Carcass wrote:
Set A comprises all 3-digit numbers that are multiples of 6. Set B comprises all 3-digit numbers that are multiples of 4 but are not multiples of 8. How many elements does (A ∪ B) comprise?
224
225
263
265
300
(A ∪ B) = A + B - (A ∩ B)
i.e. (6 ∪ 4) = (multiples of only 6) + (multiples of only 4) - (multiples of both 4 and 6)
Multiples of only 6:
First term = \(102\)
Last Term = \(996\)
Number of terms = \(\frac{(996 - 102)}{6} + 1 = 150\)
Multiples of 4:
First term = \(100\)
Last Term = \(996\)
Number of terms = \(\frac{(996 - 100)}{4} + 1 = 225\)
But, every second number in this series is also divisible by \(8\) (with starting and ending number a multiple of 4), Therefore Multiples of only \(4 = \frac{225 + 1}{2} = 113\)
Multiples of 4 and 6:
Multiples of 4 = 100, 104, 108, 112, .....
Multiples of 6 = 102, 108, 114, 120, ...
First term common = \(108\)
Last Term common = \(996\)
Common difference would be \(24\)
Applying the formula of A.P to find the number of terms:
{
NOTE: We are using this formula to find the number of terms above}
\(T_n = a + (n - 1)d\)
\(996 = 108 + (n - 1)24\)
\(n= 38\)
So, \((6 ∪ 4) = (150) + (113) - (38) = 225\)
Hence, option B