Question 1

12 men and 16 boys can do a piece of work in 5 days while 13 men and 24 boys finish it in 4 days. 5 men and 10 boys will finish it in?

Accepted Answer

(12 men + 16 boys) can do a work in 5 days. (60 men + 80 boys) can do a work in 1 day. (13 men + 24 boys) can do a work in 4 days. (52 men + 96 boys) can do a work in 1 day. ⇒ 60 men + 80 boys = 52 men + 96 boys 8 men = 16 boys and 1 man = 2 boys Now, 12 men + 16 boys = (12 x 2 + 16) boys = 40 boys can do the work in 5 days ∴ 5 men + 10 boys = [5 x 2 + 10] boys = 20 boys will take = (40 x 5)/20 = 10 days

Question 2

'A' can do a piece of work in 2 hours, 'B' can do thrice the work in 8 hours and 'C' can do the same work as 'A' in 8 hours. If all of them work together, how long it would take them to complete the work?

Accepted Answer

Efficiency of A = 50% [100/2] Efficiency of B = 37.5% [100/(8/3)] Efficiency of C = 12.5% [100/8] Combined efficiency of A,B and C = 100% So, they complete the work in 1 hour.

Question 3

If one man or two woman or three boys can finish a work in 88 days , then how many days will one man one woman and one boys together take to finish the same work?

Accepted Answer

1 man = 2 woman = 3 boys ∴ 1 man + 1 woman + 1 boy = (3 + 3/2 + 1) boys = 11/2 boys Here M1 = 3 M2 = 11/2, D1 = 88 and D2 = ? According to the formula M1D1 = M2D2 ∴ D2 = M1D1/M2 ∴ Required number of days = 3 x 2 x 88/11 = 3 x 2 x 8 = 48 days

Question 4

When A alone does a piece of work, he takes 25 days more than the time taken by (A + B) to complete that particular work, while B alone takes 49 days more than the time taken by (A + B) to finish the same work . A and B together will take what time …

Accepted Answer

According to the formula, Required time = √ab Where a = 25 and b = 49 ∴ Require time = √25 x 49 = 5 x 7 = 35 days

Question 5

X can do 1/4 of work in 10 days, Y can do 40% of the same work in 40 days and Z can do 1/3 of the work in 13 days. Who will complete the work first?

Accepted Answer

Time taken to complete the work by X =10 x 4 = 40 = 40 days Time taken to complete the work by Y = (40 x 100)/40 = 100 days Time taken to complete the work by Z = 13 x 3 = 39 days Hence, Z will complete the work first.

Question 6

P and Q can finish a work in 30 days, They work at it for 10 days and then Q left. The remaining work is done by P alone in 20 more days. How long will P take to finish the work alone?

Accepted Answer

(P + Q)'s 10 day's work= (1/30) x 10 = 1/3 Remaining work = (1 -1/3) = 2/3 2/3 work is done by P in 20 days. ∴ Whole work is done by P in 20 x (3/2) = 30 days

Question 7

A and B can do a piece of work in 60 days and 75 days, respectively. Both begin together but after a certain time A leaves off. In such case B finishes the remaining work in 30 days. After how many days did A leave?

Accepted Answer

Here A = 60 days, B = 75 days, D = 30, T = ? ∴ Required time = (B - D) x [A/(A + B)] = (75 - 30) x [60/(60 + 75)] = (45 x 60)/135 = 20 days ∴ Required time = 20 days

Question 8

20 women can complete apiece of work in 7 day. If 8 more women are put on the job. In how many days will they complete the work?

Accepted Answer

Given, M1 = 20 D1 = 7, M2 = 20 + 8 = 28 and D2 = ? Using formula, M1D1 = M2D2 ⇒ 20 x 7 = 28 x D2 ∴ D2 = (20 x 7)/28 = 5 days

Question 9

12 men can do a piece of work in 24 days. How many days are needed to go complete the work, if 8 men do this work?

Accepted Answer

Given, M1 = 12, M2 = 8 , D1 = 24 , W1 = 1, W2 = 1 According to the formula, M1D1W2 = M2D2W1 ⇒ 12 x 24 x 1= 8 x D2 x 1 ∴ D2 = (12 x 24)/8 = 12 x 3 = 36 days

Question 10

20 men can cut 30 tree in 4 h. If 4 men leave the job, how many trees will be cut in 12 h?

Accepted Answer

Given, M1 = 20, T1 = 4, W1 = 30, M2 = 20 - 4 = 16, T2 = 12 and W2 = ? According to the formula, M1T1W2 = M2T2W1 ⇒ 20 x 4 x W2 = 16 x 12 x 30 ∴ W2 = (16 x 12 x 30)/(20 x 4) = 12 x 6 = 72 trees

Question 11

40 men can build a wall 200 m long in 12 days, working 8h a day. What will be the number of days that 30 men will take to built a similar wall 300 m long, working 6h per days.

Accepted Answer

Given, M1 = 40, W1 = 200, T1 = 8, D1 = 12, M2 = 30, W2 = 300, T2 = 6 and D2 = ? Then, using formula M1T1D1W2 = M2T2D2W1 ⇒ 40 x 8 x 12 x 300 = 30 x 6 x D2 x 200 ∴ D2 = (40 x 8 x 12 x 300)/(30 x 6 x 200) = 32 days.

Question 12

If m men working m h per day, can do m units of work in m days, then n men working n h per day would be able to complete how units of work in days?

Accepted Answer

Let the required number of units of work = k According to the formula, W1= m and W2 = k M1T1D1W2 = M2T2D2W1 ⇒ m x m x m x k = n x n x n x m ∴ x= (m x n3) / m3 = n3/m2

Question 13

20 men complete one-third of a work in 20 days. How many more men should be employed to finished the rest of work in 25 more days.?

Accepted Answer

Given, M1 = 20. M2 = 20 + k W1 = 1/3, W2 = 1 - 1/3 = 2/3 D1 = 20 and D2 = 25 According to the formula, M1D1W2 = M2D2W1 ⇒ 20 x 20 x (2/3) = (20 + k) x 25 x 1/3 ⇒ 20 x 20 x (2/3) = (20 + k) x 25/3 ⇒ 16 x 2 = 20 + k ∴ k = 32 - 20 = 12

Question 14

45 people take 18 days to dig a pond. If the pond would have to be dug in 15 days, then the number of people to the employed will be?

Accepted Answer

Given, M1 = 45, D1 = 18, M2 = ? , D2 = 15 By using the formula, M1D1 = M2D2 ∴ M2 = M1D1/D2 ⇒ M2 = (45 x 18)/15 = 3 x 18 ⇒ M2 = 54

Question 15

If 3 men or 4 women can built a wall in 43 days, in how many days can 7 men and 5 women build this wall?

Accepted Answer

3 men = 4 women ∴ 1 man = 4/3 women ∴ 7 men + 5 women = 7 x (4/3) + 5 = 43/3 women ∴ M1D1W2 = M2D2W1 ∴ 4 x 43 x 1 = (43/3) x D2 x 1 ∴ D2 = 3 x 4 = 12 days

Question 16

A and B can do a piece of work in 18 days. B and C in 24 days, C and A in 36 days. Find the time in which A, B and C working together can finish the work.

Accepted Answer

(A+ B )' s 1 day ' s work = 1/18 (B +C )'s 1 day ' s work = 1/24 (C + A) 's 1 day 's work = 1/36 From Eqs. (i), (ii) and (iii), we get 2(A + B + C)' s 1 day 's work = 1/18 + 1/24 + 1/36 (A + B + C)'s 1 day's work = (4 + 3 + 2)/(72 x 2) = 9/(72 x 2) = 1/16 ∴ (A + B + C) can complete the work in 16 days

Question 17

A and B can do a piece of work in 72 days. B and C can do it in 120 days, A and C and do it in 90 days. In what time can A alone do it?

Accepted Answer

(A + B )'s 1 day ' s work = 1/72 (B + c)'s 1 day' s work = 1/120 (A + C)'s 1 day's work = 1/90 2(A + B + C)'s 1 days work = 1/72 + 1/120 + 1/90 ∴ ( A + B + C )'s 1 day's work (5 + 3 + 4)/(360 x 2) = 12/(360 x 2) = 1/60 ∴ A's 1 day's work = (A + B + C) 's 1 day's work - (B + C)'s 1 day's work = 1/60 - 1/120 = (2 - 1)/120 = 1/120 ∴ A alone can finish the work in 120 days.

Question 18

25 men reap a field in 20 days. When should 15 men leave the work, if the whole field is to be reaped in 37 1/ 2 day after they leave the work?

Accepted Answer

25 men reap the field in 20 days ∴ 10 men can reap the feild in (20 x 25)/ 10 = 50 days. When 15 men leave the work, 10 men remain and they can reap in 371/2 days = (371/2)/ 50 = 3/4 of the field Hence, all men must work till (1 - 3/4) = 1/4 of the field is reaped in 20/4 = 5 days.

Question 19

A can do a piece of work in 5 hours, B in 9 hours and C in 15 hours. If C could work with them for 1 hour only the time taken by A and B together to complete the work is?

Accepted Answer

∵ (1/5 + 1/9 + 1/15) = 17/45 work is finished in 1 hour ∴ Remaining work = 1 - 17/45 = 28/45 ⇒ (A + B)'s 1 hour's work = 1/5 + 1/9 = 14/45 14/45 work is done by (A and B) in 1 hour 28/45 work will be done by A and B in (45/14) x (28/45) = 2 hours

Question 20

A can do a piece of work in 24 days while B alone can do it in 16 days. But with the help of C, they finish the work in 8 days. C alone can do the work in?

Accepted Answer

∵ C's 1 day's work = [( A + B + C)'s 1 day's work] - [(A + B)'s 1 day's work] = [1/8 - (1/24 + 1/16)] = (1/8 - 5/48) = 1/48 ∴ C alone can do it in 48 days.

Time and Work Questions