420+ Solved Time and Work Problems with Solutions and Explanation

Q:

If 2 men and 3 women can do a piece of work in 8 days and 3 men and 2 women in 7 days. In how many days can the work be done by 5 men and 4 women working together?

A) 3 days	B) 6 days
C) 4 days	D) 2 days

Answer & Explanation Answer: C) 4 days

Explanation:

From the given data,

=> (2 M + 3W) 8 = (3M + 2W)7

=> 16M + 24W = 21M + 14 W

=> 10W = 5M

=> 2W = M

=> 14W × ? = 7W × 8

? = 4 days

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29 16037

Q:

A can do a piece of work in 30 days. He works at it for 6 days and then B finishes it in 18 days. In what time can A and B together it ?

A) 14 1/2 days	B) 11 days
C) 13 1/4 days	D) 12 6/7 days

Answer & Explanation Answer: D) 12 6/7 days

Explanation:

Let 'B' alone can do the work in 'x' days

6/30 + 18/x = 1

=> x = 22.5

1/30 + 1/22.5 = 7/90

=> 90/7 = 12 6/7 days

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29 17760

Q:

Two pipes A and B can fill a cistern in 4 minutes and 6 minutes respectively . If these pipes are turned on alternately for 1 minute each how long will it take to the cistern to fill?

Answer

As the pipes are operating alternatively, thus their 2 minutes job is = $\frac{1}{4} + \frac{1}{6} = \frac{5}{12}$

In the next 2 minutes the pipes can fill another $\frac{5}{12}$ part of cistern.

Therefore, In 4 minutes the two pipes which are operating alternatively will fill $\frac{5}{12} + \frac{5}{12} = \frac{5}{6}$ Remaining part = $1 - \frac{5}{6} = \frac{1}{6}$

Pipe A can fill $\frac{1}{4}$ of the cistern in 1 minute

Pipe A can fill $\frac{1}{6}$ of the cistern in = $\frac{2}{3}$ min

Therefore, Total time taken to fill the Cistern

4 + $\frac{2}{3}$ minutes.

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28 13388

Q:

16 women can do a piece of work in 10 days while 15 men can do the same work in 12 days. All men and all women work alternately starting with all men, then find the time taken by them to complete the whole work?

A) 10 5/2 days	B) 11 days
C) 11 2/5 days	D) 12 1/2 days

Answer & Explanation Answer: A) 10 5/2 days

Explanation:

Let efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively

15×12×m = 10×16×w

⇒ m/w = 8/9

Total work = 15 × 12 × 9 = 1620 units

In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units

So, in 10 days work done will be = 263 × 5 = 1315 units

Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days

Total days = 10 5/2 days.

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27 5057

Q:

9 men and 12 boys finish a job in 12 days, 12 men and 12 boys finish it in 10 days. 10 men and 10 boys shall finish it in how many days ?

A) 15 days	B) 11 days
C) 14 days	D) 12 days

Answer & Explanation Answer: D) 12 days

Explanation:

9M + 12B ----- 12 days ...........(1)
12M + 12B ------- 10 days........(2)
10M + 10B -------?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B............(3)

From (1) & (3)

18B + 12B = 30B ---- 12 days

20B + 10B = 30B -----? => 12 days.

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22 19786

Q:

P and Q can complete a job in 24 days working together. P alone can complete it in 32 days. Both of them worked together for 8 days and then P left. The number of days Q will take to complete the remaining work is ?

A) 56 days	B) 54 days
C) 60 days	D) 64 days

Answer & Explanation Answer: D) 64 days

Explanation:

(P+Q)'s 1 day work = 1/24

P's 1 day work = 1/32

=> Q's 1 day work = 1/24 - 1/32 = 1/96

Work done by (P+Q) in 8 days = 8/24 = 1/3

Remainining work = 1 - 1/3 = 2/3

Time taken by Q to complete the remaining work = 2/3 x 96 = 64 days.

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22 16833

Q:

6 men can complete a piece of work in 12 days. 8 women can complete the same piece of work in 18 days whereas 18 children can complete the piece of work in 10 days. 4 men, 12 women and 20 children work together for 2 days. If only men were to complete the remaining work in 1 day how many men would be required totally?

A) 38	B) 72
C) 36	D) 76

Answer & Explanation Answer: C) 36

Explanation:

Given 4men, 12 women and 20 children work for 2 days.

Workdone for 2 days by 4men, 12 women and 20 children = $\frac{4 x 2}{6 x 12} + \frac{12 x 2}{8 x 18} + \frac{20 x 2}{18 x 10} = \frac{1}{2}$

Therefore, remaining work = 1 - $\frac{1}{2}$ = $\frac{1}{2}$

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

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20 12820

Q:

Four pipes P,Q, R and S can fill a cistern in 20,25, 40 and 50 hours respectively.The first pipe P was opened at 6:00 am, Q at 8:00 am, R at 9:00 am and S at 10:00 am. when will the Cistern be full?

A) 4:18 pm	B) 3:09 pm
C) 12:15 pm	D) 11:09 am

Answer & Explanation Answer: B) 3:09 pm

Explanation:

Efficiency of P= 100/20= 5% per hour

Efficiency of Q= 100/25= 4% per hour

Efficiency of R= 100/40= 2.5% per hour

Efficiency of S=100/50= 2% per hour

Cistern filled till 10 am by P, Q and R

$\begin{matrix} Till 10.00 am Pipe P filled 20 % \\ Till 10.00 am Pipe Q filled 8 % \\ Till 10.00 am Pipe R filled 2.5 % \end{matrix}\} 30.5 %$

Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern.

Rest of cistern to be filled = 100 - 30.5 = 69.5%

Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern

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