Mo2men wrote:In a class of 100 students, 80 passed Physics, 70 passed Chemistry, and 40 passed Math. If 10 students failed in all the three subjects, at least how many of the students passed all the three subjects?
(A) 0
(B) 5
(C) 10
(D) 20
(E) 25
Total who pass = Physics + Chemistry + Math - (exactly 2 subjects) - 2(all 3 subjects).
Since 10 of 100 students pass none of the 3 subjects, the total number who pass at least one subject = 100-10 = 90.
Of these 90 students, 80 pass Physics, 70 pass Chemistry, and 40 pass Math.
Plugging these values into the equation above, we get:
90 = 80 + 70 + 40 - (exactly 2 subjects) - 2(all 3)
(exactly 2) + 2(all 3) = 100.
To MINIMIZE the number who pass all 3 subjects, we must MAXIMIZE the number who pass exactly 2 subjects.
Since 80 of the 90 students above pass Physics, the maximum who could pass only Chemistry and Math = 90-80 = 10.
Since 70 of the 90 students above pass Chemistry, the maximum who could pass only Physics and Math = 90-70 = 20.
Since 40 of the 90 students above pass Math, the maximum who could pass only Physics and Chemistry = 90-40 = 50.
Thus:
Maximum who pass exactly 2 subjects = 10+20+50 = 80.
Plugging this value into the blue equation above, we get:
80 + 2(all 3) = 100
2(all 3) = 20
all 3 = 10.
Since at most 80 students pass exactly 2 subjects, at least 10 students must pass all 3 subjects.
The correct answer is
C.
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