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.
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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