Quantitative Aptitude - Arithmetic Ability Questions

Number of shares = 14400120=120.

Face value = Rs.(100 x 120) = Rs.12000.

Annual income = Rs (5/100*12000) = Rs. 600

Given total number of cows = k
Now, 1st son share = k/2
2nd son share = k/4
3rd son share = k/5
4th son share = 9

(k) + (k/4) + (k/5) + 9 = k
=> k - (19k/20) = 9
=> (20k-19k)/20 = 9
=> k = 180.

Since the word hexagon contains 7 different letters,the number of permutations is 7P3 = 7 x 6 x 5 =210

We first count the number of committee in which

(i). Mr. Y is a member
(ii). the ones in which he is not

Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).

We can choose 1 more in5+2C1=7 ways.

Case (ii): If Mr. Y is not a member then we left with (6+4-1) people.
we can select 3 from 9 in 9C3=84 ways.

Thus, total number of ways is 7+84= 91 ways.

Smallest number of 6 digits is 100000

on dividing 100000 by 111 we get 100 as remainder 

 Number to be added = (111 -100) = 11

 Required Number  = 100011

Total of results = 50 x 9 = 450

Total of first four results = 52 x 4 = 208

Total of Last four results = 49 x 4 = 196

Hence the fifth result = 450 - (208 + 196) = 46

Cost of 1 share  =Rs.10-34+14=Rs192  

Cost of 96 shares  = Rs.192*96 = Rs. 912.

50 men can build a tank in 40 days 

Assume 1 man does 1 unit of work in 1 day

Then the total work is 50×40 = 2000 units

50 men work in the first 10 days and completes 50×10 = 500 units of work

45 men work in the next 10 days and completes 45×10 = 450 units of work

40 men work in the next 10 days and completes 40×10 = 400 units of work

35 men work in the next 10 days and completes 35×10 = 350 units of work

So far 500 + 450 + 400 + 350 = 1700 units of work is completed and

Remaining work is 2000 - 1700 = 300 units

30 men work in the next 10 days. In each day, they does 30 units of work.

Therefore, additional days required = 300/30 =10

Thus, total 10+10+10+10+10 = 50 days required.