DisciplinedPrep
Set A contains all multiples of 4 less than 500, Set B contains all odd multiples of 7 less than 500, and Set C comprises all multiples of 6 less than 500. How many elements are present in A ∪ B ∪ C?
A. 186
B. 202
C. 220
D. 228
E. 243
Assuming we are talking about positive multiples:
1 to 500
All multiples of 4 -> 500/4 = 125
The last multiple is 500 which we should not include since we need multiples less than 500.
Hence 124 multiples.
All odd multiples of 7
First, let's find all multiples of 7.
500/7 = 71
First multiple is odd, then even, then odd and so on till 71st multiple which is odd.
Hence 36 odd multiples. Since they are odd, they will have 0 overlap with multiples of 4 and/or 6.
All multiples of 6 -> 500/6 = 83
First multiple is odd, then even, then odd and so on till 83rd multiple which is odd. (e.g. 6, 12, 18, 24, 30, 36 ...)
Every odd multiple will be divisible by 2 and every even multiple will be divisible by 4. The even multiples are already counted above while counting all multiples of 4. So ignore them.
We have 42 odd multiples of 6.
A ∪ B ∪ C = 124 + 36 + 42 = 202
Answer (B)