source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38897 | & \frac 12 x & + 5 \\ \hline \end{array}$$ The ask what you have to multiply $4x^2$ by to get $4x^2$, then multiply, then subtract.....
The following is multiple choice question (with options) to answer.
Evaluate: 50 - 12÷4×2 = | [
"44",
"54",
"16",
"27"
] | A | According to order of operations, 12÷4×2 (division and multiplication) is done first from left to right
12÷4×2 = 3 × 2 = 6
Hence
50 - 12÷4×2 = 50 - 6 = 44
correct answer A)44 |
AQUA-RAT | AQUA-RAT-38898 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
I sold a book at a profit of 10%. Had I sold it for $120 more, 15% would have been gained. Find the cost price? | [
"$2000",
"$2500",
"$2400",
"$3120"
] | C | 115% of cost - 110% of cost = $120
5% of cost = $120
cost = 120*100/5 = $2400
Answer is C |
AQUA-RAT | AQUA-RAT-38899 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
In how many ways can the letters of the word ABROAD be rearranged such that the vowels always appear together? | [
"6!/2!",
"3!*3!",
"4!/2!",
"4! *(3!/2!)"
] | D | In the word ABROAD , there are 3 vowels - 2 A's and O
Number of ways the letters of word ABACUS be rearranged such that the vowels always appear together
= (4! * 3! )/2!
We can consider the the 3 vowels as a single unit and there are 3 ways to arrange them . But since 2 elements of vowel group are identical we divide by 2! .
The entire vowel group is considered as a single group .
Answer D |
AQUA-RAT | AQUA-RAT-38900 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Mohit sold an article for Rs. 18000. Had he offered a discount of 10% on the selling price, he would have earned a profit of 8%. What is the cost price of the article? | [
"15000",
"15002",
"15029",
"15329"
] | A | Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x = 18000 - 10/100(18000) = 18000 - 1800 = Rs. 16200
=> 1.08x = 16200
=> x = 15000
Answer: A |
AQUA-RAT | AQUA-RAT-38901 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A cistern is filled by pipe A in 10 hours and the full cistern can be leaked out by an exhaust pipe B in 15 hours. If both the pipes are opened, in what time the cistern is full? | [
"50hrs",
"30hrs",
"70hrs",
"80hrs"
] | B | time taken to full the cistern=(1/10-1/15)hrs
=1/30
=30hrs
ANSWER:B |
AQUA-RAT | AQUA-RAT-38902 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B together can do a piece of work in 8 days. If A alone can do the same work in 12 days, then B alone can do the same work in? | [
"25 days",
"24 days",
"22 days",
"27 days"
] | B | B
24 days
B = 1/8 – 1/2 = 1/24 => 24 days |
AQUA-RAT | AQUA-RAT-38903 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A man invests Rs. 8,000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum? | [
"Rs. 1200",
"Rs. 1300",
"Rs. 1500",
"Rs. 4000"
] | D | Explanation :
Interest on Rs.8000 at 5% per annum = ( 8000 × 5 × 1) / 100 = Rs. 400
Let his additional investment at 8% = x
Interest on Rs.x at 8% per annum = ( x × 8 × 1 ) / 100 = 2x/25.
To earn 6% per annum for the total, interest = (8000 + x) × 6 × 1/100.
=> 400 + 2x/25 = (8000 + x) × 6 × 1/100.
=> 40000 + 8x = (8000 + x) × 6.
=> 40000 + 8x = 48000 + 6x.
=> 2x=8000.
=> x=4000.
Answer : D |
AQUA-RAT | AQUA-RAT-38904 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
For a group of n people, k of whom are of the same sex, the (n-k)/n expression yields an index for a certain phenomenon in group dynamics for members of that sex. For a group that consists of 20 people, 8 of whom are females, by how much does the index for the females exceed the index for the males in the group? | [
" 0.05",
" 0.0625",
" 0.2",
" 0.25"
] | C | Index for females = (20-8)/20 = 3/5 = 0.6
Index for males = (20-12/20 = 2/5 = 0.4
Index for females exceeds males by 0.6 - 0.4 = 0.2
Answer: C |
AQUA-RAT | AQUA-RAT-38905 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
There is a 30% increase in the price of an article in the first year, a 20% decrease in the second year and a 10% increase in the next year. If the final price of the article is Rs. 2288, then what was the price of the article initially? | [
"Rs.2038",
"Rs.2073",
"Rs.2000",
"Rs.2027"
] | C | Let the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 - 20% of 130 = 130 - 26 = Rs. 104.
In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.
But present price of the article is Rs. 2288
for 114.4 ---> 100 ; 2288 ---> ?
Required price = (2288 * 100)/114.4 = 20 * 100 = Rs.2000.
Answer:C |
AQUA-RAT | AQUA-RAT-38906 | algorithms
Title: How many times can you pour milk among 3 buckets? This is a problem from usaco, which I solved, but I don't know how to estimate the number of times you can pour milk among the jugs, my solutions just uses a big enough number.
Here's the problem statement:
Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.
Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.
For example, if the input is 8, 9, 10 then the output is 1, 2, 8, 9, 10.
Another example: if the input is 2, 5, 10 then the output is 5, 6, 7, 8, 9, 10.
My approach is to recursively try all possible move sequences, keeping track of the contents of bucket C each time bucket A is empty. Currently I stop the recursion at the arbitrary depth 200. What is the correct depth to stop the recursion at? Since the total amount of water is C liters (C being the capacity of bucket C), the number of possible states is at most the number of non-negative integer solutions of $x+y+z = C$, which is $\binom{C+2}{2}$. Any sequence of moves longer than that must repeat a state. Therefore it is enough to recurse up to level $\binom{C+2}{2}$.
The following is multiple choice question (with options) to answer.
There are three foam generators in the factory, each of the first two can generate 10 liters of foam in one hour and the third can generate 18 liters in an hour. The three generators start working together at the same time and after one hour and a half one of the first generators stops working and two hours after that the third generator stops working and only one generator is left. If 5 hours after they all started to work the last generator stops working, how many liters of foam were generated? | [
"128",
"132.",
"146.",
"154."
] | A | Let the foam generators capacity be -
A = 10 lit/hr
B = 10 lit/hr
C = 18 lit/hr
Total foam generation in 1 hour will be 38 lit ( 10 + 10 + 18 ) ; since in 1 and 1/2 hr they will generate 38 + 19 => 57 litres...
Now one of the first generators stops working ( Say A stops working ) , so we have -
B = 10 lit/hr
C = 18 lit/hr
Total foam generation in 2 hour will be 56 litres {2 (10+18)}
The third generator stops working and now only B works for the remaining time 1 and 1/2 ( 5 - 1 and 1/2 - 2 )...
Foam generated by B will be 3/2 * 10 => 15 litres
So, total foam generated will be 128 Litres ( 57 + 56 + 15 ) ; hence answer will be (A) |
AQUA-RAT | AQUA-RAT-38907 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
In 1998 the profits of company N were 10 percent of revenues. In 1999, the revenues of company N fell by 30 percent, but profits were 16 percent of revenues. The profits in 1999 were what percent of the profits in 1998? | [
"80%",
"105%",
"120%",
"112%"
] | D | 0,112R = x/100*0.1R
Answer D |
AQUA-RAT | AQUA-RAT-38908 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
30 percent of Andrea's living room floor is covered by a carpet that is 4 feet by 9 feet. What is the area of her living room floor? | [
"14.4",
"120",
"50.4",
"60"
] | B | 30% of Area of the floor = 4*9 square feet = 36 square feet
i.e. 100% Area of Floor = (36/30)*100 = 120 square feet
Answer: Option B |
AQUA-RAT | AQUA-RAT-38909 | # How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.
The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.
I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.
I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.
Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\to 720/6 = 120$.
So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.
I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.
The following is multiple choice question (with options) to answer.
There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament? | [
"10",
"20",
"40",
"60"
] | A | Method 1:
Take the first amateur. He plays a game with each of the other four i.e. 4 games.
Now take the second one. He has already played a game with the first one. He plays 3 games with the rest of the 3 amateurs i.e. 3 more games are played.
Now take the third amateur. He has already played a game each with the first and the second amateur. Now he plays 2 games with the remaining 2 amateurs so 2 more games are played.
Now go on to the fourth amateur. He has already played 3 games with the first 3 amateurs. He just needs to play a game with the last one i.e. 1 more game is played.
The last amateur has already played 4 games.
Total no of games = 4+3+2+1 = 10
Method 2:
Each person is one participant of 4 games. So there are in all 4*5 = 20 instances of one participant games. But each game has 2 participants so total number of games = 20/2 = 10
Answer: A. |
AQUA-RAT | AQUA-RAT-38910 | # If $28a + 30b + 31c = 365$, then what is the value of $a +b +c$?
Question: For 3 non negative integers $a, b, c$; if $28a + 30b + 31c = 365$ what is the value of $a +b +c$ ?
How I approached it : I started immediately breaking it onto this form on seeing it :
$28(a +b +c) +2b +3c = 365 .......(1)$ $30(a +b +c) -2a +c = 365 .......(2)$ $31(a +b +c) -b -3a = 365 .......(3)$
And then I find out that
$365 = 28*13 + 1......(1')$; $365 = 30*12 + 5......(2')$ $365 = 31*11 + 24......(3')$
Now as we see (1) and (1') as well as (3) and (3') or even equations $2$ and $2'$ do not combine quiet congruently, so I meet with a dead end here.
my issue : how should I approach such problems where we are given no other equations or data? Basically I am asking what are a few ways to get a solution for this problem.
• $28\times 2 + 30\times 7 + 31\times 3 = 359\ne 365$ – Joey Zou Aug 11 '16 at 21:25
• This is a trick question, you are supposed to observe that these are the lengths of months and 365 is the length of a normal year. – f'' Aug 11 '16 at 21:31
More direct path to $a+b+c = 12$:
The following is multiple choice question (with options) to answer.
If 28a+30b+31c=425.
then a+b+c=?.a,b,c are natural numbers | [
"14",
"8",
"6",
"9"
] | A | have a look on your calender
since,we all knw a year consist of 365 days
february is d only month which has 28 days
4 months in a year has 30 days
and,rest 7 months has 31 days..
so,following d given eq. we can write 28*1 + 30*6 + 31*7..
hence values of a,b and c are 1, 6 and 7 respectively..
a+b+c=14
ANSWER:A |
AQUA-RAT | AQUA-RAT-38911 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A cistern is filled by a tap in 3 1/2 hours. Due to leak in the bottom of the cistern, it takes half an hour longer to fill the cistern. If the cistern is full how long will it take the leak to empty it? | [
"88hours",
"38hours",
"78hours",
"28hours"
] | D | 2/7 - 1/x
= 1/4
x = 28
Answer:D |
AQUA-RAT | AQUA-RAT-38912 | $$3 = (321)(-18+41k) + (123)(47-107k) \text{ for any } k$$
so for instance we also get ($k=1$):
$$3 = (321)(23) + (123)(-60)$$
-
The following is multiple choice question (with options) to answer.
What is the place value of 3 in the numeral 53468? | [
"3000",
"30",
"3",
"330"
] | A | place value of 3 = 3 * 1000 = 3000
Answer is A |
AQUA-RAT | AQUA-RAT-38913 | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
The following is multiple choice question (with options) to answer.
A child is playing with letter blocks (a playing cube in which an english letter is written). Before leaving the child to play on her own, her mon arranged 5 blocks into the word APPLE. How many different 6 letter words can the child make with these 5 blocks (Words need not have a meaning.She is just a child!) ? | [
"2!",
"3!",
"4!",
"5!"
] | C | There are 6 letter blocks to be arranged. Note that the letter P occurs twice. 2 'P' blocks can be arranged only in one way, since the two blocks are exactly the same(PP is same as PP). So we can consider PP as one letter and proceed to find the arrangement of 4 letters, A, PP, L, E. These four blocks can be arranged in 4! ways. So answer is C. 4!. |
AQUA-RAT | AQUA-RAT-38914 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
A certain company charges $6 per package to ship packages weighing less than 2 pounds each. For a package weighing 2 pounds or more, the company charges an initial fee of $6 plus $2 per pound. If the company charged $30 to ship a certain package, which of the following was the weight of the package, in pounds? | [
"6",
"8",
"10",
"12"
] | D | Let the weight of the package be X.
2X + 6= 30
X = 12
The answer is D. |
AQUA-RAT | AQUA-RAT-38915 | Q 4.Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:
If same pair of coins is tossed at random, find the probability of getting:
SOLUTION:
Number of trials = 500
Number of outcomes of two heads (HH) = 105
Number of outcomes of one head (HT or TH) = 275
Number of outcomes of no head (TT) = 120
(i) Probability of getting two heads = $\frac{Frequency\;of\;getting\;2\;heads}{Total\;No.\;of\;trails}$ = $\frac{105}{500}$ = $\frac{21}{100}$
(ii) Probability of getting one head = $\frac{Frequency\;of\;getting\;1\;heads}{Total\;No.\;of\;trails}$ = $\frac{275}{500}$ = $\frac{11}{20}$
(iii) Probability of getting no head = $\frac{Frequency\;of\;getting\;no\;heads}{Total\;No.\;of\;trails}$ = $\frac{120}{500}$ = $\frac{6}{25}$
#### Practise This Question
The lines shown below never meet at any point. So, these lines are not parallel. Say true or false.
The following is multiple choice question (with options) to answer.
If four coins are tossed, the probability of getting two heads and two tails is | [
"3/8",
"6/11",
"2/5",
"4/5"
] | A | Since four coins are tossed, sample space = 24
Getting two heads and two tails can happen in six ways.
n(E) = six ways
p(E) = 6/24 = 3/8
Answer : A |
AQUA-RAT | AQUA-RAT-38916 | Hint: You may suppose, w.l.o.g. that $$|x-2|<\frac12$$, in which case $$|x-1|>\frac12$$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$
$$\delta<\min(1,\varepsilon)/2$$ should do the job.
The following is multiple choice question (with options) to answer.
If 0.75: x :: 5:9, then x is equal to: | [
"1.12",
"1.16",
"1.2",
"1.35"
] | D | Explanation:
(x * 5) = (0.75 *9)
X=6.75/5 = 1.35
Answer: D |
AQUA-RAT | AQUA-RAT-38917 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
By investing in 16 2/3% stock at 64, one earns Rs. 1500. The investment made is: | [
"Rs. 5640",
"Rs. 5760",
"Rs. 7500",
"Rs. 8600"
] | B | To earn Rs. 50/3 , investment = Rs. 64.
To earn Rs. 1500, investment = Rs.64 x 3/50x 1500 = Rs. 5760.
ANSWER :B |
AQUA-RAT | AQUA-RAT-38918 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A dishonest dealer professes to sell goods at the cost price but uses a weight of 720 grams per kg, what is his percent? | [
"15%",
"25%",
"38.9%",
"45%"
] | C | Explanation:
720 --- 280
100 --- ? => 38.9%
Answer: C |
AQUA-RAT | AQUA-RAT-38919 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A 600 meter long train crosses a signal post in 40 seconds. How long will it take to cross a 3 kilometer long bridge, at the same speed? | [
"240",
"288",
"665",
"199"
] | A | S = 600/40 = 15 mps
S = 3600/15 = 240 sec
= 4 min
Answer: A |
AQUA-RAT | AQUA-RAT-38920 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
By selling an article at Rs.800, a shopkeeper makes a profit of 25%. At what price should he sell the article so as to make a loss of 25%? | [
"123",
"480",
"266",
"1251"
] | B | SP = 800
Profit = 25%
CP = (SP)*[100/(100+P)]
= 800 * [100/125]
= 640
Loss = 25% = 25% of 640 = Rs.160
SP = CP - Loss = 640 - 160 = Rs.480Answer: B |
AQUA-RAT | AQUA-RAT-38921 | # 3 balls are drawn from a bag contains 6 white balls and 4 red balls, what is the probability that 2 balls are white and 1 ball is red?
A bag contains 6 white balls and 4 red balls. If 3 balls are drawn one by one with replacement, then what is the probability that 2 balls are white and 1 ball is red?
$$\frac{18}{125}$$
What I did Probability of getting a white ball= $$6/10=3/5$$ Probability of getting a red ball= $$4/10=2/5$$ Probability of getting 2 balls white and 1 ball red = $$6/10*6/10*4/10=18/125$$
But the answer is $$\frac{54}{125}$$. Why are we multiplying it by $$3$$? Please someone elaborate this part
This is a gmat exam question.
• Look at it this way, if you don't multiply by three, then your answer is the probability that we pick $2$ white balls and $1$ red ball in that order. – WaveX Sep 9 '17 at 14:35
This is a binomial experiment with $P(W)=\frac{6}{10}=\frac{3}{5}$. Apply the formula : $$f(2)=C_2^3\cdot \left(\frac{3}{5}\right)^2\cdot \frac{2}{5}=\frac{54}{125}.$$
The following is multiple choice question (with options) to answer.
A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw | [
"64",
"128",
"132",
"222"
] | A | Explanation:
From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.
Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]
=1+[3×6]+[3×(6×5/2×1)]=1+18+45=64
Option A |
AQUA-RAT | AQUA-RAT-38922 | I am pretty sure that I am wrong.
Could someone help me?
Thanks
The answer is ${6 \over 25 \pi}$ ft /min according to the answer sheet
You can't take $r=12$ in $$v={1 \over 3} \pi r^2 h\\ {dv \over dm} = {2 \over 3} \pi (12){dh \over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\frac{r}{h} = \frac{12}{24} = \frac{1}{2}$, and $r=\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \over 3} \pi r^2 h$$ $$v={1 \over 3} \pi (\frac{h}{2})^2 h$$ $$v={1 \over 3} \pi \frac{h^3}{4}$$ $$\frac{dv}{dm} = \pi (r)\frac{h^2}{4}\frac{dh}{dm}$$
$$6 = \pi (\frac{h}{2})^2\frac{dh}{dm}$$ $$6 = \pi (\frac{h^2}{4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \pi (\frac{100}{4})\frac{dh}{dm}$$ We get $\frac{dh}{dm} = \frac{6}{25\pi}$.
The following is multiple choice question (with options) to answer.
A drought decreased the amount of water in city X’s reservoirs from 118 million gallons to 96 million gallons. If the reservoirs were at 79 percent of total capacity before the drought began, approximately how many million gallons were the reservoirs below total capacity after the drought? | [
"67",
"58",
"54",
"46"
] | C | let T be the total capacity of the reservoir.
Now we can equate before cases i.e 79/100(T) = 118.
Lets take approximate values for easy calculation 80/100(T) = 120 => T = 150 MG.
We need to subtract total - after drought case = 150 - 96 = 54 MG..
Option C is correct answer... |
AQUA-RAT | AQUA-RAT-38923 | c#, strings, converting
// assert
inchDimension.Millimeters.Should().Be(architecturalDimension.Millimeters);
copiedDimension.ShouldBeEquivalentTo(architecturalDimension);
}
/// <summary>
/// Tests mathmatical operators we will test the properties at the same time.
/// </summary>
[TestMethod]
public void Dimensions_Math_Operators()
{
// arrange
Dimension inchDimension = new Dimension(DimensionType.Inch, 14.1875);
Dimension architecturalDimension = new Dimension("1'2 3/16\"");
// act
Dimension subtractionDimension = inchDimension - architecturalDimension;
Dimension additionDimension = inchDimension + architecturalDimension;
// assert
subtractionDimension.Feet.Should().BeApproximately(0, .00000001, "Doubles math should get us at least this close");
additionDimension.Millimeters.Should().BeApproximately(720.725, .00000001, "Doubles math should get us at least this close");
additionDimension.Architectural.ShouldBeEquivalentTo("2'4 6/16\"");
}
/// <summary>
/// Tests Architectural string inputs.
/// </summary>
[TestMethod]
public void Dimensions_Architectural_Constructor()
{
// arrange
Dimension dimension1 = new Dimension("1'2 3/16\"");
Dimension dimension2 = new Dimension("1'");
Dimension dimension3 = new Dimension("1'2\"");
Dimension dimension4 = new Dimension("2 3/16\"");
Dimension dimension5 = new Dimension("1'2-3/16\"");
Dimension dimension6 = new Dimension("3/16\"");
Dimension dimension7 = new Dimension("121103");
Dimension dimension8 = new Dimension("-1'2\"");
The following is multiple choice question (with options) to answer.
Evaluate : 986 x 237 + 986 x 863 | [
"986000",
"968000",
"978000",
"987000"
] | A | 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.
Answer is A. |
AQUA-RAT | AQUA-RAT-38924 | # an exam has 50 multiple choice questions with 5 options
An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,
1. Compute the expected mean for the student score
2. Compute the standard deviation for the student score
3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : What is the expected success rate? How do you expect the proportion of students who will score less or equal to 20?
If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?
a. What is the expectation of this student's degree?
b. what is the standard deviation of this student's grade?
b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?
1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 21 at 0:59
• thanks I did it – Nidal May 21 at 1:29
It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.
The following is multiple choice question (with options) to answer.
The class mean score on a test was 50, and the standard deviation was 15. If Jack's score was within 2 standard deviations of the mean, what is the lowest score he could have received? | [
"20",
"31",
"45",
"90"
] | A | 1SD from the mean ranges from 35 to 65, where 65 is within SD above the mean and 35 within 1SD below the mean
2SD=15 TWICE=30 from the the mean, which is 80 to 20, where 80 is within 2 SD above the mean AND 20 is within 2 SD below the mean.
Answer = A |
AQUA-RAT | AQUA-RAT-38925 | 12n + 7 7, 19, 31, 43, 67, 79, 103, 127, 139, 151, ... A068229
12n + 11 11, 23, 47, 59, 71, 83, 107, 131, 167, 179, ... A068231
The following is multiple choice question (with options) to answer.
256, 225, 196, 169, 144, ? | [
"49",
"64",
"81",
"121"
] | D | Explanation :
The pattern is 16^2, 15^2, 14^2, 13^2, 12^2, ...
So next number is 11^2 = 121
Answer : Option D |
AQUA-RAT | AQUA-RAT-38926 | Case 1 : A pair of socks are present, hence exactly 2 draws for the socks to match. Case 2 : 2 pair of socks are present in the drawer. How?
To start with, instead of looking for a matching pair, let's find the probability that both socks are red.
The Sock Drawer: Probability and Statistics Problem A drawer contains red socks and black socks. Two drawn at random. When two socks are drawn at random, the probability that both are red is 1/2. That should be what you do first with an easy assignment as this. It would seem you can either make a pair or have a mismatched pair, and that both of those events would have equal chances, making for a 50 percent probability. (Deatsville, AL, USA). In a drawer $r$ red, $b$ blue, and $g$ green socks.
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The following is multiple choice question (with options) to answer.
In a shop, 40% socks are white and rest are black. 60% socks are made of cotton and rest are made of wool. 25% white socks are made of cotton and 20 of the black socks are made of wool. How many black socks Z are made of cotton? | [
"100",
"80",
"60",
"50"
] | A | I'll prefer to solve this with plug n play method.
As we know, White Cotton socks are only 25% and Total Cotton socks is 60% - Hence Black Cotton socks Z has to be [highlight]HIGH[/highlight]number. Thats why we've to pick high number... say 100 or 80.
100 fits perfectly after relevant calculations. Hence answer A. |
AQUA-RAT | AQUA-RAT-38927 | This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.
What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000
I see no argument of multiplying this by two.
This problem can be solved in another way and maybe this way shows that no need of multiplication:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60
Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000
Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
The same way here:
What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).
The following is multiple choice question (with options) to answer.
A certain junior class has 3000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair? | [
"1/40000",
"1/3600",
"9/2000",
"1/60"
] | A | Let's see
Pick 60/3000 first
Then we can only pick 1 other pair from the 800
So total will be 60 / 800 *3000
Simplify and you get 1/40000
Answer is A |
AQUA-RAT | AQUA-RAT-38928 | P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)]
The following is multiple choice question (with options) to answer.
A sum amounts to Rs.4410 in 2 years at the rate of 5% p.a. If interest was compounded yearly then what was the principal? | [
"Rs.4000",
"Rs.5000",
"Rs.4500",
"Rs.4800"
] | A | CI=4410,R=5,N=2
CI=P[1+R/100]^2
=P[1+5/100]^2
4410=P[21/20]^2
4410[20/21]^2
4000
ANSWER:A |
AQUA-RAT | AQUA-RAT-38929 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
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Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
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02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
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Re: Mixture problem-Can someone explain this [#permalink]
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02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containg 19% alcohol and now the percentage of alcohol was found to be 26%. What quantity of whisky is replaced ? | [
"1/3",
"2/3",
"2/5",
"3/5"
] | B | Let us assume the total original amount of whiskey = 10 ml ---> 4 ml alcohol and 6 ml non-alcohol.
Let x ml be the amount removed ---> total alcohol left = 4-0.4x
New quantity of whiskey added = x ml out of which 0.19 is the alcohol.
Thus, the final quantity of alcohol = 4-0.4x+0.19x ----> (4-0.21x)/ 10 = 0.26 ---> x = 20/3 ml.
Per the question, you need to find the x ml removed as a ratio of the initial volume ---> (20/3)/10 = 2/3.
Hence, B is the correct answer. |
AQUA-RAT | AQUA-RAT-38930 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
In store A there are 50 pairs of pants for every 40 store B has. The price ratio between the pants in store B and the pants in store A is 3:4. If all the pants were sold in both places until the stock ran out, what is the ratio between the total amount stores A earned to the total amount store B earned? | [
"3:16.",
"2:3.",
"5:3.",
"3:4."
] | C | 1st statement : ratio of pants
Store A : Store B
50x : 40x
5X:4X
Price :
4y:3y
Total revenue
20xy : 12xy
5:3
Answer is C |
AQUA-RAT | AQUA-RAT-38931 | to calculate area, perimeter or volume of a figure. Find the area of the face we are looking directly at. A (rectangular) cuboid is a closed box which comprises of 3 pairs of rectangular faces that are parallel to each other and joined at right angles. The Surface Area of a cuboid is the total of all the areas on each face added together. Then , Total surface Area of Cuboid = 2( lxb + bxh +hxl) square units. The injury is in the area of a small tarsal bone in the foot, the cuboid bone. Go to Surface Area or Volume. The cuboid bone is within the area of the mid-foot. Total surface area of cuboid is 1216 sq cm. The volume is found using the formula: Which is usually shortened to: It doesn't really matter which one is length, width or height, so long as you multiply all three together. For More Videos - Subscribe In this video, we will solve first two problem statements from class 9 NCERT, Exercise 13. Sol: Total area that can be painted = 9. Irregular Prism Volume Calculator - A prism has the same cross section all along its length. Write a Python program to calculate surface volume and area of a cylinder. 8 cm cubic and length of the two edges are 2 cm and 6 cm. You can change the Name, Class, Course, Date, Duration, etc. Some of the worksheets for this concept are Common 2 d and 3 d shapes, Geometric nets pack, Volume and surface area work, Surface area of solids, Surface area, Surface areas of prisms, Surface area of 3d shapes, Module mathematical reasoning. 6 x 3² = 54. So, total surface area of cuboid = 2(lb + bh + lb) = 2(5x × 4x + 4x × 2x + 5x × 2x) = 2(20x2 + 8x2 + 10x2) = 2(38x2) = 76x2 Total surface area of cuboid = 1216 cm2 76x2 = 1216cm2 x2 = 16 x = 4 Dimensions of cuboid are. Students learn how to find the surface area of cubes and cuboids using nets and 3D drawings. Submitted by IncludeHelp, on May 08, 2020. Answers included. It is also known as a right rectangular prism. A patient with cuboid
The following is multiple choice question (with options) to answer.
The side of a cube is 15 m, find it's surface area? | [
"1350",
"1750",
"1150",
"1450"
] | A | Surface area = 6 a(power)2 sq. units
6 a(power)2 = 6 × 225 = 1350 m(power)2
Answer is A. |
AQUA-RAT | AQUA-RAT-38932 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 220 m long is running with a speed of 59 kmph.. In what will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going | [
"8 sec",
"10 sec",
"12 sec",
"14 sec"
] | C | Explanation:
Speed of the train relative to man = (59 + 7) kmph
= 66 ×5/18 m/sec = 55/3 m/sec.
Time taken by the train to cross the man = Time taken by it to cover 220 m
at (55/3) m / sec = (220 ×3/55) sec = 12 sec
Answer: Option C |
AQUA-RAT | AQUA-RAT-38933 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 500m long takes 10 sec to cross a man walking at 5kmph in a direction opposite to that of the train. Find the speed of the train? | [
"175kmph",
"150kmph",
"162kmph",
"145kmph"
] | A | Let the speed of the train be x kmph
Speed of the train relative to man = x+5 = (x+5)*5/18 m/sec
500/[(x+5)*5/18] = 10
10(x+5) = 1800
x = 175kmph
Answer is A |
AQUA-RAT | AQUA-RAT-38934 | It would help of you reported some descriptive statistics or a histogram of your age and height data, such as mean, variance, skewness, kurtosis.
• Thank you for your help, can you tell me how to know that the certain data comes from normal distribution for example in your answer stated that age can not be from normal distribution, what about other data like height.What is the criteria that i must know.i want to learn more about this because it seems that i have misunderstand the concept since i am new to this.Thanks again. Dec 17, 2014 at 6:43
• Yet, normal distribution often is used as an approximation for such variables as age. And it is not really a problem since you can define age_centred as age - mean(age) and you have a variable with mean 0, with some standard deviation, positive and negative values. So I wouldn't be so strict about it.
– Tim
Dec 17, 2014 at 7:44
• You cannot have negative height for people either, but that wouldn't be a barrier to me to describing height as normally distributed if that was a good approximation. For that matter, why use any distribution with infinite bounds for measurements that can only be finite? As @Tim says, it is all a matter of approximations acceptable given the data and given the purpose. Dec 17, 2014 at 11:03
• I agree that normal distribution could be a good approximation for bounded data sometimes, but the question was about whether the data is from normal or not. Dec 17, 2014 at 13:35
• The age of graduating seniors from high school could potentially be normally distributed and also take on negative values if mean centered as @Tim mentioned. Dec 19, 2015 at 18:55
The following is multiple choice question (with options) to answer.
The average age of applicants for a new job is 20, with a standard deviation of 8. The hiring manager is only willing to accept applicants whose age is within one standard deviation of the average age. Assuming that all applicants' ages are integers and that the endpoints of the range are included, what is the maximum number of different ages of the applicants? | [
"8",
"16",
"17",
"18"
] | C | Minimum age = Average - 1 Standard deviation = 20 - 8 = 12
Maximum age = average + 1 Standard deviation = 20+ 8 = 28
Maximum number of different ages of the applicants = 28 - 12 + 1 = 17
Answer C |
AQUA-RAT | AQUA-RAT-38935 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning? | [
"39",
"35",
"42",
"40.5"
] | A | let the average after 16th inning =x
then total run after 16th inning=16x
then total run after 17th inning=16x+87
then average run after 17th inning=(16x+87)/17
(16x+87)/17=x+3;
x=36;
average after 17th inning =36+3=39
ANSWER:A |
AQUA-RAT | AQUA-RAT-38936 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
Set S consists of integers {8, 11, 12, 14, 15}. If integer n is included in the set, the average (arithmetic mean) of set S will increase by 25%. What is the value of integer n? | [
"28",
"32",
"36",
"40"
] | D | The average of the numbers in set S is 12.
If we increase the mean by 25%, the new mean is 15.
Thus, on average, 5 numbers increase by 5.
Therefore n = 15+25 = 40
The answer is D. |
AQUA-RAT | AQUA-RAT-38937 | In total, there are $$6! = 720$$ ways for all the people to sit on the chairs.
1. First there are $$6$$ ways for you to take a seat, $$2$$ ways for your friend to sit next to you. Now with $$4$$ people left with $$4$$ chairs, there are $$4!=24$$ ways for them to sit. So the probability would be $$\frac {24 \times 2 \times 6}{720}=\frac{2}{5}$$
2. Similar to 1. , there are $$6$$ ways for you to have the first seat, $$1$$ way for your friend to sit opposite you and $$24$$ ways for the rest to sit. The probability would be: $$\frac {24 \times 1 \times 6}{720}=\frac{1}{5}$$
3. In fact this is the complement of 1. so the probability would be $$1- \frac{2}{5}=\frac{3}{5}$$
(Sorry, I was late and English is my second language)
• hi, for number two, just assume you are already seated, there are 5 seats available, so the chance is indeed 1/5 as you calculated. Same argument can be used at the other questions. – Alucard Oct 17 '18 at 12:00
The following is multiple choice question (with options) to answer.
There are 6 people and 6 chairs. In how many different ways people can occupy the chairs? | [
"360",
"500",
"420",
"720"
] | D | number of ways =6! = 720
Answer is D |
AQUA-RAT | AQUA-RAT-38938 | to calculate area, perimeter or volume of a figure. Find the area of the face we are looking directly at. A (rectangular) cuboid is a closed box which comprises of 3 pairs of rectangular faces that are parallel to each other and joined at right angles. The Surface Area of a cuboid is the total of all the areas on each face added together. Then , Total surface Area of Cuboid = 2( lxb + bxh +hxl) square units. The injury is in the area of a small tarsal bone in the foot, the cuboid bone. Go to Surface Area or Volume. The cuboid bone is within the area of the mid-foot. Total surface area of cuboid is 1216 sq cm. The volume is found using the formula: Which is usually shortened to: It doesn't really matter which one is length, width or height, so long as you multiply all three together. For More Videos - Subscribe In this video, we will solve first two problem statements from class 9 NCERT, Exercise 13. Sol: Total area that can be painted = 9. Irregular Prism Volume Calculator - A prism has the same cross section all along its length. Write a Python program to calculate surface volume and area of a cylinder. 8 cm cubic and length of the two edges are 2 cm and 6 cm. You can change the Name, Class, Course, Date, Duration, etc. Some of the worksheets for this concept are Common 2 d and 3 d shapes, Geometric nets pack, Volume and surface area work, Surface area of solids, Surface area, Surface areas of prisms, Surface area of 3d shapes, Module mathematical reasoning. 6 x 3² = 54. So, total surface area of cuboid = 2(lb + bh + lb) = 2(5x × 4x + 4x × 2x + 5x × 2x) = 2(20x2 + 8x2 + 10x2) = 2(38x2) = 76x2 Total surface area of cuboid = 1216 cm2 76x2 = 1216cm2 x2 = 16 x = 4 Dimensions of cuboid are. Students learn how to find the surface area of cubes and cuboids using nets and 3D drawings. Submitted by IncludeHelp, on May 08, 2020. Answers included. It is also known as a right rectangular prism. A patient with cuboid
The following is multiple choice question (with options) to answer.
The edges of cuboid are 4 cm; 5 cm and 6 cm. Find its surface area? | [
"228",
"148",
"992",
"772"
] | B | 2(4*5 + 5*6 + 4*6) = 148
Answer: B |
AQUA-RAT | AQUA-RAT-38939 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
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cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
The average score in an examination of 10 students of a class is 60. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 30 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is | [
"89",
"90",
"92",
"99"
] | D | The average score in an examination of 10 students of a class is 60 --> the total score = 10*60 = 600;
The 5 smallest scores have an average of 55 --> the total score of the 5 smallest scores = 275.
From above, the total score of the 5 largest scores = 600 - 275 = 325.
Say the 5 largest scores are a, b, c, d, and e (where a<b<c<d<e, since each of the top five scorers had distinct scores). We want to maximize e. To maximize e, we need to minimize a, b, c, and d. The least value of a, is 55 (The least score of the top 5 , a, should be equal to the highest of the bottom 5 and to minimize the highest of the bottom 5, all scores from the bottom 5 should be equal). In this case the least values of b, c, and d are 56, 57, and 58 respectively:
a + b + c + d + e = 55 + 56 +57 + 58 + e = 325;
e = 99.
Answer: D. |
AQUA-RAT | AQUA-RAT-38940 | # Another probability question from my textbook
Suppose that we have a tennis tournament with 32 players. Players are matched in a completely random fashion, and we assume that each player always has probability 1/2 to win a match. What is the probability that two given players meet each other during the tournament.
-
I am trying to teach myself probability. Just to clarify, not exactly a student looking for homework solutions - but still a student – user669083 Jan 23 '12 at 22:29
Easy general answer for $n$ players in a knockout tournament (and here $n=32$):
There are $\dfrac{n(n-1)}{2}$ potential pairs for matches.
To have one winner, $n-1$ players must be knocked out, so there are $n-1$ actual matches.
So the probability that a particular pair actually have a match is $\dfrac{n-1}{{n(n-1)/2}} = \dfrac{2}{n}$.
-
Very elegant! – Rasmus Jan 24 '12 at 7:27
yes it seems to be most elegant one. – user669083 Jan 24 '12 at 16:15
Hint 1: Consider the number of other players a particular player meets with what probability: one other with probability $1/2$; two others with probability $1/4$; three others with probability $1/8$; etc.
Hint 2: What is the expected number of other players a particular player meets?
Hint 3: How does Hint 2 relate to the original question?
Answer: $$\dfrac{1 \times \dfrac{1}{2} + 2 \times \dfrac{1}{2^2} + 3 \times \dfrac{1}{2^3} + 4 \times \dfrac{1}{2^4} + 5 \times \dfrac{1}{2^4}}{31} = \dfrac{1}{16}$$
The following is multiple choice question (with options) to answer.
Let us say that a table tennis tournament was going on with knock out terms which means the one who loses the match is out of the tournament. 86 players took part in that tournament. How many matches were played? | [
"90 matches",
"95 matches",
"99 matches",
"85 matches"
] | D | D 85 matches. The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 85 matches were played. |
AQUA-RAT | AQUA-RAT-38941 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed at which a man can row a boat in still water is 15 km/hr. If he rows downstream, where the speed of current is 3 km/hr, how many seconds will he take to cover 100 meters? | [
"14",
"16",
"18",
"20"
] | D | The speed of the boat downstream = 15 + 3 = 18 km/hr
18 km/hr * 5/18 = 5 m/s
The time taken to cover 100 meters = 100/5 = 20 seconds.
The answer is D. |
AQUA-RAT | AQUA-RAT-38942 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
In a sports club with 30 members, 17 play badminton and 18 play tennis and 2 do not play either. How many members play both badminton and tennis? | [
"7",
"8",
"9",
"10"
] | A | Let x play both badminton and tennis so 17-x play only badminton and 19-x play only tennis. 2 play none and there are total 30 students. hence,
(17-x)+(18-x)+x+2=30
37-2x+x=30
37-x=30
x=7
So 7 members play both badminton and tennis.
A |
AQUA-RAT | AQUA-RAT-38943 | Since this is a quadratic equation, and the leading coefficient is $+1$, we have
$$\Delta < 0$$
Whence the equation is always strictly positive (that is, it's always $>0$).
• I think it's worth to note that we need the leading coefficient to be positive too, even if it's trivial. Jun 28 '18 at 7:56
• @Botond You're right! Jun 28 '18 at 8:00
There is also the following way.
For $x\geq-1$ we obtain $$x^2+x+1=x^2+(x+1)>0$$ and for $x<-1$ we obtain $$x^2+x+1=x(x+1)+1>0+1>0.$$
• This is also one of the most elegant proofs in reply to this OP. However, that ugly trick with the $0$ on the LHS of your last inequality is not needed. It suffices to note that in this case $x+1<0.$ Jun 28 '18 at 10:44
• @Allawonder The last inequality is true for $x<-1$. See better my post. Jun 28 '18 at 10:46
• Sorry. That was not what I meant. I have edited my comment. See above again. Jun 28 '18 at 10:51
• @Allawonder Yes, in this case $x(x+1)>0$, which gives $x(x+1)+1>0.$ If you wish you can fix my post. Jun 28 '18 at 10:55
If $x$ is positive, then $x^2+x+1$ is clearly positive.
If $x$ is negative then $x^2-x+1$ is certainly positive. Now $$(x^2+x+1)(x^2-x+1)=x^4+x^2+1$$ is certainly positive, so $x^2+x+1$ must also be positive in this case.
If $x$ is zero then $x^2+x+1=1\gt 0$
The following is multiple choice question (with options) to answer.
The positive value of x that satisfies the equation (1 + 2x)^3 = (1 + 3x)^4 is between
Bunuel, can you please explain this one? | [
"0 and 0.5",
"0.5 and 1",
"1 and 1.5",
"1.5 and 2"
] | B | Trial and error would probably be the easiest way to solve this problem. When x is large enough positive number, then because of the exponents (5>4), LHS will be more than RHS (as you increase the positive value of x the distance between the values of LHS and RHS will increase).
Try x=1 --> LHS=3^5=81*3=243 and RHS=4^4=64*4=256, so (1 + 2x)^5 < (1 + 3x)^4. As you can see LHS is still slightly less than than RHS. So, the value of x for which (1 + 2x)^5 = (1 + 3x)^4 is slightly more than 1.
Answer: B. |
AQUA-RAT | AQUA-RAT-38944 | For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
• Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider – Irishpanda Feb 18 '16 at 13:08
• haha, I had this same thought process initially – question_asker Feb 18 '16 at 16:03
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
• Very elegantly thought and stated. Best answer. – Vynce Aug 30 '17 at 2:20
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $$g_n$$ the n'th girl's secret number and $$a_n$$ the average she gave aloud.
The following is multiple choice question (with options) to answer.
The average of first five multiples of 2 is? | [
"6",
"8",
"9",
"5"
] | A | Average = 2(1 + 2 + 3 + 4 + 5)/5
= 30/5
= 6.
Answer:A |
AQUA-RAT | AQUA-RAT-38945 | Math Help - N positive integer problem with a + exponent
1. N positive integer problem with a + exponent
Here it is a problem that is driving me nuts...
Note: ^ means the exponent, in this case the exponent is "+".
If n is a positive integer, then n^+ denotes a number such that n < n^+ < n+1.
So decide which of the following options is greater (or if both ofthem are equal or if it is impossible to determine it).
Option A: 20^+ / 4^+
Option B: 5^+
What I did: 20^+ / 4^+ SO, (20/4)^+ , SO (5)^+ , so I concluded that both of them are equal, but the correct answer is " It can´t be determined which option is greater". Could anyone explain me why??
2. We have:
$20< 20^+ < 21$
$4< 4^+ < 5$
$5<5^+<6$
Clearly we cannot divide these inequalities.
$\frac{20< 20^+ < 21}{4< 4^+ < 5} = ???????$
But we can determine bounds.
How do we obtain the smallest possible value of $\frac{20^+}{4^+}?$ We minimize the numerator and maximize the denominator. The numerator's smallest value is close to $20,$ and the denominator's largest value is close to $5.$ So, the fractional value is $\frac{20^+}{4^+}> \frac{20}{5}=4$.
Now, How do we obtain the largest possible value of $\frac{20^+}{4^+}?$ We maximize the numerator and minimize the denominator. We would end up with $\frac{20^+}{4^+} < \frac{21}{4} = 5.25.$
So, $4<\frac{20^+}{4^+}<5.25,$ Which could be greater or less than $5 <5^+<6$ depending on which value $5^+$ were to take on.
The following is multiple choice question (with options) to answer.
When positive integer A is divided by positive integer B, the result is 9.44. Which of the following could be the integer B is ? | [
"11",
"44",
"24",
"25"
] | D | the remainder will be obtained from the decimal part when A is divided by B i.e. 0.44
0.44 = 44/100 = 11/25 so possible remainders are 11,22,33,44.
Only option A-11 satisfies this
for remainder 11,
A=236 and B=25
Ans - D |
AQUA-RAT | AQUA-RAT-38946 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
Three persons invested Rs.9000 in a joint business. The second person invested Rs.1000 more than the first and the third Rs.1000 more than second. After two years, they gained Rs.3600. How much third person will get? | [
"2400",
"1600",
"2980",
"2707"
] | B | First persons investment = x
Second persons investment = x + 1000
Third persons investments = x + 2000
x + x + 1000 + x + 2000 = 9000
3x = 6000
x = 2000
Ratio = 2000 : 3000 : 4000
2:3:4
4/9 * 3600 = 1600
Answer:B |
AQUA-RAT | AQUA-RAT-38947 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
If 10 crates have 98 apples each and how many apples more is required in order to make 100 apples in each crate? | [
"5",
"10",
"15",
"20"
] | D | Each crate requires 2 apples and totally there are 10 crates so required apples = 10 * 2 = 20
Answer: D |
AQUA-RAT | AQUA-RAT-38948 | However, this includes the solution when $$x_1,x_2,x_3,x_4,x_5=0$$, which is explicitly stated in the problem to be excluded. Hence our final answer is $$\boxed{6187}$$.
Questions like this are often worded ambiguously, forcing you to figure out a sensible interpretation. In this case, you aren't told whether different balls of the same color should be considered different when counting "ways". However, if we considered same-colored balls distinct, then the colors would be irrelevant (and since the number of balls is "unlimited", we wouldn't have enough information to answer the question). So we can infer that you're expected to think of two handfuls as the same if they look the same (e.g. "5 red and 4 blue"), even though multiple different sets of balls might look the same way.
The "unlimited" here just means that it's possible to get a handful where all balls are the same color (of any color). Since we're only interested in handfuls of at most 12 balls, it doesn't matter whether the box has 12 of each color or 100 of each color.
In other words, you're looking for the number of non-negative integer solutions to
$$x_1+x_2+x_3+x_4+x_5\le12$$
and the "unlimited" means there are no explicit upper limits on the variables.
The following is multiple choice question (with options) to answer.
A company that ships boxes to a total of 15 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (assume that the order of the colors in a pair does not matter) | [
"8",
"7",
"6",
"5"
] | D | Back-solving is the best way to solve this problem.
You basically need 12 combinations (including single colors)
If we start from option 1->
1=> 4C2 + 4 = 10 (Not enough)
2=> 5C2 + 5 = 15 (Enough)
Since the minimum number is asked. It should be 5.
Answer - D |
AQUA-RAT | AQUA-RAT-38949 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
What is the difference between the compound interest on Rs.12000 at 20% p.a. for one year when compounded yearly and half yearly? | [
"Rs.140",
"Rs.120",
"Rs.130",
"Rs.110"
] | B | When compounded annually, interest
= 12000[1 + 20/100]1 - 12000 = Rs.2400
When compounded semi-annually, interest
= 12000[1 + 10/100]2 - 12000 = Rs.2520
Required difference = 2520 - 2400 = Rs.120
ANSWER:B |
AQUA-RAT | AQUA-RAT-38950 | 5,224 views
In a certain town, the probability that it will rain in the afternoon is known to be $0.6$. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to $25°C$, the probability that it will rain in the afternoon is $0.4$. The temperature at noon is equally likely to be above $25°C$, or at/below $25°C$. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above $25°C$?
1. $0.4$
2. $0.6$
3. $0.8$
4. $0.9$
Answer is C) $0.8$
$P$(rain in afternoon) $= 0.5\times P($rain when temp $\leq 25) + 0.5 \times P($ rain when temp $> 25 )$
$0.6 = 0.5\times 0.4 + 0.5\times P($ rain when temp $> 25 )$
so,
$P$( rain when temp $> 25$ ) $= 0.8$
This is a question of Total Probability where after happening on one event E1, the probability of another event E2 happening or not happening is added together to get the probability of happening of Event E2.
Given P(Rain in noon) =0.6 (This is total probability given).
"The temperature at noon is equally likely to be above 25°C, or at/below 25°C."
means P(Temp less than or 25) = P(Temp >25) =0.5
P(Rain in noon) = P(Temp $\leq$ 25) * P(Rain | Temp $\leq$ 25) + P(Temp $>$ 25) * P(Rain| Temp $>$ 25)
0.6= (0.5*0.4) + (0.5*X)
X=0.8 Ans (C)
The following is multiple choice question (with options) to answer.
If the probability E of rain on any given day in city x is 50% what is the probability it with rain on exactly 3 days in a five day period? | [
"8/125",
"2/25",
"5/16",
"8/25"
] | C | 5C3 (1/2)^3(1/2)^2
= (5 * 4 * 3!)/2!3! * 1/32
E= 5/16
Answer - C |
AQUA-RAT | AQUA-RAT-38951 | meters (m), centimeters (cm) & millimeters (mm). This packet covers all you need to know about sectors of a circle. The curved part of the sector is of the circumference of the circle but to find the perimeter of the sector we must add (the radius of the circle is) so the perimeter of the sector is. A worksheet where you need to find the perimeter of sectors given the radius and angle of the arc. The perimeter of a rectangle is 40 cm. The portion of the circle's circumference bounded by the radii, the arc , is part of the sector. The perimeter of a certain sector of a circle of radius is 5.6 m and 27.2 m. Find the area of the sector. Information that I knew just from looking at the diagram that could prove useful. Find A(P-abc). If the angle is θ, then this is θ/2π the fraction of the full angle for a circle. Sector-A sector is a portion of a circle which is enclosed between its two radii and the arc adjoining them. (2) Given that r = 2√2 and that P = A, (b) show that θ = 2 1 The formula for the area of a sector is (angle / 360) x π x radius 2.The figure below illustrates the measurement: As you can easily see, it is quite similar to that of a circle, but modified to account for the fact that a sector is just a part of a circle. Videos, worksheets, 5-a-day and much more Formula: s=r x θ l=s+(2*r) Where, r = Radius of Circle θ = Central Angle s = Sector of Circle Arc Length l = Sector of Circle Perimeter Related Calculator: Perimeter of sectors. 15.8k VIEWS. Answer is: units. Simplifying expressions. A FULL LESSON on calculating the area and perimeter of a sector. Practice Question: Calculate perimeter of sector of circle for the following problems: N.B. Perimeter is the distance around a closed figure and is typically measured in millimetres (mm), centimetres (cm), metres (m) and kilometres (km). The angle subtended at the center of the circle by the arc is called the central angle. The sector to the right is a fraction of the circle to the left so the the area of
The following is multiple choice question (with options) to answer.
The sector of a circle has radius of 28 cm and central angle 90o. Find its perimeter? | [
"100 cm",
"85 cm",
"75 cm",
"95 cm"
] | A | Perimeter of the sector = length of the arc + 2(radius)
= (90/360 * 2 * 22/7 * 21) + 2(28)
= 44 + 56 = 100 cm
Answer: A |
AQUA-RAT | AQUA-RAT-38952 | $$\frac{(12)(9)}{2}$$ = 54
Answer A
_________________
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Re: A polygon has 12 edges. How many different diagonals does it have? [#permalink]
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20 Sep 2017, 15:59
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Top Contributor
1
Gnpth wrote:
A polygon has 12 edges. How many different diagonals does it have?
(A) 54
(B) 66
(C) 108
(D) 132
(E) 144
The polygon in question has 12 vertices.
Let's focus on ONE VERTEX, which we'll call vertex A
How many diagonal can be drawn from vertex A?
We'll there are 11 other vertices to connect to, HOWEVER the vertices on either side of vertex A are out of contention, since joining two adjacent points will not create a diagonal.
So, we can create a diagonal with vertex A by connecting to any of the 9 eligible vertices.
Using the same logic, we can conclude that we can create a diagonal with vertex B by connecting to any of the 9 eligible vertices.
And we can create a diagonal with vertex C by connecting to any of the 9 eligible vertices.
There are 12 vertices, so the TOTAL number of diagonals = (12)(9) = 108
HOWEVER, this is not the final answer, since we have counted each diagonal TWICE.
For example, in the first step, we counted the diagonal created by joining vertex A with vertex B, and in the second step, we counted the diagonal created by joining vertex B with vertex A
To account for this duplication, we must take 108 and divide it by 2 to get: 54
Answer:
Cheers,
Brent
_________________
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Re: A polygon has 12 edges. How many different diagonals does it have? [#permalink] 20 Sep 2017, 15:59
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A polygon has 12 edges. How many different diagonals does it have?
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
The following is multiple choice question (with options) to answer.
how many internal diagonals does a nonagon (nine sided polygon) have? | [
"25",
"27",
"40",
"56"
] | B | Number of diagonals in any polygon can be found using this formula: n(n-3)/2
Here n = 9
No. of diagonals = 9(9 - 3)/2
= 27
Ans B |
AQUA-RAT | AQUA-RAT-38953 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
Students in Class I, II and III of a school are in the ratio of 3 : 5 : 8. Had 15 more students admitted to each class, the ratio would have become 6 : 8 : 11. How many total students were there in the beginning? | [
"112",
"64",
"96",
"80"
] | D | Explanation:
Increase in ratio for 3 classes is 6 - 3 = 8 - 5 = 11 - 8 = 3.
Given, 15 more students are admitted to each class.
Therefore, 3 :: 15 ⇒ 1 :: 5
Therefore, 3 + 5 + 8 = 16 :: 16 x 5 = 80.
Hence, total students in the beginning were 80.
Correct Option: D |
AQUA-RAT | AQUA-RAT-38954 | electric-circuits, voltage
Title: Unsatisfactory explanation for the EMF measurement of a battery Experimentally, I have seen how hooking up a battery to a simple circuit just with a high-resistance voltmeter raises the voltage reading (allegedly to a level equal to the EMF of the battery).
However, I find the explanation for why the reading rises, much less to an EMF, very unconvincing. We were told that the internal resistance and the necessary potential drop is ignored, because there is no current in the said circuit, hence why the voltmeter measures an EMF. How can this make sense? There clearly must be some current, albeit very little, flowing, for the high-resistance voltmeter to even have a reading, and that little current will still experience resistive forces from the internal resistance of the cell - so the EMF should not be attainable. Or is there a mechanism by which, when there is very little current, resistors are ignored, hence no work has to be done to traverse them?
Clearly I am wrong, as experimentally I saw the voltage rise in that super simple cell. My point of view suggests that there shouldn't be a difference between the reading in said circuit and the potential difference in a circuit consisting of 3 resistors (the voltmeter in this case measures the drop between these 3 resistors, note there is no other significant source of resistance other than the internal resistance). Really, in my theoretical understanding they should produce an equal reading, but they don't.
So, to be honest, not only do I believe that the reading we saw shouldn't have been the EMF, but not even any different from the reading in a normal circuit, as described (the latter belief clearly conflicts with reality).
I an eager to know what I am thinking wrongly about. Please ask if I can help clarify anything!
Thank you very much :)! You're quite correct that there will be some current flowing, so there must be a voltage drop due to the internal resistance of the battery. The EMF measured by any voltmeter will always be less than the true EMF.
If the internal resistance of the battery is $R_b$ and the resistance of your voltmeter is $R_m$ then the voltage you measure will be:
$$ V = \frac{R_m}{R_m + R_b} E $$
The following is multiple choice question (with options) to answer.
A voltmeter is used: | [
"22",
"266",
"27",
"26"
] | C | Explanation:
Answer: Option C |
AQUA-RAT | AQUA-RAT-38955 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
Of the three-digit integers greater than 900, how many have two digits that are equal to each other and the remaining digit different from the other two? | [
" 90",
" 82",
" 80",
" 45"
] | A | Let the three digit number be represented as X Y Z.
There are 3 cases:
Case I.[ X=Y ]Z is not equal to XY :XXZorYYZ
X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways
After X is chosen, Y can be chosen in 1 way
After XY are chosen, Z can be chosen in 9 ways
Thus, possible No of digits = (3 ways) * (1 way) * (9 ways) = 27 ....(1)
[example numbers: 774,779,882,993 etc]
Case II.[ X=Z ]Y is not equal to XZ:XYXorZYZ
X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways
After X is chosen, Z can be chosen in 1 way
After XZ are chosen, Z can be chosen in 9 ways
Thus, possible No of digits = (3 ways) * (9 ways) * (1 way) = 27 ....(2)
[example numbers: 747,797,828,939 etc]
Case III.[ Y =Z ]X is not equal to YZ :XYYorXZZ
X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways
After X is chosen, Y can be chosen in 9 ways
After Y is chosen, Z can have 1 way
Thus, possible No of digits = (3 ways) * (1 way) * (9 ways) = 27 ....(3)
[example numbers: 744,799,822,933 etc]
Therefore, total numbers of possible digits [sum of Case (1), (2)(3) above] = 27 + 27 + 27 - 1 = 80
One digit is subtracted from total number of possible digits to eliminate one possibility of XYZ = 700 to satisfy the condition that digit > 900.
Answer:(A) |
AQUA-RAT | AQUA-RAT-38956 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
0, 6, 18, 36, 60, 90, ?
What number should replace the question mark? | [
"126",
"120",
"119",
"126"
] | A | A 126
add 6, 12,18,24,..... |
AQUA-RAT | AQUA-RAT-38957 | Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.
Hence your final answer is $$3!\cdot 10!\cdot \frac{4!}{2!2!}$$
Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.
Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.
So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$
For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
The following is multiple choice question (with options) to answer.
In how many ways can 7 boys be seated in a circular order? | [
"120",
"360",
"420",
"720"
] | D | Number of arrangements possible = (7-1)!
= 6!
= 6*5*4*3*2*1
=720
Ans -D |
AQUA-RAT | AQUA-RAT-38958 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A grocer has a sale of Rs 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs, 6500 ? | [
"4991",
"4992",
"2777",
"2977"
] | A | Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs.[(6500 x 6) - 34009]
= Rs. (39000 - 34009)
= Rs. 4991.
Answer: A |
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The following is multiple choice question (with options) to answer.
A and B started a business investing Rs. 10,000 and Rs 20,000 respectively. In what ratio the profit earned after 2 years be divided between A and B respectively? | [
"3:2",
"9:2",
"18:20",
"1:2"
] | D | A: B = 10000 : 20000 = 1 : 2
ANSWER:D |
AQUA-RAT | AQUA-RAT-38960 | # Making Friends around a Circular Table
I have $n$ people seated around a circular table, initially in arbitrary order. At each step, I choose two people and switch their seats. What is the minimum number of steps required such that every person has sat either to the right or to the left of everyone else?
To be specific, we consider two different cases:
1. You can only switch people who are sitting next to each other.
2. You can switch any two people, no matter where they are on the table.
The small cases are relatively simple: if we denote the answer in case 1 and 2 for a given value of $n$ as $f(n)$ and $g(n)$ respectively, then we have $f(x)=g(x)=0$ for $x=1, 2, 3$, $f(4)=g(4)=1$. I’m not sure how I would generalize to larger values, though.
(I initially claimed that $f(5)=g(5)=2$, but corrected it based on @Ryan’s comment).
If you’re interested, this question came up in a conversation with my friends when we were trying to figure out the best way for a large party of people during dinner to all get to know each other.
Edit: The table below compares the current best known value for case 2, $g(n)$, to the theoretical lower bound $\lceil{\frac{1}{8}n(n-3)}\rceil$ for a range of values of $n$. Solutions up to $n=14$ are known to be optimal, in large part due to the work of Andrew Szymczak and PeterKošinár.
The moves corresponding to the current best value are found below. Each ordered pair $(i, j)$ indicates that we switch the people in seats $(i, j)$ with each other, with the seats being labeled from $1 \ldots n$ consecutively around the table.
The following is multiple choice question (with options) to answer.
There is a circular table. It has 60 chairs. Drakula came and sit on a chair beside another person. What is the minimum no. of person will be so, that wherever he sits he always gets to sit beside a person | [
"30",
"20",
"15",
"10"
] | B | if total seats N = 5 in a circular arrangement ( _ _ M _ _ M ) where first ( _ ) is connected with last Man(M), then min no of persons required= 2.
similarly If N = 9, Man required = 3 i.e,( _ _ M _ _ M _ _ M )
If N= 12, Man required = 4 i.e, ( _ _ M _ _ M _ _ M _ _ M)
have u got any relation??
the no of man required is = Total no of seats in circle / 3
So, A/Q total seat = 60. Man required = 60/3=20
ANSWER:B |
AQUA-RAT | AQUA-RAT-38961 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
An employer pays Rs.20 for each day a worker works and for felts Rs.3 for each day is ideal at the end of sixty days a worker gets Rs.280 . for how many days did the worker remain ideal? | [
"10",
"40",
"30",
"20"
] | B | Suppose a worker remained ideal for x days then he worked for 60-x days
20*(60-x)-3x=280
1200-23x=280
23x=920
x=40
Answer is B. |
AQUA-RAT | AQUA-RAT-38962 | # Math Help - Finding the values of a and b
1. ## Finding the values of a and b
Hello everyone. This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
The keyword in this irksome question would be the word 'or'. So it denotes that that this involves quadratic formulas.
Can anyone give me a clue so that I may make a breakthrough in understanding this problem? Thank you so much!
2. The wording is a bit iffy, but they actually meant that the solutions for that equation is x = 3 AND x = -4.
3. Originally Posted by PythagorasNeophyte
This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
$x=\dfrac{-a + \sqrt{a^2 - 4 \times 1 \times b}}{2}$ and $x=\dfrac{-a - \sqrt{a^2 - 4 \times 1 \times b}}{2}$
We know that the minus squareroot usually gives us the smaller answer of x. So then we just substitute in our x answers to get:
$3=\dfrac{-a + \sqrt{a^2 - 4 \times b}}{2}$ and $-4=\dfrac{-a - \sqrt{a^2 - 4 \times b}}{2}$
Now solve for a and b using simultaneous equation.
4. Originally Posted by Educated
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
The following is multiple choice question (with options) to answer.
If a and b are the two values of t that satisfy
the equation t2 – 6t + 8 = 0, with a > b, what is the value of a – b? | [
"2",
"4",
"6",
"8"
] | A | Factor the left side of the equation:
t2 – 6t + 8 = 0
(t – 2)(t – 4) = 0
Split this equation into two equations and solve:
t – 2 = 0 t – 4 = 0
t = 2 t = 4
Thus, a = 4 and b = 2. So a – b = 4 – 2 = 2.
correct answer A)2 |
AQUA-RAT | AQUA-RAT-38963 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B can do a piece of work in 11 days. With the help of C they finish the work in 5 days. C alone can do that piece of work in? | [
"15.5 days",
"19.5 days",
"17.5 days",
"9.2 days"
] | D | C = 1/5 – 1/11 = 6/55 => 9.2 days
ANSWER:D |
AQUA-RAT | AQUA-RAT-38964 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
Rahul went to a shop and bought things worth Rs. 25, out of which 30 Paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items? | [
"Rs. 15",
"Rs. 12.10",
"Rs. 19.70",
"Rs. 16.80"
] | C | Explanation :
Total cost of the items he purchased = Rs.25
Given that out of this Rs.25, 30 Paise is given as tax
=> Total tax incurred = 30 Paise = Rs.30/100
Let the cost of the tax free items = x
Given that tax rate = 6%
∴ (25 − (30/100) − x)(6/100) = 30/100
⇒ 6(25 − 0.3 − x) = 30
⇒ (25 − 0.3 − x) = 5
⇒ x = 25 − 0.3 − 5 = 19.7
Answer : Option C |
AQUA-RAT | AQUA-RAT-38965 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
In an exam, Amar scored 64 percent, Bhavan scored 36 percent and Chetan 44 percent. The maximum score awarded in the exam is 500. Find the average mark scored by all the three boys? | [
"384",
"826",
"240",
"269"
] | C | Average mark scored by all the three boys =
[64/100 (500) + 36/100 (500) + 44/100 (500)] / 3
= 240
Answer: C |
AQUA-RAT | AQUA-RAT-38966 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
Mike took a taxi to the airport and paid $2.50 to start plus $0.25 per mile. Annie took a different route to the airport and paid $2.50 plus $5.00 in bridge toll fees plus $0.25 per mile. If each was charged exactly the same amount, and Annie's ride was 22 miles, how many miles was Mike's ride? | [
"30",
"36",
"42",
"48"
] | C | The cost of Annie's ride was 2.5+5+(0.25*22) = $13
Let x be the distance of Mike's ride.
The cost of Mike's ride is 2.5+(0.25*x) = 13
0.25*x = 10.5
x = 42 miles
The answer is C. |
AQUA-RAT | AQUA-RAT-38967 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
A dinner has to be hosted with 5 managers.Find the number of ways in which the managers may be selected from among 9 managers,if 2 managers will not attend the meeting together. | [
"1.35",
"2.None of these",
"3.12",
"4.126"
] | B | 9 managers, but two of them cannot attend the meeting together.
We can split it into two cases.
1. Meeting without these two managers in it. That would mean selecting 5, from the remaining 7 which is 7C5 = 21
2. Meeting with one of the two managers. select 1 manager from two, and then select 4 from the remaining 7, which is 2C1 X 7C4 = 70.
So, answer is 21 + 70 = 91.
Answer : B |
AQUA-RAT | AQUA-RAT-38968 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Braun invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Braun got back 6 times his original investment. What is the value of n? | [
"50 years",
"25 years",
"62 years 6 months",
"37 years 6 months"
] | C | Explanatory Answer
Let us say Braun invested $100.
Then, at the end of 'n' years he would have got back $600.
Therefore, the Simple Interest earned = 600 - 100 = $500.
We know that Simple Interest = (Principal * number of years * rate of interest) / 100
Substituting the values in the above equation we get 500 = (100 * n * 8) / 100
Or 8n = 500
Or n = 62.5 years.
correct choice is (C) |
AQUA-RAT | AQUA-RAT-38969 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The cost price of a cow and a horse is 3 lakhs. The cow is sold at 20% profit and horse at 10% loss.Overall gain is Rs.4200. What is the cost price of the cow? | [
"113000",
"114000",
"115000",
"116000"
] | B | let costs of cow and horse are x,y so x+y=300000------(1)
(x*120/100)+(y*90/100)=304200
120x+90y=30420000------(2).
solve equations (1) and (2)
x=114000
ANSWER:B |
AQUA-RAT | AQUA-RAT-38970 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
Legendary Member
Posts: 1892
Joined: 14 Oct 2017
Followed by:3 members
### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Legendary Member
Posts: 2663
Joined: 14 Jan 2015
Location: Boston, MA
Thanked: 1153 times
Followed by:127 members
GMAT Score:770
by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not? | [
"1/10",
"1/5",
"3/10",
"2/5"
] | C | We have three units - Tom (T), Mary (M) & the Rest (R).
Total ways of making a community of 3 from 6 people = 6 x 5 x 4 = 120
Now ways the community can be formed with Tom in it and Mary NOT in it:
TRR
RTR
RRT
TRR posb = 1 (way of picking Tom) x 4 (4 people left excluding Tom and Mary) x 3 (3 people left after excluding Tom, Mary and the person selected before) = 12
RTR and RRT have the possibility as TRR. Therefore possibilities the community can be formed with Tom in it and Mary NOT in it = 12 + 12 + 12 = 36
Hence probability =36/120=3/10
Option C |
AQUA-RAT | AQUA-RAT-38971 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long? | [
"40",
"28",
"26",
"27"
] | A | Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 140 = 500 m
Required time = 500 * 2/25 = 40 sec
Answer: A |
AQUA-RAT | AQUA-RAT-38972 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
If 5% more is gained by selling an article for Rs. 360 than by selling it for Rs. 340, the cost of the article is? | [
"127",
"688",
"400",
"121"
] | C | Let C.P. be Rs. x.
Then, 5% of x = 360 - 340 = 20
x/20 = 20 => x = 400
Answer: C |
AQUA-RAT | AQUA-RAT-38973 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
Difference between the length &breadth of a rectangle is 15m. If its perimeter is 206 m, then its area is? | [
"1400 m^2",
"2500 m^2",
"2520 m^2",
"1800 m^2"
] | D | Solving the two equations, we get: l = 45 and b = 40.
Area = (l x b) = (45 x 40) m2 = 1800 m^2
D |
AQUA-RAT | AQUA-RAT-38974 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
By selling 8 pens for a rupee a woman loses 25%. How many for a rupee should he sell in order to gain 25%? | [
"2",
"14",
"3",
"5"
] | D | D
75% --- 8
125% --- ?
75/125 * 8 = 5 |
AQUA-RAT | AQUA-RAT-38975 | for full-screen mode. TRIG WHEEL 1. A car wheel with a diameter of 18 inches spins at the rate of 10 revolutions per second. Interested in seeing how well you grasp a specific area of trigonometry? Take Study. 11 mm into kilometers and divided. They are going to start the chapter on graphing trigonometric functions on Monday, so this is an introductory activity to help students develop some of the vocabulary and background knowledge that. Trigonometry problems dealing with the height of two people on a ferris wheen. All we ask is that you link back to this site. For angles greater than 2π or less than −2π, simply continue to rotate around the circle, just like the ferris wheel. To the nearest inch per minute, what is the linear velocity of a point on the rim? asked by helo on July 3, 2013; trigonometry. Trig TX56 Stack NAV/COM Trig's TX56 and TX57 Nav/Com units provide the ideal platform to update legacy avionics or equip your new aircraft. Before beginning this activity, students should have been introduced to sine and cosine. WHEEL TRIG FUNCTIONS. Applications of trigonometry to waves Two-dimensional motion Suppose that a wheel of radius R is rotating anticlockwise as shown in Figure 38. This means that f(45) = 4(1) and 200. You will see that the values for sin and cos all lie on a circle of diameter ONE UNIT, centre (0. The field emerged during the 3rd century BC, from applications of geometry to astronomical studies. What is the linear speed of a point on its rim in feet per minute? Solution: Equation: 2. Chapter 13 : Trigonometric Ratios and Functions Bicycle Gears. Book title: Algebra and Trigonometry Publication date: Feb 13, 2015 Location: Houston, Texas Book. Trigonometry simply means calculations with triangles (that's where the tri comes from). Ferris Wheel Trig Problem. Trigonometry. including basic Ferris Wheel problems considering a person's position over time. The 0° angle is to the right in the "X" axis and aligned with the center of the bolt circle in the "Y" axis. It's really hard to find models and contexts for Unit Circle Trigonometry. The Trigonometry of Circles - Cool Math has free online cool
The following is multiple choice question (with options) to answer.
If the wheel is 10 cm then the number of revolutions to cover a distance of 1056 cm is? | [
"16.8",
"26",
"14",
"12"
] | A | 2 * 22/7 * 10 * x = 1056 => x
= 16.8
Answer: A |
AQUA-RAT | AQUA-RAT-38976 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
If the price of rice rises from Rs. 6 per kg to Rs. 8 per kg, to have no increase in his expenditure on rice, a person will have to reduce his consumption of sugar by | [
"25",
"62",
"27",
"19"
] | A | Explanation:
Let original consumption = 100 kg and consumption = y kg.
So, 100 x 6 = y x 8
y = 75 kg
Reduction in consumption = 25 %.
ANSWER: A |
AQUA-RAT | AQUA-RAT-38977 | I have to design a laboratory activity to answer the question, "What is the relationship between the diameter of a circle and the area of the circle?" My teacher said to use a previous activity as a templete to help me. For the
6. ### Math
Two perpendicular diamters(cutting each other at right angles) of a circle cut the circumference at four(4) points. A square is formed by joining the 4 points. If the circumference of the circle is 132cm, find the area of the
7. ### Algebra
The Circumference and area of a circle of radius r are givin by 2 [pie] r and [pie] r[2], respectively use 3.14 for the constant [pie] A. What is the circumference of a circle with a radius of 2 m? B.What is the area of a circle
8. ### Geometry
A circle with a radius of 1/2 ft is dilated by a scale factor of 8. Which statements about the new circle are true? Check all that apply. A.The length of the new radius will be 4 feet. B.The length of the new radius will be 32
9. ### Math
23.What is the approximate circumference of a circle with a radius of 6 centimeters? A.12 B.18 C.24 D.36 D 24.What is the length of the diameter if the radius is 15? A.5 B.15 C.30 D.45 C 27.What is the circumference of the circle
10. ### math
suppose you copy the circle using a size factor of 150%.what will bet he radius,diameter,circumference,and the area of the image? radius will be 1.5 times as big diameter will be 1.5 times as big but the formula for area will be
11. ### area of circle
My circle has a radius of 8ft. What is the distance of the circle? What "distance" are you talking about? Diameter? Circumference? What is the area of a circle with the radius of 32 cm????
More Similar Questions
The following is multiple choice question (with options) to answer.
If the area of circle is 616 sq cm then its circumference? | [
"22",
"88",
"99",
"66"
] | B | 22/7 r2 = 616 => r = 14
2 * 22/7 * 14 = 88.Answer:B |
AQUA-RAT | AQUA-RAT-38978 | that is y=???abc
#### Bacterius
##### Well-known member
MHB Math Helper
Re: find the value of x and y
$$x^2 \equiv x \pmod{1000} \tag{1}$$
[JUSTIFY]Clearly for $x$ coprime to $1000$, the only solutions are the trivial $x = 0$ and $x = 1$, so all nontrivial solutions for $x$ must be divisible by either $2$ or $5$, or both. This leaves us with about $600$ possible candidates. Now it should be clear that if a solution works for $1000$, it will also work for $100$, and for $10$. Let's try and find the solutions for $10$:[/JUSTIFY]
$$x^2 \equiv x \pmod{10} \tag{2}$$
[JUSTIFY]We have $5$ potential solutions here, namely $2, 4, 5, 6, 8$ (modulo $10$). Let's check each of them manually, some calculations show that only $5$ and $6$ work. So the solutions must have either $5$ or $6$ as a last digit. Let's now extend our search to $100$, armed with this information (which reduces the set of possible solutions considerably):[/JUSTIFY]
$$x^2 \equiv x \pmod{100} \tag{3}$$
[JUSTIFY]We know from the previous step that the potential solutions are $5, 6, 15, 16, \cdots$. A short exhaustive search tells us that only $25$ and $76$ work. Finally, we have narrowed down the search enough to look for the real solutions:[/JUSTIFY]
$$x^2 \equiv x \pmod{1000} \tag{4}$$
The only possible solutions are $25, 76, 125, 176, \cdots$. Again, trying those 20 potential solutions shows that only $376$ and $625$ work. Therefore, the solutions are $x \in \{ 0, \, 1, \, 376, \, 625 \}$. QED.
The following is multiple choice question (with options) to answer.
If x+y=25 and x^2y^3 + y^2x^3=25, what is the value of xy? | [
"0",
"1",
"2",
"3"
] | B | x^2*y^3 + y^2*x^3=25
=> x^2*y^2*(y+x)=25
=> x^2*y^2*25=25 [given, x+y=25]
=> (xy)^2 = 1
=> xy = 1
ANSWER:B |
AQUA-RAT | AQUA-RAT-38979 | First we choose two values, there are 13 values (2 to A), so $$13\choose2$$.
Then we want to choose two cards of the first value out of four cards, $$4\choose 2$$
Again, we want to choose two cards of the second value out of four cards, $$4\choose 2$$
And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), $${52-8\choose1} = {44\choose1}$$
So we get: $${{{13\choose2}\times{4\choose2}\times{4\choose2}\times{44\choose1} }\over{52\choose2}} = {198\over4165} ≈ 0.0475$$
The following is multiple choice question (with options) to answer.
The last time Rahul bought Diwali cards, he found that the four types of cards that he liked were priced Rs.4.50, Rs.5.50, Rs.8.00 and Rs.10.00 each. As Rahul wanted 40 cards, he took five each of two kinds and fifteen each of the other two, putting down the exact number of 10 rupees notes on the counter payment. How many notes did Rahul give? | [
"25",
"20",
"24",
"32"
] | D | Explanation :
Let the four cards be a, b ,c and d. 'x' be the number of notes.
Then, according to the question,
=> 5a + 5b + 15c + 15d = 10*x.
=> a + b + 3(c+d) = 2 * x . -----------------------------(i)
Since, a + b = 4.5+5.5 =10.
And,
=> c + d = 8+10=18
Putting these values in (i), we get :-
=> 10+ 3 * 18 = 2 * x.
=> x =32 .
Answer : D |
AQUA-RAT | AQUA-RAT-38980 | evolution, population-dynamics
Title: How many humans have been in my lineage? Is it almost the same for every human currently living? If I were to count my father, my grandfather, my great-grandfather, and so on up till, say chimps, or the most common ancestor, or whatever that suits the more accurate answer, how many humans would there have been in my direct lineage?
And would it be almost the same for every human being currently living? A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is likely to be higher. Research has shown that the great apes (chimps, gorilla, orangutan) have generation times comparatble to humans, in the range of 18-29 years (Langergraber et al. 2012).
Using 7 million years and 20 years yields an estimated 350000 ancestral generations for each living human. A more conservative estimate, using an average generation time of 28, would result in 250000 generations. However, some have argued that the human-chimp split is closer to 13 million years old, which would mean that approximately 650000 generations have passed (using a generation time of 20 years).
The exact number of ancestral generations for each human will naturally differ a bit, and some populations might have higher or lower numbers on average due to chance events or historical reasons (colonizations patterns etc). However, due to the law of large numbers my guess would be that discrepancies are likely to have averaged out. In any case, the current estimates of the human-chimp split and average historical generation times are so uncertain, so that they will swamp any other effects when trying to calculate the number of ancestoral generations.
However, this is only answering the number of ancestral generations. The number of ancestors in your full pedigree is something completely different. Since every ancestor has 2 parents, the number of ancestors will grow exponentially. Theoretically, the full pedigree of ancestors can be calculated using:
The following is multiple choice question (with options) to answer.
A family consists of grandparents, parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family? | [
"38 5/7 years",
"31 5/7 years",
"81 5/7 years",
"31 9/7 years"
] | B | Required Average
= [(67 * 2) + (35 * 2 ) + (6 * 3)]/(2 + 2 + 3)
= (134 + 70 + 18)/7 = 31 5/7 years.
Answer:B |
AQUA-RAT | AQUA-RAT-38981 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
A magician has five animals in his magic hat: 3 doves and 2 rabbits. If he pulls two animals out of the hat at random, what is the chance V that he will have a matched pair? | [
"V=2/5",
"V=3/5",
"1/5",
"1/2"
] | A | P(both doves) + P(both rabbits) = p(matched pair)
(3/5)*(2/4) + (2/5)*(1/4) = 2/5
Hi How did you get the probability 2/4 and 1/4???
You have 3 doves and 2 rabbits i.e. a total of 5 animals.
The probability that you pick a dove on your first pick is 3/5 (since there are 3 doves)
The probability that you pick a dove on your second pick too is 2/4 (because now only 2 doves are left after we picked a dove in the first pick. ALso only 4 animals are left to choose from)
Similarly,
The probability that you pick a rabbit on your first pick is 2/5 (since there are 2 rabbits)
The probability that you pick a rabbit on your second pick too is 1/4 (because now only 1 rabbit is left after we picked a rabbit in the first pick. Also only 4 animals are left to choose from)
Probability of picking a matched pair = 3/5 * 2/4 + 2/5 * 1/4 = 2/5 |
AQUA-RAT | AQUA-RAT-38982 | So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.
To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39
The following is multiple choice question (with options) to answer.
Charlene's annual bonus is calculated in the following manner: Charlene receives 20 percent of her first $11,000 in sales and 32 percent of her sales in excess of $11,000. If Charlene received a bonus of $7,300, what were her annual sales? | [
"$ 18,687",
"$ 22,000",
"$ 24,000",
"$ 26,000"
] | A | Let x be the total sales in $.
For the 1st 11,000 you get 20% ---> 0.2*11000 = 2200$
For the remaining (x-11000) $, you get 32% ---> 0.32*(x-11000)
Thus, 0.32*(x-11000) + 2200 = 7300 ----> x= 18,687 $
A is thus the correct answer. |
AQUA-RAT | AQUA-RAT-38983 | Kudos [?]: 132821 [4], given: 12378
Re: A small, rectangular park has a perimeter of 560 feet and a diagonal m [#permalink] 13 Nov 2014, 13:09
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
If the perimeter of a rectangular garden is 600 m, its length when its breadth is 200 m is? | [
"299 m",
"777 m",
"200 m",
"100 m"
] | D | 2(l + 200) = 600 => l
= 100 m
Answer:D |
AQUA-RAT | AQUA-RAT-38984 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A 140 m long train is running at a speed of 55 Km/hr. It crossed a platform of length 520 m in ? | [
"41.1 sec",
"20.2 sec",
"31.8 sec",
"43.2 sec"
] | D | Speed = 55 Km/hr
(to convert km/hr in to M/s)
= 55 x 5/18 M/s
Distance = 140 m + 520 m ( If questions is about train crossing a post you need to consider only the length of Train, )
= 660 m
Time = Distance / Speed
= 660 x 18 / (5 x 55)
= 43.2 sec
Ans is :D |
AQUA-RAT | AQUA-RAT-38985 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
If S is the sum of the reciprocals of the consecutive integers from 71 to 80, inclusive, which of the following is less than S?
I. 1/7
II. 1/8
III. 1/9
IV. 1/7>S >1/8 | [
"I",
"IV",
"II",
"III"
] | B | Since we summarize the reciprocals from 80 to 71, we can say also that we add ten numbers who are all (with one exception 1/80) greater than 1/80, so that the sum must be greater than 1/8.
On the other side we can say that we add the reciprocals from 71 to 80, so that the sum has to be less than the sum of ten times 1/81.
sum has to be greater than 1/8 and less than 1/7
B |
AQUA-RAT | AQUA-RAT-38986 | Related Rate Prob - Ships sailing in different directions
• November 5th 2010, 09:15 PM
dbakeg00
Related Rate Prob - Ships sailing in different directions
Ship A is 15 miles east of point O and moving west at 20mi/h. Ship B is 60 miles south of O and moving north at 15mi/h. a)Are they approaching or seperating after 1 hour and at what rate? b)after 3hrs?
Let D=distance between the ships at time t
$D^2=(60-15t)^2+(15-20t)^2$
$2D*\frac{dx}{dt}=(2)(-15)(60-15t)+(2)(-20)(15-20t)$
$2D*\frac{dx}{dt}=(-30)(60-15t)+(-40)(15-20t)$
$2D*\frac{dx}{dt}= 1250t-2400$
$\frac{dx}{dt}=\frac{1250t-2400}{2D}$
$D=\sqrt{5^2+45^2}=5\sqrt{82}$
$\frac{dx}{dt}=\frac{1250t-2400}{10\sqrt{82}}$
a) $\frac{dx}{dt}=\frac{1250*1-2400}{10\sqrt{82}}=approaching\ at \frac{-115}{\sqrt{82}}$ mi/h
b) $\frac{dx}{dt}=\frac{1250*3-2400}{10\sqrt{82}}=seperating\ at \frac{135}{\sqrt{82}}$ mi/h
The book agrees with me on part A, but on part B it says the answer should be $seperating\ at\ \frac{9\sqrt{10}}{2}$ mi/h. Am I missing something obvious or is the book wrong here?
• November 6th 2010, 02:24 AM
Ackbeet
The following is multiple choice question (with options) to answer.
A ship sails 4 degrees north, then 13 S. then 17 N. then 19 S. and has finally 11 degrees of south latitude. What was her latitude at starting ? | [
"STARTED WITH EQUATOR",
"STARTER WITH NORTH POLE",
"STARTER WITH SOUTH POLE",
"STARTER WITH 80 DEGREE NORTH POLE"
] | A | Let x = the latitude sought.
Then marking the northings +, and the southings -;
By the statement x + 4 - 13 + 17 - 19 = -11
And x = 0. |
AQUA-RAT | AQUA-RAT-38987 | # How many numbers made out of 5 distinct digits contain 4 and are even
This is an exercise from a book on combinatorics and I can't seem to wrap my head around it: How many numbers are there, made out of 5 distinct digits, which contain '4' and are even. The answer according to the book is 7686.
I distinguished some cases (not sure if I distinguished in a 'smart' way)
• the last digit is 4. In that case, there are 8 possibilities for the first digit (not 0, not 4). The second digit has 8 possibilities (not the first, not 4), the third has 7, the fourth has 6. The total amount is $$8\cdot 8\cdot 7\cdot 6 = 2688$$.
• the last digit isn't 4. Therefore, the last digit must be 0,2,6 or 8. We must pick 3 digits from the remaining 8 digits. This can be done in $$8 \cdot 7 = 56$$ ways. These 3 digits along with '4' have to be distributed over the first 4 places, giving a total of $$56 \cdot 4 \cdot 4! = 5376$$ ways. However, these include some invalid numbers: those starting with $$0$$. These have to be substracted. The first digit is fixed, the last digit can be $$2,6,8$$, so 3 possibilities. We have to pick 2 digits from the remaining 7 digits. This can be done in $$7 \cdot 6/2 = 21$$ ways. We then need to distribute these two digits and '4' over 3 spaces, so $$3!$$ possibilities, giving a total of $$3 \cdot 21 \cdot 3! = 378$$ invalid numbers.
The total therefore equals $$2688 + 5376 - 378 = 7686$$ ways.
This seems like a brute force solution. Anyone who can 'smoothen' this, for example by making a smarter choice of distinguished cases?
I would break the problem into three main cases instead of two. There are $$5$$ positions for the $$4$$: the first position, the middle three, or the last position.
The following is multiple choice question (with options) to answer.
The product of the digits of the four-digit number E is 36. No two digits of are identical. How many different numbers are possible values of E ? | [
"6",
"12",
"24",
"36"
] | C | The main question here is how can you arrange 4 different digits E. We don't even need to find the number. from the question itself we know that there are 4 different digits and none of them is 0 as the product is non-zero.
Therefore no of ways to arrange 4 diff. digits is 4! which is 24
hence C |
AQUA-RAT | AQUA-RAT-38988 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, there by increasing his average by 6 runs. His new average is: | [
"48 runs",
"50 runs",
"70 runs",
"75 runs"
] | A | Let average for 10 innings be x. Then,
(10x + 108)/11 = x + 6
= 11x + 66 = 10x + 108
= x = 42.
New average = (x + 6) = 48 runs.
ANSWER A |
AQUA-RAT | AQUA-RAT-38989 | RonL
Quite right!
I have made a picture of how could these streets look like.
There is also second question.
How many there are areas which are edged with streets from every side?
I have counted from picture that there are 15, I must say that I can't think another way at the moment. :D
• Apr 24th 2006, 05:50 PM
ThePerfectHacker
If you have $n$ lines than at most you can have.
$\frac{n(n-1)}{2}$ intersection points.
Thus,
$\frac{n(n-1)}{2}=21$
Thus,
$n^2-n+42=0$
Thus, $n=-6,7$
Disregard the negative which yields $n=7$
Proof,
If you have $n$ lines than to maximize the number of intersetion points in a plane you need them to be non-parallel such as no two are parallel to each other* then they need to intersect in such a way that no two pass through the same point. Now for the first line you have $n-1$ intersection points because it intersects each line EXCEPT for itself thus there are a total of $n-1$. While the second line interesects at $n-2$ because it cannot intersect with itself and not with the first because two lines intersect in only one point and it already intersected with the first line, while with the other lines it intersects with other points other than the first. The third one intersects with $n-3$ and so on. Thus you need to find the sum:
$(n-1)+(n-2)+(n-3)+...+2+1$
Which is $\frac{n(n-1)}{2}$ by the fact of the sum of an arithmetic series.
*)I did not find a rigorous demonstartion of this argument but if you think about it it is obvious. By that I of course mean the construction is possible.
• Apr 24th 2006, 05:57 PM
ThePerfectHacker
I once tried to develope a math riddle based on what you said. Never got to solving it.
-----------
The first question of the riddle I answered, i.e. the max number of distinct points.
The following is multiple choice question (with options) to answer.
In a city where all streets run east-to-west, all avenues run north-to-south, and all intersections are right angles as shown below, Jessica needs to walk from the corner of 6th Street and 3rd Avenue to the corner of 1st Street and 1st Avenue. If Jessica will randomly choose from any route that allows her to walk the fewest number of blocks, what is the probability that she walks exactly two blocks on 1st Avenue? | [
"1/21",
"1/7",
"4/21",
"10/21"
] | C | She choose a route which allows her to walk fewest number of blocks
there are 2 blocks she will cross towards east moves represented by E,E and there will be 5 blocks she will cross towards south represented by S,S,S,S,S
total number of ways 7!5!2!7!5!2!
there are 4 ways she can cross 2 blocks on av.1 as shown in attached image.
4/21 is Answer C . |
AQUA-RAT | AQUA-RAT-38990 | # Math Help - counting
1. ## counting
arrange 12 people into 4 teams of 3 people
would u choose the team first then the players: 4C1 x 12C3 first team
3C1 x 9C3 second team and carry on for third and fourth then multiply all of ur results together?
2. Originally Posted by qwerty10
arrange 12 people into 4 teams of 3 people
This is known in the trade as an unordered partition.
Suppose that we have N distinct objects and $N=j\cdot k$ then we can group them into k cells with j in each cell.
The number of ways is $\frac{N!}{(k!)^j(j!)}$
3. But how would u then extend that to allow for the restriction that 2 particular people are not allowed in the same team?
4. Originally Posted by qwerty10
But how would u then extend that to allow for the restriction that 2 particular people are not allowed in the same team?
Why did you change this question after four days?
For this new question, the answer is
$\binom{10}{2}\binom{8}{2}\frac{6!}{(3!)^2(2!)}$
5. Hello, qwerty10!
Plato is correct . . . Here is an explanation.
Arrange 12 people into 4 teams of 3 people.
Suppose we label the teams A, B, C and D.
Choose 3 of the 12 people for team A.
. . There are: . $_{12}C_3$ ways.
Choose 3 of the remaining 9 people for team B.
. . There are: . $_9C_3$ ways.
Choose 3 of the remaining 6 people for team C.
. . There are: . $_6C_3$ ways.
Choose 3 of the remaining 3 people for team D.
. . There are: . $_3C_3$ ways.
To arrange 12 people into 4 distinguishable teams of 3 players each,
. . there are: . $\left(_{12}C_3\right)\left(_9C_3\right)\left(_6C_3 \right)\left(_3C_3\right)$
The following is multiple choice question (with options) to answer.
At lovely gym class can be divided into 8 teams with an equal number of players on each team or into 12 teams with an equal number of players on each team. What is the lowest possible number of students in the class? | [
" 20",
" 24",
" 36",
" 48"
] | B | We are given that at lovely gym class can be divided into 8 teams or 12 teams, with an equal number of players on each team. Translating this into two mathematical expressions we can say, where G is the total number of students in the gym class, that:
G/8 = integer and G/12 = integer
This means that G is a multiple of both 8 and 12.
We are asked to determine the lowest number of students in the class, or the lowest value for variable “G”. Because we know that G is a multiple of 8 and of 12, we need to find the least common multiple of 8 and 12. Although there are technical ways for determining the least common multiple, the easiest method is to analyze the multiples of 8 and 12 until we find one in common.
Starting with 8, we have: 8, 16, 24, 32
For 12, we have: 12, 24
For the multiples of 12, we stopped at 24, because we see that 24 is also a multiple of 8. Thus, 24 is the least common multiple of 8 and 12, and therefore we know that the lowest possible number of students in the gym class is 24.
Answer B. |
AQUA-RAT | AQUA-RAT-38991 | Show Tags
18 Jun 2014, 06:09
Thanks much Bunuel
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
Show Tags
18 Jun 2014, 20:26
1
maggie27 wrote:
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.
Also note that area of inscribed square is always half than that of the original square
As Bunuel pointed out, if it goes less than 50, it means some of the vertex is not touching side of the original square.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
Show Tags
01 Jul 2015, 18:04
If $$x^{2}$$ is area of square, then find x, one side of the square. If square is inscribed, then diagonal is the length of larger square and therefore the diagonal is $$10$$. To determine the side, the formula also includes the area of the square, $$x^{2}$$. So, if $$2x^{2} = 100$$ then $$x^{2}=50$$
D.
Thanks,
A
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
Show Tags
31 Jul 2016, 20:43
Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible value of the area of square T ?
A. 45
B. 48
C. 49
D. 50
E. 52
The following is multiple choice question (with options) to answer.
What is the perimeter of a square with area 9p^2/25 ? | [
"3p/4",
"3p^2/4",
"12p/5",
"3p^2"
] | C | Area of square, (side)^2 = (3p/5)^2
Therefore side of the square = 3p/5
Perimeter of square = 4*side = 4* (3p/5) = 12p/5
Answer is C. |
AQUA-RAT | AQUA-RAT-38992 | # Remainder on division with $22$
What is the remainder obtained when $$14^{16}$$ is divided with $$22$$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $$22$$, except the remainder?
How should I proceed?
• $14^{16} = (22 - 8)^{16}$ – ab123 Jul 3 '18 at 10:37
• $8^{16} = 64^8 = (66 - 2)^8$ – ab123 Jul 3 '18 at 10:41
• @ab123 Why not make an answer of that? – Arthur Jul 3 '18 at 10:41
You can use binomial expansions and see that $$14^{16} = (22 - 8)^{16}$$ implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$ Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 \implies 2^8 = 256 \text{ divided by } 22 \implies \text{remainder = 14}$
Since $$14^2 \equiv -2$$ so $$14^{16} \equiv (-2)^8 \equiv 16^2\equiv (-6)^2 \equiv 14$$
or
$$14^2 = 22k -2$$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22\cdot 11+14$$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7\cdot14^{15}$ by $11$.
The following is multiple choice question (with options) to answer.
A certain no. when divided by 35leaves aremainder 25, what is the remainder if the same no.be divided by 15? | [
"1",
"3",
"4",
"6"
] | C | Explanation:
35 + 25 = 60/15 = 4 (Remainder)
C |
AQUA-RAT | AQUA-RAT-38993 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
Ritesh and Co. generated revenue of Rs. 2,500 in 2006. This was 12.5% of its gross revenue. In 2007, the gross revenue grew by Rs. 2,500. What is the percentage increase in the revenue in 2007? | [
"12.5%",
"20%",
"25%",
"50%"
] | A | Explanation :
Given, Ritesh and Co. generated revenue of Rs. 2,500 in 2006 and that this was 12.5% of the gross revenue.
Hence, if 2500 is 12.5% of the revenue, then 100% (gross revenue) is:
=>(100/12.5)×2500.
=>20,000.
Hence, the total revenue by end of 2007 is Rs. 10,000. In 2006, revenue grew by Rs. 2500. This is a growth of:
=>(2500/20000)×100.
=>12.5%.
Answer : A |
AQUA-RAT | AQUA-RAT-38994 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
Rose is two years older than Bruce who is twice as old as Chris. If the total of the age of Rose, B and Chris be 37 years, then how old is Chris ? | [
"7 years",
"10 years",
"12 years",
"13 years"
] | A | Let Chris's age be x years. Then, Bruce's age = 2x years.Rose's age = (2x + 2) years.
(2x + 2) + 2x + x = 37
5x = 35
x = 7
Hence, Chris's age =7 years.
Answer : A |
AQUA-RAT | AQUA-RAT-38995 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
A clock shows the time as 12 a.m. If the minute hand gains 5 minutes every hour, how many minutes will the clock gain by 6 p.m.? | [
"45 minutes",
"55 minutes",
"35 minutes",
"30 minutes"
] | D | there are 6 hours in between 12 a.m. to 6 p.m.
6*5=30 minutes.
ANSWER:D |
AQUA-RAT | AQUA-RAT-38996 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the effect on its area? | [
"1288",
"1299",
"1000",
"10000"
] | D | 100 * 100 = 10000
125 * 80 = 10000
Answer: D |
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