rohan2345 wrote:

In a class of 40 students, the number of students who passed the math exam is equal to half the number of students who passed the science exam. Each student in the class passed at least one of the two exams. If 5 students passed both exams, then the number of students who passed the math exam is

(A) 5

(B) 10

(C) 15

(D) 20

(E) 25

Let’s denote the number of students who passed only the math exam by x and the number of students who passed only the science exam by y. Then, since there are 5 students who passed both exams, the number of students who passed the math exam is x + 5 and the number of students who passed the science exam is y + 5.

Since the number of students who passed the math exam is equal to half the number of students who passed the science exam, we have x + 5 = (y + 5)/2, or, 2(x + 5) = y + 5. Simplifying this equation, we get y = 2x + 5.

To determine the number of students who passed the math exam, we will use the following formula:

# of students in the class = # who passed math + # who passed science - # who passed both + # who passed neither

We are given that there are 40 students in total and each student passed at least one exam; therefore, # who passed neither = 0. Then, we have:

40 = (x + 5) + (y + 5) - 5 + 0

40 = x + y + 5

x + y = 35

Let’s substitute y = 2x + 5:

x + 2x + 5 = 35

3x = 30

x = 10

We see that the number of students who passed only the math exam is 10. We must add to this number those 5 students who passed both math and science. Thus, the total number of students who passed the math exam is x + 5 = 10 + 5 = 15.

Answer: C

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