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Word Problems - EZ Solutions - Advanced Workbook

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Word Problems - EZ Solutions - Advanced Workbook

by shanice » Mon Aug 13, 2012 5:10 am
At a certain company, two-seventh of the managers attended the company's annual meeting, and one-half of them were females. If one-tenth of the managers who did not attend the annual meeting were females, what fractional part of the total number of managers were males?

Solution from the book:-

No. of managers who did attend the annual meeting = 2/7
Females = 1/2 x 2/7 = 1/7
Males = 1/2 x 2/7 = 1/7

No. of managers who did not attend the annual meeting = 1-2/7 = 5/7
Females = 1/10 x 5/7 = 1/14
Males = 9/10 x 5/7 = 9/14

Total No. of Male Managers = 1/7+9/14=2/14+9/14=11/14
So, the answer is 11/14.

My workings:-

No. of managers who did attend annual meeting = 2/7m
Females = 1/2 x 2/7m = 1/7m
Males = Since female is 1/7m so male should be 1/7

No.of managers who did not attend the annual meeting = 5/7(remaining balance as 2/7 of the managers
attended the company's annual meeting)
Females = 1/10m x 5/7 = 1/14m
Males = 13/14m(remaining balance after deducting
1/14m of females)

Total No. of male managers = 1/7+13/14=2/14+13/14=15/14
So, my answer is 15/14.

M = managers


However, my answer is wrong. What is wrong with my workings and understanding here?

Thank you, guys.
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by anuprajan5 » Mon Aug 13, 2012 6:07 am
Shanice,

You have got all the steps right except for this step:

Males = 13/14m(remaining balance after deducting

You need to deduct males from the managers who did not attend, not from 1.

Therefore - Male managers that did not attend = 5/7 - 1/14(females that did not attend) = 9/14


Therefore total male managers = 1/7 (first part that attended from your answer) + 9/14 = 11/14

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by GMATGuruNY » Mon Aug 13, 2012 12:12 pm
shanice wrote:At a certain company, two-seventh of the managers attended the company's annual meeting, and one-half of them were females. If one-tenth of the managers who did not attend the annual meeting were females, what fractional part of the total number of managers were males?
This is an EITHER/OR group question.
Every manager is EITHER male OR female.
Every manager EITHER attended OR did not.
For an EITHER/OR group problem, use a GROUP GRID to organize the data.

Let the total number of managers = the LCM of the denominators in the problem (7,2,10) = 70.
Fill in the grid step-by-step.
Two-seventh of the managers attended the company's annual meeting, and one-half of them were females.
_______________M________F_______Total

A:_____________10______10________20

DA_______________________________50

total:_____________________________70
One-tenth of the managers who did not attend the annual meeting were females.
_______________M________F_______Total

A:_____________10______10________20

DA____________ 45______5_________50

total:__________ 55_____15_________70
Males/total = 55/70 = 11/14.
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by tisrar02 » Mon Aug 13, 2012 5:51 pm
My way I did this was as follows:

2/7 attended the meeting. 1/2 of them were females. 1/10 of those who did not attend were females.

Let the total number of managers =14

4/14 attended and 2/14 were females.

Meaning that 10/14 did not attend.

1/10 of those who did not attend were female so 1 female... Total number of females would be: 1/14+2/14=3/14

Number of males then would be 14/14-3/14= 11/14

Hope this helps.

I usually look at what the fractions are and then do my workings based on them.
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