source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37497 | # Difference between revisions of "2021 AMC 10B Problems/Problem 3"
## Problem
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$
## Solution 1
Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations $$5j=2s$$$$j+s=28,$$ and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is $$\boxed{(C) \text{ } 8}.$$
## Solution 2 (Fast but Not Rigorous)
We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$. Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{C}$. ~samrocksnature
## Solution 3
The following is multiple choice question (with options) to answer.
Of the students at a certain high school, 90 percent took a course in algebra or geometry and 30 percent took courses in both. If the percent of students who took a course in algebra was 2 times the percent of students who took a course in geometry, what percent of the students took a course in algebra but not geometry? | [
"50%",
"40%",
"45%",
"35%"
] | A | You can also just use the formula
n(A or B) = n(A) + n(B) - n(A and B)
90 = 2X + X - 30
X = 40
2X = 80
This is the number of people who took algebra including those who took both. To get the number of people who took only algebra, 80 - 30 = 50%
Answer (A) |
AQUA-RAT | AQUA-RAT-37498 | 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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Kudos [?]: 79 [1] , given: 10
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.
David obtained 72, 60, 35, 62 and 84 marks (out of 100) in English, Mathematics,Physics, Chemistry and Biology What are his average marks ? | [
"75",
"62.6",
"87",
"165"
] | B | Explanation:
Average = (72 +60+ 35+ 62+84 )/ 5 = 313/5 = 62.6.
Answer:B |
AQUA-RAT | AQUA-RAT-37499 | 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.
If a = (3)b and c = 5b, then which of the following represents the average (arithmetic mean) of a, b, and c, in terms of b ? | [
"b/3",
"b/4",
"2b",
"3b"
] | D | a = (3)b and c =5b.
The mean = (a + b + c)/3 = (3b + b + 5b)/3 = 3b.
Answer is D |
AQUA-RAT | AQUA-RAT-37500 | # 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 umbrella for Rs. 300, a shop keeper gains 20%. During a clearance sale, the shopkeeper allows a discount of 10% on the marked price. His gain percent during the sale is? | [
"5",
"6",
"8",
"9"
] | C | Marked price = Rs. 300
C.P. = 100/120 * 300 = Rs. 250
Sale price = 90% of Rs. 300 = Rs. 270
Required gain % = 20/250 * 100 = 8%.
Answer:C |
AQUA-RAT | AQUA-RAT-37501 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
In a workshop there are 3 kinds of beds, 3 kinds of closets, 2 kinds of shelves and 7 kinds of chairs. In how many ways can a person decorate his room if he wants to buy in the workshop one shelf, one bed and one of the following: a chair or a closet? | [
"168.",
"80.",
"60.",
"48."
] | C | ways to select one shelf = 2
ways to select one bed = 3
ways to select one of the following: a chair or a closet = 7+3 = 10
ways to select all of them together = 2*3*10 = 60
Hence, C is the answer. |
AQUA-RAT | AQUA-RAT-37502 | $\displaystyle \frac{3}{4} \times 0.04 = 0.03$
4. Outch ! Sorry, I'll think again about it !
5. Ok, so $\displaystyle P(A/R)=\frac{P(A \cap R)}{P\color{red}(R)}$
Plus $\displaystyle P(R/A)=\frac{P(A \cap R)}{P(A)} \Longleftrightarrow P(A \cap R)=P(R/A)P(A)$
Hence : $\displaystyle {\color{blue} P(A/R)=\frac{P(R/A)P(A)}{P(R)}}$
I misread and $\displaystyle {\color{blue} P(R/A)=.04}$, not $\displaystyle P(R \cap A)$
Let's calculate P(R).
The following is multiple choice question (with options) to answer.
The simple form of the ratio 4/3:3/7 is? | [
"7:3",
"10:3",
"15:3",
"25:3"
] | A | 4/3 : 3/7 = 21:9 = 7:3
ANSWER:A |
AQUA-RAT | AQUA-RAT-37503 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
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A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
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Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
A vessel of capacity 90 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution? | [
"72.9",
"78.5",
"25.4",
"31.5"
] | A | Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T
For the given problem, T = 90, y = 9 and n = 2.
Hence, quantity of milk finally in the vessel
= [(90 - 9)/90]2 (90) = 72.9 litres.
ANSWER A |
AQUA-RAT | AQUA-RAT-37504 | history
Title: About Year Zero It is said that, in part due to a poor grasp of mathematics (the number zero to be precise) our forebears, when they developed the calendar went from 1 BC/BCE to 1 AD/CE. Year Zero is missing - it would've been, for Christians, the year Jesus Christ was born.
How do astronomers, the go-to-people when it comes to developing/maintaining/improving accurate calendars, have to say about this? If year 1 AD/CE is actually the year 0 AD/CE, does that mean this is 2021 AD and not 2022 AD?
I did a little reading on Wikipedia (it has a page on Year Zero) and it seems that astronomers have decided to treat year 1 BC/BCE as year 0 and if that's the case, this year, which is about to end in 5 days, is the year 2022. Astronomers don't use the BC/BCE or the AD/CE labels, instead preferring to use the mathematical signs - and + for them, respectively. This is of course mathematically very sound, but the problem is 1 BC (year zero) is an unremarkable year - nothing of global significance occurred in that year. Of note is the fact that there's no real consensus on the Jesus' birth year - the same source, a Dionysius Exiguus, gives conflicting dates (1 BC and 1 AD).
Is 1 BC (Year Zero) of astronomical significance - did something special (astronomically) happen that year? There is no special significance of year 0=1BC. It appears to be the result of a rough calculation done by Dionysius Exiguus.
This year is actually 2022 CE. We don't know exactly when Jesus was born (it was probably a few years before year 0. But for dates, that is irrelevant. We use the count of years as determined by Dionysius. End of.
Acually astronomers tend not to use calendar dates in scientific work anyway.
Astronomers tend to use "Julian Day". This is a simple way of measuring time. Named for Julius Scaliger (nothing to do with Julius Caesar, or the Julian Calendar). It simply counts the number of days since a remote point in the the past. It makes it possible to specify the time and date of nearly all astronomical events by simply giving a single number.
The following is multiple choice question (with options) to answer.
The calendar for the year 2001 is same for which of the following year? | [
"2005",
"2007",
"2011",
"2006"
] | B | SOLUTION :
The total number of odd days from 2001 onwards should be zero.
Now, as an ordinary year adds 1 odd day and a leap year adds 2 odd days, we have:
2001, 2002, 2003, 2005, 2006 – 1 odd day each
2004 – 2 odd days
Hence, at the end of 2006 total number of odd days = 7 or 0
Therefore, the calendar for the year 2001 is repeated in the year 2007.
ANSWER IS B |
AQUA-RAT | AQUA-RAT-37505 | The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.
As Martin Schulz pointed out in the comments (now deleted),
9165784320 is divisible by 7
• I have 1385679240 & 7165932840 neither divisible by 7 – DeNel Oct 27 '18 at 0:10
• My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that. – DeNel Oct 27 '18 at 0:11
• "The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4. – elias Oct 27 '18 at 2:08
• the correct wording would be the last 2 digits must be divisible by 4 and the last 3 digits must be divisible by 8. In fact we just need to check for divisibility by 8, since that implies divisibility by 4 as well – phuclv Oct 27 '18 at 6:24
This is a good problem to attack by computer:
It turns out that there are 7344 solutions, including 1056 to the bonus.
So I thought, how far can we go?
And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.
Code used (Python 3 IDLE):
The following is multiple choice question (with options) to answer.
A seven-digit phone number is divisible by 4. After adding the two digits of the area code the number is not divisible by 4, which of the following cannot be the area code? | [
"07.",
"06.",
"05.",
"04."
] | D | Given that 7 digit number is divisible by 4
we know any number will not be divisible by 4 if the sum of all digits of the number is not the multiple of 4.
Hence looking at the option we can say that the number that should be added to the seven digit number must be 04 as adding it will make it divisible by 4... hence answer is D |
AQUA-RAT | AQUA-RAT-37506 | Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
The following is multiple choice question (with options) to answer.
How many trucks are there if each truck carrying 70 packages and total of 490 packages? | [
"A)7",
"B)6",
"C)9",
"D)11"
] | A | Sol.
Total packages 490
Each truck carries 70 packages
= 490 / 70
= 7
Answer : A |
AQUA-RAT | AQUA-RAT-37507 | If the division "comes out even", then the remainder is zero.
So we have: . $k^2 - 8k + 15 \:=\:0 \quad\hdots\quad See?$
10. Thanks for the help! Will read on synthetic division further.
The following is multiple choice question (with options) to answer.
If R is a positive integer, and R/15 = 8.2, what is the remainder when R is divided by 15? | [
"A.1",
"B.2",
"C.3",
"D.4"
] | C | If R were a multiple of 15, the quotient R/15 would be an integer. The fact that it's a decimal tells us that 15 goes into R eight whole times and some decimal part of a time. This decimal part, 0.2 is the remainder divided by the divisor. Let B be the remainder.
B/15 = 0.2 = 1/5
B = (15)*(1/5) = 3
Answer =(C) |
AQUA-RAT | AQUA-RAT-37508 | # 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 shop contains 500 articles each of a cost of $10. He sold them at a price of $11. If the shop keeper has a order of 200 articles, how much profit he will get? | [
"A)2100",
"B)2000",
"C)1500",
"D)200"
] | D | Total money after selling articles = 200*11 = 2200
cost price = 200*10 = 2000
profit = 200
correct option is D |
AQUA-RAT | AQUA-RAT-37509 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
A man has some hens and cows. If the number of heads be 48 and the number of feet equals 144, then the number of hens will be: | [
"22",
"23",
"24",
"26"
] | C | let hens be x and cows be y
now,
feet: x*2+y*4=144
heads: x*1+y*1=48
implies, 2x+4y=144 and x+y=48
solving these two equations, we get x=24 and y=24
Therefore, hens are 26.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37510 | #### Subhotosh Khan
##### Super Moderator
Staff member
Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
Incorrect.
Two $$\displaystyle \infty$$s cannot be equated (that way algebraically)...
#### Subhotosh Khan
##### Super Moderator
Staff member
So if you see a Number like 546342 you know its not Prime since you can divide it by 2. Same with 4 and 8. Those numbers are, as far as i know, never Prime. So this is a way to see non Primes. But it dosnt help you do see if a Number is Prime. It only shows the ones who are not. And not even that many of them
That's the way to figure out whether a number is divisible by 2.
For 3 - add the digits of the number and if the sum is divisible by 3 - the original number is divisible by 3.
for 5 - the last digit of the number must 0 or 5.
There are many more divisibility rules.
#### lookagain
##### Senior Member
ers here.
Interessting way, but that would mean you have to know the Primes up to like 10.000 right ? Or at least the ones, that are near to 10.000 Squared.
No, in this case you would just need to know up to the prime number 31.
The square of the next prime, 37, puts it past 1,091.
So, you are checking divisibility by only 11 prime numbers.
(2, 3, ... , 31)
#### Cubist
##### Junior Member
If you want to find ALL the primes between 507 and 10647, using only a pen and paper, then using the "Sieve of Eratosthenes" might still be the best way? To draw a 100x100 grid you'd probably need a big piece of paper
But if you want to test a single given number within the above range then lookagain's method seems best to me. There are several other methods (algorithms) and if you are interested google "primality test". I don't understand how these methods work myself.
The following is multiple choice question (with options) to answer.
There r 4 prime numbers in ascending order. The multiplication of first 3 is 385 and that of last 3is 1001. what is the last number is | [
"10",
"12",
"11",
"13"
] | D | Explanation:
abcbcd=3851001=>ad=513
abcbcd=3851001=>ad=513
So d = 13
D |
AQUA-RAT | AQUA-RAT-37511 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
Which of the numbers below is the odd one out
1,3,7,13,24,21,31. | [
"3",
"7",
"21",
"24"
] | D | 24 is the odd man.. diff between sucessive no.s are multiples of 2
ANSWER:D |
AQUA-RAT | AQUA-RAT-37512 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.6 km? | [
"277",
"289",
"269",
"600"
] | D | Distance covered by Amar = 18/4.8 (1.6km)
= 3/8(1600)
= 600 m
Answer: D |
AQUA-RAT | AQUA-RAT-37513 | Thus,
200 x 0.3 = 60 received student loans
200 x 0.4 = 80 received scholarships
We are trying to determine what percent of those surveyed said that they had received neither student loans nor scholarships.
Let’s fill all this information into a table. Note that each row sums to create a row total, and each column sums to create a column total. These totals also sum to give us the grand total, designated by 200 at the bottom right of the table.
Statement One Alone:
25 percent of those surveyed said that they had received scholarships but no loans.
Using statement one we can determine the number of students who received scholarships but no loans.
200 x 0.25 = 50 students who received scholarships but no loans.
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.
Statement Two Alone:
We are given that 50 percent of those surveyed who said they had received loans also said that they had received scholarships. From the given information we know that 60 students received loans; thus, we can determine the number of these 60 students who also received scholarships.
60 x 0.5 = 30 students who received loans who also received scholarships
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement two is sufficient to answer the question.
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Re: In a survey of 200 college graduates, 30 percent said they [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
What will be 60 percent of a number whose 200 percent is 90 ? | [
"144",
"72",
"27",
"None of these"
] | C | Answer
∵ 200% of N = 90
⇒ N = (90 x 100) / 200 = 45
∴ 60% of N = (60 x 45) / 100 = 27
Correct Option: C |
AQUA-RAT | AQUA-RAT-37514 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
The following is multiple choice question (with options) to answer.
Find the smallest number in a GP whose sum is 38 and product 1728 | [
"12",
"20",
"8",
"All of these"
] | C | In GP :- a/r+a+ar....=38
a(1+r+r^2)=38r......eq1
Product a^3=1728
a=12
now put this value in eq1
12(1+r+r^2)=38r
r=2/3 and 3/2..so sallest number is 8
ANSWER:C |
AQUA-RAT | AQUA-RAT-37515 | 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
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
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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Posts: 6
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 30% 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"
] | A | 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 = 1/3.
Hence, A is the correct answer. |
AQUA-RAT | AQUA-RAT-37516 | # For dinner, $n$ people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so ...
Problem:
For dinner, $$n$$ ($$n \geq 4$$) people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so that Ana and Ivan do not sit next to each other and at least one of them sits next to Mark? (Note: the round table implies seating arrangements that differ only in rotation.)
My attempt:
If I have $$n$$ people sitting around circular table, the number of different arrangements are $$(n-1)!$$.
If I have $$2$$ people Mark and Ana number of arrangements that they can sit next to each other is $$2 \cdot (n-2)!$$. So the number of arrangements that Mark sit next to Ivan is also $$2 \cdot (n-2)!$$, and sitting next to Ana also $$2 \cdot (n-2)!$$.
• yes.....it is fixs
– josf
Nov 2 '19 at 14:49
• "... seating arrangements that differ only in rotation... " are what? Considered to be the same, or different? From the context, I assume they should be considered to be the same, but the language isn't clear. Nov 11 '19 at 16:28
Method 1: Seat Mark. We will use him as our reference point.
Only Ana sits next to Mark: She can be seated in two ways, to his left or to his right. That leaves $$n - 2$$ seats. Since Ivan cannot sit next to Ana or Mark, he may be seated in $$n - 4$$ ways. The remaining $$n - 3$$ people can be seated in the remaining $$n - 3$$ seats in $$(n - 3)!$$ ways as we proceed clockwise around the table relative to Mark. Hence, there are $$2(n - 4)(n - 3)!$$ such arrangements.
The following is multiple choice question (with options) to answer.
At a dinner party, 5 people are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. what is the total number E of different possible seating arrangements for the group? | [
"5",
"10",
"24",
"32"
] | C | Circular arrangements= (n-1)!
Hence (4!) = 24=E
Answer is C |
AQUA-RAT | AQUA-RAT-37517 | # A simple riddle related to addition of odd numbers
I'm not sure if this type of question can be asked here, but if it can then here goes:
Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.
PS : The inception of this question is a result of a random discussion that I was having during the break hour :)
-
Hint: adding an odd number of odd numbers will give ... (is this a real question ?) – Raymond Manzoni May 25 '13 at 8:27
Hmm..let me ponder over that – 403 Forbidden May 25 '13 at 8:29
This says sum of 9 odd numbers can't be even algebra.com/algebra/homework/word/numbers/… True? – 403 Forbidden May 25 '13 at 8:32
Yes: consider the last digit in binary if it helps. – Raymond Manzoni May 25 '13 at 8:36
A direct approach:
Any given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like $${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$ (the integers $a,b,\ldots,i$ may or may not be the same). Grouping things together, this is equal to $$2(a+b+c+d+e+f+g+h+i+4)+1.$$ Thus, the result is odd.
The following is multiple choice question (with options) to answer.
If a and b are both odd numbers, which of the following is an odd number? | [
"a + b",
"a + b + 2",
"ab+2",
"ab + 1"
] | C | Sol.
Product of two odd numbers is always odd and adding 2 gives odd.
Answer C |
AQUA-RAT | AQUA-RAT-37518 | You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.
Per statement 1:$$x=-y$$ . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: $$x^2=y^2$$ ---> $$x= \pm y$$. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
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17 Dec 2015, 06:54
statement 1 is sufficient as absolute value for -ve should give +ve
statement 2 is sufficient as
it will give either x and y both are +ve or both are -ve
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
### Show Tags
09 Mar 2018, 03:18
HarveyKlaus wrote:
Is |x| = |y|?
(1) x = -y
(2) x^2 = y^2
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong.
Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
Question:
The following is multiple choice question (with options) to answer.
If y is 70% greater than x, than x is what % less than y? | [
"18.7",
"25.5",
"33.3",
"41.2"
] | D | y = 1.7x
x = y/1.7 = 10y/17
x is 7/17 less which is about 41.2% less than y.
The answer is D. |
AQUA-RAT | AQUA-RAT-37519 | # Probability of A Specific Type of Experiment Occuring
The problem is:
An experimenter is studying the effects of temperature, pres-sure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration.
a.If any particular experimental run involves the use of a single temperature, pressure, and catalyst, how many experimental runs are possible?
b.How many experimental runs are there that involve use of the lowest temperature and two lowest pressures?
c. Suppose that five different experimental runs are to be made on the first day of experimentation. If the five are randomly selected from among all the possibilities, so that any group of five has the same probability of selection, what is the probability that a different catalyst is used on each run?
I am stuck on part c. I know that the total number of experiments is, $3 \cdot 4 \cdot 5=60$; and so, the total number of ways of choosing five of the sixty experiments would be ${{60}\choose{5}}$. What I am having trouble with is finding the total number of sequence of five experiments where a catalyst is not repeated in each sequence. I thought, for the first experiment, there would be $5 \cdot 4 \cdot 3$, and for the second experiment, $4 \cdot 4 \cdot 3$, and for the third, $3 \cdot 4 \cdot 3$, and so on. However, this isn't correct. What am I doing wrong?
-
Hmm, people just seemed to suddenly stop viewing this thread. Perhaps word got out that this is too boring of a problem. – Mack Feb 3 '13 at 18:32
The following is multiple choice question (with options) to answer.
31 of the scientists that attended a certain workshop were Wolf Prize laureates, and 14 of these 31 were also Nobel Prize laureates. Of the scientists that attended that workshop and had not received the Wolf prize, the number of scientists that had received the Nobel Prize was 3 greater than the number of scientists that had not received the Nobel Prize. If 50 of the scientists attended that workshop, how many of them were Nobel Prize laureates? | [
"A)11",
"B)18",
"C)25",
"D)29"
] | C | lets solve by creating equation..
W = 31..
Total = 50..
Not W = 50 -31= 19..
Now let people who were neither be x, so out of 19 who won Nobel = x+3..
so x+x+3=19 or x=8..
so who won Nobel but not Wolf = x+3 = 11..
But people who won BOTH W and N = 13..
so total who won N = 11+14 = 25..
C |
AQUA-RAT | AQUA-RAT-37520 | # 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.
The difference between the simple interest received from two different sources on Rs. 1500 for 3 years is Rs. 13.50. The difference between their rates of interest is? | [
"0.9%",
"0.4%",
"0.3%",
"0.2%"
] | C | (1500 * R1 * 3)/100 - (1500 * R2 * 3)/100 = 13.50 4500(R1 - R2) = 1350
R1 - R2 = 0.3%
Answer: C |
AQUA-RAT | AQUA-RAT-37521 | ### 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, B and C are partners. A receives 2/3 of profits, B and C dividing the remainder equally. A's income is increased by Rs.200 when the rate to profit rises from 5 to 7 percent. Find the Capital of B? | [
"3999",
"7799",
"2500",
"2772"
] | C | A:B:C = 2/3:1/6:1/6 = 4:1:1
x * 2/100 * 2/3 = 200
B capital = 15000*1/6 = 2500.Answer: C |
AQUA-RAT | AQUA-RAT-37522 | Note that $-169 + ( 4 \times 52 ) = 39$, so $-169 \equiv 39 \mod 52$.
Try this yourself for the second question.
• how did you get the four? – Nicole Oct 13 '14 at 23:36
• @Nicole Four is the number of times you need to add 52 to -169 to get a positive number. Note that adding any lesser multiple of 52 will leave you with a negative number. – Nick Oct 13 '14 at 23:38
The following is multiple choice question (with options) to answer.
What is the sum of the multiples of 4 between 19 and 130 inclusive? | [
"2012",
"2044",
"2060",
"2072"
] | D | the fastest way in an AP is to find the average and multiply with total integers..
between 19 and 130, the smallest multiple of 4 is 20 and largest =128..
average=(20+128)/2=148/2=74.
total numbers = (128-20)/4 + 1= 27+1=28..
sum=28*74=2072
ans D |
AQUA-RAT | AQUA-RAT-37523 | 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.
The average of first five prime numbers greater than 3 is? | [
"32.2",
"10.6",
"32.3",
"32.8"
] | B | 5 + 7 + 11 + 13 + 17
= 53/5
= 10.60
Answer = B |
AQUA-RAT | AQUA-RAT-37524 | 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.
Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually | [
"Rs 312",
"Rs 412",
"Rs 512",
"Rs 612"
] | D | Explanation:
Please apply the formula
Amount=P(1+R100)nC.I. = Amount - P
Answer: D |
AQUA-RAT | AQUA-RAT-37525 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
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A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
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01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
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Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter E of liquid A was contained by the can initially? | [
"10",
"20",
"21",
"25"
] | C | As A:B::7:5 ---> only option C is a multiple of 7 and hence it is a good place to start. Also A:B::7:5 means that , A = (712)*Total and B = (5/12)*Total
If A = 21 , B = 15 ---> remove 9 litres ---> you remove (7/12)*9 of A ---> A remaining = 21-(7/12)*9 = 63/4
Similarly, for B, you remove (5/12)*9 ---> B remaining = 15 - (5/12)*9 = 45/4 and then add 9 more litres of B ---> 9+45/4 = 81/4
Thus A/B (final ratio) = (45/4)/(81/4) = 7:9 , the same as the final ratio mentioned in the question.
Hence C is the correct answer.
A/B = 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9), where 7x and 5x are initial quantities of A and B respectively.
Thus, 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9) ---> giving you x=3. Thus A (original) E= 7*3 = 21.C |
AQUA-RAT | AQUA-RAT-37526 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
Kate is 12 years old. His age is 4 times the age of Robbie. Find the age of Robbie. | [
"1yrs",
"2yrs",
"3yrs",
"4yrs"
] | C | Let Kate age, x = 12
robbie = x/4
=12/4
=3yrs
ANSWER:C |
AQUA-RAT | AQUA-RAT-37527 | 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.
Five years ago, the average age of A, B, C and D was 45 years. With E Joining them now, the average of all the five is 51 years. The age of E is? | [
"45",
"47",
"55",
"49"
] | C | Solution
5 years ago average age of A,B,C,D = 45 years
=> 5 years ago total age of A, B, C, D = 45 x 4 = 180 years
=> Total present age of A, B, C, D = 180 + 5x4 = 200 years
If E's present age is x years = 200+x/5= 51
x=55 years. Answer C |
AQUA-RAT | AQUA-RAT-37528 | So we can split the divisors of $n$ except $\sqrt{n}$ into pairs, hence $n$ has an odd number of divisors.
• I know I'm being stupid, but I have to ask - doesn't 30 = 2 * 3 * 5 have an odd number of divisors? What am I missing? – TV's Frank Nov 17 '16 at 15:02
• @TV'sFrank: $30$ has $3$ distinct prime factors ($2,3,5$) and $8$ divisors ($1,2,3,5,6,10,15,30$). – barak manos Nov 17 '16 at 15:07
• Aha! Thanks! :) – TV's Frank Nov 17 '16 at 15:26
• How do you find out "Given an integer n, for every integer d which divides n, the integer n/d also divides n." ? It is obvious, but without this obvious start, concluding that squares are the only integers which have an odd number of divisors is not forcely straighforward – Edmondo1984 Aug 26 '18 at 20:33
If a locker number has an even number of factors, it will be alternatively opened and closed and even number of times, ending in the same configuration it started.
Square numbers have a odd number of factors.
The idea here is that it is easy to identify pairs of students who will open and close a locker (well, as easy as any interesting math problem ever is!). If any given student will open or close a locker, it means their number is a divisor of the locker's number.
Consider locker X. Let's assume there is a student Y who toggles the locker. This means we know X/Y is an integer. This also means that student X/Y will also toggle the locker.
Take locker 12, with factors 1, 2, 3, 4, 6, and 12. If I rearrange them into pairs, (1, 12) (2, 6) (3, 4) we can see that for every student who opens a locker, there will be one that closes the locker.
The following is multiple choice question (with options) to answer.
At a certain high school, a student’s popularity is determined by his or her locker number. Whoever has the locker number with the greatest number of distinct prime factors is the most popular student in the school. If Dhoni ,Kamal, Gokul, and Dinesh get lockers with the numbers 50, 400, 150, and 25, respectively, who is the most popular student? | [
"Dhoni",
"Kamal",
"Gokul",
"Dinesh"
] | C | Distinct prime factors of 50 (Dhoni) : 2,5
Distinct prime factors of 400 (Kamal): 2 and 5.
Distinct prime factors of 150 (Gokul): 2,3, and 5
Distinct prime factors of 25 (Dinesh): 5.
So most popular student - Gokul
Answer C |
AQUA-RAT | AQUA-RAT-37529 | |\
4| \
| \
|_____ \ <- theta
3
Now use the pythagorean theorem to find the final sides length and then finally calculate $\sin \theta, \cos \theta$.
EDIT: I've been corrected properly, you also need to consider a triangle with negative values on the legs, since $4/3=(-4)/(-3)$ i.e.
|\
-4| \
| \
|_____ \ <- theta
-3
and then go through the process as above to find the second set of possible values for $\sin \theta, \cos \theta$.
I'm going to go for a more visual approach to this, ignoring the triangle above. Consider a 2D cartesian graph, let's try to draw the possible triangles that the above condition represents. In order for $\tan \theta = \frac{4}{3}$ we either have a triangle in the first or third quadrant. Drawing these two triangles we either get
/|
/ |
/ |
/ |
/ | 4
/ |
theta -> /______|
3
or
-3
-------------- <- theta
| /
| /
-4 | /
| /
| /
| /
| /
| /
| /
| /
| /
| /
|/
Now we can use the pythagorean theorem (taking the hypotenuse as positive since the slope here is clearly positive) to find the last sides of the triangles and from there find $\sin \theta, \cos \theta$.
The following is multiple choice question (with options) to answer.
If 4 and 11 are the lengths of two sides of a triangular region, which of the following can be the length of the third side?
I. 5
II. 13
III. 15 | [
"I only",
"II only",
"I and II only",
"II and III only"
] | B | The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.
i.e let us assume the third side be c and other sides be a and b, then as per rule
c > a - b and c < a + b.
Let's take options
option 1. the three sides will be 5 , 4 and 11..
then 11 > 5-4 --Yes
11 < 4+9...NO
option 2... 13, 4 and 11.
then 13 > 11-4...Yes
13 < 11+4...Yes
Option 3..15 ,4 and 11.
15 < 11+4...No
15 > 11-4...Yes...
Only 13 is correct .
Answer: option B is correct.. |
AQUA-RAT | AQUA-RAT-37530 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Tim and Élan are 90 miles away from one another. They are starting to move towards each other simultaneously, Tim at a speed of 10 Mph and Élan at a speed of 5 Mph. If every hour they double their speeds, what is the distance that Tim will pass until he meets Élan? | [
"30 miles.",
"35 miles.",
"45 miles.",
"60 miles."
] | D | In 1st hour: Relative speed of the two will be 10+5 = 15 miles / hour. As they move towards each other, The distance between them reduces by 15 miles. Distance left between them: 90 - 15 = 75 miles.
In 2nd hour: Speed of Tim: 20Mph, That of Elan: 10 Mph. Relative speed of the two will be 20+ 10 = 30 miles / hour. As they move towards each other, The distance between them reduces by 30 miles. Distance left between them: 75 - 30 = 45 miles.
In 3rd hour: Speed of Tim: 40 Mph, That of Elan: 20 Mph. Relative speed of the two will be 40+ 20 = 60 miles / hour.
Now in order to cover the remaining distance of 45 miles. They will take: (45/ 60) hrs. = (3/4) hour.
Distance travelled by Tim?
In first hour: 1 hour * (10 miles per hour) = 10 miles
In second hour: 1 hour * (20 miles per hour) = 20 miles
In third hour: (3/4) hour * (40 miles per hour) = 30 miles
Total distance travelled: 10+20+30 = 60 miles.
D is the answer |
AQUA-RAT | AQUA-RAT-37531 | Shanonhaliwell April 8th, 2018 03:31 PM
Quote:
Originally Posted by romsek (Post 591335) outstanding, you seem to be getting the hang of things.
Thanks to you, I was able to do it.
All times are GMT -8. The time now is 12:30 AM.
The following is multiple choice question (with options) to answer.
Peeta asks Katniss, 'It is 10 am as per my watch right now.Do you know what will be the time 23, 999, 999 hours later?'
Katniss knows the answer. Do you ? | [
"5pm",
"6pm",
"4am",
"7am"
] | D | D
7pm
The time after 24, 000, 000 hours will be the same as it is now. We want the time 3 hours before that and thus you can deduct three hours from 9 pm. So the time will be 6 pm. |
AQUA-RAT | AQUA-RAT-37532 | # (GR. 10) 10 people are to be seated in a row. What is the total number of ways if…
Please help me! I understand what the question is asking for, but I can’t seem to get the right answer. The correct no. of ways should be $$645,120$$, though that may be incorrect. If anyone is kind enough to show me the solution, I would be very grateful.
$$10$$ people are to be seated in a row. What is the total number of ways in which this can be done if Eric and Carlos always have exactly one of the other people sitting between them?”
EDIT: Oh wow that was fast! Thank you for your kind hints! I was finally able to get the answer.
• Please show us your calculation. – saulspatz Feb 23 at 14:54
• I think it should be $8!*8*2$.I think your answer is correct. Cheers :) – Abhinav Feb 23 at 15:00
The possible positions of the two people are $$1-3,2-4,\cdots ,8-10$$ that is $$8$$ possibilities. We can swap the places, so multiply with $$2$$. Then, multiply with $$8!$$ because the other people can have $$8!$$ possible orders.
Here's a hint to get started. Suppose Alice is seated between Eric and Carlos. Then we can treat Eric-Alice-Carlos as a block to be arranged with the other $$7$$ students.
There are a total of $$10$$ people so there are $$8$$ people who could be seated between Eric and Carlos. There are $$2$$ ways of seating "Eric, other person, Carlos" or "Carlos, other person, Eric". Now treat those $$3$$ people as a single "person"- there are $$8!$$ ways to seat those $$8$$ "people". There are, then, $$8!(2)(8)= 645120$$ ways to do this. That is the same as Peter's answer.
The following is multiple choice question (with options) to answer.
There was a cycle race going on. 1/5th of the those in front of a person and 5/6th of those behind him gives the total number of participants. How many people took part in the race? | [
"30",
"31",
"32",
"33"
] | B | let total =x
1/5*(x-1)+5/6(x-1)=x by solving this we will get 31
ANSWER:B |
AQUA-RAT | AQUA-RAT-37533 | # 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 retailer sold two articles at a profit percentage of 10% each. The cost price of one article is three – fourth that of the other. Find the ratio of the selling price of the dearer article to that of the cheaper one | [
"4:3",
"3:4",
"41:31",
"51:41"
] | A | Let C.P of one of the article be X, Then C.P of the other = ¾X,
S.P 1 = 11X/10, S.P2 =3/4*11/10X,
S.P1/S.P2 = 11X/10 *40/33X = 4/3
S.P1: S.P2 = 4 : 3
ANSWER:A |
AQUA-RAT | AQUA-RAT-37534 | # 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 tradesman by means of his false balance defrauds to the extent of 20%? in buying goods as well as by selling the goods. What percent does he gain on his outlay? | [
"56%",
"44%",
"78%",
"55%"
] | B | B
g% = 20 + 20 + (20*20)/100
= 44% |
AQUA-RAT | AQUA-RAT-37535 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
A crow leaves its nest, and flies back and forth from its nest to a nearby ditch to gather worms. The distance between the nest and the ditch is 400 meters. In one and a half hours, the crow manages to bring worms to its nest 15 times. What is the speed of the crow in kilometers per hour? | [
"2",
"4",
"6",
"8"
] | D | The distance between the nest and the ditch is 400 meters.
15 times mean = A crow leaves its nest, and flies back ( going and coming back) i.e. 2 times we get total 30 rounds.
So the distance is 30 * 400 = 12000.
D = ST
12000/1.5 = T , I think we can take 12000 meters as 12km, then only we get T = 8.( 1000 meters = 1 km)
D) |
AQUA-RAT | AQUA-RAT-37536 | the largest and smallest integers in the group. This method should take three decimal arguments, and return the smallest of the three. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. It is similar to modulus – ‘%’ operator of C/C++ language. Note that this is not the modulo * operation (the result can be negative). Print its smallest integer divisor greater than 1. Find the least common multiple and greatest common divisor of the given three numbers : Find the least common multiple and the greatest common divisor of numbers 12, 24, 48, 192 and 288. LCM(Least Common Multiple) is the smallest positive number which is divisble by both the numbers. The macro below does the. During the reconstruction of the railway tracks 40 meter long pieces of rails were replaced by 15 meter long pieces. (For example: 7/3 = 3 and 10/2 = 5). Dynamically Calculate Prime Numbers. The smallest divisor (other than 1) of a composite number is a/an a) odd number b) even number c) prime number d) composite number. Thus the smallest number we can represent would be: 1E(-(MIN_EXP-digits-1)*4), eg, for digits=5, MIN_EXP=-32767, that would be 1e-131092. For example divisors of 6 are 1, 2, 3 and 6, so divisor_sum should return 12. gcd(a, b) = 5 3 · 7 2 = 6125. On a number line, we get hundredths by simply dividing each interval of one-tenth into 10 new parts. Example: The smallest weird number is 70. Write a Java program that computes the absolute value of the product of 2 numbers given by the user. Here dividend is 205 and divisor is 2 therefore remainder is 1. The last divisor will be the HCF of given numbers. In particular, we have added an entirely new Ch. The SMALL function will automatically ignore TRUE and FALSE values, so the result will be the nth smallest value from the set of actual numbers in the array. In this tutorial we will write couple of different Java programs to find out the GCD of two numbers. The first perfect number is 6, because
The following is multiple choice question (with options) to answer.
If the least common multiple of two prime numbers x and y is 10, where x > y, then the value of 2x + y is | [
"7",
"9",
"11",
"12"
] | D | LCM of two prime numbers will basically be multiple of those numbers. LCM = 10 = 2*5, both of which are prime.
Now we know x>y, so x=5 and y =2.
2x+y = 2*5 + 2 = 10 + 2 = 12.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37537 | Suppose that 5 cards are dealt from a 52-card deck. What is the probability of drawing at least two kings given that there is at least one king?
The Attempt at a Solution
Let ##B## denote the event that at least 2 kings are drawn, and ##A## the event that at least 1 king is drawn. Because ##B## is a strict subset of ##A##,
$$P(B|A) = P(A \cap B)/P(A) = P(B)/P(A)$$
Compute ##P(A)##, ##P(A^c )## denotes the probability of not drawing a single king.
$$P(A) = 1 - P(A^c) = 1 - \frac{48 \choose 5}{52 \choose 5} \approx 1 - 0.6588$$
Compute ##P(B)##, ##P(B^c)## denotes the probability of not drawing at least 2 kings, which is the sum of probabilities of drawing 1 king ##P(1)## and the probability of not drawing a single king ##P(A^c)##.
$$P(B) = 1 - P(B^c) = 1 - (P(1) - P(A^c))$$
$$P(1) = \frac{5 \times {4\choose 1} \times 48 \times 47 \times 46 \times 45 }{52 \times 51 \times 50 \times 49 \times 48} \approx 0.299$$
where the numerator is the number of ways one can have a hand of 5 containing a single king.
$$P(B) \approx 1 - (0.299 + 0.6588) \approx 0.0422$$
finally,
$$P(B|A) = P(B)/P(A) = 0.0422 / (1-0.6588) \approx 0.1237$$
The following is multiple choice question (with options) to answer.
In a board game, a stack of 46 cards, 8 of which have pictures of fruit, are shuffled and then placed face down. If the first 2 cards selected do not have pictures of fruit, what is the probability that the third card selected will have a picture of fruit? | [
"2/11",
"1/6",
"1/8",
"3/7"
] | A | After 2 cards have been selected, there are 44 cards remaining.
P(third card has a picture of fruit) = 8/44 = 2/11
The answer is A. |
AQUA-RAT | AQUA-RAT-37538 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
Jacob is now 13 years younger than Michael. If 9 years from now Michael will be twice as old as Jacob, how old will Jacob be in 4 years? | [
" 8",
" 7",
" 15",
" 21"
] | A | Jacob=x years, Michael=x+13 years
9 years from now,
2(x+9)=x+22
2x+18=x+22
x=4
x+4=8 years
Answer A |
AQUA-RAT | AQUA-RAT-37539 | ### 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.
Lakshmi invested some money in 25% stock at 500.If Usha wants to invest in an equally good 50% stock ,he must purchase a stock worth of : | [
"Rs. 1000",
"Rs. 1200",
"Rs. 1300",
"Rs. 1400"
] | A | Solution
For an income of Rs. 25, investment = Rs. 500
For an income of Rs. 50, investment = Rs.(500/25x50) = Rs. 1000
Answer A |
AQUA-RAT | AQUA-RAT-37540 | Tools » Screen Aspect Ratio & Dimension Calculator Screen Aspect Ratio & Dimension ... with a diagonal screen size of and ... ) takes up ??? Then drag the corners to create an arbitrary rectangle.Calculate the length of the diagonals.Click 'show details' to verify your answer. Mathematics - Mathematical rules and laws - numbers, areas, volumes, exponents, trigonometric functions and more ; Related Documents . Height. A = a × b A = a × b. where a a and b b are the length and width of the rectangle, respectively. Keep in mind that this is a decimal fraction; for example .5 is 1/2 inch and .75 is 3/4 inch. Length of Diagonal of Rectangle Formula: The diagonal of a rectangle is determined by the following formula. Width = 10 in Simply enter the length of a side of the square and the diagonal will be calculated quickly. Use our online diagonal of a rectangle calculator to find diagonal of rectangle by entering the width and height. Diagonal of a Square Calculator - calculate the diagonal of a square. Use this square calculator to find the side length, diagonal length, perimeter or area of a geometric square. Enter the measurement that you know (diagonal, width or height) and the other two will be calculated. Area = length x width It is necessary to follow the next steps: Enter the length and width of a rectangle in the box. Related Topics . Right triangle calculator to compute side length, angle, height, area, and perimeter of a right triangle given any 2 values. Diagonal Matrix Calculator is a free online tool that displays the result whether the given matrix is a diagonal or not for the given matrix. % of the device surface area. iForce Systems LLC, Through two sides and the angle between them, Through the radius of the inscribed circle, Through the radius of the circumscribed circle, By the diagonals and the angle between them, Through the diagonals and the angle between them, Through the sides and the angle between them, Total surface area of the regular pyramid across the height, Lateral surface area of the regular pyramid through the height, Lateral surface area of the regular pyramid through the apothem, Isosceles triangle, through side and height, Isosceles triangle, through side and angle. Calculate screen dimensions (height/width/area, in inches or
The following is multiple choice question (with options) to answer.
The size of a flat-screen television is given as the length of the screen’s diagonal. How many square inches greater is the screen of a square 24-inch flat-screen television than a square 17-inch flat-screen television? | [
"143.5",
"154",
"160",
"148.75"
] | A | If we take a square with side length x and draw a diagonal, we get two isosceles right triangles.
If we focus on one such right triangle, we see that the legs have length x.
square 24-inch flat-screen television
The diagonal (hypotenuse) = 24
So, we can apply the Pythagorean Theorem to get x² + x² = 24²
Simplify: 2x² = 24²
Divide both sides by 2 to get: x² = 24²/2
Since the area of the square = x², we can see that the area of this square is 24²/2
square 17-inch flat-screen television
The diagonal (hypotenuse) = 17
So, we can apply the Pythagorean Theorem to get x² + x² = 17²
Simplify: 2x² = 17²
Divide both sides by 2 to get: x² = 17²/2
Since the area of the square = x², we can see that the area of this square is 17²/2
DIFFERENCE IN AREAS =24²/2-17²/2
= 288 -144.5 i.e = 143.5
A |
AQUA-RAT | AQUA-RAT-37541 | • Nice approach. +1. – N.S.JOHN Sep 24 '16 at 3:26
• And are you really just 12? – N.S.JOHN Sep 24 '16 at 3:26
• @N.S.JOHN Yes, I am twelve. Thank you for your appreciation. – астон вілла олоф мэллбэрг Sep 24 '16 at 3:28
The following is multiple choice question (with options) to answer.
A man said to his son, "I was two-third of your present age when you were born". If the present age of the man is 48 years, find the present age of the son? | [
"25.7 years",
"28 years",
"29.3 years",
"28.8 years"
] | D | Present age of the son be P, he was born P years ago.
The age of the man was: (48 - P).
His age when the son was born should be equal to 2/3 of P.
(48 - P) = 2/3 P
5P = 144 => P = 28.8
ANSWER:D |
AQUA-RAT | AQUA-RAT-37542 | . Sum of the First n Terms of an Arithmetic Sequence Suppose a sequence of numbers is arithmetic (that is, it increases or decreases by a constant amount each term), and you want to find the sum of the first n terms. Sum of First 50 Odd Numbers; Sum of First 50 Even Numbers; How to Find Sum of First 50 Natural Numbers? Example: Sum 1 to 100 = 100 * 101 / 2 = 50 * 101 = 5050. Are all integers even? Though both programs are technically correct, it is better to use for loop in this case. 1729 = 1 3 + 12 3 = 10 3 + 9 3. Problem 44. . . The sum of the integers from 1 to 25 is as follows: 325 To get the answer above, you could add up all the digits like 1+2+3... +25, but there is a much easier way to do it! Consecutive integers are numbers that follow each other in order. . In elementary school in the late 1700’s, Gauss was asked to find the sum of the numbers from 1 to 100. . 200 xx (1+200)/2 = 20100 The sum of a finite arithmetic sequence is equal to the count of the number of terms multiplied by the average of the first and last terms. Put them in pairs: 1 & 500 2 & 499 3 & 498 etc There are 250 pairs, each pair = 501 250*501 = 125250-----How about only the odds: 1+3+5+7+...493+495+497+499? Therefore, 62500 is the sum of first 250 odd numbers. . Let there be n terms in this A.P. The sum of the odd numbers (from 1) up to to 500 is 62500. What is the sum of first 500 odd numbers? This is an A.P. The sum of 1+2+3+4+...496+497+498+499+500? Find the number and sum of all integer between 100 and 200, divisible by 9: ----- Numbers between 100 and 200, divisible by 9: 108 117 126 135 144 153 162 171 180 189 198 The sum : 1683 Flowchart: C++ Code Editor: Contribute your code and comments through Disqus. step 2 apply the input parameter values in the formulaSum =
The following is multiple choice question (with options) to answer.
what is the sum of the first 25 odd natural numbers | [
"300",
"400",
"625",
"600"
] | C | sum of first n odd natural nuber is = n^2
so sum of the first 25 odd natural numbers=(25)^2=625
ans 625
ANSWER:C |
AQUA-RAT | AQUA-RAT-37543 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
Find the smallest number which when divided by 13 and 16 leaves respective remainders of 2 and 5. | [
"187",
"197",
"207",
"219"
] | B | Let 'N' is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) - (common difference of divisors and remainders) = (208) - (11) = 197.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37544 | 9. Men's weights follow a normal distribution with a mean of 172 pounds and a standard deviation of 29 pounds.
1. What is the probability that a randomly selected man carrying a 20 lb bag collectively weighs more than 195 lbs.
2. If an airplane is full of 213 men (and no women or children), each with a 20 lb bag, what is the probability that the total weight is greater than 41535 lbs (the weight limit for the airplane)?
1. With the bag the mean weight is $\mu = 192$. The standard deviation remains the same. $z_{195} \doteq 0.1034$. So $P(z \gt 0.1034) \doteq 0.4588$
2. If the total weight is 41535 lbs, the average weight of the 213 men is 195 lbs. Central limit theorem applies. $\mu = 192$, $\displaystyle{\sigma = \frac{29}{\sqrt{213}} \doteq 1.9870}$. Thus $z_{195} = 2.1282$. So the probability of exceeding the weight limit is $P(z \gt 2.1282) \doteq 0.0167$.
The following is multiple choice question (with options) to answer.
Mike weighs twice as much as Mary. Mike's weight is 150% of Michelle's weight. Alan weighs 140% of Charlie's weight. Charlie weighs 180% of Mary's weight. Which of these 5 people weighs the highest? | [
"Mary",
"Mike",
"Michelle",
"Alan"
] | D | Mary < Michelle < Mike
Charlie < Alan
Let x be Mary's weight.
Mike's weight is 2x and Alan's weight is 1.4*1.8x = 2.52x.
The answer is D. |
AQUA-RAT | AQUA-RAT-37545 | 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 210 m long, running with a speed of 63 km/hr will pass a tree in? | [
"18 sec",
"16 sec",
"76 sec",
"12 sec"
] | D | Speed = 63 * 5/18 = 35/2 m/sec
Time taken = 210 * 2/35 = 12 sec
Answer:D |
AQUA-RAT | AQUA-RAT-37546 | # 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 average weight of 25 persons sitting in a boat had some value. A new person added to them whose weight was 46 kg only. Due to his arrival, the average weight of all the persons decreased by 5 kg. Find the average weight of first 25 persons? | [
"2211",
"2721",
"227",
"176"
] | D | 25x + 46 = 26(x – 5)
x = 176
Answer:D |
AQUA-RAT | AQUA-RAT-37547 | So total 6 ways we can assign project to three groups.
So total number of ways we can do the required is 15 Ways of distributing people in 3 groups X 6 ways of distributing 3 projects among the three formed groups.
= 90
Hope this helps
Probus
_________________
Probus
~You Just Can't beat the person who never gives up~ Babe Ruth
Manager
Joined: 16 Jul 2016
Posts: 54
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
05 Mar 2019, 19:03
Suppose the six students are 1,2,3,4,5,6
Now student 1 HAS to be paired with someone else. He/she can be paired with 2,3,4,5, or 6. So that's 5 possibilities.
Once that has occurred (without loss of generality suppose 1 pairs with 2. 3 can then pair with 4,5, or 6. That's 3 possibilities. Only one pair is left. 5x3=15 groups of 3.
For each group there are 3x2x1 different arrangements of the topics. 15x6=90.
Intern
Joined: 22 Oct 2017
Posts: 19
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
06 Mar 2019, 03:35
arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540
Can someone please explain to me why the red part of 15*3!, because I really can't figure out the reasoning behind the permutation of the assignments,
guaranteed kudos for accurate responses
Thanks
Director
Joined: 04 Aug 2010
Posts: 546
Schools: Dartmouth College
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
06 Mar 2019, 03:57
arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540
The following is multiple choice question (with options) to answer.
There are 4 periods in each working day of a school. In how many ways can one organize 3 subjects such that each subject is allowed at least one period? | [
"32",
"36",
"24",
"34"
] | B | 3 subjects can be arranged in 4 periods in 4P3 ways.
Remaining 1 period can be arranged in 3P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (4p3 x 3P1)/2! = 36
ANSWER:B |
AQUA-RAT | AQUA-RAT-37548 | If we use "number" in essentially any of the usual senses (integer, real number, complex number), yes, zero is a number.
-
Dear @Bongers : I don't disagree with the observation in the first line, but in almost all sources, prime numbers are defined to be nonzero, and I think that kind of trumps anything you might say about divisors. I'm not suggesting you need to change anything: I just wanted to bring this to future readers' attention. Regards! – rschwieb Oct 25 '13 at 12:46
@rschwieb I find the definition of prime numbers to be non-zero an exceptionally boring reason as to why zero is not a prime number. It implies that "zero is not a prime number because someone defined it that way", which isn't true. Zero is not a prime number because it has more than two divisors, and this allows people define it out. – user1729 Oct 28 '13 at 20:19
@user1729 Aesthetically I totally agree with you :) I was just merely stating the reality of the printed situation. Since the "real" definition of prime requires you to exclude zero, I imagine that the "nonzero" condition appears in the "gradeschool" definition as an artifact, and they mostly just don't notice this convenience possible with their definition. What annoys me more is the fact that so many geometry texts insist a trapezoid has exactly two parallel sides... – rschwieb Oct 29 '13 at 13:27
• Zero is not a prime number as prime numbers are defined for integers greater than 1.
• Zero is an even number. Definition of an even number with modular arithmetic:
$\forall x\in \mathbb{Z},\, x$ is even if and only if $x\equiv 0 \pmod 2$
As $0$ satisfies the definition, then it is an even number.
The following is multiple choice question (with options) to answer.
Which one of the following is not a prime number? | [
"2",
"3",
"5",
"9"
] | D | 9 is divisible by 3. So, it is not a prime number.
Answer:D |
AQUA-RAT | AQUA-RAT-37549 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
At a company, the average wage of male employees is $5 per hour, and the average wage of female employees is $8 per hour. If the average wage of all employees is $7.5 per hour, what is the ratio of the number of male employees to the number of female employees? | [
"1 to 5",
"2 to 3",
"3 to 2",
"5 to 3"
] | A | Average hourly wage of male employees = 5 $
Average hourly wage of female employees = 8 $
Average hourly wage of all employees = 7.5 $
Let number of male employees = m
number of female employees = f
7.5 = (5m + 8f ) / m+f
=>7.5m + 7.5f = 5m + 8f
=> 2.5m = .5f
=> m/f = 2.5/.5 = 1/5
Answer A |
AQUA-RAT | AQUA-RAT-37550 | # probability of the 3rd pick is a boy
10 students are in a class,
4 of these students are boys and 6 are girls
the teacher wanted to randomly select 3 students to represent the class in an event
what is the probability that the 3rd student is a boy?
Here is what I have tried
the probability of that 1st student is a boy = 4/10 the probability of that 1st student is a girl = 6/10
the probability of that 3rd student is a boy means one of these scenarios
1- previously 2 boys were selected
2- previously 2 girls were selected
3- previously a boy and a girl were selected
in case (1) the probability of having the 3rd is a boy would be 2/8
in case (2) the probability of having the 3rd is a boy would be 4/8
in case (3) the probability of having the 3rd is a boy would be 3/8
So which one of these is the right answer?!
• You can do it that way but there's no need to, each choice is like any other so the answer is $\frac 4{10}$. To do it the way you started, you need to weight each case by the probability of being in that scenario. – lulu Apr 28 at 22:50
• @lulu but when we select the 1st and 2nd student, number of students change and the distribution changed as well.. how come it is the same as the 1st pick? – asmgx Apr 28 at 22:53
• If you specify the first two choices then of course the answer changes, but if you don't specify them then there is no information from them, – lulu Apr 28 at 22:56
• – JMoravitz Apr 28 at 22:59
• If you were to simplify that god-awful expression however... it very simply results in $\frac{4}{10}$... which as many of the answers and comments in the linked question as well as other answers and comments here already will tell you makes sense as being "the third" person to be selected is not different enough from being "the first" person selected, so naturally the probability the third is a boy is going to be the same as the probability the first is a boy which is obviously $\frac{4}{10}$ with no tedious calculations required. – JMoravitz Apr 28 at 23:50
The following is multiple choice question (with options) to answer.
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: | [
"21/46",
"25/117",
"1/50",
"3/25"
] | A | Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= 25C3 `
=(25 x 24 x 23)/(3 x 2 x 1)
= 2300.
n(E) = (10C1 x 15C2)
=10 x((15 x 14) /(2 x 1))
= 1050.
P(E) = n(E)/n(S) =1050/2300=21/46
Option A |
AQUA-RAT | AQUA-RAT-37551 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
A person X working alone can complete a work in 6 days. A person Y completes the same amount of work in 15 days, and a person Z when working alone can complete the same amount of work in 30 days. All three people work together for 2 days but then X and Y leave. How many more days does Z need to complete the work after X and Y leave? | [
"12",
"13",
"14",
"15"
] | C | The fraction of work completed in two days is 2/6 + 2/15 + 2/30 = 16/30
The fraction of work remaining is 1 - 16/30 = 14/30
The number of days required for Z is 14 days.
The answer is C. |
AQUA-RAT | AQUA-RAT-37552 | ### 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 and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew 1/4 of his capital and B withdrew 1/5 of his capital. The gain at the end of 10 months was Rs. 760. A's share in this profit is: | [
"Rs. 330",
"Rs. 360",
"Rs. 380",
"Rs. 430"
] | A | If earlier their investmwnts were 4x and 5x for 3 months
and then
3x and 4x for next 7 months,
Then profit will be divided in ratio of (4x*3 + 3x*7) :( 5x*3+4x*7) or 33:43
Hence A's share in profit of Rs 760 = 760*33/76 = Rs 330.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37553 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
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Posts: 1892
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### 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>
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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 student committee on academic integrity has 30 ways to select a president and vice president from a group of candidates. The same person cannot be both president and vice president. How many candidates are there? | [
"7",
"8",
"9",
"6"
] | D | xC1 * (x-1)C1=30
x^2 -x -30 =0
(X-6) (X+5) = 0
X= 6 , -5
-5 can't possible.
D |
AQUA-RAT | AQUA-RAT-37554 | # Determine the maximum possible volume, excluding the volume of the legs.
A table is to be constructed by gluing together 68 cubes of dimension $$1\times 1\times 1$$. All four legs and the rectangular top will be formed by the cubes. The four legs must be the same length and must be one cube thick, and the top is one cube thick, as well. What is the maximum possible volume of the space between the table's top and the floor, excluding the space taken up by the table's legs?
My solution:
Let $$a$$ be the area of the top and $$c$$ be the length of a leg.
\begin{align} a + 4c &= 68 \cr a &= 68 - 4c \end{align} We try to find the maximum value of ac, and then subtract 4c:
$$ac = (68 - 4c)c = -4c^2 + 68c$$
Completing the square gets $$-4(c - 17/2)^2 + 289$$.
The lowest value of $$ac$$ is when $$c= 17/2$$, and the lowest value is $$289$$. Then we subtract $$289 - 4c = 289 - 34 = 255$$.
Therefore, the maximum volume when excluding the volume of the legs is $$255$$.
However, the actual answer is $$256$$ as the maximum volume when excluding the legs. Where did I go wrong?
You want to maximise the volume excluding the legs, which is $$ac-4c$$. You maximised $$ac$$ instead.
We have that $$ac-4c=-4c^2+64c$$.
Now complete the square:
$$-4c^2+64c=256-(16-2c)^2$$
The maximum of $$256$$ is at $$c=8$$.
• But maximizing ac - 4c gives is maximizing -4c^2 + 64c, and completing the square gives -4(c - 8)^2 - 64. When c = 8, the expression equals -64. What exactly do you mean by maximizing ac - 4c? – David Dong Aug 5 '19 at 17:00
The following is multiple choice question (with options) to answer.
Three table runners have a combined area of 212 square inches. By overlapping the runners to cover 80% of a table of area 175 square inches, the area that is covered by exactly two layers of runner is 24 square inches. What is the area of the table that is covered with three layers of runner? | [
" 18 square inches",
" 20 square inches",
" 24 square inches",
" 28 square inches"
] | C | Total = a + b + c - (sum of EXACTLY 2-group overlaps) - 2*(all three) + Neither
80%*175 = 212 - 24 - 2*(all three) + 0
2*(all three) = 212 - 24 - 140
all three = 24
Answer: C |
AQUA-RAT | AQUA-RAT-37555 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
The traffic lights at 3 different roads change after every 9 sec, 72 sec, 108 sec respectively. If they all change simultaneously at 9:15 AM then at what time will they again change simultaneously. | [
"9:18:36",
"9:42:15",
"10:15:24",
"9:54:12"
] | A | LCM of 9,72,108 = 216 = 3mnts 36sec
The lines will change every 3m 36sec
The light will change again after 9:15 AM is = 9:15+0:03:36 = 9:18:36
Answer is A |
AQUA-RAT | AQUA-RAT-37556 | I'm thankful for all ideas!
The answer is Yes, it is 50%.
Proof
The probability that a given coin lands tails is $\frac12$, and the probability that it doesn't is $1-\frac12=\frac12$. Therefore, using the Binomial Distribution Formula, the probability that we get exactly $r$ tails from $n$ tosses is:
$P(r) =\binom n r \left(\frac12\right)^r\left(\frac12\right)^{n-r}=\binom n r \left(\frac12\right)^n$
We must prove that $P(0)+ P(2) + P(4) + ... = \frac12$. So, we need to prove that, if $n$ is odd (and therefore if $n=491$):
$\binom n 0\left(\frac12\right)^n + \binom n 2\left(\frac12\right)^n+\binom n 4\left(\frac12\right)^n + ... +\binom{n}{n-1} \left(\frac12\right)^n = \frac12$
i.e. that:
$\binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1} = 2^{n-1}=S$, say
Now it is well known that:
$\binom n 0 + \binom n 1+\binom n 2 + ... +\binom{n}{n} = 2^{n}$ ...(1)
(The proof of this is to write down the Binomial Expansion of $(1+1)^n\;( = 2^n)$)
The following is multiple choice question (with options) to answer.
If 15% of 30% of 50% of a number is 108, then what is the number? | [
"4000",
"3050",
"4400",
"4800"
] | D | Let the number be a
Given, 15/100 * 30/100 * 50/100 * a = 108
=> 3/20 * 3/10 * 1/2 * a = 108
=> a = 10 * 20 * 10 * 2 = 4800.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37557 | 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.
A bike covers a certain distance at the speed of 64km/h in 8 hrs. If the bike was to cover the same distance in approximately 6 hrs, at what approximate speed should the bike travel? | [
"85 km/h",
"85.34 km/h",
"87.67 km/h",
"90.25 km/h"
] | B | Ans.(B)
Sol. Total distance = 64 × 8 = 512 km
Now speed = 512/6 = 85.34 km/h |
AQUA-RAT | AQUA-RAT-37558 | P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2 P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2
(1)
Show whether or not the pair {A,B}{A,B} is independent.
### Solution
P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc)P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc), and P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc)P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc).
PA = 0.4*0.7 + 0.3*0.3
PA = 0.3700
PB = 0.6*0.7 + 0.2*0.3
PB = 0.4800
PA*PB
ans = 0.1776
PAB = 0.4*0.6*0.7 + 0.3*0.2*0.3
PAB = 0.1860 % PAB not equal PA*PB; not independent
## Exercise 2
Suppose {A1,A2,A3} ci |C{A1,A2,A3} ci |C and ci |Cc ci |Cc, with P(C)=0.4P(C)=0.4, and
The following is multiple choice question (with options) to answer.
If P(A)=0.4, P(B)=0.6 and P(A ∪ B)=0.8. What is the value of P(A∩B')=? | [
"0.1",
"0.2",
"0.3",
"0.4"
] | B | Solution:
P(A∪B) = P(A) + P(B) - P(A∩B') =>
0.8=0.4 - P(A∩B) =>P(A∩B)=0.2
P(A∩B') = P(A) - P(A∩B) = 0.4 - 0.2 = 0.2
Answer B |
AQUA-RAT | AQUA-RAT-37559 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work? | [
"10",
"13",
"14",
"15"
] | B | Let the required number of days be x.
Less persons, More days (Indirect Proportion)
More working hours per day, Less days (Indirect Proportion)
Persons 30 : 39 :: 12 : x
Working hours/day 6 : 5
Therefore 30 x 6 x x = 39 x 5 x 12
=> x =(39 x 5 x 12)/(30 x 6)
=> x = 13.
Correct answer is B |
AQUA-RAT | AQUA-RAT-37560 | I don't see it...
• May 21st 2010, 12:36 PM
wonderboy1953
Hint
Look for cycles.
• May 21st 2010, 03:22 PM
Soroban
Hello, pollardrho06!
Quote:
Find the number of positive integers not exceeding 1000
that are divisible by 3 but not by 4.
Every third number is divisible by 3.
. . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3.
But every twelfth number is divisible by 3 and by 4.
. . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4.
Therefore, there are: . $333 - 83 \:=\:250$ such numbers.
• May 21st 2010, 03:46 PM
pollardrho06
Quote:
Originally Posted by Soroban
Hello, pollardrho06!
Every third number is divisible by 3.
. . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3.
But every twelfth number is divisible by 3 and by 4.
. . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4.
Therefore, there are: . $333 - 83 \:=\:250$ such numbers.
Wow!! The greatest integer function!! Gr8!! Thanks!!
The following is multiple choice question (with options) to answer.
How many three-digit integers between 310 and 400, exclusive, are divisible by 3 when the tens digit and the ones digit are switched? | [
"3",
"19",
"30",
"22"
] | C | Answer is 30
No of digits between 310 and 400 (exclusive) = 400-310-1 = 89
No Intergers divisble by 3 =~ 89/3 ~ 30
Divsiblivity rule for 3 is Sum of all digits must be divisble by 3. It doesn't matter if ones digit and tens digits are switched.
e.g 372 is divisble by 3 (becae 3+7+2= 12 (divisble by 3))
switch digits 732 also divisble by 3 (becae 7+3+2= 12 (divisble by 3))
Divsiblivity rule for 3 is Sum of all digits must be divisble by 3. It doesn't matter if ones digit and tens digits are switched.
You have right. So it is enough to proceed like usual to find the numbers in a given range that are divisible by 3. So 399-312/3 +1 =30.
Answer : C |
AQUA-RAT | AQUA-RAT-37561 | 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.
A beats B by 20metres ,B beats C by20 metres in a 100 m race .By how much A beats C | [
"35metres",
"36metres",
"37metres",
"38metres"
] | B | A : B = 100 : 80
B : C = 100 : 80
A : C = (A/B * B/C)
= (100/80 * 100/80)
= (100/64)
Therefore A beats C by = (100-64) = 36metres.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37562 | You want to end up with exactly 6 gallons in the unmarked container.
(YES: the unmarked container > 6 gallons!)
Your mission is to do it in exactly 9 moves, and using the MINIMUM number of gallons of water
from the water supply.
A "move" is any pouring, filling or emptying.
For example, filling the 9-gal from the water supply,
then pouring it to fill the 4-gal container,
then emptying it is 3 moves: fill - pour - empty.
How do you do it?
Spoiler:
(1) Fill the "9".
Code:
* *
|/////|
|/////| * *
|/////| | |
|/////| | |
* * |//9//| | 0 |
| | |/////| | |
| 0 | |/////| | |
*-----* *-----* *-----*
4 9 U
(2) Pour "9" into "4".
Code:
* *
| |
| | * *
| | | |
|/////| | |
* * |/////| | 0 |
|/////| |//5//| | |
|//4//| |/////| | |
*-----* *-----* *-----*
4 9 U
(3) Empty "4",
Code:
* *
| |
| | * *
| | | |
|/////| | |
* * |/////| | 0 |
| | |//5//| | |
| 0 | |/////| | |
*-----* *-----* *-----*
4 9 U
(4) Pour "9" into "4".
Code:
* *
| |
| | * *
| | | |
| | | |
* * | | | 0 |
|/////| | | | |
|//4//| |//1//| | |
* - - * * - - * * - - *
4 9 U
(5) Pour "9" into "U".
The following is multiple choice question (with options) to answer.
A machine fills 150 bottles of water every 8 minutes. How many minutes it takes this machine to fill 675 bottles? | [
"54 minutes",
"34 minutes",
"32 minutes",
"21 minutes"
] | C | 8 minutes are needed to fill 150 bottles. How many groups of 150 bottles are there in 675 bottles?
675 * 150 = 4.5 = 4 and 1/2
For each of these groups 8 minutes are needed. For 4 groups and 1/2
8* 4 + 4 = 32 + 4 = 36 minutes. (4 is for 1/2 a group that needs half time)
We can also find the final answer as follows
4.5 x 8 = 32 minutes correct answer C |
AQUA-RAT | AQUA-RAT-37563 | 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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Kudos [?]: 79 [1] , given: 10
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.
The average of runs of a cricket player of 10 innings was 37. How many runs must he make in his next innings so as to increase his average of runs by 4 ? | [
"76",
"81",
"85",
"87"
] | B | Explanation:
Average = total runs / no.of innings = 37
So, total = Average x no.of innings = 37 x 10 = 370.
Now increase in avg = 4runs. So, new avg = 37+4 = 41runs
Total runs = new avg x new no. of innings = 41 x 11 = 451
Runs made in the 11th inning = 451 - 370 = 81 Answer: B |
AQUA-RAT | AQUA-RAT-37564 | 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.
How long does a train 145 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length? | [
"52.66",
"53.66",
"54.66",
"58.66"
] | B | T = (660 + 145)/54 * 18/5
T = 53.66
ANSWER:B |
AQUA-RAT | AQUA-RAT-37565 | 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 100 m long can cross an electric pole in 20 sec and then find the speed of the train? | [
"19 Kmph",
"18 Kmph",
"72 Kmph",
"17 Kmph"
] | B | Length = Speed * time
Speed = L/T
S = 100/20
S = 5 M/Sec
Speed= 5*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 18 Kmph
Answer: B |
AQUA-RAT | AQUA-RAT-37566 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along theoundary of the park at the speed of 12 km/hr completes one round in 8 min, then the area of the park (in sq. m) is? | [
"128798",
"134768",
"153600",
"168799"
] | C | Perimeter = Distance covered in 8 min. = 12000 x 8 m = 1600 m.
60
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m2 = 153600 m
C |
AQUA-RAT | AQUA-RAT-37567 | Hence, the probability of selecting three balls of one color and two balls of a different color when five balls are selected from the $7 + 8 + 9 = 24$ balls in the bag is $$\frac{\dbinom{7}{3}\dbinom{8}{2} + \dbinom{7}{3}\dbinom{9}{2} + \dbinom{7}{2}\dbinom{8}{3} + \dbinom{7}{2}\dbinom{9}{3} + \dbinom{8}{3}\dbinom{9}{2} + \dbinom{8}{2}\dbinom{9}{3}}{\dbinom{24}{5}}$$
• I was wondering whether you could have any cases where you would draw 3 blue balls, 1 red, and 1 white ball? – user262291 Apr 9 '16 at 15:10
• That's a different question since choosing three of one color and two of another color when five balls are selected precludes the possibility of selecting three different colors. The probability of selecting three blue balls, one red ball, and one white ball in the first scenario is $$\frac{\binom{7}{3}\binom{7}{1}\binom{7}{1}}{\binom{21}{5}}$$ In the second scenario, the probability is $$\frac{\binom{9}{3}\binom{7}{1}\binom{8}{1}}{\binom{24}{5}}$$ The second scenario makes it clearer what I am counting, namely selections of three blue balls, one red ball, and one white ball. – N. F. Taussig Apr 9 '16 at 18:15
Whenever you can, relate the problem to a known category of familiar problems.
Here, you can treat it the way you compute poker probabilities:
We want $3-2$ of a kind, so numerator will be [Choose kinds] $\times$ [Choose balls from each kind]
The following is multiple choice question (with options) to answer.
A bag contains 7 green and 8 white balls. If two balls are drawn simultaneously, the probability that both are of the same colour is ? | [
"7/19",
"7/15",
"7/16",
"7/12"
] | B | Drawing two balls of same color from seven green balls can be done in ⁷C₂ ways.
Similarly from eight white balls two can be drawn in ⁸C₂ ways.
P = 7C2/15C2 + 8C2/15C2 = 7/15
Answer: B |
AQUA-RAT | AQUA-RAT-37568 | ### 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.
The sale price sarees listed for Rs.600 after successive discount is 20% and 5% is? | [
"227",
"358",
"456",
"787"
] | C | 600*(80/100)*(95/100)
= 456
Answer: C |
AQUA-RAT | AQUA-RAT-37569 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Joined: 02 Sep 2009
Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
It would take one machine 5 hours to complete a large production order and another machine 4 hours to complete the same order. How many hours would it take both machines, working simultaneously at their respective constant rates, to complete the order? | [
"A 9/20",
"B 20/9",
"C 7/19",
"D 5/21"
] | B | The rate of the first machine is 1/5 job per hour;
The rate of the second machine is 1/4 job per hour;
Thus, the combined rate of the machines is 1/5+1/4=9/20 job per hour, which means that it takes 1/(9/20)=20/9 hours both machines to do the job.
Answer: B. |
AQUA-RAT | AQUA-RAT-37570 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
Mixture A is 20 percent oil, and mixture B is 60 percent oil. If the two are poured together to create a 4-gallon mixture that contains 40 percent oil, approximately how many gallons of mixture A are in the mixture? | [
"1.0",
"0.8",
"0.7",
"0.6"
] | B | (60-40)/(40-20) = Qa/Qb
20/20 = Qa/Qb
1 = Qa/Qb
Qa = (1/5) * 4 = 4/5 = 0.8 approx
Answer : B |
AQUA-RAT | AQUA-RAT-37571 | Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x
Solving, xt will get cancelled and you will get:
11000 = 11x
x is 1000
sum of both investments is 2x = 2000 which is Option D
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Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
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02 Jul 2017, 07:29
1
1
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000
Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.
The following is multiple choice question (with options) to answer.
If a sum of money borrowed at 5 percent/annum SI amounts to Rs.1020 in 4 yrs, then the sum of money borrowed is? | [
"Rs. 850",
"Rs. 960",
"Rs. 1000",
"Rs. 1200"
] | A | Sol We have, A : Rs.1U2D
Let the pr xnpal be R5 x
Then mterest u — 12 : wzo — x
Therefore, by u xglurmula.
_umxx
’ Rx’: (
, _ma>< 1nzn—xv
\4Ve|\a\e,Jc— 5“
:> ~(:51DD—S*(or6x:51UU
or, J: : w 350
. The sum of money burrowed : Rs. 850
A) |
AQUA-RAT | AQUA-RAT-37572 | ### 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.
3 pig’s and 10 hens are brought for Rs.1200. If the average price of a hen be Rs.30. What is the average price of a pig. | [
"300",
"350",
"400",
"425"
] | A | Explanation:
Average price of a hen = Rs.30
total price of 10 hens = 10*30 = Rs.300
But total price of 3 pigs and 10 hens = Rs. 1200
Total price of 3 pigs is = 1200 - 300
= 900
Average price of a pig = 900/3
= Rs. 300
Answer: A |
AQUA-RAT | AQUA-RAT-37573 | momentum, velocity, linear-algebra, relative-motion
Title: A question on time taken in different frames of references Let us say , a swimmer has to swim upstream and downstream a river , between two points which are at distance "L" from each other . The swimmer can swim in still-water at velocity "Vs" , and the velocity of the river is "Vr".
Now in the frame or reference of the river bank (or,ground) , his velocity upstream is Vs-Vr, and downstream is Vs+Vr, so the total time taken is
T1=L{1/(Vs+Vr) + 1/(Vs-Vr)}.
Now if we observe the motion in the frame of reference of the river , the swimmer's velocity is Vs only , and as the distance is same , the time taken in this case is
T2=2L/Vs
But shouldn't the time come the same ,regardless of the frame of reference ? I've tried putting in some values of Vs and Vr also and they are certainly not equal .
Please help me out , and point out where am I making a mistake and what should be the correct way of solving in different frames if references .
Thank you . One way to look at it is:
In the bank’s referential, the swimmer appears to swim at a speed:
Vs + Vr
between points A and B (separated by a distance L)
In the swimmer’s referential, the banks appears to move at a speed:
V = - Vs - Vr
Or
V = - (Vs + Vr )
So, basically the length between A and B is the same in both cases and so is the module of the speed (respectively of the swimmer or of the river bank, depending upon the point of vue (referential)).
Therefore, in both cases, the time needed to swim from A to B by is the same, as common intuition suggests.
T = L / ( Vs + Vr )
The following is multiple choice question (with options) to answer.
A girl swims downstream 48 km and upstream 18 km taking 6 hours each time; what is the speed of the current? | [
"2.2",
"2.3",
"2.1",
"2.5"
] | D | 48 --- 6 DS = 8
? ---- 1
18 ---- 6 US = 3
? ---- 1 S = ?
S = (8 - 3)/2 = 2.5
Answer:D |
AQUA-RAT | AQUA-RAT-37574 | A number is divisible by 11 if and only if the difference of the sum of the odd numbered digits (the first digit, the third digit, ...) and the sum of its even numbered digits is divisible by 11. The sum of the odd numbered digits of $y$ is $250\cdot(2+7)$ and the sum of the even numbered digits of $y$ $250\cdot(7+2)$. The difference between these two quantities is $0$; so $y$ is divisible by 11.
-
This explains it very well! Thank you so much. – SNS Feb 27 '12 at 23:42
$2772=99 \times 28$
so
$277227722772\ldots277227722772 = 99 \times 28 \times 100010001\ldots000100010001$
and so is divisible by both $9$ and $11$ (and $4$ and $7$ and other numbers).
The following is multiple choice question (with options) to answer.
Which of the following numbers is divisible by each one of 3, 7, 9 and 11 ? | [
"2079",
"456",
"974",
"562"
] | A | 639 is not divisible by 7
2079 is divisible by each of 3, 7, 9, 11.
ANSWER A |
AQUA-RAT | AQUA-RAT-37575 | In solving many kinematic problems for gears, there is a key rule to key in mind: When two gears are meshing, the tangential velocities of both gears at the point of meshing are equal.
(Also, as said in EMiller's answer, gear C seems to be the smaller orange gear, and gear D the larger yellow gear, contrary to the diagram's labelling).
This rule will help provide the equations we need to solve this problem. Note that there are 2 pairs of meshing gears (A meshes with C and B meshes with D). The points of meshing for each pair are $P$ and $Q$ respectively. Therefore, this gives us 2 equations to work with. It is then a matter of relating the tangential velocities at the point of meshing with the velocity of the centre of gear C and D. I will define the following terms:
Let $v_{a,p}, v_{c,p}$ be the tangential velocities of gears A and C, respectively, at point P.
Let $v_{b,q}, v_{d,q}$ be the tangential velocities of gears A and C, respectively, at point P.
Let $V$ be the velocity of the centre of gear C and D.
(The following diagrams represents gears A and C, and then B and D, as circles that are in pure rolling with one another:)
According to the rule above, we get the following two equations:
$$v_{a,p}=v_{c,p} \qquad (Eq.1)$$ $$v_{b,q}=v_{d,q} \qquad (Eq.2)$$
$v_{a,p}$ and $v_{b,q}$ are simple enough to determine since the centres of gears A and B don't move:
$$v_{a,p}=\omega_a R_a = 0.4\text{m/s} \qquad v_{b,q}=\omega_b R_b = 0.4\text{m/s}$$
The following is multiple choice question (with options) to answer.
Two interconnected, circular gears travel at the same circumferential rate. If Gear A has a diameter of 10 centimeters and Gear B has a diameter of 50 centimeters, what is the ratio of the number of revolutions that Gear A makes per minute to the number of revolutions that Gear B makes per minute? | [
"1:5",
"9:25",
"5:1",
"25:9"
] | C | Same circumferential rate means that a point on both the gears would take same time to come back to the same position again.
Hence in other words, time taken by the point to cover the circumference of gear A = time take by point to cover the circumference of gear B
Time A = 2*pi*25/Speed A
Time B = 2*pi*5/Speed B
Since the times are same,
50pi/Speed A = 10pi/Speed B
SpeedA/Speed B = 50pi/30pi = 5/1
Correct Option: C |
AQUA-RAT | AQUA-RAT-37576 | logic
$x\{t/y\}$ is $t$ if $x=y$, and $x$ otherwise.
$(e+e')\{t/y\} = e\{t/y\} + e'\{t/y\}$ (easy case)
$(\sum_x e)\{t/y\} = ??$
The following is multiple choice question (with options) to answer.
If x/y = 5/3, then (x + y)/(x - y) = ? | [
"4",
"1/5",
"-1/6",
"-1/5"
] | A | Any x and y satisfying x/y = 5/3 should give the same value for (x + y)/(x - y).
Say x=5 and y=3, then (x + y)/(x - y) = (5 + 3)/(5-3) = 4.
Answer: A. |
AQUA-RAT | AQUA-RAT-37577 | Therefore H = (n+1)Hn n.
This is a much better approach as we need only make one approximation to find the sum: 101(ln(100) + 1/200 + 0.5772) 100 423.924; incredibly the actual sum, without using the approximation, is 423.925 (3 d.p.).
Problem ID: 209 (17 Feb 2005) Difficulty: 4 Star
Only Show Problem
The following is multiple choice question (with options) to answer.
For all even numbers n, h(n) is defined to be the sum of the even numbers between 4 and n, inclusive. What is the value of h(12)/h(10) ? | [
" 1.8",
" 3",
" 6",
" 1.5"
] | D | CONCEPT: When terms are in Arithmetic Progression (A.P.) i.e. terms are equally spaced then
Mean = Median =(First+Last)/2
and Sum = Mean*Number of terms
h(12) = [(4+12)/2]*5 = 40
h(10) = (4+10)/2]*4 = 28
h(12)/h(10) = (40) / (28) ~ 1.5
Answer : D |
AQUA-RAT | AQUA-RAT-37578 | 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.
Last year, sales at Company X were 10% greater in February than in January, 15% less in March than in Feb, 20% greater in April than in March, 10% less in May than in April, and 5% greater in June than in May. In which month were sales closest to Jan? | [
"Feb",
"Mar",
"Apr",
"May"
] | D | Jan = $100
10% greater in February than in January ---> Feb = 1.1 Jan = $110
15% less in March than in Feb ---> Mar = 0.85 Feb = $93.5
20% greater in April than in March ---> Apr = 1.2 Mar = $112
10% less in May than in April ---> May = 0.9 Apr = $101 (we have a winner)
5% greater in June than in May --->Jun = 1.05 May = $106
Ans D |
AQUA-RAT | AQUA-RAT-37579 | # 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.
A boat running downstream covers a distance of 12 km in 2 hours while for covering the same distance upstream, it takes 3 hours. What is the speed of the boat in still water? | [
"8 kmph",
"5 kmph",
"4 kmph",
"6 kmph"
] | B | Speed downstream =12/2 = 6 kmph
Speed upstream =12/3 = 4 kmph
Speed of the boat in still water = 6+4 / 2= 5 kmph
ANSWER:B |
AQUA-RAT | AQUA-RAT-37580 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
The following is multiple choice question (with options) to answer.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: | [
"101",
"107",
"111",
"185"
] | C | EXPLANATION
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Answer C |
AQUA-RAT | AQUA-RAT-37581 | # How many distinct ways to climb stairs in 1 or 2 steps at a time?
I came across an interesting puzzle:
You are climbing a stair case. It takes $n$ steps to reach to the top. Each time you can either climb $1$ or $2$ steps. In how many distinct ways can you climb to the top?
Is there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?
Let $F_n$ be the number of ways to climb $n$ stairs taking only $1$ or $2$ steps. We know that $F_1 = 1$ and $F_2 = 2$. Now, consider $F_n$ for $n\ge 3$. The final step will be of size $1$ or $2$, so $F_n$ = $F_{n-1} + F_{n-2}$. This is the Fibonacci recurrence.
The following is multiple choice question (with options) to answer.
A person walking takes 26 steps to come down on a escalator and it takes 30 seconds for him for walking. The same person while running takes 18 second and 34 steps. How many steps are there in the escalator? | [
"45 steps",
"46 steps",
"47 steps",
"48 steps"
] | B | 26+30n=18+34n
12n=8
n=2/3 substitute n valve
26+30(2/3)=46 steps
ANSWER:B |
AQUA-RAT | AQUA-RAT-37582 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row upstream at 8kmph and downstream at 13 kmph. The speed of the stream is | [
"2.5 km/hr",
"4.2 km/hr",
"5 km/hr",
"10.5 km/hr"
] | A | Sol.
Speed of stream = 1/2(13 -8) kmph = 2.5 kmph.
Answer A |
AQUA-RAT | AQUA-RAT-37583 | concentration
Title: Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%? The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).
The following is multiple choice question (with options) to answer.
How much water must be mixed in 300 ml of sugar solution which contains 40% sugar by Weight, such that it becomes a 30% sugar solution? | [
"50 ml",
"75 ml",
"90 ml",
"100 ml"
] | D | Total Quantity = 300 ml
Sugar by weight = 120 ml
Water= 180 ml
Let x be the amount of water added to the solution
Then 120/(300 + x)=30/100
=>1200 = 900 + 3x
=>x= 100 ml
ANSWER:D |
AQUA-RAT | AQUA-RAT-37584 | • geometric mean — An average obtained by calculating the nth root of a set of n numbers. For example the geometric mean of 7, 100, and 107 is %3→74 900 = 42. 15, which is considerably less than the arithmetic mean of 71. 3 … Big dictionary of business and management
• Weighted geometric mean — In statistics, given a set of data, : X = { x 1, x 2, ..., x n } and corresponding weights, : W = { w 1, w 2, ..., w n } the weighted geometric mean is calculated as: ar{x} = left(prod {i=1}^n x i^{w i} ight)^{1 / sum {i=1}^n w i} = quad exp… … Wikipedia
• Mean — This article is about the statistical concept. For other uses, see Mean (disambiguation). In statistics, mean has two related meanings: the arithmetic mean (and is distinguished from the geometric mean or harmonic mean). the expected value of a… … Wikipedia
• Geometric-harmonic mean — In mathematics, the geometric harmonic mean M( x , y ) of two positive real numbers x and y is defined as follows: we first form the geometric mean of g 0 = x and h 0 = y and call it g 1, i.e. g 1 is the square root of xy . We then form the… … Wikipedia
• arithmetic mean — arithmetic average An average obtained by adding together the individual numbers concerned and dividing the total by their number. For example, the arithmetic mean of 7, 20, 107, and 350 is 484 4 = 121. This value, however, gives no idea of the… … Big dictionary of business and management
• mean — arithmetic mean; n. the average of a group of observations calculated by adding their values and dividing by the number in the group. When one or more observations are substantially different from the rest, which can influence the arithmetic mean … The new mediacal dictionary
The following is multiple choice question (with options) to answer.
The harmonic mean of two numbers xx and yy, symbolized as h(x,y)h(x,y), is defined as 2 divided by the sum of the reciprocals of xx and yy, whereas the geometric mean g(x,y)g(x,y) is defined as the square root of the product of xx and yy (when this square root exists), and the arithmetic mean m(x,y)m(x,y) is defined as x+y2x+y2. For which of the following pairs of values for xx and yy is g(x,y)g(x,y) equal to the arithmetic mean of h(x,y)h(x,y) and m(x,y)m(x,y)? | [
"x=−2 y=−1",
"x=−1, y=2",
"x=2, y=8",
"x=8, y=8"
] | D | A. x=−2 , y=−1
B. x=−1 , y=2
C. x=2 , y=8
D. x=8 , y=8
E. x=8 , y=64
The arithmetic mean, the geometric mean and the harmonic mean of two equal numbers are equal((x+x)/2=x, sqrt(x*x)=x, (2x*x)/(x+x)=x).
So without tedious calculation, (D) should be answer. |
AQUA-RAT | AQUA-RAT-37585 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Two pipes A and B can fill a tank in 15 min and 20 min respectively. Both the pipes are opened together but after 4 min, pipe A is turned off. What is the total time required to fill the tank? | [
"14 min 49 sec",
"14 min 70 sec",
"14 min 41 sec",
"14 min 40 sec"
] | D | Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15
Remaining part = 1 - 7/15 = 8/15
Part filled by B in 1 minute = 1/20
1/20 : 8/15 :: 1 ; x
x = 8/15 * 1 * 20 = 10 2/3 min = 10 min 40 sec.
The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec.
Answer:D |
AQUA-RAT | AQUA-RAT-37586 | • $P_2$ will fly $\big[1-(d+r+y)\big]$ distance away from the airport in the counter-clockwise direction to meet up with $P_3$.
• At this point, $P_2$ will donate $z$ fuel to $P_3$.
• $P_2$ and $P_3$ will then both fly back $z$ distance, arriving at a distance of $1-d-r-y-z$ from the airport with no fuel.
• After refuelling at the airport, $P_1$ will fly the distance towards $P_2$ and $P_3$ and refund each of them for that much fuel. All three planes will then head back to the airport together.
From this, we must have
• $0 \leqslant s\leqslant d/3$: $P_1$ can fly $s$ distance forward and backwards, and refund $P_2$ for $s$ distance
• $z\geqslant 0$: cannot donate negative fuel
• $2x + 1-d-r-y \leqslant d+r+y$: $P_3$ must not run out of fuel before $P_2$ can reach it again
• $1-d-r-y - z \leqslant d/4$: $P_1$ can reach $P_2$ and $P_3$, refund them both, and the three of them will have enough fuel to head back to the airport
• $2x + 2s + 1-d-r-y - z\leqslant d+r+y + z$: $P_2$ and $P_3$ must not run out of fuel before $P_1$ can reach them again
Putting these together:
The following is multiple choice question (with options) to answer.
The first flight out of Phoenix airport had a late departure. If the next three flights departed on-time, how many subsequent flights need to depart from Phoenix on-time, for the airport's on-time departure rate to be higher than 60%?
I will see what is the quickest way to solve it then I will provide the explanation | [
"6",
"7",
"9",
"4"
] | D | The following approach might be the easiest one and less error prone.
We need on-time departure rate to be higher than 6/10, so it should be at least 7/11, which means that 7 out of 11 flights must depart on time. Since for now 3 out of 4 flights departed on time then 7-3=4 subsequent flights need to depart on-time.
Answer: D |
AQUA-RAT | AQUA-RAT-37587 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
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Joined: 22 May 2016
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
How many liters of water must be added to 20 liters of milk and water containing 10% water to make it 50% water? | [
"10",
"12",
"14",
"16"
] | D | By rule of alligation:
50% - 10% = 40%
100% - 50% = 50%
Quantity of pure water : Quantity of the mixture = 4 : 5
There are 20 liters of mixture, so we need to add 16 liters of pure water.
The answer is D. |
AQUA-RAT | AQUA-RAT-37588 | Define y = 2^{2x}. Then
y^2 - y - 2 = 0
(y - 2)(y + 1) = 0
Thus y = 2 or y = -1.
So
2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
or
2^{2x} = -1, which is impossible.
Thus x = 1/2.
-Dan
5. Originally Posted by erika
Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
In general, you don't. However there are occasional special cases where you can. For example:
Solve for x:
(2^x)*(3^x) = 216
(2*3)^x = 216
6^x = 6^3
Thus x = 3.
So keep an eye out for ones you can do.
-Dan
The following is multiple choice question (with options) to answer.
If x and y are integers and 3^(x-5) = 3^(y+2), what is y in terms of x? | [
"x - 6",
"x - 7",
"x - 2",
"x + 2"
] | B | Answer: A
3^(x-5) = 3^(y+2)
Since the base is the same, I isolated the equations and solved for y.
x-5 = y+2
y=x-7
Ans : B |
AQUA-RAT | AQUA-RAT-37589 | ### 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.
Each of three investments has a 20% of becoming worthless within a year of purchase, independently of what happens to the other two investments. If Simone invests an equal sum Q in each of these three investments on January 1, the approximate chance that by the end of the year, she loses no more than 1/3 of her original investment is | [
"90%",
"80%",
"70%",
"60%"
] | A | The problem asks for the approximate chance that no more than 1/3 of the original investment is lost. We can apply the “1 – x” technique: what’s the chance that more than 1/3 of the original investment is lost? There are two outcomes we have to separately measure:
(a) All 3 investments become worthless.
(b) 2 of the 3 investments become worthless, while 1 doesn’t.
Outcome (a): The probability is (0.2)(0.2)(0.2) = 0.008, or a little less than 1%.
Outcome (b): Call the investments X, Y, and Z. The probability that X retains value, while Y and Z become worthless, is (0.8)(0.2)(0.2) = 0.032. Now, we have to do the same thing for the specific scenarios in which Y retains value (while X and Z don’t) and in which Z retains value (while X and Y don’t). Each of those scenarios results in the same math: 0.032. Thus, we can simply multiply 0.032 by 3 to get 0.096, or a little less than 10%.
The sum of these two probabilities is 0.008 + 0.096 = 0.104, or a little more than 10%. Finally, subtracting from 100% and rounding, we find that the probability we were looking for is approximately 90%.
The correct answer is A.
This problem illustrates the power of diversification in financial investments. All else being equal, it’s less risky to hold a third of your money in three uncorrelated (independent) but otherwise equivalent investments than to put all your eggs in one of the baskets. That said, be wary of historical correlations! Housing price changes in different US cities were not so correlated—and then they became highly correlated during the recent housing crisis (they all fell together), fatally undermining spreadsheet models that assumed that these price changes were independent. |
AQUA-RAT | AQUA-RAT-37590 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160, how much money did he borrow? | [
"8000",
"80288",
"2668",
"2600"
] | A | Let the sum be Rs. x. Then,
[(x * 6 * 3)/100] + [(x * 9 * 5)/100] + [(x * 13 * 3)/100] = 8160
18x + 45x + 39x = (8160 * 100)
102x = 816000 => x = 8000.
Answer:A |
AQUA-RAT | AQUA-RAT-37591 | # 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 ratio between the sale price and the cost price of an article is 12:7. What is the ratio between the profit and the cost price of that article? | [
"2:9",
"2:5",
"3:6",
"5:7"
] | D | Let C.P. = Rs. 7x and S.P. = Rs. 12x.
Then, Gain = Rs. 5x
Required ratio = 5x : 7x = 5:7
Answer:D |
AQUA-RAT | AQUA-RAT-37592 | # Math Help - Frequency Distribution?
1. ## Frequency Distribution?
Hi all, plz help me on this..
The output in units of 20 employees during one week was as follows:
65, 69, 70, 71, 70, 68, 69, 67, 70, 68
72, 71, 69, 74, 70, 73, 71, 67, 69, 70
a) Start doing tabulation of given data. Make 5 classes by using suitable class interval.
b) Make the frequency distribution.
c) Also make the column of class mark.
d) Also find out the mean by constructing relevant column “f X”
e) Also compute relative frequency distribution and cumulative frequency distribution.
2. Originally Posted by cu4mail
Hi all, plz help me on this..
The output in units of 20 employees during one week was as follows:
65, 69, 70, 71, 70, 68, 69, 67, 70, 68
72, 71, 69, 74, 70, 73, 71, 67, 69, 70
a) Start doing tabulation of given data. Make 5 classes by using suitable class interval.
b) Make the frequency distribution.
c) Also make the column of class mark.
d) Also find out the mean by constructing relevant column “f X”
e) Also compute relative frequency distribution and cumulative frequency distribution.
Sort the date in ascending or descending order. You have 20 employees, but you would like 5 classes. These classes do not have to be the same sample size, so make their intervals equal the length of: $\frac{MAX-MIN}{5}$.
3. Originally Posted by cu4mail
Hi all, plz help me on this..
The output in units of 20 employees during one week was as follows:
65, 69, 70, 71, 70, 68, 69, 67, 70, 68
72, 71, 69, 74, 70, 73, 71, 67, 69, 70
The following is multiple choice question (with options) to answer.
For the past n days, the average (arithmetic mean) daily production at a company was 60 units. If today's production of 90 units raises the average to 62 units per day, what is the value of n ? | [
"30",
"14",
"10",
"9"
] | B | (average production for n days) * n = (total production for n days) --> 60n=(total production for n days);
(total production for n days) + 90 = (average production for n+1 days) * (n+1) -->60n + 90 = 62 * (n+1) --> n=14.
Or as 30 extra units increased the average for n+1 days by 2 units per day then 30/(n+1)=2 --> n=14.
Answer: B. |
AQUA-RAT | AQUA-RAT-37593 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
During a certain week, a seal ate 45% of the first 80 smelt it came across, and 30% of the remaining smelt it came across. If the seal ate 40% of the smelt it came across during the entire week, how many smelt did it eat? | [
"32",
"48",
"55",
"64"
] | B | Total smelt = x.
Then, 0.45*80+0.3(x-80)=0.4*x --> x=120 --> 0.4*x=48.
Answer: B. |
AQUA-RAT | AQUA-RAT-37594 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
At present, the ratio between the ages of Arun and Deepak is 4:3. After 6 years, Arun's age will be 26 years. What is the age of Deepak at present? | [
"77 years",
"15 years",
"66 years",
"88 years"
] | B | Let the present ages of Arun and Deepak be 4x and 3x years respectively.
Then, 4x + 6 = 26 => x = 5
Deepak's age = 3x = 15 years.
Answer: B |
AQUA-RAT | AQUA-RAT-37595 | 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
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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.
The mean of four integers A will not change if all the integers are multiplied by any constant. What is always true about this set of numbers?
I. The mean of the set is 0
II. The sum of the largest member and the smallest member of the set is 0
III. The set contains both positive and negative integers | [
"I only",
"II only",
"III only",
"I and II only"
] | A | If we consider a set {1,2,3,4}, when multiplied with a constant we take that constant as 1 and multiply resulting set will contain the same numbers as the original. so how can the answer be I only.....A |
AQUA-RAT | AQUA-RAT-37596 | 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
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of five numbers is 281. The average of the first two numbers is 280 and the average of last two numbers is 178.5. What is the third number? | [
"27",
"88",
"16",
"19"
] | B | Answer:B |
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