Quantitative Aptitude - Arithmetic Ability Questions

The given number series is 152, 124, 96, 69, 40

152

152 - 28 = 124

124 - 28 = 96

96 - 28 = 68 (not 69)

68 - 28 = 40

Hence, the odd man in the given number series is 69.

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.

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.

Ratio of their shares = (35000 * 8) : (42000 * 10) = 2 : 3.

Reena's share Rs. 31570 * (2 / 5) = Rs. 12628.

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.