The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is :

Solution

Maths sir · 3 steps

Count the union with inclusion–exclusion first, then remove the multiples of 48 sitting inside it.

1. Set up. Let A be the 3-digit multiples of 3 and B the 3-digit multiples of 4. Then A ∩ B is the multiples of lcm(3, 4) = 12. Count each set using the first and last multiple between 100 and 999.

   Multiples of	First	Last	Count
   3	102	999	(999-102)/3+1 = 300
   4	100	996	(996-100)/4+1 = 225
   12	108	996	(996-108)/12+1 = 75
   48	144	960	(960-144)/48+1 = 18

   n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 300 + 225 - 75 = 450

2. Remove multiples of 48. Every multiple of 48 is also a multiple of 3 and of 4, so all 18 of them lie inside these 450 numbers. Taking them out leaves 450 - 18 = 432.
3. So. There are 432 such numbers, which is option (d).

The idea to keep. For "divisible by a or b", add both counts and subtract the multiples of lcm(a, b). Before subtracting an excluded set, check that it lies wholly inside the union.

Answer from the NTA final answer key. Solution by PrepWiser.