Time & Work - Question and Answer!

A, B and C can do a piece of work in 12, 18 and 24 days, respectively. They all begin together. A work continuously till it is finished, B leaves the work 2 days before its completion and C leaves the work 4 days before its completion. In what approximate time is the work finished?

Let the work is finished in x days

Work done by A in 1 day =	1
	12

Work done by B in 1 day =	1
	18

Work done by C in 1 day =	1
	24

x	+	x – 2	+	x – 4	= 1
12		18		24

⇒	x	+	x	–	2	+	x	–	4	= 1
	12		18		18		24		24

⇒	x	+	x	+	x	= 1 +	1	+	1
	12		18		24		9		6

⇒	6x + 4x +3x	=	18 + 2 + 3
	72		18

⇒	13x	=	23
	72		18

⇒ x = 7.07 ≈ 7 days

Rohan takes twice the time taken by Sohan to complete a piece of work and half of the time taken by Mohan to complete the work. If all of them working together can complete the work in 12 days, in how many days Rohan and Sohan working together can complete that work?

Let the time taken by Sohan to complete the work = x days

Then, time taken by Rohan to complete the work = 2x days and

time taken by Mohan to complete the work = 4x days

According to the question:  

12	(	1	+	1	+	1	)	= 1
		x		2x		4x

⇒	4 + 2 + 1	=	1
	4x		12

⇒	7	=	1
	4x		12

⇒ x = 21 days

Hence,

time taken by Sohan to complete the work = x = 21 days

time taken by Rohan to complete the work = 2x = 2 x 21 = 42 days

time taken by Mohan to complete the work = 4x = 4 x 21 = 84 days

let the no of days taken by Rohan and Sohan to complete the work = p days

According to the question:
 

p	(	1	+	1	)	= 1
		21		42

⇒ p	1 + 2	= 1
	42

⇒ p = 14 days

Two persons Aman and Bhanu can dig a pit in 20 days and 25 days respectively and a third person Cheenu can fill that pit in 50 days. All of the three persons start their work and after sometime Cheenu leaves the work. If total time taken to dig the pit from the beginning is 13 days, find after how many days Cheenu left his work?

Let total work is LCM of 20, 25 and 50 days = 100 units
 

1 day work of Aman , Bhanu and Cheenu is 5 units, 4 units and 2 units. But the nature of work of Cheenu is opposite to that of Aman and Bhanu.
 

Let Cheenu work for ‘x’ days and remaining days in which only Aman and Bhanu work is (13 – x) days.
 

According to question-
 

⇒ x × (5 + 4 – 2) + (13 – x) × (5 + 4) = 100
 

⇒ x = 8.5 days
 

Hence Cheenu left the work after 8.5 days.

Mahesh can complete a job in 5 days. Mahesh is twice as fast as Akhilesh while Akhilesh is thrice as fast as Nimesh. If all of them work together, in how many days would the job get completed?

As per the question,

Mahesh can do a piece of work in 5 days, Akhilesh can do the same work in 10 days, Nimesh can do the same work in 30 days.

So together they can do the piece of work in 1 day
 

=	(	1	+	1	+	1	)
		5		10		30

=	(	12 + 6 + 2	)	+	20	=	1
		60			60		3

In a day they finsih 1/3rd of the work.

Hence, together they will finish the piece of work in 3 days.

100 Men were employed to finish a work in 180 days. After 60 days it was found that only 1/5 of the work was done. How many more men must be employed to finish the work in the stipulated time ?

By product constancy

M1 × D1	=	M2 × D2
W1		W2

100 × 60	=	a × 120
1 5		1 –  1 5

⇒ a = 200

⇒ Total workers  =  200 

⇒ Extra men required = 200 - 100  = 100

If A can do 1/4 of a work in 3 days and B can do 1/6 of the same work in 4 days, how much will A get if both work together and are paid Rs.180 in all?

A’s one day work =	1	×	1	=	1
	4		3		12

B’s one day work =	1	×	1	=	1
	6		4		24

A’s wages: B’s wages = A’s 1 day’s work : B’s 1 day’s work

=	1	:	1	= 2 : 1
	12		24

⇒ A’s share =	2	× 180 = Rs.120
	3

20 women can complete a work in 14 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?

20 women can complete the work in 14 days 
 

Let the work completed by 1 women in 1 day = w
 

So work done by 20 women in 1 day = 20 × w
 

We know that work = M × D × H
 

Here,
 

M = no. of men 
 

D = no of days
 

H = hours
 

∴ Work done by 20 women in 14 days = 20 × w × 14
 

Let the work completed by 1 child in 1 day = c
 

So, work completed by 10 children in 1 day = 10 × c

So, work done by 10 children in 14 days = 10 × c × 14
 

∵ Work is equal 
 

∴ 20 × w × 14 = 10 × c × 14
 

 2 × w = c
 

Let 5 women and 10 children complete the work in = d days
 

∴ Complete work = (5 × w + 10 × c) × d
 

∵ 1 child = 2 women 
 

∴ 10 children = 20 women 
 

(5 women + 20 women) = 25 women 
 

Again work is equal 25 × w × d = 20 × w × 14

 

So, d =	56	days
	5

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