420+ Solved Time and Work Problems with Solutions and Explanation

Q:

Amit can do a piece of work in 45 days, but Bharath can do the same work in 5 days less, than Amit, when working alone. Amit and Bharath both started the work together but Bharath left after some days and Amit finished the remaining work in 56 days with half of his efficiency but he did the work with Bharath with his complete efficiency. For how many days they had worked together?

A) 6	B) 8
C) 9	D) 12

Answer & Explanation Answer: B) 8

Explanation:

$\begin{matrix} A m i t & B h a r a t \\ N o . o f D a y s & 45 & 40 \\ E f f i c i e n c y & 2.22 P e r c e n t (= \frac{1}{45}) & 2.5 P e r c e n t (= \frac{1}{40}) \end{matrix}$

Amit did the work in 56 days = $56 \times \frac{1}{45 \times 2} = \frac{28}{45}$

Therefore, Rest work 17/45 was done by Amit and Bharath = $\frac{\frac{17}{45}}{\frac{17}{360}}$ = 8 days

( since Amit and Bharath do the work in one day = $\frac{1}{45} + \frac{1}{40} = \frac{17}{360}$ )

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30 15580

Q:

Five men and nine women can do a piece of work in 10 days. Six men and twelve women can do the same work in 8 days. In how many days can three men and three women do the work ?

A) 18 days	B) 20 days
C) 16 days	D) 14 days

Answer & Explanation Answer: B) 20 days

Explanation:

(5m + 9w)10 = (6m + 12w)8

=> 50m + 90w = 48w + 96 w => 1m = 3w

5m + 9w = 5m + 3m = 8m

8 men can do the work in 10 days.

3m +3w = 3m + 1w = 4m

So, 4 men can do the work in (10 x 8)/4 = 20 days.

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9 15275

Q:

A is 3 times faster than B. If A can finish a work 32 days less than that of B, find the number of days need to finish the work if both are working together?

A) 12 days	B) 24 days
C) 32 days	D) 16 days

Answer & Explanation Answer: A) 12 days

Explanation:

It is given that efficiency ratio =3:1

so time ratio will be 1:3 (since work is same)

Also given that time diff = 32 days. ratio difference = 3-1 =2

2 ratio = 32 days

1 ratio = 16 days.

So A will alone finish it in 16 days and B will finish it in 16*3 = 48 days.

Total work = LCM of 16 and 48 = 48.

Total time = Total work/Total efficiency

ie; 48/4= 12 days.

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8 15206

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 14082

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 13957

Q:

A garrison of 2000 men has provisions for 54 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement ?

A) 1900	B) 2100
C) 1700	D) 2000

Answer & Explanation Answer: A) 1900

Explanation:

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

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Q:

Kaushalya can do a work in 20 days, while kaikeyi can do the same work in 25 days. They started the work jointly.Few days later Sumitra also joined them and thus all of them completed the whole work in 10 days. All of them were paid total Rs.700. What is the Share of Sumitra?

A) Rs.130	B) Rs.185
C) Rs.70	D) can't be determined

Answer & Explanation Answer: C) Rs.70

Explanation:

Efficiency of kaushalya = 5%

Efficiency of kaikeyi = 4%

Thus, in 10 days working together they will complete only 90% of the work.

[(5+4)*10] =90

Hence, the remaining work will surely done by sumitra, which is 10%.

Thus, sumitra will get 10% of Rs. 700, which is Rs.70

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Q:

12 men complete a work in 9 days. After they have worked for 6 days, 4 more men join them. How many days will they take to complete the remaining work ?

A) 2 days	B) 2.5 days
C) 2.25 days	D) 3 days

Answer & Explanation Answer: C) 2.25 days

Explanation:

1 man's 1 day work = 1/108
12 men's 6 day's work = 1/9 x 6 = 2/3

Remaining work = 1 - 2/3 = 1/3

16 men's 1 day work = 1/108 x 16 = 4/27
4/27 work is done by them in 1 day.

1/3 work is done by them in 27/4 x 1/3 = 9/4 days.

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