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Here W is the work. For e.g., if 5 men are cutting 10 trees in 2 days, working 4 hours per day. Then,

M = 5, D = 2, H = 4 and W = 10.

**1)**

H1 = 6

D1 = 18

D2 = 12

H2 = ?

We know, H1 * D1 = H2 * D2

6 * 18 = H2 * 12

H2 = 9 hours**Answer : (B)**

**2)**

M1 = 15

D1 = 20

H1 = 8

M2 = 20

D2 = 12

H2 = ?

We know, M1 * D1 * H1 = M2 * D2 * H2

15*20*8 = 20*12*H2

H2 = 10 hours

**Answer : (B)**

**3)**

In this question, Taps = Men

Number of taps required = 20*9/15 = 12

**Answer : (B)**

**4)**

Let there be X number of men.

X men can finish a piece of work in 100 days. Hence total work = 100X

If there were (X – 10) men, it would have taken 110 days to finish the work. Total work in this case = 110(X – 10)

Total work remains the same. Hence,

100X = 110(X – 10)

X = 110**Answer : (D)**

**5)**

Efficiency of Subhash = 50/10 = 5 per hour

Efficiency of Subhash and Prakash = 300/40 = 7.5 per hour

Efficiency of Prakash = (Efficiency of Subhash and Prakash) – (Efficiency of Subhash) = 7.5 – 5 = 2.5

So Prakash can copy 2.5 pages per hour. To copy 30 pages, he would require 30/2.5 or 12 hours.**Answer : (D)**

**6)**

40 men can finish a work in 60 days. Hence, total work = 40*60 = 2400

Let the 10 men left after X days.

For X days, all the 40 men worked. Total work performed = 40X

Now when 10 men quit, only 30 men were left to do the work and they took (70 – X) more days to finish it.

Total work done by 30 men = 30*(70 – X)

Now, 40X + 30*(70 – X) = 2400

X = 30 days

**Answer : (C)**

**7)**

Let the total work = 360 units

Efficiency of A = 360/45 = 8 units

Efficiency of B = 360/40 = 9 units

Efficiency of A + B = 17 units

Let A left after X days.

For X days, both A and B worked. Hence work performed = 17X

B worked for 23 days. Hence work performed by B = 23*9 = 207 units

Now, 17X + 207 = 360

X = 9 days

**Answer : (D)**

**8)**

B and C together do 8/23 of the work, hence A does (1 – 8/23) or 15/23 of the work.

A should be paid = 15*5290/23 = Rs. 3450

**Answer : (D)**

**Note :** In this question, they have asked the wages of A. Had they asked the wages of B, firstly you would have calculated the work performed by B with the formula-

Work done by B = (Portion of work done by A and B) + (Portion of work done by B and C) – 1

Work done by B = 19/23 + 8/23 – 1 = 4/23

Wages of B = 4*5290/23 = Rs. 920

**9)**

This is a very famous question. A company employed 200 workers to complete a certain work in 150 days. Here the total work is not 200*150 because 200 workers and 150 days was only a plan. In reality, only 1/4th of the work has been done in 50 days. So if they go with the same pace, 200 workers will take 200 days to complete the work.

So total work = 200*200 units

200 workers have worked for 50 days. Hence they have finished 200*50 units of work.

Remaining work = 200*200 – 200*50 = 200*150

Let the number of additional workers required = X.

Now (200+X) workers will work for 100 days to finish the work as per the schedule.

Work they need to perform = (200 + X)*100

Now, (200 + X)*100 = 200*150

X = 100

**Answer : (C)**

**Q. 10) A contractor undertook to finish a certain work in 124 days and employed 120 men.After 64 days,he found that he had already done 2/3 of work. How many men can be discharged now so that the work may finish in time?**

**A) 56 B) 44 C) 50 D) 60**

120 workers finish 2/3 of the work in 64 days. So to complete the whole work, workers will take 64*3/2 or 96 days.

Total work to be performed = 120*96

Now the workers have already finished 2/3 of the work and only 1/3 work has to be performed.

Remaining work = 120*96/3 = 120*32

Let the contractor discharges X men. Remaining workers = 120-X. These workers will continue the work for (124-64) or 60 days. Hence,

120*32 = (120-X)*60

X = 56**Answer : (A)**

__Method 2__

M1 = 120, D1 = 64, W1 = 2/3

M2 = 120 – x, D2 = 60, W2 = 1/3

(M1*D1)/W1 = (M2*D2)/W2

120*64*3/2 = (120 – x)*60*3

x = 56

**11)**

Let the second pipe fills the pool in X hours. Then first pipe takes (X+5) hours and the third pipe takes (X-4) hours to fill the pool. Now, 1st and 2nd pipe together take the same time to the fill the pool as the 3rd pipe alone. Hence,

1/(X+5) + 1/(X) = 1/(X – 4)

Solve this quadratic equation and you will get X = 10 hours

That means second pipe takes 10 hours to fill the pool while the third pipe takes 6 hours. Together they will take 10*6/(10 + 6) hours to fill the pool.

**Answer: (B)**

**I have covered almost all the possible questions from Time and Work that are asked by SSC. If you have any doubt in this topic, please drop a comment.**

**Keep Reading 🙂**

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