Percents- 700-800

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Percents- 700-800

by lavinia » Tue Oct 05, 2010 11:18 am
Currently, y percent of the members on the finance committee are women and next month, z percent of the men on the finance committee will resign. If no other personnel changes occur, then after the resignations next month, the men who remain on the finance committee will represent what percent of the total finance committee members?

How do you solve it in less than or around 2 minutes?

[spoiler]Answer: (100)(100 - z)(100 - y)/ (100)(100)-z(100-y)[/spoiler]

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by limestone » Tue Oct 05, 2010 6:35 pm
Just thoroughly understand percentage concept, and solve it as quick as you can.

Percentage of Women y/100
then that of men 100/100 - y/100 = (100-y)/100
z% resign then percentage of men will resign: (100-y)/100* z/100
that of remaining men:(100-y)/100* (100-z)/100

Try to remember some quick rules: women = y/100 then men = (100-y)/100
Here we have a product of percentage of men and percentage of not resigning

(100-y)/100_______ X ___________ (100-z)/100

The rule give you 2 bold phrases quickly.

Now come to the remaining members of the committee:

100/100 - (100-y)/100* z/100 = 100- z*(100-y)/100

Ratio of remaining men / remaining members
= (100-y)/100* (100-z)/100____/____(100- z*(100-y)/100)
= (100-y)*(100-z)/100_____/_____(100-z*(100-y))
= (100-y)*(100-z)____/_____100*(100-z*(100-y))

Change from ratio to percentage, multiply 100:
100____*___(100-y)*(100-z)____/_____100*(100-z*(100-y))
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by lavinia » Wed Oct 06, 2010 8:54 am
Hi Limestone,

First of all thank you for your answer. I woud like to mention that your answer choice is not the correct one. If you see it is:100(100-z)(100-y)/ 100 squared -z(100-y)

My explanation is similar to yours, but I feel that your answer is not the correct one or I'm wrong:)
Current W: y/100
Current M: (100-y)/100
Resign M: [z/100*(100-y)]/100=z(100-y)/ 100 squared
Remain M: Current M - Resign M= (100-y)/100- [z(100-y)]/100squared= (100-y)(100-z)/ 100 squared
Total people remain: 100/100- Resign M= 100/100- z(100-y)/100 squared= [100squared-z(100-y)] /100squared

Part=?% * Whole
(Remain M/ Total people remain)*100=?
[100(100-y)(100-z)]/ 100 squared * 100squared/ 100 squared-z(100-y)

Answer: 100(100-y)(100-z)/ 100 squared- z(100-y)

Is it a strategy that it is not time consuming for this kind of problem or we just solvit it? In my case, it takes aroud 3 minutes.

Thanks again Limestone!!!
limestone wrote:Just thoroughly understand percentage concept, and solve it as quick as you can.

Percentage of Women y/100
then that of men 100/100 - y/100 = (100-y)/100
z% resign then percentage of men will resign: (100-y)/100* z/100
that of remaining men:(100-y)/100* (100-z)/100

Try to remember some quick rules: women = y/100 then men = (100-y)/100
Here we have a product of percentage of men and percentage of not resigning

(100-y)/100_______ X ___________ (100-z)/100

The rule give you 2 bold phrases quickly.

Now come to the remaining members of the committee:

100/100 - (100-y)/100* z/100 = 100- z*(100-y)/100

Ratio of remaining men / remaining members
= (100-y)/100* (100-z)/100____/____(100- z*(100-y)/100)
= (100-y)*(100-z)/100_____/_____(100-z*(100-y))
= (100-y)*(100-z)____/_____100*(100-z*(100-y))

Change from ratio to percentage, multiply 100:
100____*___(100-y)*(100-z)____/_____100*(100-z*(100-y))

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by GMATGuruNY » Wed Oct 06, 2010 9:47 am
lavinia wrote:Currently, y percent of the members on the finance committee are women and next month, z percent of the men on the finance committee will resign. If no other personnel changes occur, then after the resignations next month, the men who remain on the finance committee will represent what percent of the total finance committee members?

How do you solve it in less than or around 2 minutes?

[spoiler]Answer: (100)(100 - z)(100 - y)/ (100)(100)-z(100-y)[/spoiler]
I would plug in values:

100 members
y = 20
20 women, 80 men
z = 50
So .5*80 = 40 men resign
80-40 = 40 men left
100-40 = 60 members left
40/60 = 66 2/3%

Correct answer: (100)(100 - z)(100 - y) / (100)(100)-z(100-y)
= 100(100-50)(100-20) / (100)(100)-50(100-20)
= 400000/6000
= 400/6 = 66 2/3
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by limestone » Wed Oct 06, 2010 9:54 am
Thanks Lavinia

I made a mistake from this step:
Now come to the remaining members of the committee:

100/100 - (100-y)/100* z/100 = 100- z*(100-y)/100
Actually, it should be

100/100 - (100-y)/100* z/100 = 100^2- z*(100-y)/100

Then the answer should be: (after I have redone it again)

100*(100-y)*(100-z)/(100^2-z*(100-y))

So many No. and x/y confused me. Lol.

I think only frequent practice can help us solve this problem both quickly and precisely. It also took me more than 2 minutes ( In the real test, hope I can solve other questions quickly to spare the time for such time consuming questions like this)

Thanks again buddy.

[/quote]
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by limestone » Wed Oct 06, 2010 9:58 am
Thanks Mitch for your nice plug-in method. It's quick, less time consuming and it really helps.
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by lavinia » Wed Oct 06, 2010 11:57 am
Thanks GMATGuruNY!

For this kind of question- how do I choose the smart numbers? Are some rules for picking the numbers that I can apply them for other similar problems?
GMATGuruNY wrote:
lavinia wrote:Currently, y percent of the members on the finance committee are women and next month, z percent of the men on the finance committee will resign. If no other personnel changes occur, then after the resignations next month, the men who remain on the finance committee will represent what percent of the total finance committee members?

How do you solve it in less than or around 2 minutes?

[spoiler]Answer: (100)(100 - z)(100 - y)/ (100)(100)-z(100-y)[/spoiler]
I would plug in values:

100 members
y = 20
20 women, 80 men
z = 50
So .5*80 = 40 men resign
80-40 = 40 men left
100-40 = 60 members left
40/60 = 66 2/3%

Correct answer: (100)(100 - z)(100 - y) / (100)(100)-z(100-y)
= 100(100-50)(100-20) / (100)(100)-50(100-20)
= 400000/6000
= 400/6 = 66 2/3

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by makkays » Thu Sep 05, 2013 3:50 am
How did you reach the values of Women 20 and Men 80. Would appreciate your help
Regards
GMATGuruNY wrote:
lavinia wrote:Currently, y percent of the members on the finance committee are women and next month, z percent of the men on the finance committee will resign. If no other personnel changes occur, then after the resignations next month, the men who remain on the finance committee will represent what percent of the total finance committee members?

How do you solve it in less than or around 2 minutes?

[spoiler]Answer: (100)(100 - z)(100 - y)/ (100)(100)-z(100-y)[/spoiler]
I would plug in values:

100 members
y = 20
20 women, 80 men
z = 50
So .5*80 = 40 men resign
80-40 = 40 men left
100-40 = 60 members left
40/60 = 66 2/3%

Correct answer: (100)(100 - z)(100 - y) / (100)(100)-z(100-y)
= 100(100-50)(100-20) / (100)(100)-50(100-20)
= 400000/6000
= 400/6 = 66 2/3

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by [email protected] » Thu Sep 05, 2013 12:47 pm
Hi lavinia,

The TESTING VALUES approach is great for these types of questions because it allows you to do quick basic calculations and avoid long-winded algebra. Round numbers tend to be best in these situations (the word "percent" tips you off that 100 would probably be a good choice)).

You did say something in your original post that worth talking about though. You shouldn't expect to spend 2 minutes (or less) on each question. That's unrealistic. Certain questions on Test Day are designed to take 3 minutes to solve and that's IF you know what what you're doing. Others require less than a minute to solve. If this question takes you 3 minutes to solve, then that's fine; by TESTING VALUES though, you should be able to do it faster.

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