Question:

% of students of a class took Statistics and 45% took Mathematics. If each student took Statistics or Mathematics and 40 took both, the total number of students in the class was:

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For two sets $A$ and $B$, $|A\cup B|=|A|+|B|-|A\cap B|$. If “everyone chose at least one,” then $|A\cup B|=N$.
Updated On: Jul 15, 2026
  • 160
  • 180
  • 200
  • 225
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The Correct Option is A

Approach Solution - 1

Let the class size be $N$. Using inclusion–exclusion and the fact that everyone took at least one of the two subjects: \[ N=\underbrace{0.80N}_{\text{Statistics}}+\underbrace{0.45N}_{\text{Mathematics}}-\underbrace{40}_{\text{both}}. \] So $N=1.25N-40$ which gives $0.25N=40$ and hence $N=160$.
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Approach Solution -2

80 percent of the students in a class took Statistics and 45 percent took Mathematics, every student took at least one of the two subjects, and 40 students took both. We need the total class size. Using each option as a trial value, we can check which one makes the overlap come out to exactly 40.

  1. 160: Students taking Statistics = \( 0.80 \times 160=128 \). Students taking Mathematics = \( 0.45 \times 160=72 \). Since every student takes at least one subject, the overlap is \( 128+72-160=40 \), which matches exactly.
  2. 180: Students taking Statistics = \( 0.80 \times 180=144 \). Students taking Mathematics = \( 0.45 \times 180=81 \). The overlap works out to \( 144+81-180=45 \), more than the given 40.
  3. 200: Students taking Statistics = \( 0.80 \times 200=160 \). Students taking Mathematics = \( 0.45 \times 200=90 \). The overlap works out to \( 160+90-200=50 \), far more than 40.
  4. 225: Students taking Mathematics would be \( 0.45 \times 225=101.25 \), which is not even a whole number of students, so this total cannot be correct.

Only a total of 160 students gives whole numbers for both subjects and an overlap of exactly 40 students.

Therefore, the correct answer is 160.

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Approach Solution -3

Every student falls into exactly one of three groups: only Statistics, only Mathematics, or both, and these three groups must add up to the whole class. Taking the overlap as a fraction \( b=40/N \) of the class, the group taking only Statistics is \( 0.80-b \) of the class and the group taking only Mathematics is \( 0.45-b \). We can test each option by checking whether these three groups sum to the whole class.

  1. 160: Here \( b=40/160=0.25 \). The three groups sum to \( (0.80-0.25)+(0.45-0.25)+0.25=0.55+0.20+0.25=1.00 \), exactly the whole class.
  2. 180: Here \( b=40/180 \approx 0.222 \). The three groups sum to \( (0.80-0.222)+(0.45-0.222)+0.222 \approx 1.028 \), more than the whole class.
  3. 200: Here \( b=40/200=0.20 \). The three groups sum to \( 0.60+0.25+0.20=1.05 \), more than the whole class.
  4. 225: Here \( b=40/225 \approx 0.178 \). The three groups sum to approximately \( 0.622+0.272+0.178 \approx 1.072 \), more than the whole class.

Only a class of 160 students makes the only-Statistics, only-Mathematics, and both groups add up to exactly the whole class.

Therefore, the correct answer is 160.

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