420+ Solved Time and Work Problems with Solutions and Explanation

Q:

4 men can repair a road in 7 hours. How many men are required to repair the road in 2 hours ?

A) 17 men	B) 14 men
C) 13 men	D) 16 men

Answer & Explanation Answer: B) 14 men

Explanation:

M x T / W = Constant
where, M= Men (no. of men)
T= Time taken
W= Work load
So, here we apply
M1 x T1/ W1 = M2 x T2 / W2
Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?
Note that here, W1 = W2 = 1 road, ie. equal work load.
Clearly, substituting in the above equation we get, M2 = 14 men.

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Q:

10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?

A) 215 days	B) 225 days
C) 235 days	D) 240 days

Answer & Explanation Answer: B) 225 days

Explanation:

Given that

(10M + 15W) x 6 days = 1M x 100 days

=> 60M + 90W = 100M

=> 40M = 90W

=> 4M = 9W.

From the given data,

1M can do the work in 100 days

=> 4M can do the same work in 100/4= 25 days.

=> 9W can do the same work in 25 days.

=> 1W can do the same work in 25 x 9 = 225 days.

Hence, 1 woman can do the same work in 225 days.

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Q:

A works twice as fast as B.If B can complete a work in 18 days independently,the number of days in which A and B can together finish the work is:

A) 4 days	B) 6 days
C) 8 days	D) 10 days

Answer & Explanation Answer: B) 6 days

Explanation:

Ratio of rates of working of A and B =2:1. So, ratio of times taken =1:2

Therefore, A's 1 day's work=1/9

B's 1 day's work=1/18

(A+B)'s 1 day's work= 1/9 + 1/18 = 1/6

so, A and B together can finish the work in 6 days

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Q:

A and B can do a piece of work in 30 days , while B and C can do the same work in 24 days and C and A in 20 days . They all work together for 10 days when B and C leave. How many days more will A take to finish the work?

A) 18 days	B) 24 days
C) 30 days	D) 36 days

Answer & Explanation Answer: A) 18 days

Explanation:

2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8

=>(A+B+C)'s 1 day's work= 1/16

work done by A,B and C in 10 days=10/16 = 5/8

Remaining work= 3/8

A's 1 day's work= $(\frac{1}{16} - \frac{1}{24}) = \frac{1}{48}$

Now, 1/48 work is done by A in 1 day.

So, 3/8 work wil be done by A in =48 x (3/8) = 18 days

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138 32727

Q:

(x-2) men can do a piece of work in x days and (x+7) men can do 75% of the same work in (x-10)days. Then in how many days can (x+10) men finish the work?

A) 27 days	B) 12 days
C) 25 days	D) 18 days

Answer & Explanation Answer: B) 12 days

Explanation:

$\frac{3}{4} \times (x - 2) x = (x + 7) (x - 10)$

$\Rightarrow x^{2} - 6 x - 280 = 0$

=> x= 20 and x=-14

so, the acceptable values is x=20

Therefore, Total work =(x-2)x = 18 x 20 =360 unit

Now 360 = 30 x k

=> k=12 days

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Q:

The ratio of efficiency of A is to C is 5:3. The ratio of number of days taken by B is to C is 2:3. A takes 6 days less than C, when A and C completes the work individually. B and C started the work and left after 2 days. The number of days taken by A to finish the remaining work is:

A) 4.5	B) 5
C) 6	D) 9 1/3

Answer & Explanation Answer: C) 6

Explanation:

A : C

Efficiency 5 : 3

No of days 3x : 5x

Given that, 5x-6 =3x => x = 3

Number of days taken by A = 9

Number of days taken by C = 15

B : C

Days 2 : 3

Therefore, Number of days taken by B = 10

Work done by B and C in initial 2 days = $2 [\frac{1}{10} + \frac{1}{15}]$ = 1/3

Thus, Rest work =2/3

Number of days required by A to finish 2/3 work = (2/3) x 9 = 6 days

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81 30486

Q:

A Contractor employed a certain number of workers to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realised that the work would get delayed by three-fourth of the scheduled time, so he at once doubled the no of workers and thus he managed to finish the road on the scheduled time. How much work he had been completed, before increasing the number of workers?

A) 10 %	B) 14 ( 2/7 )%
C) 20 %	D) Can't be determined

Answer & Explanation Answer: B) 14 ( 2/7 )%

Explanation:

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.

D * x +(100- D) * 2x= 175x

=> D= 25 days

Now , the work done in 25 days = 25x

Total work = 175x

Therefore, workdone before increasing the no of workers = $\frac{25 x}{175 x} \times 100$ % = $14 \frac{2}{7} %$

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Q:

A and B can do a piece of work in 40 and 50 days. If they work at it an alternate days with A beginning in how many days, the work will be finished ?

Answer

(A+B)'s two days work = $\frac{1}{40} + \frac{1}{50} = \frac{9}{200}$

Evidently, the work done by A and B duing 22 pairs of days

i.e in 44 days = $22 \times \frac{9}{200} = \frac{198}{200}$

Remaining work = $1 - \frac{198}{200}$ = 1/100

Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in $40 \times \frac{1}{100} = \frac{2}{5}$

Therefore, Total time taken = 44 $\frac{2}{5}$ daya

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