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Students opting for both activities

CAT 2024 Slot 3 · Quantitative Ability · Medium · Arithmetic

This is a medium Quantitative Ability question from the CAT 2024 Slot 3 paper. It tests Arithmetic. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are

  1. A.

    72 and 80, respectively

  2. B.

    75 and 90, respectively

  3. C.

    72 and 88, respectively

  4. D.

    75 and 96, respectively

Answer

A

Explanation

Let total students =250= 250. Number of girls GG satisfies 0.44×250≤G≤0.60×250  ⟹  110≤G≤1500.44 \times 250 \le G \le 0.60 \times 250 \implies 110 \le G \le 150. Number of boys B=250−GB = 250 - G.

For boys: SB=0.5BS_B = 0.5B, RB=0.7BR_B = 0.7B. Since every student opted for at least one activity, (S∩R)B=SB+RB−B=0.5B+0.7B−B=0.2B(S \cap R)_B = S_B + R_B - B = 0.5B + 0.7B - B = 0.2B.

For girls: SG=0.8GS_G = 0.8G, RG=0.6GR_G = 0.6G. (S∩R)G=0.8G+0.6G−G=0.4G(S \cap R)_G = 0.8G + 0.6G - G = 0.4G.

Total students opting for both: Both=0.2B+0.4G=0.2(250−G)+0.4G=50+0.2G\text{Both} = 0.2B + 0.4G = 0.2(250 - G) + 0.4G = 50 + 0.2G

  • Minimum value (when G=110G = 110): 50+0.2(110)=7250 + 0.2(110) = 72.
  • Maximum value (when G=150G = 150): 50+0.2(150)=8050 + 0.2(150) = 80.

Thus, the minimum and maximum numbers are 72 and 80, respectively.

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