Problems on Work Done in a Given Period of Time

Problems on Work Done in a Given Period of Time are given in this article. Every problem has answers along with explanations. Start your practice without any delay to get good marks in the exam. You can easily score good results in the exam by getting the grip on the complete concept. You have to follow the best way to solve problems. We have also included tips and tricks to solve problems on Work Done in a Given Period of Time. Therefore, make your practice easy and get complete knowledge of the concept.

Work Done = Time x Rate of Work.
If A person can complete the work in ‘x’ days, then the work done by ‘A’ in one day is equal to (\(\frac{1}{x}\)).

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Examples on Calculating Work Done in a Given Period of Time

Solve the different examples of work done in a given period of time by using the different formulas.

Example 1.

Amar and Brusly can do the work in 25 days. Brusly and Catherin can do the work in 35 days. Catherin and Amar can do the work in 30 days. In how many days, Amar, Brusly, and Catherin complete the work by working together?

Solution:

As per the details, Amar and Brusly can take time to complete the work = a = 25 days.
So, Amar and Brusly can complete the work in one day = \(\frac{1}{a}\) = \(\frac{1}{25}\)th part of the work.
Brusly and Catherin can take time to complete the work = b = 35 days.
So, Brusly and Catherin both can complete the work in 1 day = \(\frac{1}{b}\) = \(\frac{1}{35}\)th part of the work.
Catherin and Amar can take time to complete the work = c = 30 days.
Catherin and Amar can complete the work in 1 day = \(\frac{1}{c}\) = \(\frac{1}{30}\)th part of the work.
The work done by Amar, Brusly, and Catherin together in one day = \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\).
= \(\frac{1}{25}\) + \(\frac{1}{35}\) + \(\frac{1}{30}\).
L.C.M of 25, 35, and 30 is 1050.
= \(\frac{42 + 30 + 35}{1050}\) = \(\frac{107}{1050}\).
Total Work done by Amar, Brusly, and Catherin in \(\frac{107}{1050}\) days = 9.8 days.

Therefore, Total work is done by Amar, Brusly, and Catherin in 9.8 days.

Example 2.

A and B can make the toys in 14 days. B and C can make the same toys in 24 days. C and A can make the toys in 20 days. Calculate work done by each person individually?

Solution:

As per the given information, A and B can make the toys in 14 days. That is a = 14 days.
A and B can complete the work in 1 day = \(\frac{1}{a}\) = \(\frac{1}{14}\).
B and C can make the same amount of toys in 24 days. That is b = 24 days.
B and C can complete the work in one day = \(\frac{1}{b}\) = \(\frac{1}{24}\).
C and A can make the toys in 20 days. That is c = 20 days.
C and A can complete the work in 1 day = \(\frac{1}{c}\) = \(\frac{1}{20}\).
(A and B) + (B and C) + (C and A) one day’s work = \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) = \(\frac{1}{14}\) + \(\frac{1}{24}\) + \(\frac{1}{20}\).
L.C.M of 14, 24, 20 is 840.
=\(\frac{60 + 35 + 42}{840}\).
2(A + B + C)’s one day’s work = \(\frac{137}{840}\).
A, B, and C one day work =\(\frac{137}{2(840)}\)= \(\frac{137}{1680}\).
A, B, and C can complete the total work in = \(\frac{1680}{137}\) = 12 days.
Work done by ‘A’ alone = Work done by A, B, and C – Work done by B and C.
= 12 – 1/24
= \(\frac{288 – 1}{24}\) = \(\frac{287}{24}\)
Therefore, A alone can do the work in \(\frac{287}{24}\) days.
Work done by ‘B’ alone = Work done by A, B, and C – Work done by A and C.
= 12 – 1/20.
= \(\frac{240 – 1}{20}\).
= \(\frac{239}{20}\).
Therefore, B alone can complete the work in\(\frac{239}{20}\)days.
Work done by ‘C’ alone = Work done by A,B, and C – Work done by A and B.
= 12 – 1/14.
= \(\frac{168 – 1}{14}\).
= \(\frac{167}{14}\).

Therefore, C alone can complete the work in \(\frac{167}{14}\) days.

Example 3.

Ira and Mihira started work on the same project and they can finish the work in 20 days. Ira worked for 10 days and then Mihira completed the remaining work in 14 days. Find the time to complete the project by Mihira alone?

Solution:

As per the given details, Ira and Mihira both can complete the project in 20 days.
Ira and Mihira both can complete the project work in one day = \(\frac{1}{20}\)th part of the day.
Let us consider the work done by Ira in one day = p and Work done by Mihira in one day = q.
Work done by Ira and Mihira in one day = p + q = 1/20\(\frac{1}{20}\). —–(1)
Ira worked for 10 days and Mihira completed the remaining work in 14 days. That is
10p + 14q = 1 —-(2).
By comparing the equation (1) and (2), We will get Multiply the equation (1) with 10 on both sides. That is
10p + 10q = 10/20 = \(\frac{1}{2}\).
10p + 14q = 1
-4q = 1/2 – 1 = (1-2)/2 = –\(\frac{1}{2}\).
q= \(\frac{1}{8}\).
p + \(\frac{1}{8}\)= \(\frac{1}{20}\).
p=\(\frac{1}{20}\) – \(\frac{1}{8}\)
= \(\frac{2-5}{40}\)
= \(\frac{3}{40}\).
Therefore, Work done by Mihira in 8 days.

Example 4.

10 men and 10 women can complete the work in 12 days and 15 days respectively. In how many days 10 men and 10 women together can complete the work?

Solution:

As per the given information, 10 men can take time to complete the work = a = 12 days.
1 day work done by 10 men = \(\frac{1}{a}\) = \(\frac{1}{12}\).
10 women can finish the work in 15 days. That is b = 15 days.
Work done by 10 women in 1 day = \(\frac{1}{b}\) = \(\frac{1}{15}\).
10 men and 10 women together can work in one day = \(\frac{1}{12}\) + \(\frac{1}{15}\).
=\(\frac{5+4}{60}\) = \(\frac{9}{60}\).
Total work done by 10 men and 10 women together \(\frac{60}{9}\) = \(\frac{3}{20}\).

Therefore, total work is completed by 10 men and 10 women in \(\frac{3}{20}\)days.

Example 5.

In 35 days 30 men complete work by working together. After starting the work together. Every 10th day, 5 men left from the work. In how many days, the work will complete?

Solution:

As per the given information, Total work is completed in 35 × 30 = 1050 days.
1st 10 days = Total number of days are 10: the number of days × number of men working = 10 × 30 = 300 days.
2nd 10 days = Total number of days are 20: the number of days × number of men working = 10 × 25 = 250 days.
So, total work completed 300 + 250 = 550 days.
3rd 10 days = Total number of days are 30:number of days × number of men working = 10 × 20 = 200 days.
Total work completed in 550 + 200 = 750 days.
4th 10 days = Total number of days are 40: number of days × number of men working = 10 × 15 = 150 days.
Total number of days = 750 + 150 = 900 days
5th 10 days = Total number of days are 50: number of days × number of men working = 10 × 10 = 100.
Total number of days = 900 + 100 = 1000.
6th 10 days = Total number of days is 60: number of days × number of men working = 10 × 5 = 50.
Total number of days = 1000 + 50 = 1050.

Therefore, the total work is completed in 60 days.

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