TheCATExam
Sign in

CAT 2022 Slot 3 — QA Question 6

Weighted AverageEasy

In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

Answer & solution

Answer: 63

Solution

Easy

Let section B have xx students and A have x−10x-10. The combined average must be an integer; simplify the weighted-average expression until you get a remainder term, then force (x−5)(x-5) to divide 7070.

1

Write the combined average (A: x−10x-10 students at 3232; B: xx students at 6060):

(x−10)⋅32+x⋅60(x−10)+x=92x−3202x−10=46x−160x−5\frac{(x-10)\cdot32+x\cdot60}{(x-10)+x}=\frac{92x-320}{2x-10}=\frac{46x-160}{x-5}
2

Split off the integer part to expose the remainder:

46x−160x−5=46(x−5)+70x−5=46+70x−5\frac{46x-160}{x-5}=\frac{46(x-5)+70}{x-5}=46+\frac{70}{x-5}
3

Condition: the average is an integer   ⟺  70x−5\iff \dfrac{70}{x-5} is an integer   ⟺  (x−5)\iff (x-5) is a positive divisor of 7070.

Also A must have at least 11 student: x−10≥1⇒x>10⇒x−5>5x-10\ge1\Rightarrow x>10\Rightarrow x-5>5.

4

Pick the divisors of 7070 that exceed 55: 70=2×5×770=2\times5\times7, divisors {1,2,5,7,10,14,35,70}\{1,2,5,7,10,14,35,70\}. Those >5>5 are 7,10,14,35,707,10,14,35,70.

max: x−5=70⇒x=75⇒A=65min: x−5=7⇒x=12⇒A=2difference in A=65−2=63\begin{aligned} &\text{max: } x-5=70\Rightarrow x=75\Rightarrow A=65\\ &\text{min: } x-5=7\Rightarrow x=12\Rightarrow A=2\\ &\text{difference in A}=65-2=63 \end{aligned}
Difference=63\text{Difference}=\mathbf{63}

Note the question asks for the difference in section A's count. Since A=x−10A=x-10, the difference equals the difference in xx: 75−12=6375-12=63.

CAT 2022 Slot 3 QA Q6: In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number o — Solution | TheCATExam