Quantitative Aptitude - Arithmetic Ability Questions

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

=>Required number = (90 x 4) + 4   = 364.

The given number series 11, 13, 30, 96, ? follows a pattern that,

11

11 x 1 + 2 = 13

13 x 2 + 4 = 30

30 x 3 + 6 = 96

96 x 4 + 8 = 392

Hence, the next number in the given series is 392.

504/42 × 7 - 63 + 28 = 84 – 63 + 28 = 49

Hence option 5 is correct.