Find the smallest number which when increased by 20 is completely divisible by 12, 20, 15, 18 and 24.

Correct option is C

Given:
Divisors = 12, 15, 18, 20, 24
The number is increased by 20 to be perfectly divisible.
Formula Used:
Required number = LCM(a, b, c, ...) - Increment
Solution:
Find the Lowest Common Multiple (LCM) of 12, 15, 18, 20, and 24.
Prime factorization:
12 = $$2^2 \times 3$$​​
15 = $$3 \times 5$$​​
18 =  $$2 \times 3^2$$​​
20 = $$2^2 \times 5$$​​
24 = $$2^3 \times 3$$​​
LCM = $$2^3 \times 3^2 \times 5 = 8 \times 9 \times 5$$​ = 360
The smallest number completely divisible by all given divisors is 360.
Since the number is increased by 20 to get 360, subtract 20 from the LCM:
360 - 20 = 340
Final Answer
So the correct answer is (c)