Time and Work have good scope in govt exams. Here we learn to solve the tricky questions based on time & work. The explanations are given for the practice questions.

1).** A and B can complete a job in the 24 days working together. A alone can complete it in 32 days. Both of them worked together for 8 days and then A left. The number of days B will take to complete the remaining job is ?**

**Solution: **

Let B will take N days to complete the remaining job: so

According to the question,

*1*A + *1*B = *1*24

and *1*A = *1*32

∴ *1*B = *1*24 – *1*32 = *1*96

⇒ B = 96 days

According to the question,

8[*1*A + *1*B] + N x *1*B = 1

⇒ 8 x *1*24 + *N*96 = 1

⇒ *1*3 + *N*96 = 1

⇒ *N*96 = 1 – *1*3

∴ N = (2 x 96)/3 = 64

Hence, B complete the remaining job in 64 days

2). **A and B can do a piece of work in 9 days and 18 days, respectively. As they were ill, they could do 90% and 72% of their efficiency, respectively. How many days will they take to complete the work together?**

**Solution:**

A 1 day’s work = 90% of *1*9 = *1*10

B 1 day’s work = 72% of *1*18 = *1*25

∴ (A + B)’s 1 day’s work

= *1*10 + *1*25

= *(5 + 2)*50

= *7*50

Hence, time taken by them to complete the work = 50/7 days

**3). A, Band C can do a piece of work individually in 8, 12 and 15 days, respectively. A and B start working but A quits after working for 2 days. After this, C joins B till the completion of work. In how many days will the work be completed ?**

**Solution:**

Work done by A and B in 1 day

= *1*8 + *1*24 = *5*24

2 day’s work of A and B = *10*24

After 2 day’s A left the work

∴ Remaining work = 1 – *10*24 = *14*24

One day work of B and C together = *1*12 + *1*15 = *9*60

So, the number of days required by B and C to finish work

= (14/24) / (9/60) = (14/24) x (60/9) = 35/9

∴ Total days to complete the work = 2 + (35/9) = 53/9

**4). 90 men are engaged to do piece of work in 40 days but it is found that in 25 days, 2/3 work is complete. How many men should be allowed to go off, so that the work finished in time ?**

**Solution:**

Let N men are allowed,

M1= 90, D1 = 25, D2 = 15

W1 = *2*3, W2 = 1 – *2*3

= *1*3

M2 = 90 – N

According to the formula,

**M1D1W2 = M2D2W1**

⇒ (90 x 25) *1*3 = (90 – N) x 15 x *2*3

⇒ 90 x 25 x *1*3 = 10(90 – N)

⇒ 75 = 90 – N

∴ N = 90 – 75 = 15

**5). 24 men can complete a work in 16 days. 32 women can complete the same work in 34 days. 16 men and 16 women started working and worked for 12 days. How many more men are to be added to complete the remaining work in 2 days ?**

**Solution: **

24 men complete the work in 16 days

⇒ Work that are 16 men complete in 12 days = *16*24 x *12*16 = *1*2 part of work in 12 days.

32 women complete the work in 24 days

∴ Work that are 16 women complete in 14 days = *16*32 x *14*24 = *7*27 part of work in (12 + 2) = 14 days.

So, the remaining part of the work which done by (16 men + 16 women) and required additional no. of men in 2 days = 1 – (1/2 + 7/24) = 1/2 – 7/24 = 5/24

Now, in 2 days 5/24 part of the work is done by (24 x 16)/ 2 x (5/ 24) = 40 men

∴ Required additional no. of men = 40 – 16 =24.

**6). A piece of work was to be completed in 40 days, a number of men employed upon it did only half the work in 24 days, 16 more men were then set on and the work was completed in the specified time, how many men were employed at first?**

**Solution:**

Let N men were employed at first

∵ N men do 1/2 of the work in 24 days

∴ 1 man do the whole work in 24 x 2 x N = 48N days

Now, from the question,

(N + 16) men do the remaining work (1 – 1/2) = 1/2 in (40 – 24) = 16 days

∴ 1 man do the whole work in 16 x 2 x (N + 16) days

∴ 48N = 32(N + 16)

∴ N = 32 men

**7). A and B can do a piece of work in 18 days; B and C can do it in 24 days while C and A can finish it in 36 days. If A, B, C works together, in how many days will they finish the work?**

**Solution:**

Time taken by (A + B) to finish the work = 18 days.

(A + B)’s 1 day’s work = *1*18

Time taken by (B + C) to finish the work = 24 days.

(B + C)’s 1 day’s work =*1*24

Time taken by (C + A) to finish the work = 36 days.

(C + A)’s 1 day’s work = *1*36

Therefore, 2(A + B + C)’s 1 day’s work = *1*18 + *1*24 + *1*36

= *(4 + 3 + 2)*72 = *9*72

= *1*8

⇒ (A + B + C)’s 1 day’s work = *1*2 × *1*8

= *1*16

Therefore, A, B, C together can finish the work in 16 days.

**8). A and B can do a piece of work in 12 days; B and C can do it in 15 days while C and A can finish it in 20 days. If A, B, C works together, in how many days will they finish the work? In how many days will each one of them finish it, working alone? **

**Solution: **

Time taken by (A + B) to finish the work = 12 days.

(A + B)’s 1 day’s work = *1*12

Time taken by (B +C) to finish the work = 15 days.

(B + C)’s 1 day’s work = *1*15

Time taken by (C + A) to finish the work = 20 days.

(C + A)’s 1 day’s work = *1*20

Therefore, 2(A + B + C)’s 1 day’s work = *1*12 + *1*15 + *1*20

= *12*60 = *1*5

⇒ (A + B + C)’s 1 day’s work = *1*2 × *1*5

= *1*10

Therefore, A, B, C together can finish the work in 10 days.

Now, A’s 1 day’s work

= {(A + B + C)’s 1 day’s work} – {(B + C)’s 1 day’s work}

= (*1*10 – *1*15) = *1*30

Hence,

A alone can finish the work in 30 days.

B’s 1 day’s work

{(A + B + C)’s 1 day’s work} – {(C + A)’s 1 day’s work}

*1*10 – *1*20 = *1*20 Hence,

B alone can finish the work in 20 days.

C’s 1 days work

= {(A + B + C)’s 1 day’s work} – {(A + B)’s 1 day’s work}

= *1*10 – *1*12 = *1*60

Hence, C alone can finish the work in 60 days.

**9). Three pipes A,B and C are connected to a tank. These pipes can fill the tank separately in 5 hr, 10 hr and 15hr respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at (34)th of their normal rates for the 1st hour after which they supplied water at normal rate. Pipe C supplied water at (23)th of its normal rate for 1st 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled?**

**Solution:**

The part of the tank filled by A and B in first two hour

⇒*3*4 × {(*1*5 + *1*10)+(*1*5 + *1*10)}

The part of the tank filled by C

In the first two hours = 2 × *11*30 × *1*15

Remaining part = *139*360

In 1 hour,

(After 2 hours have been passed and pipes are working at a normal rate),

All the three pipes together fill = *11*30

Hence,

The time is taken to fill the remaining tank = *139*360 × *11*30

= 1.0530 hour

Thus,

The total time is taken to fill the remaining tank = 3.05 hour.

**10). A and B undertake to do a piece of work for Rs 600. An alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they can finish it in 3 days, Find the share of C?**

**Solution:**

C’s one day’s work,

= *1*3 − (*1*6 + *1*8)

= *1*24

Therefore,

A:B:C=Ratio of their one day’s work

And,

= *1*6 : *1*8 : *1*24

= 4 : 3 : 1

A’s share,

=Rs. 600 × *4*8 = 300

B’s share,

=Rs. 600 × *3*8 = 225

C’s share,

=Rs. [600−(300+225)]=Rs. 75

I hope, this article will help you a lot to understand the **Time And Work Problems**. If you still have any doubts and problems with any topic of mathematics you can ask your problem in the Ask Question section. You will get a reply shortly.