Question

Among 50 people, 25 like Lux, 28 like Fiama and 30 like Dettol and all the customers like at least one of the three soaps. The number of customers who like exactly one is greater than the number of customers who like all three soaps but less than the number of customers who like at least 2 out of the three soaps.
What is the maximum number of customers who could like exactly two out of the three soaps?

• A. 31
• B. 33
• C. 29
• D. 23

Answer 1

The correct option is B 33

Given, Number of people liking at least one soap = x = I + II + III = 50 - - -(i)
Sum of all circles = I + 2 II + 3 III = 83 - - - (ii)
Subtracting i) from ii), we get II + 2 III = 33
II = 33 - 2 III
For maximum value of II, III = 0
$\therefore$ $I{I}_{max}$ = 33

Alternatively


Given a + b + c > x
a + b + c < p + q + r + x
a + b + c + p + q + r + x = 50 --------------- 1)
Also, a + b + c + 2(p + q + r) + 3x = 25 + 28 + 30 = 83 ------------ 2)
From 1) & 2), we get
p + q + r + 2x = 33
(p + q + r) = 33 - 2x
Maximum (p + q + r) happens when x = 0 which means
Maximum value of (p + q + r) = 33