Problem Solving

In a corporation, 50 percent of the male employees and 40 percent of the female employees are at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?

A. 3/5
B. 2/3
C. 3/4
D. 4/5
E. 5/6

is the answer A?

0.5x + 0.4y = 0.42(x+y)
0.08x = 0.02 y
4x = y

The ration we are looking for is y:(x+y)
= 4x/x+4x
=4/5

Option D

Agree D

Total employees be m+ f

Male emp be m
Feamle be F

ttal number of 35 year old in organization : 0.42 (m+f)


0.42 (m+f) = 0.5 m + 0.4 f
0.02 f = 0.08 m
f = 4m

Females in organization:
f/m+f = 4m/5m = 4/5

gmatinjuly wrote:Agree D

Total employees be m+ f

Male emp be m
Feamle be F

ttal number of 35 year old in organization : 0.42 (m+f)


0.42 (m+f) = 0.5 m + 0.4 f
0.02 f = 0.08 m
f = 4m


Females in organization:
f/m+f = 4m/5m = 4/5

Do you guys typically use a grid for these types of problems? I'm a bit algebraically challenged and I find it's much easier to compute using a grid- I know we post the grid on the thread (or I can't), but just curious how you guys attack these.

Hey yaa i used to use Grids but i at times feel more comfortasble reading the problem objectively....

Say in this case if one understands that number of people under 35 of total is actually 40% of total male plus 50% of females ...then it make me easier to understand it.....and i can put in an equation for that.

If you are more comfortable with Grid approach then you should use that.....but i get confused with that at times so I avoid using it

This is a weighted average question. Very easy to solve using a number line and remembering a quick rule.

Females ---2--- overall average ----------8---------- Males

(the numbers are the distance from individual group averages to the overall average... i.e. fem=40%, overall = 42%, males = 50%)

To find the weight of each group, set up the following ratio:

f/m = (distance of males to average)/(distance of females to average)
f/m = 8/2 = 4/1

Therefore, 4 out of every 5 is female: choose (d).

* * *

A more general statement of the rule:

Avg group A ---- x----- Overall average --------y-------- Avg group B

A/B = y/x

For example:

A class writes a math test and the overall average is 75%. If the girls average 85% and the boys average 70%, what fraction of the class is boys?

boys ----- 5------ overall average---------10----------girls

b/g = 10/5 = 2/1

So, the class is 2/3 boys.

Hi Stuart,

That is a very novel approach! Me likey.

BUT the relationship you offered, that is, A/B = y/x, is not inherently understood... Not by me, at least.

Is there a link to a proof you can show? Or, if you dare, would you write it yourself?

I hate memorizing tricks like that as they don't really stick unless I get the to the root of understanding it.

Perhaps I am not alone here, but under the duress of GMAT, I would likely default to the more secure, but alas, brute force method of setting up multiple equations, like those above.

Someone needs to save me from those brute force methods. They are indeed ugly.

Stuart Kovinsky wrote: For example:

A class writes a math test and the overall average is 75%. If the girls average 85% and the boys average 70%, what fraction of the class is boys?

boys ----- 5------ overall average---------10----------girls

b/g = 10/5 = 2/1

So, the class is 2/3 boys.

Hi Stuart,

Sorry for the basic question ... how did you get to the 2/3 in the above example. Not sure I quite follow what you are saying here.

Thanks.

II wrote:

Stuart Kovinsky wrote: For example:

A class writes a math test and the overall average is 75%. If the girls average 85% and the boys average 70%, what fraction of the class is boys?

boys ----- 5------ overall average---------10----------girls

b/g = 10/5 = 2/1

So, the class is 2/3 boys.

Hi Stuart,

Sorry for the basic question ... how did you get to the 2/3 in the above example. Not sure I quite follow what you are saying here.

Thanks.

We have the part:part ratio boys:girls = 2:1

We have the part:whole ratio boys:(boys+girls) = 2:2+1 = 2:3

So, boys/(boys+girls) = 2/3

Hi Stuart,

Sorry but I am not able to understand this approach. How did you get 5 and 10 for boys and girls?

Are you picking numbers here?

Please explain.

Thanks,
Vikas Mehra

Stuart Kovinsky wrote:

II wrote:

Stuart Kovinsky wrote: For example:

A class writes a math test and the overall average is 75%. If the girls average 85% and the boys average 70%, what fraction of the class is boys?

boys ----- 5------ overall average---------10----------girls

b/g = 10/5 = 2/1

So, the class is 2/3 boys.

Hi Stuart,

Sorry for the basic question ... how did you get to the 2/3 in the above example. Not sure I quite follow what you are saying here.

Thanks.

We have the part:part ratio boys:girls = 2:1

We have the part:whole ratio boys:(boys+girls) = 2:2+1 = 2:3

So, boys/(boys+girls) = 2/3

Hi,

I didn't get 5 for boys and 10 for girls.

I noted that the boys' average was 5 marks away from that of the class and the girls' average was 10 marks away from that of the class.

Since the overall average is twice as close to the boys' average as the girls', there must be twice as many boys.

Thanks again Stuart.

Just one last doubt

You are calculating the ratio as
b/g = (distance of girls to average)/(distance of boys to average)

i.e. b/g = 10/5 = 2/1

the left hand side i.e. b/g is it the ratio of number of boys / number of girls

if that's correct then the total ratio would be


number of boys : number of girls = distance of girls to average : distance of boys to average

Stuart Kovinsky wrote:Hi,

I didn't get 5 for boys and 10 for girls.

I noted that the boys' average was 5 marks away from that of the class and the girls' average was 10 marks away from that of the class.

Since the overall average is twice as close to the boys' average as the girls', there must be twice as many boys.

netigen wrote:In a corporation, 50 percent of the male employees and 40 percent of the female employees are at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?

A. 3/5
B. 2/3
C. 3/4
D. 4/5
E. 5/6

this is a like a mixture problem.

set up is:

((1/2)M + (2/5)F)/(M+F)=42/100=21/50

cross multiply and simplify you will get:

f/m=4/1

so ratio of female to total is 4/5 -->D

you got this man!

mehravikas wrote:Thanks again Stuart.

Just one last doubt

You are calculating the ratio as
b/g = (distance of girls to average)/(distance of boys to average)

i.e. b/g = 10/5 = 2/1

the left hand side i.e. b/g is it the ratio of number of boys / number of girls

if that's correct then the total ratio would be


number of boys : number of girls = distance of girls to average : distance of boys to average

Correct (see the post near the top which reiterates this general rule)!