UNITARY METHOD TIME AND WORK

Unitary Method Definition and Example :

Definition :

Unitary-method is all about finding value to a single unit.

Example :

If 8 men can complete a work in 6 days,

1 men can complete in = 8 x 6 = 48 days.

Tricks to solve time and work problems using unitary method :

(i) If A can complete a work in "m" days, then part of the work completed by A in 1 day is 1/m.

(i) If A can complete 1/m part of the work in 1 day, then A will complete the work in "m" days.

Solved Problems

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

(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