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by Mo2men » Sat Apr 07, 2018 10:07 am
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

OA: C
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by MartyMurray » Sat Apr 07, 2018 11:12 am
The first thing that we can do is to eliminate all the students who failed all three.

100 students - 10 students who failed all three = 90 students who passed at least one test.

Now we have 90 students who passed at least one, but we have 80 who passed Physics, 70 who passed Chemistry, and 40 who passed Math, for a total of 190 passes.

So there must be some overlap, meaning that some students passed more than one test in order that 90 students could account for 190 passes.

Since the max is 90 students, and we have 80 who passed Physics and 70 who passed chemistry, for a total of 150 passes, there must be a total of at least 60 students who passed both chemistry and Physics.

In that case we would have:

60 passed Physics and chemistry.

20 passed Physics.

10 passed Chemistry.

Now we have 40 math passes to account for.

We already have 90 students. So all the math passes have to be allotted to students who have already passed something.

We want least possible number of triple passes. So, we will allot the most math passes we can to people who so far have passed only Chemistry or only Physics, but not both.

So we will have:

20 passed Physics and Math.

10 passed Chemistry and Math.

We are left with 10 who must have passed all three.

The correct answer is C. $$$$
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by GMATGuruNY » Sat Apr 07, 2018 12:01 pm
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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by Keith@ThePrincetonReview » Sat Apr 07, 2018 2:22 pm
For a different approach, think about the number of students who fail. Then, simply plug in the answers. Of 100 students, 20 fail physics, 30 fail chemistry, 60 fail math, and 10 fail all three subjects.

total students = (fail P) + (fail C) + (fail M) - (fail 2 subjects) - 2(fail 3 subjects) + (pass 3 subjects)
100 = 20 + 30 + 60 - (fail 2 subjects) −2(10) + (pass 3 subjects)
100 = 90 - (fail 2 subjects) + (pass 3 subjects)
(fail 2 subjects) = (pass 3 subjects) - 10

Now, plug the answers into this equation one at a time. The question asks for the least number of students who passed all 3 subjects, so begin with choice A.

Choice A:
(fail 2 subjects) = (0) - 10 = −10
Eliminate choice A (can't have negative number of students)

Choice B:
(fail 2 subjects) = (5) - 10 = −5
Eliminate choice B (can't have negative number of students)

Choice C - 10:
(fail 2 subjects) = (10) - 10 = 0
Choice C is the least answer that works.

The correct answer is C
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by Scott@TargetTestPrep » Wed Apr 11, 2018 5:56 am
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
We can create the equation:

Total = Physics + Chemistry + Math - (exactly 2 subjects) - 2(all 3 subjects) + (none of 3 subjects)

100 = 80 + 70 + 40 - (exactly 2 subjects) - 2(all 3 subjects) + 10

100 = 200 - (exactly 2 subjects) - 2(all 3 subjects)

100 = -(exactly 2 subjects) - 2(all 3 subjects)

100 = (exactly 2 subjects) + 2(all 3 subjects)

Since we want to minimize (all 3 subjects), we want to maximize (exactly 2 subjects). That is, we want to find the maximum students who passed exactly Physics and Chemistry, those who passed exactly Physics and Math, and those who passed exactly Chemistry and Math.

Notice that since 10 of the 100 students failed all 3 subjects, the number of students who passed at least one subject is 90. Since 80 students passed Physics, the maximum number of students who could pass exactly Chemistry and Math is 90 - 80 = 10.

Similarly, since 70 students passed Chemistry, the maximum number of students who could pass exactly Physics and Math is 90 - 70 = 20.

Finally, since 40 students passed Math, the maximum number of students who could pass exactly Physics and Chemistry is 90 - 40 = 50.

Thus, we have:

100 = (10 + 20 + 50) + 2(all 3 subjects)

100 = 80 + 2(all 3 subjects)

20 = 2(all 3 subjects)

10 = all 3 subjects

Answer: C

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