Quantitative Aptitude - Arithmetic Ability Questions

After day 1, A finishes 1/9 of the work.

After day 2, B finishes 1/7 more of the total work. (1/9) + (1/7) is finished.

After day 3, C finishes 1/5 more of total work. Total finished is 143/315.

So, after day 6, total work finished is 286/315.

Now remaining work =  29 /315

On day 7, A will work again 

Work will be completed on day 7 when A is working. He must finish 29/315 of total remaining work.

Since he takes 9 days to finish the total task, he will need 261/315 of the day. 

Total days required is 6 + (261/315) days.

This is the number of permutations of five things taken all at a time.

Therefore, answer = 5P5 = 120 ways

Choose 2 juniors and 2 seniors.

14C2*23C2 = 23023

Here the LCM of 54, 60 is

=> 2 x 3 x 3 x 3 x 10

= 6 x 9 x 10

= 54 x 10

= 540.

The given series follow the rule that :

Multiplying consecutive prime numbers
2,3,5,7,11,13,17,19....
2x3 = 6
3x5 = 15
5x7 = 35
7x11 = 77
11x13 = 143
13x17 = 221
So, the wrong term is 165.

25 Jan, 1981 = (1980 years + Period from 1.1.1981 to 25.1.1981)

Odd days in 1600 years = 0

Odd days in 300 years = 1

80 years = (60 ordinary years + 20 leap year) = (60 x 1 + 20 x 2)= 2 odd days

Jan = 25 days = (3 weeks + 4 days) = 4 odd days.

Total number of odd days = (0 + 1 + 2 + 4) = 7= 0 odd day.

Given day is Sunday.

K x 3/8 = 141 => K = 376

376 x 32.08/100 = 120