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## Comment on

Combined Test Scores## Hello Brent,

Can this question be solved using weighted average?

Thank you.

## You bet!

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: https://www.gmatprepnow.com/module/gmat-statistics/video/805

## Alternatively,

One can plug ratio's starting with "C" (as Brent advised in a previous module):

2(B):1(G)

1/3 (144)+2/3(120)=128

(144+240)/3=128

384=384 (Sucess !)

Now my question is: Is it risky to systematically follow that pattern ?

## That's a great approach! It's

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

## Yes, that approach (aka

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.

Thanks

## Statement 1 is sufficient.

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

144------------128-----------120

(B)----(16)---(mean)----(8)---(G)

Ratio is 16:8 = 2:1

## Can you explain how you would

## For each answer choice, you'd

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

ELIMINATE A

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

Perfect!!

Answer: C

Cheers,

Brent

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