Quantitative Aptitude - Arithmetic Ability Questions

B covers 28 meters in 7sec. So, B's time over the course = (7/28)*100 =250 sec.

Whereas A's time over the course = 250 -7 = 243 sec.

i.e A's time over the course is 4min , 3 sec.

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 = 

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.