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- Most Important Multiple Choice Questions
- Online Time and Work Exercise with Correct Answer Key and Solutions
- Useful for all Competitive Exams

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- Question 1 of 18
##### 1. Question

After working for 8 days, Anil finds that only ⅓ of the work has been done. He employs Rakesh who is 60 % efficient as Anil. How many more days will Anil take to complete the job?

##### Hint

In 8 days, Anil does = ⅓rd work.

∴ in 1 day, he does = ¹⁄₂₄th work

∴ Rakesh’s one day’s work

= 60% of ¹⁄₂₄ = ¹⁄₄₀ th work.

Remaining work

(Anil and Rakesh)’s one day’s work

=

= ¹⁄₁₅th work

Now, ¹⁄₁₅th work is done by them in one day.

∴ ⅔rd work is done by them in 15 × ⅔ = 10 days

- Question 2 of 18
##### 2. Question

A sum of Rs. 25 was paid for a work which A can do in 32 days, B in 20 days, B and C in 12 days and D in 24 days. How much did C receive if all the four work together?

##### Hint

A’s one day’s work = ¹⁄₃₂

B’s one day’s work = ¹⁄₂₀

(B + C)’s one day’s work =¹⁄₁₂

∴ C’s one day’s work

=

D’s one day’s work = ¹⁄₂₄

∴ (A + B + C + D)’s one day’s work

=

=

∴ Out of ⁵⁄₃₂ of work done, ¹⁄₃₀ of the work is done by C.

⇒ Out of Rs. 25 paid for the work, C will receive

Rs.

i.e.

i.e. Rs.¹⁶⁄₃

- Question 3 of 18
##### 3. Question

A and B can do a job in 15 days and 10 days, respectively. They began the work together but A leaves after some days and B finished the remaining job in 5 days. After how many days did A leave?

##### Hint

A’s one day’s work = ¹⁄₁₅th work.

B’s one day’s work = ¹⁄₁₀th work.

(A + B)’s one day’s work

work.

Let A left after x days.

∴ (A +B)’s x days’ work = ¹⁄₆th work.

Remaining work

work.

Now, in 5 days, work done by

.

∴ In 1 day work done by B

and

∴ x = 3 days

- Question 4 of 18
##### 4. Question

Mr. Suresh is on tour and he has Rs 360 for his expenses. If he exceeds his tour by 4 days he must cut down daily expenses by Rs 3. The number of days of Mr. Suresh’s tour programme is :

##### Hint

Let Suresh undertakes a tour of x days.

Then, expenses for each day = 360/x

Now,

⇒ = 3

⇒ x² + 4x – 480 = 0

⇒ x = – 24 or x = 20

Since, x ≠ -24 we have x = 20

- Question 5 of 18
##### 5. Question

A can do a job in 3 days less time than B. A works at it alone for 4 days and then B takes over and completes it. If altogether 14 days were required to finish the job, then in how many days would each of them take alone to finish it?

##### Hint

Let B can finish the work in x days.

Then A can finish the work in (x – 3) days.

B’s one day’s work

A’s one day’s work work

A’s 4 days’ work =

Remaining work

The remaining work done by B in 14 – 4

= 10 days.

Now, in 10 days, work done by

th work

∴ In 1 day, work done by B

= work

and

⇒ x = 15 days

∴ B will finish in 15 days and A will finish in 12 days

- Question 6 of 18
##### 6. Question

Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked ⅓ as efficiently as he actually did, the work would have completed in 3 days. Find the time for A to complete the job alone.

##### Hint

(A + B)’s one day’s work = ⅕th work

Let A can do job in x days. Then,

A’s one day’s work

and B’s one day’s work

Now ,

- Question 7 of 18
##### 7. Question

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

##### Hint

Let 1 man’s 1 day’s work = x and

1 boy’s 1 day’s work = y.

Then, 6x + 8y = ¹⁄₁₀

and 26x + 48y = ½

Solving these two equations, we get :

∴ (15 men + 20 boys)’s 1 day’s work

=

∴ 15 men and 20 boys can do the work in 4 days.

- Question 8 of 18
##### 8. Question

39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?

##### Hint

Let the required number of days be x.

Less persons, More days (Indirect Proportion)

More working hrs per day, Less days (Indirect Proportion)

- Question 9 of 18
##### 9. Question

A certain number of persons can dig a trench 100 m long, 50 m broad and 10 m deep in 10 days. The same number of persons can dig another trench 20 m broad and 15 m deep in 30 days. The length of the second trench is :

##### Hint

Let the required length be x metres.

More breadth, Less length (Indirect Proportion)

More depth, Less length (Indirect Proportion)

More days, More length (Direct Proportion)

∴ 20 × 15 × 10 × x = 50 × 10 × 30 × 100

- Question 10 of 18
##### 10. Question

If 8 men or 17 boys can do a piece of work in 26 days, how many days will it take for 4 men and 24 boys to do a piece of work 50 × 0.9 times as great?

##### Hint

Let the required number of days be x.

8 men = 17 boys

⇒ 4 men

∴ 4 men and 24 boys

Now, More boys , less days (Indirect Proportion)

But work → 50 × 0.9 times

∴ Required days

- Question 11 of 18
##### 11. Question

If 18 men working 5 hours a day for 8 days can complete a job, how many men working 8 hours a day for 6 days will be needed?

##### Hint

Let the required men be x.

More hours, less men (Indirect proportion)

More days, less men (Indirect proportion)

∴ 5 × 8 × 18 = 8 × 6 × x

- Question 12 of 18
##### 12. Question

4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats would be woven by 8 mat-weavers in 8 days?

##### Hint

Let the required number of mats be x.

More weavers, More mats (Direct Proportion)

More days, More mats (Direct Proportion)

- Question 13 of 18
##### 13. Question

In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?

##### Hint

Let the required number of days be x.

Less cows, More days (Indirect Proportion)

Less bags, Less days (Direct Proportion)

∴ 1 × 40 × x = 40 × 1 × 40

⇒ x = 40

- Question 14 of 18
##### 14. Question

Two coal loading machines each working 12 hours per day for 8 days handles 9,000 tonnes of coal with an efficiency of 90%. While 3 other coal loading machines at an efficiency of 80% set to handle 12,000 tonnes of coal in 6 days. Find how many hours per day each should work.

##### Hint

More machines, less hours (Indirect Proportion)

Less days, more hours (Indirect Proportion)

More amount of coal, more hours (Direct Proportion)

Less efficiency, more hours (Indirect Proportion)

⇒ 3 × 6 × 9,000 × 0.8 × x

= 2 × 8 × 12,000 × 0.9× 12

⇒ x = 16 hrs

- Question 15 of 18
##### 15. Question

A man and a boy together can do a certain amount of digging in 40 days. Their skills in digging are in the ratio of 8 : 5. How many days will the boy take, if engaged alone.

##### Hint

Let M denotes man and B denotes boy.

(M + B)’s 1 day’s work = ¹⁄₄₀

i.e.

Ratio of their skill = ⁸⁄₅

i.e.

Let efficiency of a man of 1 days work = x

i.e.

Now,

Now,

⇒ M = 65 and

- Question 16 of 18
##### 16. Question

A contractor undertakes to do a piece of work in 40 days. He engages 100 men and after 35 days, he engaged an additional 100 men and completes the work. How many days behind the schedule would the work have been, if he had not engaged the additional men?

##### Hint

(100 × 35+ 200 × 5) men can finish the work in 1 day.

i.e., 4500 men can finish the work in 1 day

∴ 100 men can finish the work in 45 days

∴ The work would be 5 days behind the schedule.

- Question 17 of 18
##### 17. Question

A contract is to be completed in 56 days and 104 men were set to work, each working 8 hours a day. After 30 days, ⅖ of the work is finished. How many additional men may be employed so that work may be completed on time, each man now working 9 hours per day?

##### Hint

Remaining work

Remaining time = 56 – 30 = 26 days

More work, more men (Direct Proportion)

Less days, more men (Indirect Proportion)

More hours, Less men (Indirect Proportion)

⇒ x = 160

∴ Additional men to be employed = 160 – 104 = 56 men

- Question 18 of 18
##### 18. Question

A tyre has two punctures. The first puncture along would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat?

##### Hint

1 minute’s work of both the punctures

=

So, both the punctures will make the tyre flat in

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