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.

Total profit = Rs. 880

A's share for managing the business i.e 

 %  =  

Remaining profit of A and B as per their capital = 880 - 110 = Rs. 770

Ratio of amounts = 5000 : 6000 = 5 : 6

Sum of ratios = 5 + 6 = 11

A's share = 

A's total share = 350 + 110 = Rs. 460

B's share = 

The given series follows a pattern that,

$$\begin{gathered} 4 + 32 = 4 + 9 = 13 \\ 13 + 52 = 13 + 25 = 38 \\ 38 + 72 = 38 + 49 = 87 \\ 87 + 92 = 87 + 81 = 198 \end{gathered}$$ 

Hence, the odd man in the given number series is 12.

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.