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# m25#18

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26 Sep 2010, 15:22
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There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village?

1. There are 40 more pupils in the second school than there are in the first.
2. There are three times as many pupils in the second school as there are in the first.

[Reveal] Spoiler: OA
B

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[Reveal] Spoiler: OE
Statement (1) by itself is insufficient. Denote $$X$$ and $$Y$$ the number of pupils in school 1 and school 2 respectively. By definition of average, we have to calculate $$\frac{12.2X + 13.1Y}{X + Y}$$ . S1 is not sufficient because one of the unknowns cannot be found.

Statement (2) by itself is sufficient. S2 says that $$Y = 3X$$ , so $$X$$ cancels out of the fraction and we can calculate the result.

Just a little confused : according to statement 1. can't we make the equation Y=X+40 ?

Thanks
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26 Sep 2010, 21:07
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shrive555 wrote:
Quote:
There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village?

1. There are 40 more pupils in the second school than there are in the first.
2. There are three times as many pupils in the second school as there are in the first.

Statement (1) by itself is insufficient. Denote $$X$$ and $$Y$$ the number of pupils in school 1 and school 2 respectively. By definition of average, we have to calculate $$\frac{12.2X + 13.1Y}{X + Y}$$ . S1 is not sufficient because one of the unknowns cannot be found.

Statement (2) by itself is sufficient. S2 says that $$Y = 3X$$ , so $$X$$ cancels out of the fraction and we can calculate the result.

Just a little confused : according to statement 1. can't we make the equation Y=X+40 ?

Thanks

Hi shrive, you are right. We can make the equation $$Y=X+40$$, but by using this equation, can we find the required avg?? I think No. Check this out.
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27 Sep 2010, 08:57
Thanks Hussain15:
well i assumed " zero " on the right hand side of the equation, which is wrong.
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02 Dec 2010, 06:26
we cant find answer by stmnt 1 but we can solve it by statement 2.
because the common term will cancel out.
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02 Dec 2010, 08:24
It just says the answer is "B", but I don't see what B is. I got 12.875 as the average age. Is this correct?
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02 Dec 2010, 08:34
B.
Average * #items = SUM
12.2 * x = 12.2x School 1
13.1 * 3x = 39.3x School 2
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? * 4x = 51.5x Total

? = 51.5x/4x doesn't matter outcome. Xs cancel and yes we get the average.

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02 Dec 2010, 11:18
Zanini wrote:
It just says the answer is "B", but I don't see what B is. I got 12.875 as the average age. Is this correct?

yes..
i solve this question using the statement 2 and got the same result.
its 12.875
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03 Dec 2010, 06:11
hi wayxi,

I dont understand. Can you explain in more detail how you came your equation below. please explain using statement 1 and 2

12.2 * x = 12.2x School 1
13.1 * 3x = 39.3x School 2
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? * 4x = 51.5x Total

Thanks
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03 Dec 2010, 07:21
B.

S1 : assume x for first school so for second school = x + 40 the number of pupils
x*avg1 + (x+40)*avg2 / 2x + 40 is not sufficient

s2 : x, 3x : x*avg1 + 3x*avg2 / 4x --> x get cancelled and hence new avg can be determined
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21 Dec 2010, 15:13
Maryann,

The formula for average goes:
Average * # Items = SUM

i.e. Average of 6 different values is 62. Then sum of the six different values are:
62 * 6 = 372

Statement 1:
x : number of students in School 1
x + 40 : number of students in School 2

SCHOOL 1: 12.1 * x = 12.1x
SCHOOL 2: 13.1 * (x+40) = 13.1x + 524
We want to the get the average of BOTH schools so we need to divide the Total SUM by Total Items.
Adding the SUM of the two schools we get 25.2x + 524.
Adding the # Items we get 2x+40
AVERAGE: ( 25.2x + 524 ) / ( 2x+40 ) . We can't work this out so its, INSUFFICENT

Statement 2:
x: number of students to School 1
3x: number of students in School 2
SCHOOL 1: 12.1 * x = 12.1x
SCHOOL 2: 13.1 * 3x = 39.3x
Adding the SUM of the schools : 12.1x + 39.3x = 51.4x
Adding #Items: x +3x = 4x
AVERAGE: (51.4x) / (4x )

The Xs cancel so we get a value. It doesn't matter what the value is as long as we're able to solve this. Its 12.85 btw. B is SUFFICENT
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07 Dec 2011, 09:45
@Wayxi

Sorry, I need to brush up on math fundamentals, but is the reason we cannot solve statement 1 because by trying to eliminate the variable in the denominator, we end up creating a quadratic equation with two potential answers? If not, whats the reasoning?
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07 Dec 2011, 11:24
What would the difficulty level of this question be?
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26 Mar 2012, 08:09
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Expert's post
shrive555 wrote:
There are two schools in the village. The average age of pupils in the first school is 12.2 years; the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village?

1. There are 40 more pupils in the second school than there are in the first.
2. There are three times as many pupils in the second school as there are in the first.

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

[Reveal] Spoiler: OE
Statement (1) by itself is insufficient. Denote $$X$$ and $$Y$$ the number of pupils in school 1 and school 2 respectively. By definition of average, we have to calculate $$\frac{12.2X + 13.1Y}{X + Y}$$ . S1 is not sufficient because one of the unknowns cannot be found.

Statement (2) by itself is sufficient. S2 says that $$Y = 3X$$ , so $$X$$ cancels out of the fraction and we can calculate the result.

Just a little confused : according to statement 1. can't we make the equation Y=X+40 ?

Thanks

There are two schools in the village. The average age of pupils in the first school is 12.2 years and the average age of pupils in the second school is 13.1 years. What is the average age of all school pupils in the village?

This is a weighted average question. Say $$x$$ and $$y$$ are the number of pupils in the first and the second schools in the village, respectively: $$average=\frac{12.2x+13.1y}{x+y}$$.

(1) There are 40 more pupils in the second school than there are in the first --> $$x+40=y$$. Not sufficient to find the average.

(2) There are three times as many pupils in the second school as there are in the first --> $$3x=y$$ --> $$average=\frac{12.2x+13.1*3x}{x+3x}=\frac{12.2+13.1*3}{4}\approx{12.9}$$. Sufficient.

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07 Dec 2012, 01:17
Hi,
Only B is sufficient to answer the question, cannot be solved by A .

If we go by A, we end up in an unknown variable that doesnot cancel out.
If we go by B, we clearly can get the answer as Sum of age of pupils (second school) can be expressed in form of Sum of Age of Pupils (First School)
and same applies to No Of students as well.

Thus, B can solved the problem alone.

Thanks,
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07 Dec 2012, 07:25
B for me because only average is needed, not the exact no of students.
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03 Dec 2013, 18:23
All,
I have a different take on this:
Total age cannot be fraction (age of any student is a whole number).
Stem1--> S - F = 40 (eqn-1)

average age of students in first school = 12.2 (hence number of students = 5, 10, 15, 20, 25, ..)
average age of students in second school = 13.1 (hence no of students = 10, 20, 30, 40, 50, ..)
In order to satisfy stem1 (equation 1) and question stem:
Number of students in school1 = 80
No of students in school 2 = 40

sufficient ( A or D)

Stem 2:
to satisfy question stem and S = 3F

S = 30, 60, 90...
F = 10, 20, 30
Not sufficient

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04 Dec 2013, 02:56
abhayrai wrote:
All,
I have a different take on this:
Total age cannot be fraction (age of any student is a whole number).
Stem1--> S - F = 40 (eqn-1)

average age of students in first school = 12.2 (hence number of students = 5, 10, 15, 20, 25, ..)
average age of students in second school = 13.1 (hence no of students = 10, 20, 30, 40, 50, ..)
In order to satisfy stem1 (equation 1) and question stem:
Number of students in school1 = 80
No of students in school 2 = 40

sufficient ( A or D)

Stem 2:
to satisfy question stem and S = 3F

S = 30, 60, 90...
F = 10, 20, 30
Not sufficient

Consider x=20 and y=60 for (1). The correct answer is B, not A. Check here: m25-101757.html#p1065083

Hope it helps.
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