Minimum Initial Class Strength

CAT 2025 Slot 1 · Quantitative Ability · Medium · Arithmetic

This is a medium Quantitative Ability question from the CAT 2025 Slot 1 paper. It tests Arithmetic. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was

Answer

35

Explanation

Let GG be the initial number of girls and BB be the initial number of boys, with B>10B > 10. After 40% girls leave, remaining girls =0.6G= 0.6G. After 60% boys leave, remaining boys =0.4B= 0.4B. We are given: 0.6G=0.4B+8    3G=2B+40    2B=3G400.6G = 0.4B + 8 \implies 3G = 2B + 40 \implies 2B = 3G - 40 Since B>10B > 10, 2B>20    3G40>20    3G>60    G>202B > 20 \implies 3G - 40 > 20 \implies 3G > 60 \implies G > 20. Also, GG must be an even integer for BB to be an integer (since 2B=3G402B = 3G - 40). To minimize total initial students G+BG + B, choose the smallest even integer G>20G > 20: If G=22G = 22: 2B=3(22)40=6640=26    B=132B = 3(22) - 40 = 66 - 40 = 26 \implies B = 13 Since B=13>10B = 13 > 10, this condition is satisfied. Minimum initial number of students =G+B=22+13=35= G + B = 22 + 13 = 35.

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