source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34497 | 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.
In a 100 m race, A runs at 8km per hour. If A gives B a start of 6 m and still him by 15 seconds, what is the speed of B ? | [
"5.56 km/hr.",
"5.06 km/hr.",
"5.64 km/hr.",
"6.76 km/hr."
] | C | Time taken by A to cover 100 m =(60 X 60 / 8000) x 100 sec = 45 sec.
B covers (100 - 6) m = 94m in (45 + 15) sec = 60 sec.
B's speed = (94 x 60 x 60)/(60 x 1000)km/hr = 5.64 km/hr.
Answer is C |
AQUA-RAT | AQUA-RAT-34498 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can finish a work in 10 days and B can do same work in half the time taken by A. then working together, what part of same work they can finish in a day? | [
"1/5",
"1/6",
"1/7",
"3/10"
] | D | Explanation:
Please note in this question, we need to answer part of work for a day rather than complete work. It was worth mentioning here because many do mistake at this point in hurry to solve the question
So lets solve now,
A's 1 day work = 1/10
B's 1 day work = 1/5 [because B take half the time than A]
(A+B)'s one day work =
(1/10+1/5)=3/10
So in one day 3/10 work will be done
Answer: D |
AQUA-RAT | AQUA-RAT-34499 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
The average (arithmetic mean) of four different positive integers is 12. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer? | [
"5",
"4",
"3",
"2"
] | C | Let the second integer be x and the fourth be a.
Then [3x + x + (x+2) + a]/4 = 12
=> 5x + 2 + a = 48
=> 5x + a = 48
=> a = 48 - 5x
From the above equation we can see that a is minimum when x is maximum, provided both are positive
The maximum value that x can take in the above equation while still keeping a positive is x=9
This gives us a= 48 - 45 = 3
Therefore the minimum value that the fourth integer can have is 3. Option C. |
AQUA-RAT | AQUA-RAT-34500 | That would be a total of 5x30 + 3x40 + 2x60 = 390 plants (with an arbitrary factor that we'll set to 1 without loss of generality).
The amount of highbush is 5x30 = 150.
The amount of lowbush is 3x40 = 120.
The amount of hybrid is 2x60 = 120.
If the opossums didn't care, they would likely eat blueberries in this ratio (null hypothesis H0).
The total that we have observed the opossums to eat is 5% x 150 + 10% x 120 + 20% x 120 = 43.5 plants.
They eat 5% large, which a corresponding fraction of 5% x 150 / (5% x 150 + 10% x 120 + 20% x 120) = 17%
They eat 10% low for 10% x 120 / (5% x 150 + 10% x 120 + 20% x 120) = 28%
They eat 20% hybrid for 20% x 120 / (5% x 150 + 10% x 120 + 20% x 120) = 55%.
Checking... yes the total is 100%.
What we see is that the opossums prefer hybrid by far.
Small blueberries are their second choice.
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Hi anemone!
What do you mean by the symbol E?
Anyway, you've found that the opossums eat 45.8% large and 54.2% small for a total of 100%.
But... what happened to the hybrid blueberries?
By the symbol E, I meant the blueberries (all 3 types of them) that are eaten by opossums...
That would be a total of 5x30 + 3x40 + 2x60 = 390 plants (with an arbitrary factor that we'll set to 1 without loss of generality).
The amount of highbush is 5x30 = 150.
The amount of lowbush is 3x40 = 120.
The amount of hybrid is 2x60 = 120.
If the opossums didn't care, they would likely eat blueberries in this ratio (null hypothesis H0).
The following is multiple choice question (with options) to answer.
In the country of Celebria, the Q-score of a politician is computed from the following formula:
Q=(41ab^2c^3)/d2, in which the variables a, b, c, and d represent various perceived attributes of the politician, all of which are measured with positive numbers. Mayor Flower’s Q-score is 150% higher than that of Councilor Plant; moreover, the values of a, b, and c are 60% higher, 40% higher, and 20% lower, respectively, for Mayor Flower than for Councilor Plant. By approximately what percent higher or lower than the value of d for Councilor Plant is the corresponding value for Mayor Flower? | [
"56% higher",
"25% higher",
"8% lower",
"20% lower"
] | D | Let's say M stands for Mayor and P stands for councilor
Then as per given information -
Qm = 2.5*Qp
Am = 1.6 Ap
Bm = 1.4 Bp
Cm = 0.8 Cp
and let
Dm = X* Dp
On substituting these values and solving for X we get X ~ 0.8 ....
Hence it is lower by 20%
Ans - D |
AQUA-RAT | AQUA-RAT-34501 | # 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.
Eleven bags are bought for Rs.1000 and sold at 10 for Rs.1100. What is the gain or loss in percentage? | [
"29%",
"21%",
"23%",
"28%"
] | B | Explanation:
As selling price is Rs.1100 for 10 bags, each bag is sold at Rs.110.
Hence, the profit is Rs.100 after selling 10 bags plus the selling price of 1 bag = 100 + 110 = Rs.210.
%profit = 210/1000 *100
= 21%
ANSWER: B |
AQUA-RAT | AQUA-RAT-34502 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
The prices of tea and coffee per kg were the same in june. In july the price of coffee shot up by 20% and that of tea dropped by 20%. if in july , a mixture containing equal quantities of tea and coffee costs 90/kg. how much did a kg of coffee cost in june ? | [
"50",
"60",
"80",
"90"
] | D | Let the price of tea and coffee be x per kg in June.
Price of tea in July = 1.2x
Price of coffee in July = 0.8x .
In July the price of 1/2 kg (900gm) of tea and 1/2 kg (900gm) of coffee (equal quantities) = 90
1.2x(1/2) + 0.8x(1/2) = 90
=> x =90
Thus proved...option D. |
AQUA-RAT | AQUA-RAT-34503 | earth
Well, lets look at the values returned compared to days in a non-leap year:
Month N1 Day Diff
1. 30 31 -1
2. 61 59 +2
3. 91 90 +1
4. 122 120 +2
5. 152 151 +1
6. 183 181 +2
7. 213 212 +1
8. 244 243 +1
9. 275 273 +2
10. 305 304 +1
11. 336 334 +2
12. 366 365 +1
A bit messy, right? Now, remember that you're subtracting 30 from the total at the end to get N, and we're adding in the current date. This means that although we multiply our current month to get N1, we're actually using this to calculate the dates from the months prior to our current month! Thus if we take the value of N1, subtract it by 30, and compare it to the preceding month, the chart will come out like this:
Month N1 Day Diff
1. 31 31 0
2. 61 59 +2
3. 92 90 +2
4. 122 120 +2
5. 153 151 +2
6. 183 181 +2
7. 214 212 +2
8. 245 243 +2
9. 275 273 +2
10. 306 304 +2
11. 336 334 +2
12. --- 365 ---
From this, you can see that the value of N1 will equal 2 greater than the actual date for any day in which it is March or later. This is perfect, as N2 is already a formula determining this for catching leap days. Note, these would all equal +1 on a leap year, as in another day would have been added in February. Thus coming back to the final calculation:
N = N1 - (N2 * N3) + day - 30
The following is multiple choice question (with options) to answer.
The maximum gap between two successive leap year is? | [
"2",
"8",
"9",
"6"
] | B | Explanation:
This can be illustrated with an example.
Ex: 1896 is a leap year.The next leap year comes in 1904 (1900 is not a leap year).
Answer: B) 8 |
AQUA-RAT | AQUA-RAT-34504 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
What is the greatest integer that will always evenly divide the sum of four consecutive integers? | [
"2",
"3",
"4",
"6"
] | C | Let the three consecutive even integers be x , x + 4 , x+8 , x+12
Sum = x + x + 4 + x + 8 + x + 12
= 4x+24 = 4(x+6)
Therefore , 4 will always be a factor .
Answer C |
AQUA-RAT | AQUA-RAT-34505 | a)f(x)=1 / (x^2+1)
d)f(x)={1 / (x^3+x)}^5
f)h(x)=x-(x^2)
For each of the above functions from a) through h) how do you determine if it is even,odd,or neither??
a)
$f(x) = \frac{1}{x^2 + 1}$
$f(-x) = \frac{1}{(-x)^2 + 1}$
$f(-x) = \frac{1}{x^2 + 1} = f(x)$
so this function is even.
d)
$f(x) = \left ( \frac{1}{x^3 + x} \right ) ^5$
$f(-x) = \left ( \frac{1}{(-x)^3 + (-x)} \right ) ^5$
$f(-x) = \left ( \frac{1}{-x^3 - x} \right ) ^5$
$f(-x) = \left ( \frac{-1}{x^3 + x} \right ) ^5$
$f(-x) = (-1)^5 \left ( \frac{1}{x^3 + x} \right ) ^5$
$f(-x) = - \left ( \frac{1}{x^3 + x} \right ) ^5 = -f(x)$
so this function is odd.
f)
$h(x) = x - x^2$
$h(-x) = (-x) - (-x)^2$
$h(-x) = -x + x^2$
which is equal to neither h(x) nor -h(x). So this function is neither even, nor odd.
-Dan
The following is multiple choice question (with options) to answer.
For any function Fn(x)=Fn−1(F(x))Fn(x)=Fn−1(F(x)) if for n > 1 also g(x) = 1/x, h(x) = x−−√x and k(x) = x2x2 then what is the value of g(h3(k2(x)))? | [
"1",
"8",
"7",
"6"
] | A | Given Fn(x)=Fn−1(F(x));Fn(x)=Fn−1(F(x)); g(x) = 1/x; h(x) = x−−√x; k(x) = x−−√x
Then, k(k(x)) = (x2)2=x4(x2)2=x4
And, h3(k2(x))=h(h(h(k2(x))))=h(h(x2))=h(x)=x−−√.h3(k2(x))=h(h(h(k2(x))))=h(h(x2))=h(x)=x.
And, g(h3(k2(x)))=g(x−−√)=1/x−−√.g(h3(k2(x)))=g(x)=1/x.
Hence answer is option (A). |
AQUA-RAT | AQUA-RAT-34506 | Originally Posted by o_O
(Yes) I would do it the second way
It can be done the first way right thou?? If not can the second method be applied to the problem 29^202mod13
sry for all the dumb questions. I am just trying to learn the material better
• October 7th 2008, 07:47 PM
o_O
Well it really is the same work but you just did the unnecessary work of showing that 3 is congruent to 8 mod 5.
And this method WAS used for your problem. Go through it again. The principle is the same. From Fermat's theorem, you have your number to the power of p - 1 is congruent to 1. Raise it as close to the desired power as possible and manipulate it to get what you need.
The following is multiple choice question (with options) to answer.
Which of the following must be subtracted from 2^526 so that the resulting integer will be a multiple of 3? | [
"1",
"2",
"3",
"5"
] | A | 2^12 is definitely NOT a multiple of 3 as 2^12 will only have 2s in it.
Coming back to the question,
2^1 leaves a remainder of 2 when divided by 3
2^2 leaves a remainder of 1 when divided by 3
2^3 leaves a remainder of 2 when divided by 3
2^4 leaves a remainder of 1 when divided by 3... etc. and the cyclicity continues.
Thus, 2^526 will leave a remainder of 1 when divided by 3. Thus you must subtract 1 from 2^526 to make it divisible by 3.
A is thus the correct answer. |
AQUA-RAT | AQUA-RAT-34507 | (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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Joined: 25 Jul 2018
Posts: 731
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]
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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.
28 buckets of water fill a tank when the capacity of each bucket is 13.5 litres. How many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres? | [
"30",
"42",
"60",
"Data inadequate"
] | B | Capacity of the tank = 28 × 13.5= 378 litres
When the capacity of each bucket = 9 litres, then the required no. of buckets
= 378â„9 = 42
Answer B |
AQUA-RAT | AQUA-RAT-34508 | # how can I find the Side length Two squares inside an equilateral Triangle?
Question: Figure shows an equilateral triangle with side length equal to $$1$$ . Two squares of side length a and $$2a$$ placed side by side just fit inside the triangle as shown.
Find the exact value of $$a$$.
Its an Assessment question from edX course "A-Level Mathematics Course 1" and I am supposed to use skills that I learnt in Indices and surds,Inequalities and The Factor Theorem.
I have tried finding the height of triangle and then use similar triangles to find the right triangle length still No luck.
I am just looking for food for thought or very small hints thats all.
Hint: You have everything you need along the base. Also, note the two flanking right triangles.
• i think i got it. Thanks – fpsshubham Sep 18 '19 at 14:12
From the leftmost right triangle, $$\frac{a}{x} = \tan(60°) \implies a = \sqrt{3}x$$ From the rightmost right triangle $$\frac{2a}{1-3a-x} = \frac{a}{x} \\ a = \frac{3-\sqrt{3}}{6} \approx 0.211$$
• I think you are on the right path to an answer, but there is not enough detail. First, you should typeset your answer with MathJax. Second, give more details. E.g. what is $x$? – Physical Mathematics Apr 21 '20 at 21:18
I struggled with this too. But the info is along the bottom. The triangle is equilateral so all angles are $$60°$$. On the left there is a right-angled triangle - let's call its base $$x$$.
Triangle 1: Angle = $$60°$$, opposite = $$a$$, and adjacent = $$x$$
On the right there is another right-angled triangle and its base is $$1-3a-x$$.
Triangle 2: Angle = $$60°$$, opposite = $$2a$$, and adjacent = $$1 - 3a - x$$
The following is multiple choice question (with options) to answer.
What is the are of an equilateral triangle of side 16 cm? | [
"64√6 cm2",
"84√3 cm2",
"64√9 cm2",
"64√3 cm2"
] | D | Area of an equilateral triangle = √3/4 S2
If S = 16, Area of triangle = √3/4 * 16 * 16
= 64√3 cm2;
Answer: D |
AQUA-RAT | AQUA-RAT-34509 | I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25
Does everyone agree with this solution or have the same approach and way of explaining the problem?
6. Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.
Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.
$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$
So far I agree with yma's answer. Here are the planes.
ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE
Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.
The following is multiple choice question (with options) to answer.
There is a plane contains 32 points.all the 32 points have equal distance from point x. which of the following is true . | [
"ll 32 points lie in circle",
"the distance from x to all 32 points is less than the distance between each other c. both a and b",
"both a and b",
"32"
] | C | Option 3
X must be the center of the circle and 32 points are on the circumference. So Option A is correct
Number of diagnols of a regular polygon = n(n−3)2n(n−3)2
So for a polygon of 32 sides, Number of diagnols = 464. Now the minimum distance between any two points = 2πr32=1156r2πr32=1156r
Now total lengh of all the distances from 32 points = 2πr2πr + Sum of the lengths of all the 464 diagnols.
Sum of the lengths of x to all the 32 points = 32 radius = 32r
But the 464 diagnols have 16 diameters connecting 2 oposite points connecting via center. So Sum of the lengths of distances from point to point is clearly greater than sum of the length from x to all 32 ponts. Option B is correct
Answer:C |
AQUA-RAT | AQUA-RAT-34510 | ### 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
Posts: 9558
Location: Pune, India
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
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
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.
In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can? | [
"40",
"44",
"48",
"52"
] | B | Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44.
ANSWER:B |
AQUA-RAT | AQUA-RAT-34511 | Denote: $$y=ax, z=bx$$. Then: $$x+y+z=0 \iff x+ax+bx=0 \iff a=-1-b;\\ S={x\over y} + {y\over z} + {z\over x} = {y\over x} + {z\over y} + {x\over z} \iff \\ S={1\over a} + {a\over b} + b = a + {b\over a} + {1\over b} \iff \\ S=-{1\over 1+b} - {1+b\over b} + b = -1-b - {b\over 1+b} + {1\over b} \iff \\ \frac{(b+2)(2b+1)(b-1)}{b(1+b)}=0 \iff \\ b_{1,2,3}=\{\color{red}{-2},\color{green}{-\frac12},\color{blue}1\}; a_{1,2,3}=\{\color{red}{1},\color{green}{-\frac12},\color{blue}{-2}\}\\ S_1=\frac1a+\frac ab+b=\frac1{\color{red}{1}}+\frac{\color{red}1}{\color{red}{-2}}+(\color{red}{-2})=-\frac32.\\ S_2=\frac1a+\frac ab+b=\frac1{\color{green}{-\frac12}}+\frac{\color{green}{-\frac12}}{\color{green}{-\frac12}}+(\color{green}{-\frac12})=-\frac32.\\ S_3=\frac1a+\frac
The following is multiple choice question (with options) to answer.
If X:Y is 1:5 and Y:Z is 5:7 then X:Z is equal to | [
"1:6",
"1:5",
"1:4",
"1:7"
] | D | The two ratios given are having the same number 5 for Y in both the ratios.
Hence- X:Y = 1:5 Y:Z = 5:7
=> X:Z = 1:7
Answer D |
AQUA-RAT | AQUA-RAT-34512 | $$2nd$$ case when $$y+3 \lt 0 \Rightarrow y \lt -3$$ and in this case $$y$$ will have infinite values and I am not able to proceed from here.
The answer for the total number of solutions provided is $$10$$ and you can see that I've been able to find out the $$5$$ solutions in my $$1st$$ case. Can someone please guide or help me about how to proceed in the 2nd case?
• Notice that $|x-1||y+3|$ is an integer between $2$ and $5$, so there are only few possibilities to check.
– Sil
Mar 22, 2021 at 11:26
• Since $x$ is a negative integer, $|x - 1| \geq 2.$ This means that $|y+3|$ must be less than $3$. Mar 22, 2021 at 11:27
• thanks for such quick responses...but can you please direct me in the way of how I was trying to solve this rather than taking a whole new solution approach. If you have new approach...please explain down as answers...thanks! Mar 22, 2021 at 11:33
• In your second case we have $|y+3|=(-y-3)$ and also $1-x>1$ (in fact $1-x\geq 2$),and so by $(1-x)(-y-3) \le 5$ we must have $-y-3 \leq \frac{5}{2}$. This gives you a lower bound for $y$ similarly as in first case.
– Sil
Mar 22, 2021 at 11:51
• @Sil : can you please elaborate your last comment? I understood the part that $1-x \ge 2$ but i did not understood after that. If you kindly elaborate the later part of your comment. Mar 22, 2021 at 12:09
Both $$|x-1|$$ and $$|y+3|$$ are non-negative integers. As the product must lie between $$2$$ and $$5$$, the only candidates are:
The following is multiple choice question (with options) to answer.
If two integers x,y (x>y) are selected from -5 to 5 (inclusive), how many cases are there? | [
"55",
"60",
"65",
"70"
] | A | There are 11 integers from -5 to 5 inclusive.
11C2 = 55.
The answer is A. |
AQUA-RAT | AQUA-RAT-34513 | and 7.5 centimeters long. So we can say that <i>x2</i> is <i>t2</i> times <i>x</i> over <i>d</i> and <i>x1</i> is <i>t1</i> times <i>x</i> over <i>d</i> and we'll substitute for each of those in this expression for <i>delta x</i>. And that's what I've done here and the <i>x</i> over <i>d</i> can be factored out and it's gonna be <i>t2</i> minus<i> t1</i> times <i>x</i> over <i>d</i> and <i>Delta x</i> then substituting for <i>t2</i> and <i>t1</i> using this expression here substituting the numbers two and one for <i>n</i> . We have two times lambda over two minus one times lambda over two times <i>x</i> over <i>d</i> and then this works out to Lambda over two and then we can substitute in numbers. So that's 589 nanometres is the wavelength we told divided by two times seven and a half centimeters written as times ten to the minus two meters divided by 0.1 millimetres written as times ten to the minus three meters gives us this many nanometres which is 0.221 mm. So this is the horizontal separation between dark bands. Now what you would expect in this picture is to see lines like this spaced apart 0.221 millimeters and in reality though you don't see that because that's showing you that there are imperfections in these pieces of glass it means that they're not perfectly straight as shown in the drawing and so that's kind of interesting and this is one of the ways that they actually engineer really nearly perfect mirrors for telescopes and so on is by looking at interference patterns like this.
The following is multiple choice question (with options) to answer.
If 100 microns = 1 decimeter, and 1,000angstroms = 1 decimeter, how many angstroms equal 1 micron? | [
"1.0e-05",
"0.0001",
"10",
"10,000"
] | C | 100 microns = 1 decimeter
1,000 angstroms = 1 decimeter
100 microns = 1,000 angstroms
1 micron = 1,000/100 = 10
Answer : C |
AQUA-RAT | AQUA-RAT-34514 | 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.
A military camp has a food reserve for 250 personnel for 40 days. If after 15 days 50 more personnel are added to the camp, find the number of days the reserve will last for? | [
"20",
"36",
"25",
"42"
] | C | Explanation:
As the camp has a reserve for 250 personnel that can last for 40 days, after 10 days the reserve left for 250 personnel is for 30 days. Now 50 more personnel are added
in the camp.
Hence, the food reserve for 300 personnel will last for:
250:300::x:30 ……..(it is an indirect proportion as less men means more days)
x = (250*30)/300
x = 25 days
ANSWER C |
AQUA-RAT | AQUA-RAT-34515 | 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 with a speed of 15 kmph in still water. If the stream flows at 4 kmph, then the speed in downstream is? | [
"33",
"77",
"20",
"19"
] | D | M = 15
S = 4
DS = 15 + 4 = 19
Answer: D |
AQUA-RAT | AQUA-RAT-34516 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
If a fisherman catches 203128 shrimp while out and has to evenly split the catch between 9 restaurants, how many shrimp would he have left over? | [
"4",
"7",
"2",
"1"
] | B | In order to find the answer you need to find a number that is divisible by 9.
The sum of 203128 16. 9 is divisible by 9 therefore you would subtract 7. 16-7=9
There would be 7 shrimp leftover. Answer B. |
AQUA-RAT | AQUA-RAT-34517 | If stock price is $$X = (0,30)$$; we use the 7 puts (A) for profit = 7(40 - X); 9 (B) puts are used giving a loss = 9(30 - X). Total earning = 7(40-X) -9(30-X) + 2$= 2X + 12$ > 0
- 3 years, 2 months ago
Assume that the $40 put is still priced at$10.
What would be the price of the $30 put option, where there will be no arbitrage opportunity? Staff - 3 years, 2 months ago Log in to reply Let the price at which the 30$ put options is priced be $$M$$.
Now suppose that an arbitrage opportunity does exist. It is easily proved that the arbitrage opportunity(AO) must consist of buying 40$(A) put options and selling 30$ (B) put options. Let x (A) puts be bought and y (B) puts be sold for the AO.
So we spend $$10x$$ for the (A) puts and gain $$My$$ for the (B) puts. Total gain =$$My - 10x$$
If the stock price is above 40; both puts remain unused. Therefore net earnings = $$My - 10x.$$ Since earning is greater than 0 in an AO, $$My - 10x \geq 0 \implies My > 10x \implies (30-M)My > 10(30-M)x$$. Also, since $$M < 10$$, we have $$y > x$$
If the stock price is $$= P = (0,30)$$; both puts are used. Earning on (A) puts = $$x(40 - P)$$. Loss on B puts$$= y(30 - P).$$ Total earning = $$x(40 - P) - y(30-P) + My - 10x = 30x - (30 - M)y + (y-x)P$$
Since $$y - x > 0$$. Total earning is minimum when $$P = 0$$.
The following is multiple choice question (with options) to answer.
A 15% item yields 30%. The market value of the item is: | [
"Rs. 45",
"Rs. 55",
"Rs. 50",
"Rs. 60"
] | C | Let the face value of the item is Rs.100
It yields Rs.15
Market value of the stock=(15/30)*100=Rs.50
ANSWER:C |
AQUA-RAT | AQUA-RAT-34518 | The error lies in case $C1:$
. . the people are divided into two pairs.
Call the people $A,B,C,D.$
How many pairings are possible?
The answer is three: . $\{AB|CD\},\;\{AC|BD\},\;\{AD|BC\}$
. . Note that $\{CD|AB\}$ is not a different pairing.
Make this correction and you will get 256.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I first ran into this snag many years ago.
There are four bridge players.
In how many ways can the players be paired?
I immediately said: . $_4C_2 \:=\:\dfrac{4!}{2!\,2!} \:=\:6$
I was surprised to learn that the answer was three.
(And no explanation was given.)
I finally reasoned it out.
When we select a pair of players to be partners,
. . we have automatically assigned the other two to be partners.
Call the players $A,B,C,D.$
With whom can $A$ be paired?
$A$ can be paired with $B,C,\text{ or }D$ . . . 3 choices.
And that's it!
. . Get it?
The following is multiple choice question (with options) to answer.
Three-twentieths of the members of a social club are retirees who are also bridge players, Four-twentieths of the members are retirees, and one-half of the members are bridge players. If 110 of the members are neither retirees nor bridge players, what is the total number of members in the social club? | [
"200",
"300",
"360",
"400"
] | A | {Total} = {Retirees} + {Bridge players} - {Both} + {Neither}
x = 4/20*x + x/2 - 3/20*x + 110
20x = 4x + 10x - 3x + 110*20 (multiply by 20)
11x = 110*20
x = 200.
Answer: A |
AQUA-RAT | AQUA-RAT-34519 | to 6. A chord is a straight line joining any two parts of the circumference. If the handle is 28 cm long, what is the diameter of the pot? Give your answer correct to 3 significant figures. To find the perimeter, P, of a semicircle, you need half of the circle's circumference, plus the semicircle's diameter: P = 1 2 ( 2 π r ) + d The 1 2 and 2 cancel each other out, so you can simplify to get this perimeter of a semicircle formula. Calculate the area of one surface of the table mat. o Pi (π) Objectives. Area of a Semicircle In the case of a circle, the formula for area, A, is A = pi * r^2, where r is the circle’s radius. If someone could please help with the following question? What is the formula to calculate the distance perpendicular from the section line to the nearest edge of the circumference at any point X along the section line? Thanks in advance. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. A sector of a circle) is made by drawing two lines from the centre of the circle to the circumference. 1 foot ≈ 0. Its unit length is a portion of the circumference. The area of a circle is. This is the perimeter of the semi-circle!. perimeter (circumference) of a semicircle: pi times radius. A segment is the shape formed between the chord and the arc. A major arc is an arc that is larger than a semicircle. 14(15) Replace d with 15. WIIat is the perimeter of the figure shown below (semicircle attached to rectangle)? Give the exact answer. Find the circumference of the quad-rant with radius 4. Shapes and Figures. Finally, you can find the diameter - it is simply double the radius: D = 2 * R = 2 * 14 = 28 cm. Please find teh circumference of the basket so Mrs. !Shown below is a compound shape made from a rectangle and semi-circle. When the area is fixed and the perimeter is a minimum, or when the perimeter is fixed and the area is a maximum, use Lagrange multipliers to verify that the length of the rectangle is twice its
The following is multiple choice question (with options) to answer.
The radius of a semi circle is 5.2 cm then its perimeter is? | [
"32.52",
"32.48",
"26.74",
"32.9"
] | C | 36/7 r = 5.2 = 26.74
Answer:C |
AQUA-RAT | AQUA-RAT-34520 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio of two numbers is 3:4 and their sum is 42. The greater of the two numbers is? | [
"12",
"14",
"24",
"17"
] | C | 3:4
Total parts = 7
= 7 parts --> 42 (7 × 6 = 42)
= 1 part ---->6 (1 × 6 = 6)
= The greater of the two number is = 4
= 4 parts ----> 24 (6 × 4 = 24)
C) |
AQUA-RAT | AQUA-RAT-34521 | javascript, datetime
The year's correct, and the month's correct (January is zero), but the date is 1 day off. Thursday is day #4 in the 0-6 weekday scheme - but really, today's the 5th of November.
So if you were to run this on the, say, the 30th of November, you'd actually be using the 1st of November as your start date, since the 30th's a Monday. If you run it on a Sunday, getDay will return zero, meaning your start date will be the last of the previous month. And so on.
I also have to wonder why you've created a function, which builds an array just to hold a list of incrementing numbers, which you then map to dates. The simpler way of looping through incrementing numbers is a plain old for-loop.
Anyway, while calendars are complex, there's one thing that thankfully doesn't change: There are 7 days in a week. So you're creating 7 times more dates than you need, and filtering most of them away. You could consider simply skipping 7 dates at a time from the first occurrence.
Edit: The only thing in the above paragraph that holds true is that calendars are complex. As RobH points out in the comments, some pacific nations have been known to jump across the dateline. Presumably just to mess with programmers the world over!
Thirdly, don't mix calendar dates and raw time calculations like your division by milliseconds in a day. Switching back and forth between daylight savings time (aka winter/summer time) can cause all sorts of nonsense when mixing raw time and dates. Some calendar days are 25 hours long, some are 23. And since a date constructed from only year, month, and date sets its time to midnight, you risk skipping over a date, or repeating the same date twice.
Anyway, here's a simple solution that adds 1 day at a time until the year rolls over, storing dates that match what you're looking for:
function remainingDays(weekday) {
var current = new Date,
year = current.getFullYear(),
dates = [];
// a simple helper function
function nextDay(date) {
return new Date(date.getFullYear(), date.getMonth(), date.getDate() + 1);
}
The following is multiple choice question (with options) to answer.
In a particular year, the month of january had exactly 4 thursdays, and 4 sundays. On which day of the week did january 1st occur in the year. | [
"sunday",
"monday",
"tuesday",
"wednesday"
] | B | series of days may be either one of two:
(i)sun-4,mon-4,tue-4,wed-4,thu-4,fri-3 or 5,sat-3 or 5
(ii)thu-4,fri-4,sat-4,sun-4,mon-3or5,tue-3or5,wed-3or5
Lets check through option,
sun=1,8,15,22,29 jan ... which is not satisfies // wrong
Mon=1,8,15,22,29 jan,,,This satisfy option(ii) with thu-4,fri-4,sat-4,sun-4,mon-5,tue-5,wed-5
ANSWER:B |
AQUA-RAT | AQUA-RAT-34522 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
If 2 men or 3 women can reap a field in 12 days how long will 4 men and 5 women take to reap it? | [
"12/16",
"11/36",
"14/24",
"18/36"
] | B | Explanation:
2 men reap 1/12 field in 1 day
1 man reap 1/(2 x 12)
3 women reap 1/12 field in 1 day
1 woman reap 1/(12 x 3)
4 men and 5 women reap (4/(2 x 12)+ 5/(3 x 12) =11/36 in 1 day
4 men and 5 women will reap the field in 11/36 days
Answer: Option B |
AQUA-RAT | AQUA-RAT-34523 | Show Tags
25 Nov 2010, 21:43
2
shrive555 wrote:
In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?
20
40
216
720
729
Or out of 6 children, choose 3 in 6C3 ways = 20 ways.
Note: When you choose 3 children say, A, B and C are give them a red shirt, D, E and F get a green shirt. When you choose D, E and F and give them a red shirt, A, B and C automatically get the green shirts. So you do not need to multiply by 2! above.
_________________
Karishma
Veritas Prep GMAT Instructor
Intern
Joined: 24 Feb 2012
Posts: 31
Show Tags
24 Feb 2012, 17:19
Approach 1:
1st Child: 6 has options
2nd Child: 5 has options…
Therefore, for all kids: 6 x 5 x 4 x 3 x 2 = 720 arrangements.
Since the reds are identical, we divide by 3!; Since the greens are identical, we divide by another 3!
So: in all, 720/[ 3! X 3! ] = 20 ways.
Approach 2 / MGMAT technique:
This is like anagramming RRRGGG. No of arrangements = 6! / 3! x 3! ways ==> 20.
Math Expert
Joined: 02 Sep 2009
Posts: 53771
Re: In how many different ways can 3 identical green shirts and [#permalink]
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24 Feb 2012, 22:59
3
1
shrive555 wrote:
In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?
A. 20
B. 40
C. 216
D. 720
E. 729
1-2-3-4-5-6 (children)
B-B-B-G-G-G (shirts)
G-B-B-G-G-B
G-G-B-G-B-B
....
The following is multiple choice question (with options) to answer.
A boy has 11 trousers and 12 shirts. In how many different ways can he select a trouser and a shirt? | [
"120ways",
"138ways",
"140ways",
"132ways"
] | D | The boy can select one trouser in 11 ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 11 * 12
= 132 ways.
Answer: D |
AQUA-RAT | AQUA-RAT-34524 | • Thank you for taking the time to help me!! How can you say that $x$ and $y$ are co-prime? – user342661 Aug 6 '16 at 2:46
• Suppose $x$ and $y$ are not co-prime. Then let $x = gw$ and $y=gz$, where $g \neq 1$. Then note that $c=xd=gwd$,and $m = yd = gzd$. Note that $gd$ divides both $x$ and $y$ and is greater than $d$ because $g>1$. This contradicts the fact that $d$ is the gcd of $m$ and $c$. Hence $x$ and $y$ are co-prime. – астон вілла олоф мэллбэрг Aug 6 '16 at 2:51
• You are welcome. – астон вілла олоф мэллбэрг Aug 6 '16 at 4:16
Assuming integers for all variables, and $c' = c/d$, $m' = m/d$ :
\begin{align} ac \equiv bc \pmod m &\iff \exists k ~~ ac - bc = km \\ % &\iff \exists k ~~ ac'd - bc'd \equiv km \\ % &\iff \exists k ~~ a - b \equiv (k/c')(m/d) \end{align}
So we have to establish that $k / c'$ is an integer. From $\gcd(c, m) = d$ we can infer $\gcd(c', m') = 1$, so
$$ac - bc = km$$ $$ac'd - bc'd \equiv k(m'd)$$ $$c'(a - b) = km'$$
So $c'$ divides $k$, so $k/c'$ is an integer.
The following is multiple choice question (with options) to answer.
If y = 30p, and p is prime, what is the greatest common factor of y and 12p, in terms of p? | [
"p",
"5p",
"6p",
"7p"
] | C | Y=30p
other number is 12p
then GCF(30p,12P)=6P; C is the correct answer |
AQUA-RAT | AQUA-RAT-34525 | javascript, algorithm, programming-challenge, ecmascript-6, palindrome
You can get the unit value eg 423 is 100 or 256378 is 100000 by raising 10 to the power of the number of digits minus one. Well not for powers of 10
eg
unit = 10 ** (Math.ceil(Math.log10(13526)) -1); // 10000
unit = 10 ** (Math.ceil(Math.log10(10000)) -1); // 1000 wrong for power of 10 number
To get the value we want we need to floor the log first
unit = 10 ** Math.floor(Math.log10(10000)); // 10000
unit = 10 ** Math.floor(Math.log10(13526)); // 10000 correct
or
unit = 10 ** (Math.log10(10000) | 0); // 10000
unit = 10 ** (Math.log10(13526) | 0); // 10000
Get digit at position of positive integer
To get the digit at any position in a number divide it by 10 raised to the power of the digit position get the remainder of that divided by 10 and floor it.
const digitAt = (val, digit) => Math.floor(val / 10 ** digit % 10);
or
const digitAt = (val, digit) => val / 10 ** digit % 10 | 0;
// Note brackets added only to clarify order and are not needed
// ((val / (10 ** digit)) % 10) | 0;
digitAt(567, 0); // 7
digitAt(567, 1); // 6
digitAt(567, 2); // 5
The following is multiple choice question (with options) to answer.
The unit digit in the product (445 * 534 * 999 * 234) is: | [
"2",
"0",
"6",
"5"
] | B | Explanation:
Unit digit in the given product = Unit Digit in (5*4*9*4) = 0
ANSWER: B |
AQUA-RAT | AQUA-RAT-34526 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A pair of articles was bought for $360 at a discount of 10%. What must be the marked price of each of the article? | [
"$200",
"$500",
"$350",
"$400"
] | A | S.P. of each of the article = 360/2 = $180
Let M.P = $x
90% of x = 360
x = 180*100/90 =$200
Answer is A |
AQUA-RAT | AQUA-RAT-34527 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
Jim bought edging to go around a circular garden with a radius of 5 feet. Later he decided to
double the diameter of the garden. How many more feet of edging must he buy? | [
"28.84' additional edging",
"48.84' additional edging",
"31.40' additional edging",
"18.84' additional edging"
] | C | circumference of small garden = 2 x 3.14 x 5 = 31.40'
double of the circumference of small garden = 2 x 31.40' = 62.80'
More feet to be buy = 62.80 - 31.40 = 31.40
Answer : C |
AQUA-RAT | AQUA-RAT-34528 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
The difference between the local value and the face value of 7 in the numeral 32675149 is | [
"75142",
"64851",
"5149",
"69993"
] | D | (Local value of 7) - (Face value of 7) = (70000 - 7) = 69993
Answer: Option D |
AQUA-RAT | AQUA-RAT-34529 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
The average age of boys in a class is 16 years and that of the girls is 15 years. What is the average age for the whole class? | [
"15",
"16",
"15.5",
"Insufficient Data"
] | D | Explanation:
We do not have the number of boys and girls. Hence we cannot find out the answer.
Answer: Option D |
AQUA-RAT | AQUA-RAT-34530 | ### 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 start a business with investments of Rs. 5000 and Rs. 4500 respectively. After 4 months, A takes out half of his capital. After two more months, B takes out one-third of his capital while C joins them with a capital of Rs. 7000. At the end of a year, they earn a profit of Rs. 5080. Find the share of each member in the profit. | [
"A - Rs. 1600, B - Rs. 1800, C - Rs. 1680",
"A - Rs. 1680, B - Rs. 1600, C - Rs. 1800",
"A - R,s. 1800, B - Rs. 1500, C - Rs. 1780",
"A - Rs. 1400, B - Rs. 1900, C - Rs. 1780"
] | A | Explanation:
A:B:CA:B:C =(5000×4+2500×8)=(5000×4+2500×8) :(4500×6+3000×6):(4500×6+3000×6) :(7000×6):(7000×6)
=40000:45000:42000=40000:45000:42000 =40:45:42=40:45:42.
A's share = Rs. 5080×401275080×40127 = Rs. 1600.
B's share = Rs. 5080×451275080×45127 = Rs. 1800.
C's share = Rs. 5080×421275080×42127 = Rs. 1680.
ANSWER IS A |
AQUA-RAT | AQUA-RAT-34531 | I will be grateful for any help, or any reading material reference.
-
You are right about the probability of guessing $3$. Your method can be adapted to solve the other problems. But my preferred way of doing it involves the binomial coefficients, which in general count the number of ways of choosing $r$ objects from $n$ objects, order irrelevant. Maybe also look up combinations. Note that if you can handle exactly $0$, $1$, $2$, $3$ you can handle the "at least" stuff. For example the probability of at least $2$ is prob. of exactly $2$ plus prob. of exactly $3$. – André Nicolas Jun 2 '11 at 11:12
@user6312: Thank you for your comment I now understand (I better say you've reminded me) about the "at least" stuff. Please take a look at the answer provided by Henry and my response to that answer and see if you can maybe shed some light on my problem. – Nikola Jun 2 '11 at 12:34
You need to be clear whether order matters. Henry's answer assumes that guessing (2,5,8) and getting (2,8,5) is three correct, not one. – Ross Millikan Jun 2 '11 at 13:08
@Ross: yes, he is right, so if I say that numbers 2,5,8 will be n that 10 numbers that are drawn then the order of them actually "coming" out is not important. – Nikola Jun 2 '11 at 13:16
If 10 out of 20 are drawn and you guess 3, then the probabilities are
The following is multiple choice question (with options) to answer.
Four different objects 1,2,3,4 are distributed at random in four places marked 1,2,3,4. What is the probability that none of the objects occupy the place corresponding to its number? | [
"17/24",
"3/8",
"1/2",
"5/8"
] | C | Explanation :
Let a particular number (say) number 2 occupies position 1.
Then all possible arrangement are given as:
=>(2,1,3,4), (2,1,4,3), (2,3,4,1), (2,4,1,4), (2,4,1,3), (2,4,3,1).
Out of these six, three (2,1,3,4), (2,3,1,4), (2,4,3,1) are not acceptable because numbers 3 and 4 occupy the correct positions.
Hence, the required probability = 3/6 = 1/2.
Answer : C |
AQUA-RAT | AQUA-RAT-34532 | ### 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.
Of the families in City X in 1991, 60 percent owned a personal computer. The number of families in City X owning a computer in 1993 was 20 percent greater than it was in 1991, and the total number of families in City X was 9 percent greater in 1993 than it was in 1991. what percent of the families in City X owned a personal computer in 1993? | [
"59%",
"55%",
"63%",
"66%"
] | D | Say a 100 families existed in 1991 then the number of families owning a computer in 1991 - 60
Number of families owning computer in 1993 = 60 * 120/100 = 72
Number of families in 1993 = 109
The percentage = 72/109 * 100 = 66%.
ANSWER:D |
AQUA-RAT | AQUA-RAT-34533 | We're asked for the price at which the TOTAL PROFIT from the sale of these units = $42,000. Since there are 600 units, then each unit has to bring in$42,000/600 = $70 in profit. The COST of each unit is$90, so to hit that total profit, we just need to increase that $90 by$70 on each unit.
$90 +$70 = $160 Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ SVP Joined: 03 Jun 2019 Posts: 1739 Location: India Re: A television manufacturer produces 600 units of a certain model each [#permalink] ### Show Tags 16 Sep 2019, 09:23 RSOHAL wrote: A television manufacturer produces 600 units of a certain model each month at a cost to the manufacturer of$90 per unit and all of the produced units are sold each month. What is the minimum selling price per unit that will ensure that the monthly profit (revenue from sales minus the manufacturer's cost to produce) on the sales of these units will be at least $42,000? A$110
B$120 C$140
D$160 E$180
A television manufacturer produces 600 units of a certain model each month at a cost to the manufacturer of $90 per unit and all of the produced units are sold each month. What is the minimum selling price per unit that will ensure that the monthly profit (revenue from sales minus the manufacturer's cost to produce) on the sales of these units will be at least$42,000?
Let the selling price per unit be $x 600 (x - 90) >= 42,000 x -90 >= 70 x>=$160
IMO D
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."
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The following is multiple choice question (with options) to answer.
A person purchased a TV set for Rs. 16000 and a DVD player for Rs. 6250. He sold both the items together for Rs. 35600
. What percentage of profit did he make? | [
"80%",
"49%",
"40%",
"60%"
] | D | The total CP = Rs. 16000 + Rs. 6250
= Rs. 22250 and SP = Rs. 35600
Profit(%) = (35600 - 22250)/22250 * 100
= 60%
Answer:D |
AQUA-RAT | AQUA-RAT-34534 | ### Show Tags
08 Feb 2012, 09:00
# of selections of books with no condition = 8C4 = 70
# of selections of books with no paperback book = 6C4 = 15
# of selections of books with at least one paperback book = 70 -15 = 55
Manager
Status: Sky is the limit
Affiliations: CIPS
Joined: 01 Apr 2012
Posts: 68
Location: United Arab Emirates
Concentration: General Management, Strategy
GMAT 1: 720 Q50 V38
WE: Supply Chain Management (Energy and Utilities)
Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
### Show Tags
20 May 2012, 03:36
3
The selection should at least contain one PAPERBACK book. There are a total of 2 PB and 6 HB books are available.
The combinations that at least one PB book will come out are: PHHH & PPHH
1. PHHH = 2C1 * 6C3 = 2 * (5*4) = 40
2. PPHH = 2C2 * 6C2 = 1 * ((6*5) / 2) = 15
In total 40+15 = 55 Ways
Manager
Joined: 07 Feb 2011
Posts: 89
Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
### Show Tags
25 Jan 2013, 07:20
Hmm....so when I did this I made an error that might have carried over from using combinatorics from probability. I found the total as 8C4 and to find the amount to subtract from it I had (6C4)(2C0). Why don't we multiply by the books not chosen and what's would it mean if we did?
I'm trying to understand a fundamental flaw I made here
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52385
Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
John had a stock of 1200 books in his bookshop. He sold 75 on Monday, 50 on Tuesday, 64 on Wednesday, 78 on Thursday and 135 on Friday. What percentage of the books were not sold? | [
"36.5%",
"46.5%",
"66.5%",
"76.5%"
] | C | Let N be the total number of books sold. Hence
N = 75 + 50 + 64 + 78 + 135 = 402
Let M be the books NOT sold
M = 1200 - N = 1200 - 402 = 798
Percentage
Books not sold / total number of books = 798/1200 = 0.665 = 66.5%
correct answer C |
AQUA-RAT | AQUA-RAT-34535 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B can do a piece of work in 7 days. With the help of C they finish the work in 4 days. C alone can do that piece of work in? | [
"33",
"9 1/3",
"30",
"88"
] | B | C = 1/4 – 1/7 = 3/28 => 28/3 = 9 1/3 days
Answer: B |
AQUA-RAT | AQUA-RAT-34536 | # Finding the minimum number of students
There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
It is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?
$1) \quad pq - \binom{q}{2}$
$2) \quad pq - \binom{p}{2}$
$3) \quad (p-1)(q-1)$
-
Something is missing. Is every student supposed to be on a committee? – JavaMan Aug 31 '11 at 16:24
@DJC:Not mentioned in the question,I guess we may have to consider that to get a solution. – Quixotic Aug 31 '11 at 16:28
@DJC: For the minimum number of students this does not matter. – TMM Aug 31 '11 at 16:30
@Thijs Laarhoven:Yes you are right but as the problem also asked for maximum number I have considered it in my solution. – Quixotic Aug 31 '11 at 16:31
@Thijs, FoolForMath, I guess my question is, should the minimum answer be in terms of $p$ and $q$? – JavaMan Aug 31 '11 at 16:31
For $1\leq i\leq p$, let $C_i$ be the set of students on the $i$th committee. Then by inclusion-exclusion, or more accurately Boole's inequalities, we have $$\sum_i|C_i|-\sum_{i<j}|C_i C_j|\leq |C_1\cup C_2\cup\cdots \cup C_p|\leq \sum_i |C_i|.$$
From the constraints of the problem, this means $$pq-{p\choose 2}\leq \#\mbox{ students}\leq pq.$$
The following is multiple choice question (with options) to answer.
A committee of three students has to be formed. There are six candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible? | [
"3",
"4",
"5",
"6"
] | D | Let's try a different problem with your approach. There are five people: A, B, C, D, E. Need to chose 3 for a committee. A and B cannot be chosen together. B and C cannot be chosen together. How many options?
Your approach: total 10 options, 5c3.
Now, assume the wrong scenario where A and B are chosen together. There are three such scenarios. (A and B are chosen, just need one more person.) So we have to subtract the three wrong options. Similarly, there are three wrong scenarios where B and C are chosen together.
This gives us 10-3-3=4 as the answer.
Yet this answer is wrong. There are five possibilities: ACD, ACE, ADE, BDE, CDE.
D |
AQUA-RAT | AQUA-RAT-34537 | ## 5 Feb 2019 Enter the compounding period and stated interest rate into the effective interest rate formula, which is: r = (1 + i/n)^n-1. Where: r = The effective
1 Apr 2019 Based on the method of calculation, interest rates are classified as nominal interest rate, effective interest rate and annual percentage yield 10 Jan 2018 The simple interest rate is the interest rate that the bank charges you for taking the loan. It is also commonly known as the flat rate, nominal rate or 10 Apr 2019 The advertised rate (also known as nominal rate) is the interest the bank charges you on the sum you borrow. Note that there are different ways to 10 Feb 2019 TaxTips.ca - The effective rate of interest depends on the frequency of compounding. (e.g. 6% compounded monthly), the stated rate is the nominal rate. Interest earned on chequing and savings accounts is usually 3 Oct 2017 In this situation, with an effective interest rate of 17.2737 percent, there is very little margin for missing out on making an amortization payment. 1) If I'm given a 7% semi-annual nominal rate, does that mean the annual nominal rate is simply 14%?. No. 7% semi-annual is 3.5% every six months. So annual 27 Nov 2016 Going further, since a nominal APR of 12% corresponds to a daily interest rate of about 0.0328%, we can calculate the effective APR if this
### 19 Apr 2013 The interest rate per annum is only the nominal interest rate. This nominal rate is equal to the effective rate when a loan is on annual-rest basis
The following is multiple choice question (with options) to answer.
The simple interest on Rs. 10 for 4 months at the rate of 3 paise per rupeeper month is | [
"1.2",
"1.4",
"2.25",
"3.21"
] | A | Sol.
S.I. = Rs. [10 * 3/100 * 4] = Rs. 1.20
Answer A |
AQUA-RAT | AQUA-RAT-34538 | n\leq 5$$ ...(2)From (1) & (2) $$n\in (-\sqrt{21},\sqrt{21}),n\in Z.$$$$n=-4,-3,-2,-1,0,1,2,3,4$$$$\Rightarrow$$ Number of values of $$n$$ are $$9$$Mathematics
The following is multiple choice question (with options) to answer.
If n is a positive integer, which of the following is a possible value of |36 - 5n|? | [
"7",
"14",
"12",
"15"
] | B | When 36>5n, then |36- 5n| = 71- 5n = (36-5n) + 1 = {multiple of 5} + 1
When 36<=5n, then |36- 5n| = -(71- 5n) = (36-70) - 1 = {multiple of 5} - 1
So, the correct answer must be 1 greater or 1 less than a multiple of 5. Only B fits.
Answer: B |
AQUA-RAT | AQUA-RAT-34539 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
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The following is multiple choice question (with options) to answer.
TOM traveled the entire 50 miles trip. If he did the first 15 miles of at a constant rate 20 miles per hour and the remaining trip of at a constant rate 45 miles per hour, what is the his average speed, in miles per hour? | [
"60.40mph",
"60.20mph",
"55.32mph",
"65.50mph"
] | A | avg speed = total distance/ total time = (d1+d2)/(t1+t2) = (15+35) / ((15/20)+(35/45)) = 50*180/149 = 60.40mph
A |
AQUA-RAT | AQUA-RAT-34540 | 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.
Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is? | [
"10",
"11",
"12",
"10.8"
] | D | Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec.
Answer: Option D |
AQUA-RAT | AQUA-RAT-34541 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
A train traveled the first d miles of its journey it an average speed of 60 miles per hour, the next d miles of its journey at an average speed of y miles per hour, and the final d miles of its journey at an average speed of 160 miles per hour. If the train’s average speed over the total distance was 90 miles per hour, what is the value of y? | [
" 68",
" 84",
" 90",
" 110"
] | D | Average speed=Total distance traveled /Total time taken
3d/d/60+d/y+d/160=90
Solving for d and y,
15y=11y+480
4y=440
y=110
Answer D |
AQUA-RAT | AQUA-RAT-34542 | 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 television screen is given as the length of the screen's diagonal. If the screens were flat, then the area of a square 20-inch screen would be how many square inches greater than the area of a square 18-inch screen? | [
"2",
"4",
"16",
"38"
] | D | Pythogoras will help here!
let the sides be x and diagonal be d
Then d ^2 = 2x^2
and
Area = x^2
Now plug in the given diagonal values to find x values
and then subtract the areas
Ans will be 20^2/2 - 18^2 /2 = 76 /2 = 38
Ans D. |
AQUA-RAT | AQUA-RAT-34543 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
how many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? | [
"5",
"10",
"15",
"20"
] | D | Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
ANSWER D 20 |
AQUA-RAT | AQUA-RAT-34544 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
Two cars are 500 cm apart. each is moving forward for 100 cm at a velocity of 50 cm/s and receding back for 50 cm at 25cm/s at what time they will collide with each other. | [
"11 sec",
"14 sec",
"12 sec",
"13 sec"
] | B | each car is moving forward for 100 cm with velocity 50cm/s
so time to cover 100cm is=100/50=2 sec
again receding back for 50 cm at 25cm/s velocity
so time to cover 50 cm is=50/25=2 sec
actually they individually cover (100-50)=50 cm in (2+2)=4 sec
so in 12 sec they together cover (150+150)=300 cm
and after that both move 100 cm in 2 sec
so after (12+2)=14 sec they cover (300+200)=500cm
so they collide after 14 sec
ANSWER:B |
AQUA-RAT | AQUA-RAT-34545 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
What is the % change in the area of a rectangle when its length increases by 30% and its width decreases by30%? | [
"0%",
"20% increase",
"20% decrease",
"9% decrease"
] | D | (13/10)*(7/10) =91/100 of original area
91/100 is a 9% decrease from 100/100 ->D |
AQUA-RAT | AQUA-RAT-34546 | inscribed in a circle. K = (1/2)(5)(8) K = 20 The area of A is 20 square units. opposite sides equal and parallel; all angles 90. Find the area of the shaded region. I do not know any of the angles and presume that none of them are right angles. Proving a quadrilateral is a parallelogram 8. A trapezium has one pair of opposite sides parallel. Problem 4 of the United States Mathematical Olympiad. These worksheets will inspire them through innovative approaches to symmetry, transformations, plane figures, and more. If you already know that the shape is a parallelogram, you will only have to show that one of the angles is a right angle and then it would follow that all of the angles are right angles. Here are four different examples of it in action. Also presume that none of the sides are parallel to each other. The steps for finding this area can be done by performing these steps: Divide the figure into two triangles by drawing a diagonal. Before going into the calculation of area, let us define what is a quadrilateral. Find the area of this quadrilateral with the given vertices, A(-8,6) B(-5,8) C(-2,6) D(-5,0). Surface Area. There are two basic kite area formulas, which can be used depending on which information you have: If you know two diagonals, you can calculate the area of a kite as:. • An isosceles trapezoid is a quadrilateral with exactly two. Angle sum of a convex quadrilateral = (4 – 2) × 180° = 2 × 180° = 360° Since, quadrilateral, which is not convex, i. To show that the figure obtained by joining the mid-points of consecutive sides of the quadrilateral is a parallelogram. Another method to calculate the surface area of a trapezium is to divide the trapezium into a rectangle and two triangles, to measure their sides and to determine separately the surface areas of the rectangle and the two triangles (see Fig. If a quadrilateral is cyclic, then the exterior angle is equal to the interior opposite angle. ] When the groups are finished, discuss each figure as a class. Isosceles: A triangle with at two equal sides. If you know
The following is multiple choice question (with options) to answer.
Find the area of the quadrilateral of one of its diagonals is 20 cm and its off sets 9 cm and 6 cm? | [
"150 cm2",
"196 cm2",
"189 cm2",
"146 cm2"
] | A | 1/2 * 20(9 + 6)
= 150 cm2
Answer:A |
AQUA-RAT | AQUA-RAT-34547 | (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.
If the cost of 12 apples is $20. What would be the cost of 20 apples? | [
"25.5",
"29.36",
"33.33",
"35.56"
] | B | The cost of one apple=20/12=1.67
The cost of twenty apples= 1.67x20= $ 33.33
Answer: B |
AQUA-RAT | AQUA-RAT-34548 | ### 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.
Income and expenditure of a person are in the ratio 10 : 8. If the income of the person is Rs. 10000, then find his savings? | [
"2003",
"2002",
"2000",
"2001"
] | C | Let the income and the expenditure of the person be Rs. 10x and Rs. 8x respectively.
Income, 10x = 10000 => x = 1000
Savings = Income - expenditure = 10x - 8x = 2x = 2(1000)
So, savings = Rs. 2000.
Answer:C |
AQUA-RAT | AQUA-RAT-34549 | This series of cash flows will yield exactly 10 % is $56.07 scenario... Stan also wants his son to be paid or received in the future amount that you expect receive! Is extremely important in many financial calculations 510.68 ; discount rate. discount rate is investment! Money received in the discussion above, we looked at one investment over the course of year! Now knows all three variables for the first offer suggests the value of$ 100 today or can... For the four discount rates trend, the amount $100 being$... Would not have realized a future sum the applicable discount rate. a present value formula shown! Costs, inflation will cause the price they pay for an investment might earn example, future... ; discount rate or the interest rate ” is used in the future cashflows expected from an investment might.... That with an initial investment of exactly $100 today or I can pay you back 100! Periods interest rates rise and the CoStar product suite often used as the present value provides a basis assessing... Today, you can buy goods at today 's prices, i.e must... Earned on the funds over the next five years time refers to future value and a... If you receive money today, you can buy goods at today 's prices, i.e from now 1 4... Aone-Size-Fits-All approach to determining the appropriate discount rate is used when referring to present. Discount lost business profits to a present value, the future value of cash flows will yield exactly %... Than$ 1,000 five years time value takes the future receiving $1,000 five years discount rate present value not. The concept that states an amount for any timeframe other than one.! More Answers money worth in today ’ s lost earnings the idea of net present value money...: present value becomes equal to the present value of money that is expected to arrive at a time. This Table are from partnerships from which investopedia receives compensation states that an amount of money that expected! Financial planning formula given below PV = CF / ( 1 + r ) t 1 of. Can pay you back$ 100 today or I can pay you $110 year. Bob knows the future amount that you expect to earn a rate return... 5,000 lump sum payment in five years from now % ) 3 2 earn. Bob gets up and says, “ I
The following is multiple choice question (with options) to answer.
The banker's gain on a bill due 1 year hence at 10% per annum is Rs. 20. What is the true discount? | [
"Rs. 200",
"Rs. 100",
"Rs. 150",
"Rs. 250"
] | A | Explanation :
TD = (BG ×100)/TR = (20×100)/(1×10) = Rs. 200 Answer : Option A |
AQUA-RAT | AQUA-RAT-34550 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
The first train leaves the station at 6:00 am with a speed of 60 km/hr. The second train leaves at 7:00 am, in the same direction from the same station, with a speed of 75 km/hr. What time will the second train catch the first train? | [
"10:00 am",
"11:00 am",
"12:00 noon",
"1:00 pm"
] | B | In sixty minutes, the first train travels 60 km.
The second train catches the first train with a relative speed of 15 km/h.
The second train will catch the first train in 4 hours.
The second train will catch the first train at 11:00 am.
The answer is B. |
AQUA-RAT | AQUA-RAT-34551 | Show Tags
18 Jun 2014, 06:09
Thanks much Bunuel
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
Show Tags
18 Jun 2014, 20:26
1
maggie27 wrote:
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.
Also note that area of inscribed square is always half than that of the original square
As Bunuel pointed out, if it goes less than 50, it means some of the vertex is not touching side of the original square.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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01 Jul 2015, 18:04
If $$x^{2}$$ is area of square, then find x, one side of the square. If square is inscribed, then diagonal is the length of larger square and therefore the diagonal is $$10$$. To determine the side, the formula also includes the area of the square, $$x^{2}$$. So, if $$2x^{2} = 100$$ then $$x^{2}=50$$
D.
Thanks,
A
Director
Joined: 04 Jun 2016
Posts: 568
GMAT 1: 750 Q49 V43
Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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31 Jul 2016, 20:43
Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible value of the area of square T ?
A. 45
B. 48
C. 49
D. 50
E. 52
The following is multiple choice question (with options) to answer.
The area of a square field is A square feet and the perimeter is p feet. If 6A=6(2p+9), what is the perimeter of the field, in feet? (Here A=a^2) | [
"28",
"36",
"40",
"56"
] | B | You can also solve this via using the given numbers in the answer choices!
Of course you need to be aware of the basic properties as outlined by the other posts above (a = x^2 and p = 4x)
Starting with D you will notice that x=14 is way too big for your area (14^2) and will not satisfy: 6A=6(2p+9) or a=2p+9
--> Eliminate D and E
Now pick B (its either too big, then its A, or too small then you know its C or it is B itsself)
And picking B indeed solves the problem! (36/4 --> 9; a= 9^2 = 81 and 81=2x36+9) |
AQUA-RAT | AQUA-RAT-34552 | 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.
A is two years older than B who is twice as old as C.If the total of the ages of A,B nd C be 27,then how old is B ? | [
"7",
"8",
"9",
"10"
] | D | Solution
Let C's age be x years .Then,B's age =2x years.A's age =(2x +2) years.
∴ (2x +2) +2x + x =27 ⇔ 5x =25 ⇔ x =5.
Hence,B's age =2x = 10 years. Answer D |
AQUA-RAT | AQUA-RAT-34553 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
if 12401 is divided by any no. then quotient is 76 and remainder is 13.what is divisor? | [
"154",
"124",
"153",
"163"
] | D | divisor=(dividend-remainder)/quotient=(12401-13)/76=12388/76=163
divisor = 163
answer D |
AQUA-RAT | AQUA-RAT-34554 | #### Opalg
##### MHB Oldtimer
Staff member
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
First, the minimum value of $xy$ must be positive, because if $xy\leqslant 0$ then $(x+y)z\geqslant 4$. So $(x+y)^2\geqslant\dfrac{16}{z^2}$, and $$7-z^2 = x^2+y^2 = (x+y)^2 - 2xy \geqslant (x+y)^2 \geqslant \frac{16}{z^2}.$$ Thus $z^2 + \dfrac{16}{z^2} \leqslant 7$. But that cannot happen, because the minimum value of $z^2 + \dfrac{16}{z^2}$ is $8$ (occurring when $z^2 = 4$).
So we may assume that $xy>0$. Let $u = \sqrt{xy}$ and $v = x+y$. Then we can write the equations as $v^2-2u^2 + z^2 = 7$, $vz+u^2=4$. Therefore $v^2 - 2(4-vz) + z^2 = 7,$ so $(v+z)^2 = 15$. But $vz\leqslant\bigl(\frac12(v+z)\bigr)^2 = \frac{15}4.$ Therefore $xy = u^2 \geqslant 4-\frac{15}4 = \frac14$.
The following is multiple choice question (with options) to answer.
Find the value of x from the below equation:
x^2−7x+10=0 | [
"5 or 2",
"3 or 5",
"3 or 7",
"4 or 6"
] | A | Here we need to find out a and b such that a + b = -7 and ab = +10
a = -5 and b = -2 satisfies the above condition.
Hence
x^2−7x+10=(x−5)(x−2)
x2−7x+10=(x−5)(x−2)
x^2−7x+10
=0
⇒(x−5)(x−2)
=0
x2−7x+10=0⇒(x−5)(x−2)=0
Step 3: Equate each factor to 0 and solve the equations
(x−5)(x−2)
=0
⇒(x−5)
=0or (x−2)=0
⇒x= 5 or 2
A |
AQUA-RAT | AQUA-RAT-34555 | 8. ## Re: Red marbles, white marbles, in a bag.
Did I get the answers right?
What do you mean I changed the question midstream?
If you mean why I changed the number of marbles, it was to see if I could to the calculation.
Apparently not?
Please could you make your reply more useful. I really don't have any idea what you mean.
The following is multiple choice question (with options) to answer.
A jar of 180 marbles is divided equally among a group of marble-players today. If 2 people joined the group in the future, each person would receive 1 marble less. How many people are there in the group today? | [
"16",
"18",
"20",
"22"
] | B | 180 = 18*10 = 20*9
There are 18 people in the group today.
The answer is B. |
AQUA-RAT | AQUA-RAT-34556 | Circular motion question
1. Nov 10, 2005
donjt81
This is the question...
A small wheel of radius 1.4cm drives a large wheel of radius 15cm by having their circumferences pressed together. If the small wheel turns at 407 rad/s, how fast does the larger one turn? Answer in rad/s
This is what I was thinking...
radius of smaller wheel = .014m
radius of larger wheel = .15m
circumference of smaller wheel = 2*pi*r = 2*3.14*.014 = .08792
angular velocity of smaller wheel (given) = 407 rad/s
angular velocity = circumference/time
time = circumference/angular velocity
=.08792/407 = .000216s
circumference of larger wheel = 2*pi*r = 2*3.14*.15 = .942
angular velocity = circumference/time
=.942/.000216 = 4361.11 rad/s
Does this approach look right?
2. Nov 10, 2005
BerryBoy
I disagree... By intuition, you can predict that the larger wheel is going to turn more slowly.. Try another approach..
Hint: Consider the fact that the speeds of circumferences are equal.
Eq: speed = w x r
w = angular velocity
r = radius
Does this help?
Sam
Last edited: Nov 10, 2005
3. Nov 10, 2005
donjt81
You are right... the larger wheel should go slower.
but since the speed of smaller wheel is 407 rad/s wont the larger wheel speed be the same?
so is the answer to the problem 407 rad/s for the larger wheel? but that doesnt make sense because the larger wheel is supposed to go slower...
I am confused...
4. Nov 10, 2005
BerryBoy
OK, so if the speed at the circumfrence is:
v = w x r (as I stated above).
If the wheels are in contact this speed is equal on both wheels (not the angular velocity). Therefore:
wsmall x rsmall = wlarge x rlarge
I can't give you anymore hints without doing it now.
The following is multiple choice question (with options) to answer.
A bicyclist's wheel has a circumference of 9 meters. if the bicyclist covers 180 meters in 5 hours at a constant speed, how many rotations does the wheel make per hour at that speed? | [
"2",
"3",
"4",
"5"
] | C | The bicyclist covers y meters in t hrs. So he covers y/t meters in one hour.
To cover y/t meters, the bicycle's wheel went round and round that is, multiple times, the wheel completed one circumference on the road. How many times did it complete one full rotation. Since the circumference is x, the number of times it completes the circumference is y/xt.=180/9*5=4
Answer (C) |
AQUA-RAT | AQUA-RAT-34557 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
If a particular is never chosen, in how many ways can a cricketbe chosen out of 15 players? | [
"239",
"259",
"364",
"370"
] | C | A particular players is never chosen, it means that 11 players are selected out of 14 players.
=> Required number of ways = 14C11
= 14!/11!x3! = 364
C |
AQUA-RAT | AQUA-RAT-34558 | If we are considering all 26 letters of the alphabet,
. . we simply append a permutation factor.
Quote:
With five letters, consider the first three letters.
3 different letters: . $ABC\;\;\Rightarrow\;\;ABCBA$ . . . one way
There are $26$ choices for the " $A$",
. . $25$ choices for the " $B$",
. . and $24$ choices for the " $C$".
Hence, there are: . $26\cdot25\cdot24 \times 1 \:=\:15,600$ ways.
Quote:
2 different letters: . $\begin{array}{ccc}ABA\;\;\Rightarrow\;\;ABABA \\ AAB\;\;\Rightarrow\;\;AABAA \\ ABB\;\;\Rightarrow\;\;ABBBA\end{array}$ . . . three ways
There are $26$ choices for the " $A$"
. . and $25$ choices for the " $B$".
Hence, there are: . $26\cdot25 \times 3 \:=\:1950$ways.
Quote:
Only one letter: . $AAA\;\;\Rightarrow\;\;AAAAA$ . . . one way
There are $26$ choices for the " $A$".
Hence, there are: . $26 \times 1 \:=\:26$ ways.
Therefore, there are: . $15,600 + 1950 + 26 \:=\:\boxed{17,576\text{ ways.}}$
• Aug 30th 2006, 06:48 PM
Quick
just to clear confusion: $\underbrace{\overbrace{15600+1950+26}^{\text{Sorob }\!\!\text{an's Way}}=17576=\overbrace{26^3}^{\text{my way}}}_{\text{both ways give the same answer}}$
The following is multiple choice question (with options) to answer.
Which of the following leads to the correct mathematical solution for the number of ways that the letters of the word APPLE could be arranged to create a five-letter code? | [
"5!",
"5! − (2!)",
"5! − (3! × 2!)",
"5!/(2!)"
] | D | APPLE - five letters can be arranged in 5! ways
since 'P' repeats 2 times , we need to divide the 5! ways by 2! to adjust repetition.
5!/(2!)
Ans. D) 5!/(2!) |
AQUA-RAT | AQUA-RAT-34559 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
The cost price of an book is 64% of the marked price. Calculate the gain percent after allowing a discount of 12%? | [
"33%",
"28.02%",
"37.5%",
"30.5%"
] | C | C
37.5%
Let marked price = $100.
Then, C.P. = $64, S.P. = $88
Gain % = 24/64 * 100 = 37.5%. |
AQUA-RAT | AQUA-RAT-34560 | You could also see what it takes to get the linear term or the constant term to match, but you will get the same answer. In other words, if you require that
$$k(x-r)(x-s)$$ and $$ax^2+bx+c$$ are the same expression for all $x$ (where $a,b,c, r,s$ are known and $k$ is to be determined), then equating coefficients on both sides gives you $$k=a$$ $$-k(r+s)=b$$ $$krs=c$$ which gives you $k$ in three apparently different ways, but they will all be the same number: $$k=a$$ $$k=-\frac{b}{r+s}$$ $$k=\frac{c}{rs}$$ (watch out if you have division by zero in the two final versions, you can't do that).
Note that if $k$ will clear a fraction in $r$ or $s$, it is often distributed over one of the linear factors instead of leaving it out front (this is the case with your problem). Similarly, you could distribute two numbers whose product is $k$ over each of the linear factors separately.
The following is multiple choice question (with options) to answer.
If x^2 − 2x − 15 = (x + r)( x + s) for all values of x, and if r and s are constants, then which of the following is a possible value of 2(r − s)? | [
"8",
"16",
"− 2",
"− 3"
] | B | We know that given ax^2 + bx + c = 0, Sum of the roots = -b/a and product of the roots = c/a.
The roots here are -r and -s.
-r - s = -(-2)/1 = r + s = -2
(-r)*(-s) = -15/1 = rs
So one of r and s is -5 and the other is 3. So 2(r - s) could be 16 or -16.
Answer (B) |
AQUA-RAT | AQUA-RAT-34561 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
The length of rectangle is thrice its breadth and its perimeter is 72 m, find the area of the rectangle? | [
"432",
"376",
"299",
"324"
] | D | 2(3x + x) = 72
l = 36 b = 9
lb = 36 * 9
= 324
D |
AQUA-RAT | AQUA-RAT-34562 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
Tom, working alone, can paint a room in 12 hours. Peter and John, working independently, can paint the same room in 6 hours and 2 hours, respectively. Tom starts painting the room and works on his own for two hour. He is then joined by Peter and they work together for two hour. Finally, John joins them and the three of them work together to finish the room, each one working at his respective rate. What fraction of the whole job was done by Peter? | [
"5/3",
"1/3",
"7/3",
"4/9"
] | D | Let the time when all three were working together be t hours. Then:
Tom worked for t+4 hour and has done 1/12*(t+4) part of the job;
Peter worked for t+2 hour and has done 1/6*(t+2) part of the job;
John worked for t hours and has done 1/2*t part of the job:
1/12*(t+4)+1/6*(t+2)+1/2*t=1 --> multiply by 12 --> (t+4)+(2t+2)+6t=12 --> t=2/3;
Hence Peter has done 1/6*(2/3+2)=1/6 *8/3=8/18=4/9
Answer: D |
AQUA-RAT | AQUA-RAT-34563 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
Rs.1500 is divided into two parts such that if one part is invested at 6% and the other at 5% the whole annual interest from both the sum is Rs.85. How much was lent at 5%? | [
"298",
"287",
"202",
"500"
] | D | (x*5*1)/100 + [(1500 - x)*6*1]/100 = 85
5x/100 + 90 – 6x/100 = 85
x/100 = 5
=> x =500
Answer:D |
AQUA-RAT | AQUA-RAT-34564 | # 2010 AMC 10B Problems/Problem 25
## Problem
Let $a > 0$, and let $P(x)$ be a polynomial with integer coefficients such that
$P(1) = P(3) = P(5) = P(7) = a$, and
$P(2) = P(4) = P(6) = P(8) = -a$.
What is the smallest possible value of $a$?
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$
## Solution
We observe that because $P(1) = P(3) = P(5) = P(7) = a$, if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$, $R(x)$ has roots when $P(x) = a$; namely, when $x=1,3,5,7$.
Thus since $R(x)$ has roots when $x=1,3,5,7$, we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$.
Then, plugging in values of $2,4,6,8,$ we get
The following is multiple choice question (with options) to answer.
If both 7^2 and 3^4 are factors of the number a*4^3*6^2*13^11, then what is the smallest possible value of a? | [
"120",
"256",
"343",
"441"
] | D | The number a must include at least 3^2*7^2 = 441
The answer is D. |
AQUA-RAT | AQUA-RAT-34565 | PRIME NUMBERS:
1. 1 is not a prime, since it only has one divisor, namely 1.
2. Only positive numbers can be primes.
3. There are infinitely many prime numbers.
4. the only even prime number is 2. Also 2 is the smallest prime.
5. All prime numbers except 2 and 5 end in 1, 3, 7 or 9.
PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);
4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.
IRRATIONAL NUMBERS
1. An irrational number is any real number that cannot be expressed as a ratio of integers.
2. The square root of any positive integer is either an integer or an irrational number. So, $$\sqrt{x}=\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).
This week's PS question
This week's DS Question
The following is multiple choice question (with options) to answer.
The average of first twelve prime numbers which are odd is? | [
"15.5",
"19.7",
"15.2",
"15.8"
] | B | Sum of first 12 prime no. which are odd = 236
Average = 236/12 = 19.7
Answer:B |
AQUA-RAT | AQUA-RAT-34566 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
The difference between two numbers is 1365. When the larger number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is : | [
"240",
"270",
"295",
"360"
] | B | Solution
Let the numbers be x and (x + 1365).
Then, x + 1365 = 6x + 15 ⇔ 5x = 1350 ⇔ x = 270.
Answer B |
AQUA-RAT | AQUA-RAT-34567 | => 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.
Of the 3,600 employees of Company X, 12/25 are clerical. If the clerical staff were to be reduced by 1/4, what percent of the total number of the remaining employees would then be clerical? | [
"40%",
"22.2%",
"20%",
"12.5%"
] | A | Let's see, the way I did it was
12/25 are clerical out of 3600 so 1728 are clerical
1728 reduced by 1/4 is 1728*1/4
so it reduced 432 people , so there is 1296 clerical people left
but since 432 people left, it also reduced from the total of 3600 so there are 3168 people total
since 1296 clerical left / 3168 people total
you get (A) 40% |
AQUA-RAT | AQUA-RAT-34568 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Train W leaves New York at 7:00 am traveling to Boston at 80mph. Train B leaves Boston at 7:45 am traveling to New York at 70 mph on a parallel track. If the distance between New York and Boston is 210 miles, at what time will the two trains pass each other? | [
"8:15 am",
"8:45 am",
"9:00 am",
"9:30 am"
] | B | train W will cover 60 kms in 45 min at 7:45, the time when train B starts...
distance left to cover is 210-60=150..
combined speed=80+70=150..
so the trains meet in 1 hour i.e. 7:45 + 1 hr=8:45
B |
AQUA-RAT | AQUA-RAT-34569 | 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.
A car covers a distance of 10km in 12 minutes. If its speed is decreased by 5km/hr, the time taken by the car to cover the same distance will be? | [
"10min 12sec",
"15min 45sec",
"13min 20sec",
"20min 4sec"
] | C | Speed = 10*60/12 = 50km/hr
New speed = 50-5 = 45 km/hr
Time taken = 10/45 = 60*2/9 = 13min 20sec
Answer is C |
AQUA-RAT | AQUA-RAT-34570 | kinematics, speed, distance
Title: How to calculate the speeds two objects need to move at in order to reach different destinations at the same time For example: On a graph there are two points. Each point has a different destination. The distance between Point A and its destination is 50, and the distance between Point B is 100. The points will begin moving at the same time at a constant speed until they both reach their destination. What speed does each point need to move in order to reach their destination at the same time as the other point if given a time, like 5 seconds. Intuitively, if one object travels twice the distance as another in the same amount of time, then first one is moving with twice the speed as the other.
You could show this mathematically by using the equations $$x_1=x_{01}+v_1t\\ x_2=x_{02}+v_2t$$ where $x_{01}$, $x_{02}$ are the initial positions of each object and $v_1$, $v_2$ are their respective speeds and $t$ is the time it takes both objects to reach their destinations.
Since they both start at the same position, then $x_{01}=x_{02}$ and so $$50=v_1t\\ 100=v_2t$$ since one travels a distance $50$ and the other $100$. Solving both these equations by dividing one into the other will give $$\frac{100}{50}=\frac{v_2 t}{v_1 t}$$ so that cancelling the $t$'s will give $$v_2=2v_1$$
The following is multiple choice question (with options) to answer.
Two cars cover the same distance at the speed of 50 and 62 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car. | [
"250 km",
"214 km",
"224 km",
"216 km"
] | A | 50(x + 1) = 62x
X = 4.16
60 * 4.16 = 250 km
ANSWER:A |
AQUA-RAT | AQUA-RAT-34571 | rhombus can be found, also knowing its diagonal. How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If we use Heron’s formula then we find area of one triangle made by two sides and a diagonal then twice of this area is area of rhombus. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Answered Formula for side of rhombus when diagonals are given 2 1. Perimeter = 4 × 12 cm = 48 cm. Thus, the total perimeter is the sum of all sides. Yes, because a square is just a rhombus where the angles are all right angles. The "base times height" method First pick one side to be the base. P = 4s P = 4(10) = 40 A rhombus is often called as a diamond or diamond-shaped. Here at Vedantu you will learn how to find the area of rhombus and also get free study materials to help you to score good marks in your exams. Since a rhombus is also a parallelogram, we can use the formula for the area of a parallelogram: A = b×h. If one of its diagonal is 8 cm long, find the length of the other diagonal. Ask your question. This is because both shapes, by definition, have equivalent sides. Any isosceles triangle, if that side's equal to that side, if you drop an altitude, these two triangles are going to be symmetric, and you will have bisected the opposite side. Given the length of diagonal ‘d1’ of a rhombus and a side ‘a’, the task is to find the area of that rhombus. Its diagonals perpendicularly bisect each other. The formula to calculate the area of a rhombus is: A = ½ x d 1 x d 2. where... A = area of rhombus; d 1 = diagonal1 (first diagonal in rhombus, as indicated by red line) d 2 = diagonal2 (second diagonal in rhombus, as indicated by purple line) Home List of all formulas of the site; Geometry. Area Of […] where b is the
The following is multiple choice question (with options) to answer.
The side of a rhombus is 26 m and length of one of its diagonals is 20 m. The area of the rhombus is? | [
"289",
"269",
"207",
"480"
] | D | 262 – 102 = 242
d1 = 20 d2 = 48
1/2 * 20 * 48 = 480
Answer: D |
AQUA-RAT | AQUA-RAT-34572 | • #### 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.
A man walking at the speed of 4 kmph crosses a square field diagonally in 3 minutes. The area of the field is: | [
"18000m2",
"20000m2",
"19000m2",
"25000m2"
] | B | Length of the diagonal= Distance covered in 3 min. at 4 km/hr.
= (4000/ 60 *3)= 200m.
Therefore, Area of the field= 1/2 * diagonal2
= ½ * 200*200 = 20000 m2
ANSWER:B |
AQUA-RAT | AQUA-RAT-34573 | 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.
During 2003, a company produced an average of 1,000 products per month. How many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 200% over its 2003 average? | [
"132,000",
"235,000",
"175,000",
"200,000"
] | A | Company produced 12*1000 = 12,000 products in 2003. If company produces X products from 2004 to 2007, then total amount of product produced in 4 years (2003 through 2007) is X+12,000. The gives the average of (X+12,000)/4.
This average needs to be 200% higher than that in 2003. In math terms, 12,000+200%(12,000) = 36,000. So:
(X+12,000)/4 = 36,000
X+12,000 = 144,000
X = 132,000
The answer is A. |
AQUA-RAT | AQUA-RAT-34574 | classical-mechanics, scaling, dimensional-analysis
$Q \propto g^{1/2} (a-d)^{5/2}$
Now, this becomes invalid when $H$ and $D$ are of the same order, but we can get an idea of the time to drain $T$ by saying that as the flow rate does not depend of $H$, $Q T = V = 4 \pi (D/2)^2 H = \pi D^2 H$ so, dropping the $\pi$ we get
$T \propto g^{1/2} \, \frac{(a-d)^{5/2}}{D^2 \, H}$
and thus for everything but the size of sand grains $d$ constant
$\frac{T(d_1)}{T(d_2)} = \left( \frac{a - d_1}{a - d_2} \right)^{5/2}$
The following is multiple choice question (with options) to answer.
In digging a pond 20 m * 10 m * 5 m the volumes of the soil extracted will be? | [
"3387",
"1000",
"2866",
"2787"
] | B | 20 * 10 * 5 = 1000
Answer: B |
AQUA-RAT | AQUA-RAT-34575 | (1) $$a^2+b^2>16$$
Doesn't tell us about the value of a. a could be 2 or 10 or many other values.
(2) a=|b|+5
a is 5 or greater since |b| is at least 0. This tells us that 'a' does not lie between -4 and 4. Hence this statement is sufficient to answer the question.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
The following is multiple choice question (with options) to answer.
If b does not equal zero, and ab = b/5, what is the value of a? | [
"A)1/5",
"B)1/4",
"C)1/3",
"D)1/2"
] | A | Explanation:
To solve for a, divide both sides of the equation by b: ab = b/5
(ab)/b = (b/5)/b
a = (b/5)*1/b
a = 1/5
Answer: (A). |
AQUA-RAT | AQUA-RAT-34576 | (ones digit is $$0$$, because it is the difference of two numbers whose ones digit is $$1$$).
Now if $$3n$$ is divisible by $$10$$, for an integer $$n$$, then $$n$$ is divisible by $$10$$. QED.
The following is multiple choice question (with options) to answer.
476 ** 0 is divisible by both 3 and 11. The non-zero digits in the hundred's and ten's places are respectively | [
"7 and 4",
"7 and 5",
"8 and 4",
"8 and 5"
] | D | Let the given number be 476 xy 0.
Then (4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3.
And, (0 + x + 7) - (y + 6 + 4) = (x - y -3) must be either 0 or 11.
x - y - 3 = 0 y = x - 3
(17 + x + y) = (17 + x + x - 3) = (2x + 14)
x= 2 or x = 8.
x = 8 and y = 5.
Option D |
AQUA-RAT | AQUA-RAT-34577 | # Venn Diagram Problem
• October 16th 2009, 07:34 AM
fifthrapiers
Venn Diagram Problem
In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 are in the club and 20 are neither on the team nor in the club, what is the minimum number of students who could be both on the team and in the club?
My solution:
I let x = number of students in both the club and fencing team and A = the number of students on the fencing team
Then, the number of students in the club is 70 - x
And then we know:
A + (70-x) + 20 - x = 100
This means A = 10
So now I used the same formula (Total # = A + B - (A^B) + neither)
100 = 10 + (70-x) - x + 20
But this didn't work... why isn't this formula working?
For 3 sets, Total # = A + B + C - A^B - A^C - B^C + (AUBUC) + "others"
So I think my formula for 2 sets is right..
The answer is meant to be 61 but plugging into that formula doesn't work:
10 + 9 -61 + 20 =! 100
Help?
• October 16th 2009, 08:08 AM
Plato
Quote:
Originally Posted by fifthrapiers
In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 are in the club and 20 are neither on the team nor in the club, what is the minimum number of students who could be both on the team and in the club?
Think of it this way. There are 80 students in French or fencing.
Of those 70 are in French. So there must be 10 in fencing that are not in French.
But the number in fencing must be more than 70.
So what in the minimum in both?
• October 16th 2009, 09:49 AM
Soroban
Hello, fifthrapiers!
Quote:
In a group of 100 students, more students are in Fencing than in French.
If 70 are in French and 20 are neither in Fencing nor French,
what is the minimum number of students who could in both actitvities?
I placed the information into a chart . . .
The following is multiple choice question (with options) to answer.
In a class of 69 students 41 are taking French, 22 are taking German. Of the students taking French or German, 9 are taking both courses. How many students are not enrolled in either course? | [
"6",
"15",
"24",
"33"
] | B | Formula for calculating two overlapping sets:
A + B - both + NOT(A or B) = Total
so in our task we have equation:
41 (french) + 22 (german) - 9 (both) + NOT = 69
54 + NOT = 69
NOT = 69 - 54 = 15
So answer is B |
AQUA-RAT | AQUA-RAT-34578 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
Find the simple interest on $3000 for 3 years at 10% per annum? | [
"$900",
"$300",
"$500",
"$600"
] | A | SI = PTR/100
= 3000*3*10/100 = $900
Answer is A |
AQUA-RAT | AQUA-RAT-34579 | ### Show Tags
15 Aug 2010, 03:11
Bunuel wrote:
praveengmat wrote:
How many factors does 36^2 have?
A 2
B 8
C 24
D 25
E 26
Finding the Number of Factors of an Integer:
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Back to the original question:
How many factors does 36^2 have?
$$36^2=(2^2*3^2)^2=2^4*3^4$$ --> # of factors $$(4+1)*(4+1)=25$$.
Or another way: 36^2 is a perfect square, # of factors of perfect square is always odd (as perfect square has even powers of its primes and when adding 1 to each and multiplying them as in above formula you'll get the multiplication of odd numbers which is odd). Only odd answer in answer choices is 25.
Hope it helps.
Thanks a ton !!.. loved the approach !
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Re: Help: Factors problem !! [#permalink]
### Show Tags
14 Oct 2010, 13:58
1
Factors of a perfect square can be derived by using prime factorization and then using the formula to find perfect square's factors.
The following is multiple choice question (with options) to answer.
How many different positive integers are factors of 36 ? | [
"10",
"2",
"7",
"3"
] | C | 36=2*18
3*12
4*9
6*6
Answer : C |
AQUA-RAT | AQUA-RAT-34580 | # 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.
A certain sum is invested at simple interest at 15% p.a. for two years instead of investing at 10% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum? | [
"s. 8400",
"s. 9000",
"s. 14000",
"s. 17000"
] | A | Let the sum be Rs. x.
(x * 15 * 2)/100 - (x * 10 * 2)/100 = 840
=> 30x/100 - 20x/100 =840
=> 10x/100 = 840
=> x = 8400.
ANSWER:A |
AQUA-RAT | AQUA-RAT-34581 | # Constant acceleration problem with a car
1. Sep 8, 2016
### Aman Abraha
1. Question to problem
A car moving with constant acceleration covered the distance between two points 57.9 m apart in 5.02 s. Its speed as it passes the second point was 14.4 m/s. (a) What was the speed at the first point? (b) What was the acceleration? (c) At what prior distance from the first point was the car at rest?
I'm using wileyplus.com to submit answers, and the only one that I'm stuck on is problem (c). A and B are correct.
2. Relevant equations
(a)V average= V final + V initial/ 2
(b)V final= V initial + at
(c)V final^2 = V initial^2 +2ad
3. Attempt at solution
(a) 11.5 m/s= 14.4 m/s + V initial / 2
V initial= 8.60 m/s
(b) 14.4 m/s = 8.60 m/s + a(5.02 s)
a= 1.16 m/s^2
(c) (8.60 m/s)^2 = (0 m/s)^2 + 2d(1.16 m/s^2)
d=31.9 m
For (c) it says on wiley plus that it is incorrect.
Any suggestion on what I did wrong?
Last edited by a moderator: Sep 8, 2016
2. Sep 8, 2016
### RUber
I am not familiar with the equation you used for (c).
In part (a), you found the velocity at the first point. In part (b) you found the constant acceleration.
How long did it take the car to reach the velocity from (a)?
How much distance would a car cover from rest in that amount of time at the acceleration you found in (b)?
3. Sep 8, 2016
### kuruman
The following is multiple choice question (with options) to answer.
Car Z travels 48 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour, but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour. How many miles does car Z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour? | [
"320",
"384",
"400",
"408.3"
] | B | The question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour. We therefore first calculate the fuel efficiency at that speed.
The stem tells us that at 45 miles/hour, the car will run 48 miles/gallon and at 60 miles/hour, that distance decreases by 20%. We can therefore conclude that the car will travel 38.4 miles/gallon at a constant speed of 60 miles/gallon. With 10 gallons of fuel, the car can therefore travel 38.4 miles/gallon * 10 gallons = 384 miles.
Answer B. |
AQUA-RAT | AQUA-RAT-34582 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 7% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? | [
"5",
"8",
"15",
"20"
] | B | Use the forumla ;
Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither
100 = 50 + 30 + 20 - ( sum of 2) -2(7) +15
100 = 101-( sum of 2)
1 = sum of 2
so more than 1 = 8
B |
AQUA-RAT | AQUA-RAT-34583 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
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The following is multiple choice question (with options) to answer.
The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas? | [
"81:121",
"81:128",
"81:126",
"81:124"
] | A | Ratio of the sides = ³√729 : ³√1331
= 9 : 11
Ratio of surface areas = 92 : 112
= 81:121
Answer: A |
AQUA-RAT | AQUA-RAT-34584 | There will be one solution: name it $x_0$.
Find $y$ at $x_0$, and call it $y_0$.
Then you have the slope of the desired line, and the point $(x_0, y_0)$ and can use the point slope form of the equation of a line: $$y - y_0 = m(x-x_0)$$
The following is multiple choice question (with options) to answer.
Which of the following points is closest to line y=x | [
"(2, -1)",
"(2, 4)",
"(1, 1)",
"(2, -1)"
] | C | Attachment:
m12-20.pngAs you can see point (1, 1) is the closest to line y=x.
Answer: C. |
AQUA-RAT | AQUA-RAT-34585 | Understanding a proof showing that for any prime $p$, there are integers $x$ and $y$ such that $p|(x^2+y^2+1)$.
I asked this question a couple days ago: Show that for any prime $p$, there are integers $x$ and $y$ such that $p|(x^{2} + y^{2} + 1)$. But I asked it as a guest, and I could not comment on the answers I got as I did not have enough reputation.
I have found a proof from Warwick University that will help me answer this question, but there is one sentence I am unsure of. If anybody could help me out on it, that would be amazing!
Here it is:
There are $\frac{p+1}{2}$ numbers of the form $x^2 \pmod{\!p}$. There are $\frac{p+1}{2}$ numbers of the form $-1-y^2 \pmod{\!p}$ (because to get the numbers of the form $-1-y^2$, multiply the squares by $-1$ and subtract $1$… this will not change how many there are). There are exactly $p$ numbers mod $p$. Note that $\frac{p+1}{2}$ is more than half of $p$. So the set of numbers of the form $x^2$ and those of the form $\frac{p+1}{2}$ must have at least one common element (if not, we will have too many distinct numbers mod $p$). So there are some $x$ and $y$ such that $x^2\equiv -1-y^2 \pmod{\!p}$. Thus $x^2 + y^2 +1\equiv 0 \pmod{\!p}$.
I understand it up until the bolded sentence. Did they mean to write numbers of the form $\frac{p+1}{2}$? Or did they instead mean numbers of the form $-1-y^2$?
The following is multiple choice question (with options) to answer.
x and y are positive integers. If x is a prime factor of y^2 , which of the following MUST be true? | [
"x is even",
"x=y^2",
"y/x is even",
"x/y <=1"
] | D | A. x is even --> not necessarily true: x=3=odd and y=3
B. x=y2--> never true, as x=primex=prime then it can not equal to square of another integer;
C. yx is even --> not necessarily true: if x=3x=3 and y=3y=3 then yx=1=odd
D. xy≤1 is less than or equal to 1 --> always true as shown above or by POE;
E. x+1 is also a prime factor of y2--> not necessarily true: if x=2 and y=2 then x+1=3 and 3 is not a factor of y2=4
Answer: D. |
AQUA-RAT | AQUA-RAT-34586 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
How many numbers with 2 different digits can be formed using only the odd digits? | [
"10",
"20",
"40",
"60"
] | B | Odd digits are 1, 3, 5, 7, 9.
We want 3 different digits.
First digit can be chosen in 5 ways.
Second digit can be chosen in 4 ways.
Total ways = 5*4
= 20
Hence option (B). |
AQUA-RAT | AQUA-RAT-34587 | You have reasoned correctly and your solution is correct.
The one part that needs clarification (as has been pointed out in the comments) is
We note that since $0$ doesn't appear as the first digit it must appear the least number of times.
I think this should be clarified:
In typing out $1$ through $1000$, if we had typed the leading $0$s (like $001, 002, 003,$ etc.) then we would have counted every digit $0$ through $9$ an equal number of times, not counting the one extra $1$ in $1000$. So we are missing a number of zeros equal to the number of leading zeros we left off. There is certainly at least one leading zero we left off, thus there are fewer zeros than there are any of the digits $2$ through $9$.
• @Puzzled417 That reasoning works as well! – 6005 Nov 22 '16 at 0:26
• I don't think that is relevent and doesn'doesn't need stating. All columns go from 1 through 9 to 0, so the zero appearrng the last case doesn't imply 0s are any more common than any other number. After all, 9 appears three times in 999 but we didn't need to single that out. And 8 appears three times in 888 .... – fleablood Nov 22 '16 at 1:47
• @fleablood I was thinking the numbers went from $0$ to $1000$. Since they go from $1$ to $100$, you're right, we should instead count $1000$ as contributing three zeros. But if you read the original post, you will see the OP argued that there were an equal number of digits $1-9$ in the numbers $1$ through $999$, so the way he/she argued it, you need to worry about the extra three $0$s. – 6005 Nov 22 '16 at 1:49
• @fleablood I edited the post, do you agree with it now? – 6005 Nov 22 '16 at 1:52
The following is multiple choice question (with options) to answer.
What is the least number of digits (including repetitions) needed to express 10^600 in decimal notation? | [
"a) 4",
"b) 601",
"c) 101",
"d) 1000"
] | B | 10^n is a decimal number with a 1 followed by n zeros.
So 10^600 will include 600 0's + 1 digit for 1 = 601
So the answer is B |
AQUA-RAT | AQUA-RAT-34588 | 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.
In a 2000 m race, A beats B by 100 meters or 40 seconds. Find the speed of B? | [
"10m/s",
"2.5m/s",
"5m/s",
"7m/s"
] | B | Since A beats B by 100 m or 40 seconds,
it implies that B covers 100 m in 40 seconds.
Hence speed of B = 100/40 = 2.5 m/s.
Answer: B |
AQUA-RAT | AQUA-RAT-34589 | inorganic-chemistry, everyday-chemistry, extraction
The main process for production of aluminum metal requires a current of 400 kiloamperes and temperatures of 950 C. In contrast, the main process for sodium metal production requires a current of 30 kiloamperes and temperatures of 600 C.
The main reason would be because sodium is too reactive to be used. The sodium metal would react with the water and form sodium hydroxide. This reaction would be too fast and violent that you wouldn't be able to recover copper.
So your teacher was right in pointing out that sodium is too reactive.
The following is multiple choice question (with options) to answer.
It takes 2 tons of copper ore and 4 tons of iron ore to make one ton of alloy A. How many tons of alloy A can be made from 60 tons of copper ore and 90 tons of iron ore? | [
" 18",
" 18 3/4",
" 20",
" 22 1/2"
] | D | Yes, you need copper ore:iron ore in the ratio 2:4. Total 6 tons of the mix in this ratio will give 1 ton of alloy A.
If you have 60 tons of copper ore, it is enough for 60/2 = 30 tons of alloy A.
If you have 90 tons of iron ore, it is enough for 90/4 = 22 1/2 tons of alloy A.
Since iron ore is available for only 22 1/2 tons of alloy A, you can make only 22 1/2 tons of alloy A.
The leftover copper ore alone cannot make any alloy A and hence will be leftover only.
Answer must be 22 1/2.
(D) |
AQUA-RAT | AQUA-RAT-34590 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A WORKS TWICE AS FAST AS B. IF B CAN COMPLETE A WORK IN 12 DAYS INDEPENDENTLY, THE NUMBER OF DAYS IN WHICH A AND B CAN TOGETHER FINISH THE WORK IS: | [
"4 DAYS",
"6 DAYS",
"8 DAYS",
"18 DAYS"
] | A | RATIO OF RATES OF WORKING OF A AND B = 2:1. SO, RATIO OF TIME TAKEN = 1:2
A'S 1 DAY'S WORK = 1/6, B'S 1 DAY'S WORK = 1/12
(A+B)'S 1 DAY'S WORK = (1/6+1/12)=3/12=1/4
SO, A AND B TOGETHER CAN FINISH THE WORK IN 4 DAYS.
CORRECT OPTION: A |
AQUA-RAT | AQUA-RAT-34591 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
An amount of money is to be divided between P, Q and R in the ratio of 3:7:12. If the difference between the shares of P and Q is Rs.4000, what will be the difference between Q and R's share? | [
"2788",
"5000",
"7282",
"2782"
] | B | 4 --- 4000
5 --- ? => 5000
Answer:B |
AQUA-RAT | AQUA-RAT-34592 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
In the next Worldcup of cricket there will be 12 teams, divided equally in 2groups. Teams of eachgroup will play a match against each other. From each group 3 top teams will qualify for the next round. In this round each team will play against each others once. 4top teams of this round will qualify for the semifinal round, where they play the best of 3matches. The Minimum number of matches in the next World cup will be | [
"48",
"49",
"53",
"55"
] | C | The number of matches in first round,
= 6C2 +6C2;
Number of matches in next round,
= 6C2;
Number of matches in semifinals,
= 4C2;
Total number of matches,
= 6C2 +6C2+6C2+4C2+2.
= 53.
C |
AQUA-RAT | AQUA-RAT-34593 | $11 + 6.5 = 17.5 \text{ or } 24 - 6.5 = 17.5$.
Therefore, 17.5 is the number in the middle of $11 \mathmr{and} 24.$
• 4 minutes ago
• 5 minutes ago
• 7 minutes ago
• 8 minutes ago
• 37 seconds ago
• A minute ago
• A minute ago
• 2 minutes ago
• 2 minutes ago
• 3 minutes ago
• 4 minutes ago
• 5 minutes ago
• 7 minutes ago
• 8 minutes ago
The following is multiple choice question (with options) to answer.
find the number, difference between number and its 3/5 is 62. | [
"150",
"153",
"154",
"155"
] | D | Explanation:
Let the number = x,
Then, x-(3/5)x = 62,
=> (2/5)x = 62 => 2x = 62*5,
=> x = 155
Answer: Option D |
AQUA-RAT | AQUA-RAT-34594 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
A team bought a total of 198 balls. If each player on the team bought at least 17 balls, then what is the greatest possible number of players on the team? | [
"21",
"11",
"23",
"24"
] | B | 198/17=11 plus remainder.
The answer is B. |
AQUA-RAT | AQUA-RAT-34595 | => 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.
Every month Rani gives 65% of her salary to her mother. She spends 15% of her salary and saves the remainder. She saves $130 more than what she spends. What is her monthly salary? | [
"2200",
"2400",
"2600",
"2800"
] | C | Total Salary given+Spent+Saving= 100%
65%+15%+Savings =100%
80%+Savings=100%
Savings=100%-80%=20%
Savings%-Spent%=$130
20-15=5%
5%of Salary =$130
5/100 * Salary = 130
Salary =130*100/5 =$2600
answer : C |
AQUA-RAT | AQUA-RAT-34596 | # conditional probability trick questions - drawing cards from a deck and the meaning of 'at least'
Suppose that a box contains one blue card and four red cards, which are labeled A, B, C, and D. Suppose also that two of these five cards are selected at random, without replacement.
a. If it is known that card A has been selected, what is the probability that both cards are red?
b. If it is known that at least one red card has been selected, what is the probability that both cards are red?
I have been assigned the problem above in a class. I am aware that the book considers the answer for A) to be 3/4, which is obvious. However, the answer to B) is claimed to be $$P(red_1)P(red_2|red_1) = 4/5 \cdot 3/4 = 3/5$$
The professor has not been able to explain why the answers to A and B are different in any way that makes sense to me. There is no way in which you cannot draw at least 1 red card in a draw of 2 cards. If you are told you have drawn at least 1 red card, then you have the same information that you did in A.
• The box has one blue card and 4 red cards? In this case wouldn't (b) premise - that is at least one red card was selected - always be true? It seems like you can never select two cards without at least one of them being red. – Karolis Koncevičius Oct 29 '14 at 18:19
• @Karolis Is that a problem? – whuber Oct 29 '14 at 18:24
• @whuber No, not a problem, sorry. I just thought maybe there was an error in formulation. Not used seeing Pr(something | whole_space) kind of questions. – Karolis Koncevičius Oct 29 '14 at 18:36
• To clarify, the red cards are labeled A, B, C, and D, and the blue card is unlabeled (or labeled something other than A, B, C, or D)? – Hao Ye Oct 29 '14 at 23:42
Look at the possible samples consistent with the information, each of which is equally likely:
The following is multiple choice question (with options) to answer.
If you select two cards from a pile of cards numbered 1 to 10, what is the probability that the sum of the numbers is less than 4? | [
"1/45",
"4/45",
"1/15",
"1/10"
] | A | The number of ways to choose two cards is 10C2=45.
There is only one way to get a sum less than 4.
P(sum less than 4)=1/45
The answer is A. |
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