Question: 21 -

A can finish a work in 18 days and B can do the same work in half the time taken by A. So if they working together, what part of the same work can finished in a day?

1/7

5/6

6

1/6

Answer:

1/6

Solution:

First find the 1 day work of both (A & B)
A's 1 day's work = 1/18
and 
B's 1 day's work = 1/9 (B can do work in half time)
(A + B)'s 1 day's work = (1/18+1/9) 
= (1+2)/18 = 3/18 = 1/6 
so A & B together can do 1/6 of work in 1 day

Question: 22 -

P, Q and R are three typists who working simultaneously can type 216 pages in 4 hours. In one hour, R can type as many pages more than Q as Q can type more than P. During a period of five hours, R can type as many pages as P can during seven hours. How many pages does each of them type per hour ?

14,17,20

15,17,22

16,18,22

15, 18, 21 

Answer:

15, 18, 21 

Solution:

Let's the number of pages typed in one hour by P, Q and R be p, q and r respectively. Then,
P,Q and R typed page in 1 hrs = 216/4
=> p + q + r = 216/4
=> p + q + r = 54 ...(i)
r - q = q - p => 2p = q + r ...(ii)
5r = 7p => p = 5/7 r ...(iii)
By Solving above (i), (ii) and (iii) equations
=> p = 15, q = 18, q = 21

Question: 23 -

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days what is time taken by 15 men and 20 boys?

7 days

6 days

5 days

4 days

Answer:

4 days

Solution:

Given that 
6 men and 8 boys can do a piece of work in 10 days 
26 men and 48 boys can do the same in 2 days
As the work done is equal,
10(6M + 8B) = 2(26M + 48B)
60M + 80B = 52M + 96B
=> M = 2B
=> B = M/2 ……(1)
Now Put (1) in 15M + 20B
=> 15M + 10M = 25M
Now, 6M + 8B in 10 days
=> (6M + 4M) 10 = 100M
Then D(25M) = 100M
=> D = 4 days.

Question: 24 -

Kim can do a work in 3 days while David can do the same work in 2 days. Both of them finish the work together and get Rs. 150. What is the share of Kim ?

Rs.80

Rs.60

Rs.75

Rs.65

Answer:

Rs.60

Solution:

Kim 1 day work = 1/3
David 1 day work = 1/2
Kim's 1 day's work : David's 1 day's work=(1/3)/(1/2) = 2 : 3.

Kim's share = Rs. (2/5?150)=Rs. 60

Question: 25 -

A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?

5(1/8)

4(1/4)

(3/5)

3(3/5)

Answer:

3(3/5)

Solution:

1 minute's Work by both punctures = (1/9+1/6) = (2/18+3/18) = 5/18
So both punctures can make tyre flat = 18/5 = 3(3/5)