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Arithmetic percents

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Arithmetic percents

by datonman » Fri Nov 13, 2015 10:35 am
Of the 50 researchers in a workgroup, 40 percent will be assigned to Team A and the remaining 60 percent to Team B. However, 70 percent of the researchers prefer Team A and 30 percent prefer Team B. What is the lowest possible number of researchers who will NOT be assigned to the team they prefer?

(A) 15
(B) 17
(C) 20
(D) 25
(E) 30
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by DavidG@VeritasPrep » Fri Nov 13, 2015 10:46 am
datonman wrote:Of the 50 researchers in a workgroup, 40 percent will be assigned to Team A and the remaining 60 percent to Team B. However, 70 percent of the researchers prefer Team A and 30 percent prefer Team B. What is the lowest possible number of researchers who will NOT be assigned to the team they prefer?

(A) 15
(B) 17
(C) 20
(D) 25
(E) 30
Matrix! We know that 40% of the 50 are in group A, so that's 20 in A. We know that the other 60% are in group B, so let's 30 in B. We know that 70% prefer A, so that's .75*50 = 35 who prefer A and the other 15 prefer B. So our initial matrix will look like this.

Image

Now we want to minimize the unhappy people. So let's say that all 20 people in group A want to be there. Well, if there are 35 people who prefer A, and only 20 who are in group A, the remaining 15 are stuck in Group B, and are unhappy. So the answer is A

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by Brent@GMATPrepNow » Fri Nov 13, 2015 10:55 am
I thought I'd point out that Dave's matrix approach (aka Double Matrix Method) can be used for most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of researchers, and the two characteristics are:
- assigned to Team A or assigned to Team B
- prefer Team A or prefer Team B

This question type is VERY COMMON on the GMAT, so be sure to master the technique.

To learn more about the Double Matrix Method, watch our free video: https://www.gmatprepnow.com/module/gmat- ... ems?id=919

Once you're familiar with this technique, you can attempt these additional practice questions:

Easy Problem Solving questions
- https://www.beatthegmat.com/the-aam-aadm ... 72242.html
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Difficult Problem Solving questions
- https://www.gmatprepnow.com/module/gmat- ... /video/946
- https://www.beatthegmat.com/ratio-problem-t268339.html
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- https://www.beatthegmat.com/mba/2011/05/ ... question-2

Easy Data Sufficiency questions
- https://www.gmatprepnow.com/module/gmat- ... /video/943
- https://www.beatthegmat.com/for-what-per ... 70596.html
- https://www.beatthegmat.com/ds-quest-t187706.html

Medium Data Sufficiency questions
- https://www.beatthegmat.com/sets-matrix-ds-t271914.html
- https://www.beatthegmat.com/each-of-peop ... 71375.html
- https://www.beatthegmat.com/a-manufacturer-t270331.html
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Difficult Data Sufficiency questions
- https://youtu.be/dsCeqF9Kbk8
- https://www.beatthegmat.com/double-set-m ... 71423.html
- https://youtu.be/dOZ9KM1m5Hs
- https://www.beatthegmat.com/sets-t269449.html
- https://www.beatthegmat.com/mba/2011/05/ ... question-3

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Jim@StratusPrep » Fri Nov 13, 2015 1:16 pm
I have attached another diagram that is pretty helpful for this question.
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by Matt@VeritasPrep » Fri Nov 13, 2015 2:44 pm
Arithmetic is an option here too.

We know that Team A has 20 people and Team B has 30 people.

If we put everybody on Team B who wants to be on Team B, that's 15 people. The rest of team B will be unwilling members, so that's also 15 people.

Since everyone left wants to be on Team A, we'll give them those spots, meaning there are no unhappy people on Team A.

Hence our 15 people who were forced to be on Team B are the only miserable ones, giving us a minimum of 15, or A.
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