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CAT 2022 Slot 2 — QA Question 14

Simple EquationsEasy

Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is

Answer & solution

  • A

    24

  • 20

  • C

    21

  • D

    22

Solution

Easy

Total of 5 distinct integer scores is 5×38=1905\times 38=190. Exactly 3 scored above 32, so 2 (including Amit, the lowest) scored ≤32\le 32. Push the others to extremes: one extreme minimises Amit's score, the other maximises it.

Five distinct integers in [0,50][0,50]; sum =190=190; exactly three exceed 32; Amit is the smallest.

1

Minimise Amit by maximising everyone else. The three high scorers take 50,49,4850,49,48 and the other non-Amit student (must be ≤32\le 32) takes 3232:

Amitmin⁡=190−(50+49+48+32)=190−179=11\text{Amit}_{\min}=190-(50+49+48+32)=190-179=11
2

Maximise Amit. Amit must be the lowest and ≤32\le 32, with another distinct student also ≤32\le 32. So the largest Amit can be is 3131 (leaving 3232 for that other student):

Amitmax⁡=31\text{Amit}_{\max}=31

(This is feasible: e.g. scores 31,32,?,?,?31,32,?,?,? with three above 32 summing to 190−63=127190-63=127, like 40,43,4440,43,44.)

3

Difference:

31−11=2031-11=20
Difference=20\text{Difference}=\mathbf{20}
CAT 2022 Slot 2 QA Q14: Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50 — Solution | TheCATExam