Problem Solving

[GMAT math practice question]

Out of 75 students, 17 students enrolled in a Physics class, 28 students enrolled in a Chemistry class, and 39 students enrolled in a Biology class. 5 students enrolled in both the Physics and Chemistry classes, 7 students enrolled in both the Chemistry and Biology classes, and 6 students enrolled in both the Biology and Physics classes. If 4 students enrolled in Physics, Chemistry, and Biology, how many students did not enroll in any of the three science classes?

A. 2
B. 3
C. 4
D. 5
E. 6

Max@Math Revolution wrote:[GMAT math practice question]

Out of 75 students, 17 students enrolled in a Physics class, 28 students enrolled in a Chemistry class, and 39 students enrolled in a Biology class. 5 students enrolled in both the Physics and Chemistry classes, 7 students enrolled in both the Chemistry and Biology classes, and 6 students enrolled in both the Biology and Physics classes. If 4 students enrolled in Physics, Chemistry, and Biology, how many students did not enroll in any of the three science classes?

A. 2
B. 3
C. 4
D. 5
E. 6

We can use the following formula:
Total = Group 1 + Group 2 + Group 3 - (at least 2 groups) + (all 3 groups) + None.

In the problem above:
Total = 75.
Group 1 = physics = 17.
Group 2 = chemistry = 28.
Group 3 = biology = 39.
At least 2 groups = (physics and biology) + (chemistry and biology) + (biology and physics) = 5+7+6 = 18.
All 3 groups = 4.
None = N.

Plugging the values above into the formula, we get:
75 = 17 + 28 + 39 - 18 + 4 + N
75 = 70 + N
N = 5.

The correct answer is D.

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The total number of students is given by
a + b + c + d + e + f + g + h = 75.
Of these,
a + d + f + g = 17 are enrolled in Physics,
b + d + e + g = 28 are enrolled in Chemistry, and
c + e + f + g = 39 are enrolled in Biology.
Adding these three equations gives ( a + b + c ) + 2( d + e + f ) + 3g = 84.
We also know that
d + g = 5 study both Physics and Chemistry,
e + g = 7 study both Chemistry and Biology, and
f + g = 6 study both Physics and Biology.
Adding these three questions yields ( d + e + f ) + 3g = 18.
Since 4 students are enrolled in all three classes, we have g = 4,

Plugging this value for g into the above equation yields
(d + e + f) + 3g = 18
(d + e + f) + 12 = 18
d + e + f = 6.

So,
( a + b + c ) + 2( d + e + f ) + 3g = 84 yields
(a + b + c) + 2(6) + 3(4) = 84
(a + b + c) + 12 + 12 = 84
a + b + c = 60.

Finally, using the equation, a + b + c + d + e + f + g + h = 75, we see that
(a + b + c) + (d + e + f) + g + h = 75
60 + 6 + 4 + h = 75
h = 5.

Therefore, the answer is D.