Averages

[tanyasethi]

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.

I thought the answer to this question would be C but the answer is B

Could someone please explain the solution of this problem to me?

[vineeshp]

No. Since the equation for the total average will be

12.2X + 13.1Y / X + Y.

From statement 2, all Y will get replaced by 3X thus allowing us to remove X from both denominator and numberator. And directly give us the total average.

[amit2k9]

a (12.2m+13.1n)/(m+n)
m-n=40

thus 12.2(n+40)+13.1n/2n+40. hence cannot be solved.not sufficient.

b n/m=3 hence
12.2m+13.1*3m/4m hence m gets cancelled as common factor. sufficient.

B it is.

[GMATGuruNY]

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.

I thought the answer to this question would be C but the answer is B

Could someone please explain the solution of this problem to me?

This is a DS weighted average question. Very common.
We can answer this sort of question very quickly without doing any math.
A weighted average question has 3 key elements:

1: The average of each element.
2: The ratio of the elements when they are combined.
3: The average of the final mixture.

If we know the average of each element and the ratio, we have sufficient information to determine the average of the final mixture.
If we know the average of each element and average of the final mixture, we can determine the ratio of the elements in the mixture.

In the DS above, let F = first school and S = second school:
Average of F = 12.2.
Average of S = 13.1.

To determine the average of the whole village, we need to know F:S.

Statement 1: There are 40 more pupils in the second school than there are in the first.
S = F+40.
Doesn't give us F:S.
Insufficient.

Statement 2: There are three times as many pupils in the second school as there are in the first.
Thus, F:S = 1:3.
Sufficient.

The correct answer is B.

To illustrate:
Let F=1 and S=3.
Sum of ages = 1*12.2 + 3*13.1 = 51.5.
Average = 51.5/4 = 12.875.

Let's double the elements in the ratio.
Let F=2 and S=6.
Sum of ages = 2*12.2 + 6*13.1 = 103.
Average = 103/8 = 12.875.

Notice that the average stays the same.
As long as F:S = 1:3, the average for the village will be 12.875.

We should hope for this sort of question on the GMAT because it can be answered very quickly without doing any math.

[tanyasethi]

Hey thanks a ton for your reply! That strategy should really help speed up things for me!