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

Let us look at the next problem on "Unitary method time and work"

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. 

Let us look at the next problem on "Unitary method time and work"

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.

Let us look at the next problem on "Unitary method time and work"

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.

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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. 

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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

Let us look at the next problem on "Unitary method time and work"

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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