Question: Combined Test Scores

Comment on Combined Test Scores

Hello Brent,

Can this question be solved using weighted average?

Thank you.
gmat-admin's picture

You bet!

Weighted average = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ..

In this case, the formula is...
Average score = (proportion of girls)(girls' average score) + (proportion of boys)(boys' average score)

Let's examine the proportions first.

Let G = # of girls
Let B = # of boys

So, G + B = TOTAL number of students
And G/(G + B) = PROPORTION of girls
And B/(G + B) = PROPORTION of boys

Now we'll plug in all of the values...
128 = (G/G+B)(144) + (B/G+B)(120)
Multiply both sides by (G+B) to get: 128G + 128B = 144G + 120B
Simplify: 8B = 16G
Divide both sides by 8 to get: B = 2G

We want to find ratio of boys to girls (i.e., B/G)
Take B = 2G and divide both sides by G to get: B/G = 2, which is the same as B/G = 2/1

So, the ratio of boys to girls is 2 to 1

Answer: C

More on weighted averages:

One can plug ratio's starting with "C" (as Brent advised in a previous module):
1/3 (144)+2/3(120)=128
384=384 (Sucess !)
Now my question is: Is it risky to systematically follow that pattern ?
gmat-admin's picture

That's a great approach! It's only risky if it takes you a lot of time to check each answer choice. But the way you have set up the equation, it shouldn't take long to determine the correct answer.

I use the method of distances from the average. If the average is 128. The girls avg score is 144 and the boys average score is 120, which are 16 more and 8 less than the average score respectively. therefore, the ration between 16 and 8 is 2:1 and now we just need to ask ourselves- "Who is closer to the average? Who "weighs" more? In this case it is the boys (120) - so they get the larger unit of the ration - meaning 2. So the answer is 2:1 - C. This way we save almost all calculations.
gmat-admin's picture

Yes, that approach (aka "allegation") works too. I typically don't teach that approach, since it only works on certain types of mixture questions.

Hi Brent ,

Please help me in this question .

Q: There is a set of consecutive even integers. What is the standard deviation of the set?
(1) There are 39 elements in the set.
(2) the mean of the set is 382.

gmat-admin's picture

Statement 1 is sufficient.
We know this because ANY set of 39 consecutive even integers will have the same dispersion. For example, the numbers {2, 4, 6, 8} have the same dispersion as {24, 26, 28, 30}. Every number is 2 greater than the number before it.

Statement 2 is not sufficient.
The set could be {4} in which case the standard deviation = 0, or the set could be {4, 6} in which case the standard deviation does not equal 0

I found this approach to be faster since there is no formula to solve.
Ratio is 16:8 = 2:1

Can you explain how you would plug in ratios 3:2 or 4:3 if we were just doing guess and check?
gmat-admin's picture

For each answer choice, you'd need to assign a "nice" value to the student population, and then go from there.

For example...

(A) 4 to 1
If the ratio of boys to girls is 4 to 1, then there could be 4 boys and 1 girl
So, the 1 girl had a test score of 144.
If the 4 boys had an average score of 120, then the SUM of the boys' scores = (4)(120) = 480
Does this yield a school average of 128?
Let's find out...
(144 + 480)/5 = 624/5 = 124.8
No good. We want a school average of 128

As you can see, this process will take a while! But it works!!

(C) 2 to 1
So, there could be 2 boys and 1 girl
So, the 1 girl had a test score of 144.
If the 2 boys had an average score of 120, then the SUM of the boys' scores = (2)(120) = 240
Does this yield a school average of 128?
Let's find out...
(144 + 240)/3 = 384/3 = 128

Answer: C


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