Prep Exam 1 - #8

[Guest]

Hi,

Can someone please walk me through how to solve this problem using only choice 2?

Thanks

[GMAT 2007]

Here you go:

Avg(arithematic mean) for Males = 9.8
Avg(arithematic mean) for females = 9.1

Statement (2) - Combined arithematic mean = 9.3

If no. of males is 'x' and no. of females is 'y' then from above we know

9.8 = (x1+x2+x3....) / x where x1, x2.... are males
or 9.8x = (x1+x2+x3....)--------- (1)

9.1 = (y1+y2+y3....) / y where y1, y2 ..are females
or 9.1y = (y1+y2+y3....)---(2)

and 9.3 = ((x1+x2+x3..+) + (y1+y2+y3...)) / (x+y) -----(3) (Combined males & females)

substituting (x1+x2+x3..+) & (y1+y2+y3...)) from (1) & (2) in (3) and solving

x/y = 1/6. Hence (2) is sufficient

GMAT 2007

[Guest]

GMAT2007,

Thanks for the quick response! I appreciate your help.

[RonPurewal]

Gmat2007 has most of the setup right, but his/her answer is wrong.

If you do the substitution, you get
(9.8x + 9.1y)/(x + y) = 9.3
multiply by x + y --> 9.8x + 9.1y = 9.3x + 9.3y
subtract 9.3x and 9.1y from both sides --> 0.5x = 0.2y
therefore x = (0.2/0.5)y
--> x = (2/5)y
x/y = 2/5

Notice that the answer given by GMAT2007 (many more males than females) doesn't make sense. Everybody's average is much closer to the women's average than to the men's average, meaning that there must be substantially more women than men employees.

Incidentally, you can memorize the following fact: If you have the averages for two components of a group, and also a weighted average of the whole group, then you can find the ratio of the two components. This knowledge will in itself be sufficient to address many data sufficiency problems involving ratios and averages, without the rather extensive algebraic twists and turns shown here.

[GMAT 2007]

Thanks for correcting the calculation mistake Ron. Yes - the correct answer is x/y = 2/5. I guess I did mistake at the last step.

GMAT 2007