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Among 100 students in a class, 57 like football and 68 like

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by Max@Math Revolution » Fri Sep 13, 2019 6:44 am

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[GMAT math practice question]

Among 100 students in a class, 57 like football and 68 like basketball. The maximum possible number of students who like both football and basketball is a and the minimum possible number of students who like both football and basketball is b. What is the value of a-b?

A. 12
B. 19
C. 23
D. 28
E. 32
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by Max@Math Revolution » Sun Sep 15, 2019 5:37 pm
=>

Define n(X) to be the number of elements of the set X.
Let U be the set of all 100 students, F be the set of students who like football, and B be the set of students who like basketball.
Then n(U)=100, n(F)=57 and n(B)=68.

The maximum size of F ∩ B occurs when the set F is included in the set B. In this case, n(F ∩ B) = n(F). So a = n(F) =57.

The minimum size of F ∩ B occurs when U=F∪B. In this case, b = n(F ∩ B)=n(F)+n(B)-n(F ∪ B)=57+68-100=25 since n(F ∪ B) = n(F) + n(B) - n(F ∩ B).

Thus, a-b = 57-25 = 32.

Therefore, E is the answer.
Answer: E
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by deloitte247 » Thu Sep 19, 2019 10:43 am
Total number of students = 100
Maximum number of who likes football and basketball = a
Maximum number of students who likes football and basketball = b
The maximum number of students who likes football and basketball will be that number of students who likes football as this means they did not like football alone so all the students who like football likes basketball (This is the maximum number possible.
Therefore, a = 57
The maximum number of students who likes football and basketball = (Students who liked basketball + students who likes football ) = Total students
$$b=\left(68+57\right)-100$$
$$b=25$$
$$Hence\ a-b=57-25$$
$$=32$$

$$Answer\ is\ Option\ E$$
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