Question 1

A takes twice as much time as B, and C take thrice as much time as B to finish a work. Working together, they can finish the work in 12 days. Find the number of days needed for A to do the work alone.

Accepted Answer

Let time taken by B = N Then, time taken by A = 2N and time taken by C = 3N According to the question, (1/N) + (1/2N) + (1/3N) = 1/12 ⇒ 1 + (1/2) + (1/3) = N/12 ⇒ (6 + 3 + 2)/6 = N/12 ⇒ 11 = N/2 ∴ N = 22 ∴ Required number of days = 2N = 2 x 22 = 44 days

Question 2

10 men and 8 women together can complete a work in 5 days. Work done by one women in day is equal to half the work done by man in 1 day. How many days will it take for 4 men and 6 women to complete that work?

Accepted Answer

According to the question, 1 man = 2 women ∴ 10 men + 8 women = (20 + 8) women = 28 women 4 men + 6 women = (8 + 6) women = 14 women Given, M1 = 28, M2 = 14, D1 = 5, D2 = ? and W1 = W2 = 1 According to the formula, M1D1W2= M2D2W1 ⇒ 28 x 5 x 1 = 14 x D2 x 1 ∴ D2 = (28 x 5)/14 = 10 days

Question 3

5 men started working to complete a work in 15 days. After 5 days, 10 women are accompanied by to them to complete the work in next 5 days. If the work is to be done by women only, when could the work be over, if 10 women have started it?

Accepted Answer

According to the question, 5 men's 1 day's work = 1/15 .....(i) ∴ 5 men's 5 day's work = (1/15) x 5 = 1/3 ∴ Remaining work = 1 - (1/3) = 2/3 5 men + 10 women can do 2/3 of the work in 5 days. ∴ 5 men + 10 women can do the whole of the work in 15/2 days. ∴ (5 men + 10 women's) 1 day's work = 2/15 .......(ii) 10 women 1 days work = (2/15) - 5 men's 1 day's work = (2/15) - (1/15) = 1/15 ∴ 10 women can finish the work in 15 days

Question 4

A contract is to be complete in 50 days and 105 men were set to work, each working 8 h a day. After 25 days, 2/5th of the work is finished. How many additional men be employed, so that the work may be completed on time, each man now working 9 h a da…

Accepted Answer

According to the formula M1D1T1/W1= M2D2T2/W2 Given M1 = 105 D1 = 25 T1 = 8 W1 = 2/5 Now let the additional men be N Then, M2 = 105 + N, T2 = 9, D2 = 25 and W2 =1 - (2/5) = 3/5 On putting these value in the above formula (105 x 25 x 8) / (2/5) = [(105 + N) x 25 x 9] / (3/5) ⇒ (105 x 8) / 2 = (105 + N) x (9/3) ⇒ 105 x 4 = (105 + N) x 3 ⇒ 105 x 4 = 105 x 3 +3N ⇒ 3N = 105 ∴ N = 35 men

Question 5

If 14 men and 12 boys can finish work in 4 days, while 8 men and 16 boys can finish the same work in 5 days. Compare the 1 days work of 1 man and 1 boy.

Accepted Answer

Here, a1 = 14, b1 = 12, p = 4, a2 = 8, b2 = 16 and q = 5 One days work of 1 man/One days of 1 boy = (qb2 - pb1) / ( pa1 - qa2) = [5 x 16 - 4 x 12] / [ 4 x 14 - 5 x 8] = (80 - 48) / (56 - 40) = 32/16 = 2

Question 6

In a factory, there are equal number of women and children. Women work for 6 h a days and children for 4 h a day. During festival time, the work load goes up by 50%. The government rule does not allow children to work for more then 6 h a day. If the…

Accepted Answer

Let extra hours a day are N. According to the formula, (M1D1T1) / W1 = (M2D2T2) / W2 ⇒ [1 x 1 x (6 + 4)] / 1 = [1 x 1 x (6 + 6 + N)] / 11/2 ⇒ (3/2) x 10 = 12 + N ⇒ 15 = 12 + N ∴ N = 15 - 12 = 3 Extra hours of works everyday are 3.

Question 7

3 women and 18 children together take 2 days to complete a piece of work. How many days will 9 children alone take to complete the piece of work, if 6 women alone can complete the piece of work in 3 days?

Accepted Answer

Since, 3 women + 18 children complete work in 2 days. Therefore, (3 x 2) women + (18 x 2) children complete work in 1 days i.e, 6 women + 36 children complete work in 1 day. Work of 36 children for 1 day = 1 - (1/3) = 2/3 [∵ work of 6 women for 1 day = 1/3] ∴ 36 children do 2/3 part of the work in 1 day. 36 children can do the work in 3/2 days. 9 children can do the the work in (3/2) x 4 = 6 days

Question 8

A contractor undertook to finish a certain work in 124 days and employed 120 men. After 64 days, he found that he had already done 2/3 of the work. How many men can be discharge now, so that the work finish in time?

Accepted Answer

Given, M1 = 120, M2 = 120 - n, D1 = 64, D2 = 60 W1= 2/3 and W2 = 1/3 According to the formula, (M1D1) / W1 = (M2D2) / W2 ⇒ [120 x (64/2)] / 3 = [(120 - n) x 60] / (1/3) ⇒ (120 x 64) / (2 x 60) = (120 - n) ⇒ (120 - n) = 64 ∴ n = 120 - 64 = 56 Now, 56 men can be discharged to finish the work in time.

Question 9

A contract is to be complete in 92 days and 234 men were set to work, each working 16 h a days. After 66 days, 4/7 of the work is completed. How many addition men may be employed, so that the work may be completed in time, each man now working 18 h …

Accepted Answer

Remaining work = 1 - (4/7) = 3/7 Remaining period = (92 - 66) = 26 days Let the number of addition men = N Given, M1 = 234, D1 = 66, T1 = 16, W1= 4/7, M2 = (234 + x), D2 = 26, T2 = 18, W2 = 3/7 According to the question, M1W2T1D1= M2W1T2D2 ⇒ 234 x (3/7) x 16 x 66 = (234 x N) x (4/7) x 18 x 26 ⇒ 234 + N = (3 x 66 x 16 x 234) / (4 x 26 x 18) ⇒ 234 + N = 36 x 11 = 396 ∴ N = 396 - 234 = 162 Additional men to be employed = 162

Question 10

A, B and C can do a piece of work in 16, 32 and 48 days, respectively. They all begin together. A work continuously till it is finished, C leaves the work 2 days before its completion and B leaves the work in 1 days before its complete. In what time…

Accepted Answer

Let the work be finished in N days. Then, A's N day's work + B's (N - 1) days + work + C's (N - 2) day's work = 1 ⇒ N/16 + (N - 1/32) + (N - 2/48) = 1 ⇒ N/1 + N - 1/2 + N - 2/3 = 16 ⇒ (6N + 3N - 3 + 2N - 4)/6 = 16 ⇒ 11N - 7 = 96 ⇒ 11N = 103 ∴ N = 103/11 = 94/11 days

Question 11

Rashmi can do a piece of work in 16 days. Ravina can do the same work in 12 4/ 5 days, while Gitika can do it in 32 days. All of them started to work together but Rashmi leaves after 4 days. Ravina leaves the job 3 days before the completion of the …

Accepted Answer

Let the work lasted for N days, Then, Rashmi's 4 day's work + Ravina (N - 3) day's + Gitika's N day's work = 1 ⇒ (4/16) + (N - 3) / (124/5) + N/32 = 1 ⇒ 5(N - 3)/64 + N/32 = 1 - 1/4 ⇒ [5(N - 3) + 2N] / 64 = 3/4 ⇒ 7N - 15 = 48 ∴ N= (48 + 15)/7 = 63/7 = 9 days

Question 12

A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):

Accepted Answer

A's 5 days work = 50% B's 5 days work = 33.33% C's 2 days work = 16.66% [100- (50+33.33)] Ratio of contribution of work of A, B and C = 50 : 33 1 3 : 16 2 3 = 3 : 2 : 1 A's total share = Rs. 1500 B's total share = Rs. 1000 C's total share = Rs. 500 A's one day's earning = Rs.300 B's one day's earning = Rs.200 C's one day's earning = Rs.250

Question 13

A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?

Accepted Answer

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10. C leaves 4 days before completion of the work, which means only A and B work during the last 4 days. Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10. Remaining Work = 7/10, which was done by A,B and C in the initial number of days. Number of days required for this initial work = 7 days. Thus, the total numbers of days required = 4 + 7 = 11 days.

Question 14

12 men can complete a work in 8 days. 16 women can complete the same work in 12 days. 8 men and 8 women started working and worked for 6 days. How many more men are to be added to complete the remaining work in 1 day?

Accepted Answer

1 man's 1 day work =1/96 ; 1 woman's 1 day work = 1/192 Work done in 6 days= 6 8 96 + 8 192 = 6 × 1 8 = 3 4 Remaining work = 1/4 (8 men +8 women)'s 1 day work = 1 8 96 + 8 192 =1/8 Remaining work =1/4 - 1/8 = 1/8 1/96 work is done in 1 day by 1 man Therefore, 1/8 work will be done in 1 day by 96 x (1/8) =12 men

Question 15

P can complete a work in 12 days working 8 hours a day.Q can complete the same work in 8 days working 10 hours a day. If both p and Q work together,working 8 hours a day,in how many days can they complete the work?

Accepted Answer

P can complete the work in (12 x 8) hrs = 96 hrs Q can complete the work in (8 x 10) hrs=80 hrs Therefore, P's 1 hour work=1/96 and Q's 1 hour work= 1/80 (P+Q)'s 1 hour's work =(1/96) + (1/80) = 11/480. So both P and Q will finish the work in 480/11 hrs Therefore, Number of days of 8 hours each = (480/11) x (1/8) = 60/11

Question 16

A and B can do a piece of work in 30 days , while B and C can do the same work in 24 days and C and A in 20 days . They all work together for 10 days when B and C leave. How many days more will A take to finish the work?

Accepted Answer

2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8 =>(A+B+C)'s 1 day's work= 1/16 work done by A,B and C in 10 days=10/16 = 5/8 Remaining work= 3/8 A's 1 day's work= 1 16 - 1 24 = 1 48 Now, 1/48 work is done by A in 1 day. So, 3/8 work wil be done by A in =48 x (3/8) = 18 days

Question 17

A can do a certain work in the same time in which B and C together can do it.If A and B together could do it in 20 days and C alone in 60 days ,then B alone could do it in:

Accepted Answer

(A+B)'s 1 day's work=1/20 C's 1 day work=1/60 (A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15 Also A's 1 day's work =(B+C)'s 1 day's work Therefore, we get: 2 x (A's 1 day 's work)=1/15 =>A's 1 day's work=1/30 Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60 So, B alone could do the work in 60 days.

Question 18

A works twice as fast as B.If B can complete a work in 18 days independently,the number of days in which A and B can together finish the work is:

Accepted Answer

Ratio of rates of working of A and B =2:1. So, ratio of times taken =1:2 Therefore, A's 1 day's work=1/9 B's 1 day's work=1/18 (A+B)'s 1 day's work= 1/9 + 1/18 = 1/6 so, A and B together can finish the work in 6 days

Question 19

A Contractor employed a certain number of workers to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realised that the work would get delayed by three-fourth of the scheduled time, s…

Accepted Answer

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days. D * x +(100- D) * 2x= 175x => D= 25 days Now , the work done in 25 days = 25x Total work = 175x Therefore, workdone before increasing the no of workers = 25 x 175 x × 100 % = 14 2 7 %

Question 20

A,B,C together can do a piece of work in 10 days.All the three started workingat it together and after 4 days,A left.Then,B and C together completed the work in 10 more days.In how many days can complete a work alone ?

Accepted Answer

(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.

Time and Work Questions