In a class...!

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In a class...!

by chaitanya.bhansali » Thu Aug 07, 2014 10:16 am
In a class of 80 students, 40% of the students play basketball, 50% play football and 30% play table tennis. If 5 students play both football and table tennis but not basketball, what is the maximum possible number of students that play both basketball and tennis?
A) 11 B) 19 C) 21 D) 24 E) 32

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by Jim@StratusPrep » Thu Aug 07, 2014 11:21 am
Here are the number of students in each category:

Basketball: 32
Football: 40
Table Tennis: 24

If 5 play football and table tennis your new breakdown would be:

Basketball: 32
Football: 35
Table Tennis: 19
Football and Table Tennis: 5

This give you a total of 91 students. You can have an overlap of 11 between Basketball and Table Tennis to reach a total of 80:

Basketball: 21
Football: 35
Table Tennis: 8
Football and Table Tennis: 5
Basketball and Table Tennis: 11

A
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by GMATGuruNY » Thu Aug 07, 2014 11:28 am
chaitanya.bhansali wrote:In a class of 80 students, 40% of the students play basketball, 50% play football and 30% play table tennis. If 5 students play both football and table tennis but not basketball, what is the maximum possible number of students that play both basketball and tennis?
A) 11 B) 19 C) 21 D) 24 E) 32
We can use the following formula:

T = B + F + T - (BF + BT + FT) - 2(BFT)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in B, everyone in F, and everyone in T:
Those in exactly 2 of the groups (BF + BT + FT) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (BFT) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

The following values are given:
T = 80.
B = 40% of 80 = 32.
F = 50% of 80 = 40.
T = 30% of 80 = 24.
FT = 5.

Plugging these values into the formula above, we get:
80 = 32 + 40 + 24 - (BF + BT + 5) - 2(BFT)
80 = 91 - BF - BT - 2(BFT).
BT = 11 - BF - 2(BFT).

The value of BT will be maximized if BF=0 and BFT=0, in which case BT=11.

The correct answer is A.

For an official problem about 3 overlapping groups, check here:
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by feedrom » Fri Aug 08, 2014 12:55 pm
GMATGuruNY wrote:

T = B + F + T - (BF + BT + FT) - 2(BFT)
Mitch,

I got confused with the formula. I think it should be "+2(BFT) because when we subtract 3 groups of "both", we subtract the group of "three" twice.

Could you explain? Thanks.