UNITARY METHOD TIME AND WORK

Problem 1 :

A can do a piece of work in 15 days while B can do it in 10 days. How long will they take together to do it?

Solution :

A’s 1 day work  =  1/15   

B’s 1 day work  =  1/10

(A + B)’s  1 day work  =  1/15 + 1/10

L.C.M of (15, 10)  =  30. 

Then, we have

(A + B)’s  1 day work  =  2/30 + 3/30

(A + B)’s  1 day work  =  5/30

(A + B)’s  1 day work  =  1/6

So, A and B can together finish the work in 6 Days.

Problem 2 :

A, B and C can do a piece of work in 12 days, 15 days and 10 days respectively. In what time will they all together finish it ?

Solution :

A’s 1 day work  =  1/12   

B’s 1 day work  =  1/15

C’s 1 day work  =  1/10

(A + B + C)’s  1 day work  =  1/12 + 1/15 + 1/10

L.C.M of (12, 15, 10)  =  60. 

Then, we have

(A + B + C)’s  1 day work  =  5/60 + 4/60 + 6/60

(A + B + C)’s  1 day work  =  15/60

(A + B + C)’s  1 day work  =  1/4

So, A, B and C can together finish the work in 4 days.

Problem 3 :

5 workers can complete a work in 20 days. In how many days can 10 workers complete the same work ?

Solution :

Given : 5 workers can complete the work in 20 days

Number of days taken by one worker to complete the same work is 

=  5 ⋅ 20 days

=  100 days

Number of days taken by 10 workers to complete the same work is 

=  100 / 10 days

=  10 days

So, 10 workers can complete the work in 10 days. 

Problem 4 :

A and B together can do a piece of work in 35 days, while A alone can do it in 60 days. How long would B alone take to do it ?

Solution :

Let x be the no. of days taken by B to complete the work. 

A’s 1 day work  =  1/60   

B’s 1 day work  =  1/x

(A + B)’s  1 day work  =  1/60 + 1/x

L.C.M of (60, x)  =  60x. 

Then, we have

(A + B)’s  1 day work  =  (x/60x) + (60/60x)

(A + B)’s  1 day work  =  (x + 60)/60x ------(1)

Given : A and B together can do the work in 35 days. 

So, we have 

(A + B)’s  1 day work  =  1/35 ------(2)

From (1) and (2), we get

(x + 60)/60x  =  1/35

35(x + 60)  =  1 ⋅ 60x

35x + 2100  =  60x

2100  =  25x

84  =  x

So, B alone can complete the work in 84 Days.

Problem 5 :

A can do a piece of work in 20 days while B can do it in 15 days. With the help of C, they finish the work in 5 days. In what time would C alone do it ?

Solution :

Let x be the no. of days taken by C to complete the work. 

A’s 1 day work  =  1/20   

B’s 1 day work  =  1/15

C’s 1 day work  =  1/x

(A + B + C)’s  1 day work  =  1/20 + 1/15 + 1/x

L.C.M of (20, 15, x)  =  60x. 

Then, we have

(A + B + C)’s 1 day work =  (3x/60x) + (4x/60x) + (60/60x)

(A + B + C)’s  1 day work  =  (3x + 4x + 60)/60x

(A + B + C)’s  1 day work  =  (7x + 60)/60x ------(1)

Given : A, B and C together can do the work in 5 days. 

So, we have 

(A + B + C)’s  1 day work  =  1/5 ------(2)

From (1) and (2), we get

(7x + 60)/60x  =  1/5

5(7x + 60)  =  1 ⋅ 60x

35x + 300  =  60x

300  =  25x

12  =  x

So, C alone can complete the work in 12 Days.

Problem 6 :

A can do a piece of work in 12 days and B alone can do it in 16 days. They worked together on it for 3 days and then A left. How long would B take to finish the remaining work ?

Solution :

A’s 1 day work  =  1/12   

B’s 1 day work  =  1/16

(A + B)’s  1 day work  =  1/12 + 1/16

L.C.M of (12, 16)  =  48. 

Then, we have

(A + B)’s 1 day work  =  4/48 + 3/48

(A + B)’s 1 day work  =  7/48

Then, the amount of work completed by A and B together in 3 days is

 =  3 ⋅ 7/48

=  7/16

Amount of work left for B to complete is

=  9/16

Number of days that B will take to finish the work is

=  Amount of work / part of the work done in 1 day

=  (9/16) / (1/16)

=  (9/16) ⋅ (16/1)

=  9

So, no. of days taken by B to finish the remaining work is 9 days.

Problem 7 :

A can do 1/4 part of a work in 5 days, while B can do 1/5 part of the work in 6 days. In how many days can both do it together ?

Solution :

A’s 1 day work  =  (1/4) / 5  =  1/20    

B’s 1 day work  =  (1/5) / 6  =  1/30

(A + B)’s  1 day work  =  1/20 + 1/30

L.C.M of (20, 30)  =  60. 

Then, we have

(A + B)’s 1 day work  =  3/60 + 2/60

(A + B)’s 1 day work  =  5/60

(A + B)’s 1 day work  =  1/60

So, A, and B can together finish the work in 60 days.

Problem 8

A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?

Solution :

A’s 1 minute work  =  1/10    

B’s 1 minute work  =  1/6

(A + B)’s  1 minute work  =  1/10 - 1/6

L.C.M of (10, 6)  =  30. 

Then, we have

(A + B)’s 1 minute work  =  3/30 - 5/30

(A + B)’s 1 minute work  =  - 1/15

(A + B)’s 1 minute work  =  - 1/15

(Here, negative sign indicates that when both the pipers are opened together, 1/15 part of the tank will be emptied in 1 minute)

Given : 2/5 part of the tank is full 

When both the pipes are opened, time taken to empty 2/5 part of the tank is

=  Amount of water in tank / water emptied in 1 minute

=  (2/5) / (1/15)

=  (2/5) ⋅ (15/1)

=  6 minutes

So, when the tank is two-fifth full, if both the pipes are open, it will take 6 minutes to empty. 

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