**Question 1: ** Ram and Shyam undertake a piece of work for Rs 300. Ram can do it in 20 days and Shaym can do it in 60 days. With the help of Radha, they finish it in 10 days. How much should Radha be paid for her contribution?

**Solution: ** Ram alone takes 20 days and Shyam alone takes 60 days to finish the work.

All together can finish in 10 days.

Let the total work done is LCM(10, 20, 60) = 60

Ram’s efficiency = 60/20 = 3

Shyam’s efficiency = 60/60 = 1

Combined Efficiency = 60/10= 6

Radha’s efficiency = 6 – 3 -1 = 2

Radha contributed 1/3 of the total work done.

So, she should be paid 1/3 of the total amount = 300/3 = **Rs 100**

**Question 2: ** A can do a piece of work in 30 days. He works only for 5 days and left, then B finishes the remaining work in 15 more days. In how many days will A and B together finish the work?

**Solution: ** 25 days work of A completed by B in 15 days.

25A=15B

A/B=3/5

A’s efficiency = 3

B’s efficiency = 5

Total work done= 3 * 30 = 90

Work done by A and B together=total work done/total efficiency

= 90/8

= **11.25 days**

**Question 3: **Twenty five employees can finish a project in 40 days.After how many days should 10 employees leave the job so that the project is finished in 50 days.

**Solution: ** Let n be the number of days when 10 employees left the project.

Total work done= 25*40= 1000

25x + (50 – x)15 = 1000

25x + 750 – 15x = 1000

10x = 250

x = 25

Hence, after **25 days **10 employees left the job.

**Question 4: ** A can type 85 pages in 10 hours. A and B together can type 500 pages in 40 hours.How much time B will take to type 80 pages.

**Solution: ** A can type 85 pages in 10 hours

Then, A can type 340 pages in 40 hours.

A and B together can type 500 pages in 40 hours

So, B can type number of pages= 500 – 340 = 160 pages in 40 hours.

Hence, B can type 80 pages in **20 hours**.

**Question 5: ** A, B and C can do a piece of work in 10, 12 and 15 days respectively.They all start the work together but A leaves after the 2 days of work and B leaves 3 days before the work is completed.Find the number of days the work completed.

**Solution: ** Total work done is LCM(10, 12, 15)=60 unit

A’s efficiency = 60/10= 6

B’s efficiency = 60/12= 5

C’s efficiency = 60/15= 4

First two days all work together

So, the work completed in first two days= 15 x 2 = 30 unit

Remaining work= 60 – 30 = 30 unit

If B completes 3 day work also = 3 x 5 = 15 unit

Total work remaining= 30 + 15 = 45 unit

Number of days B and C works= 45/9=5

Total number of days to complete the work = 2 + 5 = **7 days.**

**Question 6: ** A can do 1/6 work in 4 days and B can do 1/5 of the same work in 6 days. In how many days both will finish the work?

**Solution: ** A can complete the work in 6*4 = 24 days

B can complete the work in 5*6 = 30 days

Total work done LCM(24, 30)= 120 unit

A’s efficiency = 120/24= 5

B’s efficiency = 120/30= 4

Total time taken = total work done/ total efficiency

= 120/9

=** 40/3 days**

**Question 7: ** A company undertake a project to build 2000 m long bridge in 400 days and hire 50 men for the project. After 100 days, he finds only 400 m of bridge has been completed.Find the (approx )number of extra men he hire to complete the project on time.

**Solution: ** Use here ** M _{1} D_{1} / W_{1} = M_{2} D_{2} / W_{2}**

50 x 100/ 400 = [(50 + x) 300]/ 1600

4 x 5000 = 15000 + 300x

20000 – 15000 = 300x

3x = 50

x= 16.66

x =

**17 men**to hire to complete the project on time.

**Question 8: ** In a hostel mess there was sufficient food for 400 students for 31 days of a month. After 26 days 150 students go to their home. For how many extra days will the rest of food last for the remaining students.

**Solution: **After 26 days food left in mess= 5 * 400

Students remaining in hostel mess =400 – 150 = 250

Let x be the extra days.

5 * 400 = (5 + x ) * 250

2000 = 1250 + 250x

250x = 750

x = 3 days

Hence, food will last for **3 extra days**.

**Question 9: ** Two candles A and B of same height can burn completely in 6 hours and 8 hours respectively. If both start at same time with their respective constant speed, then calculate after how much time the ratio of their height will become 3:4.

**Solution: ** Total time is LCM (6, 8) =24

A’s efficiency = 24/6 = 4

B’s efficiency = 24/8= 3

After x time height will become 3:4

So, (24 – 4x) /( 24 – 3x) = 3/4

96 – 16x = 72 – 9x

7x = 24

x = 24/7 = **3.42 hours**

**Question 10: ** A and B men can build a wall in 20 and 30 hours respectively but if they work together they use 220 less bricks per hour and build the wall in 15 hours. Find the number of bricks in the wall.

**Solution: ** Work done is LCM(20, 30) = 60

A’s efficiency= 60/20 = 3

B’s efficiency= 60/30 = 2

If they work together their efficiency will be 5.

Working together their efficiency 60/15 = 4

Efficiency is less than by 5 – 4 = 1

1 -> 220 less bricks

60 -> 220 x 60 = **13200 bricks** in the wall.

**Question 11: ** A + B and B + C can do a work in 12 days and 15 days respectively. If A work for 4 days and B work for 7 days then C complete the remain work in next 10 days. Then calculate in how much time C would complete the whole work?

**Solution: ** Total work done is LCM(12, 15)=60

A+B ‘s efficiency = 60/12 = 5

B+C ‘s efficiency = 60/15 = 4

A work for 4 days, B work for 7 days then A and B work together for 4 days.

B work for 7 days, C work for 10 days then B and C work together for 3 days.

A+B B+C C 4 days 3 days 7 days |_{x5}|_{x4}| 20 12 60-20-12=28

C’s efficiency = 28/7 = 4

Hence, C can complete the work in 60/4 = **15 days**.

**Question 12: ** In a company there are three shifts for a day during the three shifts the average working efficiency of the workers is 80%, 70% and 50% respectively. A work is complete in 60 days by the group working in the first shift only. If the work is done in all the three shift per day then how many less days are required to complete the work.

**Solution: **

Shifts I II III 80% 70% 50% 8 7 5

Total work done obtain by using I shift only = 60 x 8 = 480 unit

Total efficiency = 8 + 7 + 5 = 20

Total days required = 480/20 = 24 days

Total less required days = 60 – 24 = **36 days**

**Question 13: ** In a factory same number of women and children are present. Women works for 6 hours in a day and children work 4 hours in a day.In festival season workload increases by 60% and government does not allow children to work more than 6 hours per day.If their efficiency are equal and remain work is done by women then how many extra hours/day increased by women?

S**olution.** Shortcut

Let they earn 1 Rs/hr.

Woman Child Earns 6 + 4 = 10 | | |_{60%}__ max 6 = 16

Workload increases by 60% from 10 to 16.

Children can work maximum 6 hours

Then women work per day 16 – 6 = 10

So, it increases by** 4 hours/day** extra.

**Question 14: ** A start a work and left after 2 days and remaining work is done by B in 9 days. If A left after 3 days then B complete the remaining work in 6 days. Then in how many days A and B can complete the work individually.

**Solution.**

Work done is equal in both cases.

2A + 9B = 3A + 6B

A=3B

A/B=3/1

Total work done =3 x 2 + 9 x 1

= 15

A alone take 15/3 = **5 days**.

B alone take 15/1 = **15 days.**

**Question 15: ** A and B can complete a work together in 30 days. They start work together and after 23 days B left the work and the whole work is completed in 33 days. Find the work completed by A alone.

**Solution: ** A’s completed the 7 days work of B in 3 days.

3A = 7B

A’s efficiency = 7

B’s efficiency = 3

Total work done = ( 7 + 3 ) * 30 = 300 unit

A alone can complete in = **300/7 days.**

**Question 16: ** A alone would take 64 hours more to complete a work then A + B work together. B take 4 hours more to complete a work alone than A and B work together.Find in how much time A alone complete the work.

**Solution: ** **First method**

Let A and B take x hours to complete a work together.

A alone would take (x + 64) and B alone would take (x + 4)hours to complete the work.

A( x + 64) = x (A + B)

64A =x B …………(1)

B(x + 4)= x(A + B)

4B = x A……………(2)

**from (1)and (2)**

64A = x * x A/4

x^{2} = 256

x = 16

A alone = 16 + 64 = **60 hours**

**Shortcut method –**

x^{2} = more of A * more of B

x^{2}= 64 * 4

x^{2}= 256

x= 16

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