Question: Combined Test Scores

Comment on Combined Test Scores

Also, we can apply another weighted average to solve this particular question. Here's how:
Firstly , consider the total number of students in the school be 'T', of which there are 'B' number of boys and 'G' number of girls. One can also consider the value of number of girls to be 'G' = T - B.
Now, Applying weighted average formula: We have,

Weighted average = (proportion)(boy's average test score) + (proportion)(girl's average test score)

Substituting corresponding value in the equation yields,

128 = (120)(B/T) + (144)((T-B)/T) [since. G = T-B]

Multiplying both sides by T, we have,

128T = 120B + 144T - 144B

Upon rearranging we get,
24B = 16T
Thus, B = (2/3)T
and therefore, G = T -B = (1/3)T

As we are asked to determine ratio of the number of boys to girls:

Boys : Girls = (2/3)T : (1/3)T = 2 : 1

Hence, the correct answer choice, here, is still option C.

greenlight-admin's picture

Perfect!

How does (2/3)T : (1/3)T equal 2 to 1?
greenlight-admin's picture

Take: (2/3)T : (1/3)T
Divide both sides by T to get the EQUIVALENT ratio 2/3 : 1/3
Multiply both sides by 3 to get the equivalent ratio 2 : 1

Cheers,
Brent

Sum1 = 144 (G) => sum of score of G number of girls

Sum2 = 120 (B) => sum of score of B number if boys

Sum1 + Sum2 = 128 (G+B) => sum of score of total number of students ( G + B)

substitute Sum1 and Sum2 in 3rd equation

144(G) + 120(B) = 128(G) + 128(B)
16(G) = 8(B)
B/G = 2/1 there C

I did it using Tug and War method.
No. of Girls = 144
No. of Boys = 120.
If equal number of girls and boys are at each end, the average would be 132. Equal distance (12) from both boys and girls

B-------(avg)-------G
120 132 144

However, now we are given average of 128. So we know that we do not have equal number of boys and girls. The average is closer to number of boys than girls (128- 120 = 8 vs 144-128 = 16). So we can say more number of boys are there than the girls.


B-------(given avg)-----------------G
120 (8) 128 (16) 144

Using proportion, we can say that B:G = 16:8 => 2:1
greenlight-admin's picture

Nice work. That method is also called "Alligation."

Does the "Alligation" method works in all given problems like this one?
greenlight-admin's picture

I have chosen not to teach the alligation method because it is not easily modified to handle more advanced question.

That said, this question CAN be solved using alligation.

Cheers,
Brent

I'm getting the result but somewhat in a different form.
Assume: a = girls b = boys

(a/a+b)*144 + (b/a+b)* 120 = 128

after solving the solution is : 16a : 8b or 2:1
which states that the ratio of girls to boys is 2:1 therefore boys to girls should be 1:2.

What am I doing wrong?
greenlight-admin's picture

Great idea to apply the weighted averages formula!

This part is perfect: (a/a+b)*144 + (b/a+b)* 120 = 128

When we simplify this equation, we get 16a = 8b

Our goal is to find the ratio of boys to girls.
In other words, we want to find the value of b/a

Take 16a = 8b
Divide both sides by 8 to get: 2a = b
Divide both sides by a to get: 2 = b/a
In other words, 2/1 = b/a
We can also write this as b : a = 2 : 1

Does that help?

Cheers,
Brent

Hey Brent,

Very confused on when I get 16G = 8B, why I can’t divide both sides by 8 resulting in 2G:1B.
greenlight-admin's picture

The EQUATION 16G = 8B, is different from the RATIO 16G : 8B

Let's take the EQUATION 16G = 8B, and find a pair of values (a G-value and a B-value) that satisfies this equation.

One possible solution is G = 1 and B = 2
In other words, for 1 girl, we have 2 boys.
So, the RATIO of girls to boys is 1 : 2

Also note that the ratio 2G : 1B doesn't make any sense.
We might say the ratio of girls (G) to boys (B) is 2 : 1
Or we can say G : B = 2 : 1

Conversely, the ratio 2G : 1B doesn't tell us what's happening.

Does that help?

Cheers,
Brent

PS: It might be useful to review the introductory lesson on ratios: https://www.greenlighttestprep.com/module/gre-arithmetic/video/1086

You are hands down the best GRE instructor.
greenlight-admin's picture

Thanks aferro1989!

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