Can do a piece of work in 10 days and b in 20 days they work together but 2 days before the completion of the work a leaves. in how many days was the work?

A's one day's work =`1/10`

B's one day's work =`1/12`

C's one day's work =`1/15`

A, B and C's one day's work =`1/10+1/12+1/15`

`=(6+5+4)/60=15/60=1/4`

Let the work completed in x days

∴ A's 2 days work + B's (x − 3) days work + C's x days work = 1

`⇒2xx1/10+("x"-3)xx1/12+"x"xx1/15=1`

`⇒1/5+("x"-3)/12+"x"/15=1`

`(12+5"x"-15+4"x"=60)/60` ...(L.C.M of 5, 12, 15 = 60)

⇒ 9x = 60 − 12 + 15

⇒ 9x = 63

`⇒"x"=63/9=7`

∴ Work will last for 7 days

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Correct Answer:

Description for Correct answer:

Let the required days be x.

A works for (x - 2) days, while B works for x days.

According to the question,

\( \Large\frac{x - 2}{10} + \frac{x}{20} = 1 \)

2x- 4 + x = 20

3x = 24

x = 8 days


Part of solved Time and work questions and answers : >> Aptitude >> Time and work

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Similar Questions

The completion of the work would have been scheduled assuming that A & B will work together for completing the work.

Let us suppose it to be x days.

but,

B worked for x - 2 days.

Let the total work = L. C. M (10, 15)

= 30 units.

A completes → \(\dfrac{30}{10}=3\) units/day

B completes → \(\dfrac{30}{15}=2\) units /day

\(x=\left(\frac{10×15}{10+15}\right)=\frac{150}{25}=6 \) days

For 4 days, A & B worked together

Work done = 4 × 5 = 20 units.

Remaining work = 10 units. 

We need to find how much time A will take to complete 10 units of work. 

A completes 3 units in 1 day

A completes 10 units in 10/3 = 3.33 days. 

Total days → \(4+3\frac{1}{3}=7\frac{1}{3}\) days

 Alternate Method

Formula:

W = RT

Where W is work and T is time

Given, A complete a work in 10 days

A complete a work in 15 days

Let, 1 work = 30 unit (for simplification: LCM of 10, 15)

Person

R

T

W

A

R = W/T = 3

10

30

B

R = W/T = 2

15

30

If A & B together do a work then, R = R(A) + R(B) = 3 + 2 = 5

A + B

5

T = W/R = 6

30


Hence, both together can complete a work in 6 days.

If 'B' leaves 2 days before the scheduled completion of the work then A alone will do the work for the last 2 days.

Person

R

T

W

A + B

5

4

W = RT = 20


Now remaining work is (30 – 20 = 10), which completed alone by A

Person

R

T

W

A

3

T = W/R = 10/3

10


Now total time required to complete work by both if 'B' leaves 2 days before the scheduled completion

Time = \(4+\frac{10}{3}=7\frac{1}{3}\) days