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

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

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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)

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.

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

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