Practice Questions on Time and Work with Multiple Methods and Tricks

Time and Work

Example 1: If 3 men or 4 women can reap a field in 43 days, how long will 7 men and 5 women take to reap it?

Solution:  Method 1:  3 men in 43 days reap = 1 field
   ⇒ 3 men can reap in 1 day = 1/43 of the field
   ⇒ 1 men can reap in 1 day = 1/(43x3) of the field
Similarly, 4 women in 43 days reap = 1 field
   ⇒ 4 women can reap in 1 day = 1/43 of the field
   ⇒ 1 women can reap in 1 day = 1/(43x4) of the field
Therefore, 7 men and 5 women can reap in 1 day = 7 x 1/(43x3) + 5 x 1/(43x4) = 1/12th of the field
   ⇒ 7 men and 5 women will reap the whole field in = 12 days

 Method 2:  3 men = 4 women
   ⇒ 1 man = 4/3 women
   ⇒ 7 men = 7 x 4/3 women = 28/3 women
   ⇒ 7 men + 5 women = (28/3 + 5) women = 43/3 women
Now the question is: If 4 women can reap a field in 43 days, how long will 43/3 women take to reap it?
Now using basic formula: (M1 x D1) = (M2 x D2)
   ⇒ 4 x 43 = 43/3 x D2
   ⇒ D2 = 12 days

Example 2: A is twice as good a workman as B. Together, they finish the work in 4 days. In how many days can it be done by each separately?

Solution:  Method 1:  Let B finish the work in 2x days. Since A is twice as good as B, so A will take half the time taken by B. So A will finish the work in x days.
(A + B)'s one day work = 1/x  + 1/2x = 3x/2x2
   ⇒ Total days taken to finish the work = 2x2/3x = 14
   ⇒ x = 21
   ⇒ A finishes the work in 21 days and B finishes the work in 21 x 2 = 42 days

 Method 2:  Twice + one time = Thrice active person does the work in 14 days.
   ⇒ One time active person (B) will do it in = 14 x 3 = 42
   ⇒ Twice active person (A) will do it in  = 42/2 = 21 days

Example 3: 10 men can finish a piece of work in 10 days, whereas it takes 12 women to finish it in 10 days. If 15 men and 6 women undertake to complete the work, how many days will they take to complete it?

Solution:  Method 1:  10 men x 10 days = 12 women x 10 days
   ⇒ 10 men = 12 women
   ⇒ 15 men  = 12/10 x 15 = 18 women
   ⇒ 15 men + 6 women = 24 women

Now the question is: If 12 women can finish a piece of work in 10 days, how long will 24 women take to finish it?
   ⇒ 12 women x 10 days = 24 women x D2 days
   ⇒ D2 = 5 days

 Method 2:  We know that: (M1 x D1) = (M2 x D2)
   ⇒ 10 men x 10 days = 15 men x N1 days
   ⇒ N1 = 20/3 days
Similarly, 12 women x 10 days = 6 women x N2 days
   ⇒ N2 = 20 days
One day work of 15 men and 6 women = 1/N1 + 1/N2 = 3/20 + 1/20 = 1/5
   ⇒ Total number of days taken by 15 men and 6 women = 5 days

Example 4: A builder decided to build a farmhouse in 40 days. He employed 100 men in the beginning and 100 more after 35 days and completed the construction in stipulated time. If he had not employed the additional men, how many days behind schedule would it have been finished?

Solution:  Method 1:  Work done by 100 men in 35 days = 35 x 100 = 3500
Work done by 200 men in last 5 days = 200 x 5 = 1000
   ⇒ Total work = 3500 + 1000 = 4500
If he had not employed the additional men then, total work done by 100 men in 40 days = 40 x 100 = 4000
Difference = 4500 - 4000 = 500
   ⇒ To do this work, 100 men will take = 500/100 = 5 days more
The work would have lasted 5 days behind the schedule time.

 Method 2:  200 men  can do the rest of the work in 5 days.
   ⇒ 100 men can do the rest of the work in = 5 x 200/100 = 10 days
If he had not employed the additional men, 100 men would take (10 - 5) = 5 days more

Example 5: A team of 30 men is supposed to do a work in 38 days. After 25 days, 5 more men were employed and the work was finished one day earlier. How many days would it have been delayed if 5 more men were not employed?

Solution: After 25 days, 5 more men were employed and work finished 1 day earlier (37 days)
Given, 35 men done the rest of the job in 12 days
   ⇒ 30 men can do the rest of the job in = 12 x 35/30 = 14 days
   ⇒ Work would have finished in (25 + 14) = 39 days if 5 more men were not employed.
   ⇒ Work would have been delayed for (39 - 38) = 1 day

Example 6: 38 men, working 6 hours a day can do a piece of work in 12 days. Find the number of days in which 57 men working 8 hours a day can do twice the work. Assume that 2 men of the first group do as much worked in 1 hour as 3 men of the second group do in 1.5 hours.

Solution: 2 men x 1 hour = 3 men x 1.5 hours
   ⇒ 2 men of first group = 4.5 men of the second group
   ⇒ Efficiency of persons in first group to the second group = 4.5 : 2

Using the Formula: (M1 x D1 x T1 X E1)/W1= (M2 x D2 x T2 x E2)/W2
   ⇒ (38 x 6 x 12 x 4.5)/1 = (57 x D2 x 8 x 2)/2
   ⇒ D2 = 27 days

Example 7: A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in how many days?

Solution:  Method 1:  Basic: Work done by A in one day = 1/15
Work done by B in one day = 1/10
They worked together for 2 days, Work done by them = 2(1/15 + 1/10) = 1/3
   ⇒ Remaining work = 1 - 1/3 =  2/3
A can do one work in = 15 days
   ⇒ A can do 2/3 work in = 15 x 2/3 = 10 days
   ⇒ Whole work completed in = 2 + 10 = 12 days

 Method 2:  Direct Formula: A's one day work = 1/15
B's one day work = 1/10
B work for 2 days, Let A works for x days.
Formula: (No of days A worked) x (A's one day work) + (No of days B worked) x (B's one day work) = Total work (1)
   ⇒ x/15 + 2/10 = 1
   ⇒ x = 12 days

 Method 3:  A can do a work in 15 days and B can do it in 10 days.Let total work = LCM of (15 & 10) = 30 units
   ⇒ A's one day efficiency = 30/15 = 2 units per day
   ⇒ B's one day efficiency = 30/10 = 3 units per day
   ⇒ (A + B)'s one day efficiency = 2 + 3 = 5 units per day

A and B worked for 2 days, so (A + B)'s2 days work = 5 x 2 = 10 units
   ⇒ Remaining work = 30 - 10 = 20 units
   ⇒ Number of days taken by A to complete the remaining work = 20/2 = 10 days
   ⇒ Total no. of days = 2 + 10 = 12 days

Example 8: A and B can do a work in 12 days, B and C in 15 days and C and A in 20 days. If A, B and C work together, they will complete the work in?

Solution: (A + B)'s one day work = 1/12
                       (B + C)'s one day work = 1/15
                       (C + A)'s one day work = 1/20
Adding all three equations, we will get: 2 (A + B + C)'s one day work = 1/12 + 1/15 + 1/20 = 1/5
   ⇒ (A + B + C)'s one day work = 1/10
   ⇒ A, B and C together can finish in 10 days

Example 9: A and B working separately can do a piece of work in 9 and 15 days respectively. If they work for a day alternatively, with A beginning, then the work will be completed in how many days?

Solution: A's one day work = 1/9
                       B's one day work = 1/15
Work done in first Two days = 1/9 + 1/15 = 8/45
   ⇒ Work done in 10 days = 5 x 8/45 = 8/9
Remaining work = 1 - 8/9 = 1/9
Now its A's turn to do work 11th day, time taken by A to complete 1/9  work = 9 x 1/9 = 1 day
   ⇒ Work will be completed in = 10 + 1 = 11 days

Example 10: A, B and C can do a piece of work in 30, 20 and 10 days respectively. A is assisted by B on one day and by C on the next day, alternately. How long would the work take to finish?

Solution: A's one day work = 1/30
                       B's one day work = 1/20
                       C's one day work = 1/10
On first day A and B worked together, second day A and C worked together.
   ⇒ Work done in first two days = 2/30 + 1/20 + 1/10 = 13/60
   ⇒ Work done in 8 days = 4 x 13/60 = 52/60
Remaining work = 1 - 52/60 = 2/15
Now, on 9th day, A and B will work together.
   ⇒ (A + B)'s one day work = 1/30 + 1/20 = 1/12
Remaining work = 2/15 - 1/12  = 1/20
   ⇒ (A + C)'s one day work = 1/30 + 1/10 = 2/15
Time taken to complete the remaining work = 1/20 x 15/2 = 3/8 days
    ⇒ Total time taken = 9 + 3/8 = 93/8 days

Example 11: A can do a piece of work in 8 days which B can destroy in 3 days. A has worked for 6 days, during the last 2 of which B has been destroying; how many days must A now work alone to complete the work?

Solution: Work done by A in 6 days = 6/8
                       Work destroyed by B in 2 days = 2/3
   ⇒ Work remained after destruction = 6/8 - 2/3 = 1/12
   ⇒ Now time taken by A to complete the remaining work (11/12) = 8 x 11/12 = 71/3 days

Example 12: A man, a woman and a boy can complete a job in 3, 4 and 12 days respectively. How many boys assist 1 man and 1 woman to complete the job in 1/4 of a day?

Solution: 1 man's one day work = 1/3
                       1 woman's one day work = 1/4
                       1 boy's one day work = 1/12
   ⇒ (1 man + 1 woman)'s 1/4 day's work = 1/4 x (1/3 + 1/4) = 7/48   
   ⇒ Remaining work = 1 - 7/48 = 41/48
1 boy's 1/4 day's work = 1/12 x 1/4 = 1/48
   ⇒ 41/48 work will be done by = 41/48 x 48 = 41 boys

Example 13: A can do a piece of work in 70 days and B is 40% more efficient than A. The number of days taken by B to do the same work is?

Solution: Efficiency of A : Efficiency of B = No of Days taken by B : No of Days taken by A
   ⇒ 100 : 140 =  D2 : 70
   ⇒ D2 = 50 days

Example 14: A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs 3200. With the help of C they completed the work in 3 days. How much is to be paid to C?

Solution: A's one day work = 1/6
                      B's one day work = 1/8
                     (A + B + C)'s one day work = 1/3
   ⇒ C's one day work = 1/3 - (1/6 + 1/8) = 1/24
Ratio of their one day's work = 1/6 : 1/8 : 1/24 = 4:3:1
Sum of the ratios = 4 + 3 + 1 = 8
   ⇒ C's share = 1/8 x 3200 = Rs 400

Example 15: A man and a boy recieved Rs 800 as wages for 5 days for the work they did together. The man's efficiency in the work was three times that of the boy. What are the daily wages of the boy?

Solution: Ratio of Efficiency = 3:1
   ⇒ Boy's Share = 1/4 x 800 = Rs 200
   ⇒ Daily wages of Boy = 200/5 = Rs 40