UNITARY METHOD TIME AND WORK
About "Unitary method time and work"
On this web page "Unitary method time and work", we will learn to solve time and work problems using unitary method
Unitary method definition and example :
Definition :
Unitary-method is all about finding value to a single unit.
Unitary-method can be used to calculate cost, measurements like liters and time.
Example :
If 8 men can complete a work in 6 days,
1 men can complete in = 8 x 6 = 48 days.
Unitary method time and work - Practice problems
To have better understanding on unitary method time and work, let us look at some practice problems.
Problem 1 :
7 men can complete a work in 52 days. In how many days will 13 men finish the same work?
Solution :
7 men can complete a work in = 52 days
Then, one can complete in = 7 x 52 = 364 days
13 men can complete in = 364 / 13 = 28 days
Hence, 13 men can complete the work in 28 days
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Problem 2 :
A truck covers a particular distance in 3 hours with the speed of 60 miles per hour. If the speed is increased by 30 miles per hour, find the time taken by the truck to cover the same distance
Solution :
Given : Time = 3 hours and Speed = 60 mph
Then, Distance = Time x Speed
Distance = 3 x 60 = 180 miles
If the given speed 60 mph is increased by 30 mph,
then the new speed = 90 mph
Then, Time = Distance / Speed
Time = 180 / 90 = 2 hous
Hence, if the speed is increased by 30 mph, time taken by the truck is 2 hours.
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Problem 3 :
David can complete a work in 6 days working 8 hours per day. If he works 6 hours per day, how many days will he take to complete the work ?
Solution :
8 hours per day --------> 6 days to complete the work
1 hour per day ---------> 8 x 6 = 48 days
6 hours per day ---------> 48 / 6 = 8 days
Hence, David can complete the work in 8 days working 6 hours per day.
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Problem 4 :
Alex takes 15 days to reduce 30 kilograms of his weight by doing 30 minutes exercise per day. If he does exercise for 1 hour 30 minutes per day, how many days will he take to reduce the same weight ?
Solution :
Given : Minutes per day = 30 and No. of days = 15
Total minutes in 15 days = 30 x 15 = 450 minutes
So, 450 minutes of exercise required to reduce 30 kilograms weight.
1 hour 30 minutes = 90 minutes
Then, No. of days required = 450 / 90 = 5 days
Hence, if Alex does exercise for 1 hour 30 minutes per day, it will take 5 days to reduce 30 kilograms of weight.
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Problem 5 :
If 5 men can paint a house in 18 hours, how many men will be able to paint it in 10 hours ?
Solution :
In 18 hours, the house can be painted by 5 men
In 1 hour, the house will be painted by = 18 x 5 = 90 men
In 10 hours, the house can be painted by = 90 / 10 = 9 men
Hence, 9 men will be able to paint the house in 10 hours
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Problem 6 :
In a fort, 360 men have provisions for 21 days. If 60 more men join them, how long will the provision last ?
Solution :
360 men have provisions for 21 days
1 man has provisions for = 360 x 21 = 7560 days
If 60 more men join, the total no. of men = 360 + 60 = 420
420 men have provisions for = 7560 / 420 = 18 days
Hence, if 60 more men join, provision will last for 18 days
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Problem 7 :
A man can type 9 pages of a book everyday and completes it in 50 days. How many days will he take to complete it, if he types 15 pages everyday ?
Solution :
9 pages per day ------> 50 days
1 page per day --------> 9 x 50 = 450 days
15 pages per day ------> 450 / 15 = 30 days
Hence, the man will complete the book in 30 days, if he types 15 pages per day.
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Problem 8 :
A can do a piece of work in 8 days. B can do the same in 14 days. In how many days can the work be completed if A and B work together?
Solution :
Let us find LCM for the given no. of days "8" and "14".
L.C.M of (8, 14) = 56
Therefore, total work = 56 units
A can do = 56 / 8 = 7 units/day
B can do = 56 / 14 = 4 units/day
(A + B) can do = 11 units per day
No. of days taken by (A+B) to complete the same work
= 56 / 11 days
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Problem 9 :
A and B together can do a piece of work in 12 days and A alone can complete the work in 21 days. How long will B alone to complete the same work?
Solution :
Let us find LCM for the given no. of days "12" and "21".
L.C.M of (12, 21) = 84
Therefore, total work = 84 units.
A can do = 84 / 21 = 4 units/day
(A+B) can do = 84 / 12 = 7 units/day
B can do = (A+B) - A = 7 - 4 = 3 units/day
No. of days taken by B alone to complete the same work
= 84 / 3
= 28 days
Let us look at the next problem on "Unitary method time and work"
Problem 10 :
A and B together can do a piece of work in 110 days. B and C can do it in 99 days. C and A can do the same work in 90 days. How long would each take to complete the work ?
Solution :
Let us find LCM for the given no. of days "110", "99" and "90".
L.C.M of (110, 99, 90) = 990
Therefore, total work = 990 units.
(A + B) = 990/110 = 9 units/day --------->(1)
(B + C) = 990/99 = 10 units/day --------->(2)
(A + C) = 990/90 = 11 units/day --------->(3)
By adding (1), (2) & (3), we get,
2A + 2B + 2C = 30 units/day
2(A + B + C) = 30 units/day
A + B + C = 15 units/day --------->(4)
(4) - (1) ====> (A+B+C) - (A+B) = 15 - 9 = 6 units
C can do = 6 units/day , C will take = 990/6 = 165 days
(4) - (2)====> (A+B+C) - (B+C) = 15 - 10 = 5 units
A can do = 5 units/day, A will take = 990/5 = 198 days
(4) - (3) ====> (A+B+C) - (A+C) = 15 - 11 = 4 units
B can do = 4 units/day, B will take = 990/4 = 247.5 days
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Problem 11 :
A and B can do a work in 15 days. B and C can do it in 30 days. C and A can do the same work in 18 days. They all work together for 9 days and then A left. In how many days can B and C finish remaining work?
Solution :
Let us find LCM for the given no. of days "15", "30" and "18".
L.C.M of (15, 30, 18) = 90 units
Therefore, total work = 90 units.
(A + B) = 90/15 = 6 units/day --------->(1)
(B + C) = 90/30 = 3 units/day --------->(2)
(A + C) = 90/18 = 5 units/day --------->(3)
By adding (1), (2) & (3), we get,
2A + 2B + 2C = 14 units/day
2(A + B + C) = 14 units/day
A + B + C = 7 units/day --------->(4)
A, B and C all work together for 9 days.
No. of units completed in these 9 days = 7x9 = 63 units
Remaining work to be completed by B and C = 90 - 63 = 27 units
B and C will take = 27/3 = 9 days [ Because (B+C) = 3 units/day ]
Hence B and C will take 9 days to complete the remaining work.
Let us look at the next problem on "Unitary method time and work"
Problem 12 :
A and B each working alone can do a work in 20 days and 15 days respectively. They started the work together, but B left after sometime and A finished the remaining work in 6 days. After how many days from the start, did B leave?
Solution :
Let us find LCM for the given no. of days "20" and "15".
L.C.M of (20, 15) = 60 units
Therefore, total work = 60 units.
A can do = 60/20 = 3 units/day
B can do = 60/15 = 4 units/day
(A + B) can do = 7 units/day
The work done by A alone in 6 days = 6x3 = 18 units
Then the work done by (A+B) = 60 - 18 = 42 units
Initially, no. of days worked by A and B together
= 42/7 = 6 days
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Problem 13 :
A is 3 times as fast as B and is able to complete the work in 30 days less than B. Find the time in which they can complete the work together.
Solution :
A & B working capability ratio = 3 : 1
A & B time taken ratio = 1 : 3
From the ratio, time taken by A = k and time taken by B = 3k
From "A takes 30 days less than B", we have
3k - k = 30
2k = 30 ===> k = 15
Time (A) = 15 days, Time (B) = 3x15 = 45 days
LCM (15, 45) = 45
Total work = 45 units
A can do = 45 / 15 = 3 units/day
B can do = 45 / 45 = 1 unit/day
(A + B) can do = 4 units per day
No. of days taken by (A+B) to complete the same work
= 45 / 4 = 11 1/4 days
Let us look at the next problem on "Unitary method time and work"
Problem14 :
A and B working separately can do a piece of work in 10 and 8 days respectively. They work on alternate days starting with A on the first day. In how many days will the work be completed ?
Solution :
Let us find LCM of the given no. of days "10" and "8"
LCM of (10, 8) = 40
Total work = 40 units
A can do = 40/10 = 4 units/day
B can do = 40/8 = 5 units/day
On the first two days,
A can do 4 units on the first day and B can do 5 units on the second day. (Because they are working on alternate days)
Total units completed in the 1st day and 2nd day = 9 units ----(1)
Total units completed in the 3rd day and 4th day = 9 units ----(2)
Total units completed in the 5th day and 6th day = 9 units ----(3)
Total units completed in the 7th day and 8th day = 9 units ----(4)
By adding (1),(2),(3) & (4), we get 36 units.
That is, in 8 days 36 units of the work completed.
Remaining work = 40 - 36 = 4 units
These units will be completed by A on the 9th day.
Hence the work will be completed in 9 days.
Let us look at the next problem on "Unitary method time and work"
Problem 15 :
Two pipes A and B can fill a tank in 16 minutes and 20 minutes respectively. If both the pipes are opened simultaneously, how long will it take to complete fill the tank ?
Solution :
Let us find LCM of the given no. of minutes "16" and "20"
LCM of (16, 20) = 80
Total work = 80 units
A can fill = 80/16 = 5 units/min
B can fill = 80/20 = 4 units/min
(A+B) can fill = 9 units/min
No. of minutes taken by (A+B) to fill the tank
= 80/9 = 8 8/9 minutes
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Problem 16 :
Pipe A can fill a tank in 10 minutes. Pipe B can fill the same tank in 6 minutes. Pipe C can empty the tank in 12 minutes. If all of them work together, find the time taken to fill the empty tank.
Solution :
Let us find LCM of the given no. of minutes "10", "6" and "12"
LCM of (10,6, 12) = 60
Total work = 60 units
A can fill = 60/10 = 6 units/min
B can fill = 60/6 = 10 units/min
(A+B) can fill = 16 units/min
C can empty = 60/12 = 5 units/min
If all of them work together,
(A+B) can fill = 9 units/day
No. of minutes taken by (A+B) to fill the tank
= 80/9 = 8 8/9 minutes
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Problem 17 :
A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?
Solution :
Let us find LCM of the given no. of minutes "10" and "6".
LCM of (10, 6) = 30
Total work = 30 units (to fill the empty tank)
Already tank is two-fifth full.
So, the work completed already = (2/5)x30 = 12 units
Out of 30 units, now the tank is 12 units full.
A can fill = 30/10 = 3 units/min (filling)
B can empty = 30/6 = 5 units/min (emptying)
If both the pipes are open, the tank will be emptied
2 units/minute
No of minutes taken to empty the tank (already two-fifth filled)
= 12/2 = 6 minutes
Let us look at the next problem on "Unitary method time and work"
Problem 18 :
Who is better in earning,
David earns $57.60 for 8 hours of work
or
John earns $90 for 12 hours of work ?
Solution :
To compare the given measures, convert them in to unit rates.
|
David Earning in 8 hrs = $57.60 Earning in 1 hr = 57.60 / 8 Earning in 1 hr = $7.20 |
John Earning in 12 hrs = $90 Earning in 1 hr = 90 / 12 Earning in 1 hr = $7.50 |
From the above unit rates, John earns more than David per hour.
Hence, John is earning better
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Problem 19 :
Who is better in working,
Alex completes 120 units of work in 3 hours
or
Jose completes 84 units of work in 2 hours ?
Solution :
To compare the given measures, convert them in to unit rates.
|
Alex Work in 3 hrs = 120 units Work in 1 hr = 120 / 3 Work in 1 hr = 40 units |
Jose Work in 2 hrs = 84 units Work in 1 hr = 84 / 2 Work in 1 hr = 42 units |
From the above unit rates, Jose completes more units of work than David per hour.
Hence, Jose is better in working
Let us look at the next problem on "Unitary method time and work"
Problem 20 :
Who is better,
Lily can prepare 10.4 gallons of juice in 4 days
or
Rosy can prepare 7.5 gallons of juice in 3 days ?
Solution :
To compare the given measures, convert them in to unit rates.
|
Lily No.gallons in 2 days = 5.2 No.of gallons in 1 day = 5.2/2 No.of gallons in 1 day = 2.6 |
Rosy No. gallons in 3 days = 7.5 No. of gallons in 1 day = 7.5/3 No.of gallons in 1 day = 2.5 |
From the above unit rates, Lily prepares more gallons than day.
Hence, Lily is better
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