Question

Consider the set Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3, ………., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ………..)?

• A. 80
• B. 81
• C. 82
• D. 83

Answer 1

The correct option is A 80

Option (a)

Calculating “n” which does not satisfy the requisite condition (i.e. the value of “n” which will not yield subsets with an integral multiple of 6).

The terms will fall in an AP with n =1 as the first term. The general form of the AP is 6k+1.

The last term will be 91. Thus, total number of sets which do not satisfy the condition= $\frac{90}{6}$ = 15+1 = 16.

Total number of terms = 96.

Thus, number of sub sets possible = 96 - 16 = 80. Option (a).