Question

A can do a piece of work in 30 days and B can do it in 20 days. They work . together for 4 days and then B leaves the job. In how many days, will A finish the remaining work?​

Answer 1

Step-by-step explanation:

$\because \: time \: taken \: to \: finish \: the \: work \: by \: \: \: a = \: 30 \: days \\ \therefore \: a \: \: s\: 1 \: day \: work = \frac{1}{30} part \\ \because \: time \: taken \: to \: finish \: the \: work \: by \: \: \: b = \: 20 \: days \\ \therefore \: b \: \: s\: 1 \: day \: work = \frac{1}{20} part \\(a + b) \: \: s \: 1 \: day \: work = \frac{1}{30} + \frac{1}{20} part \\ = \frac{2 + 3}{60} = \frac{5}{60} = \frac{1}{12} \: part \\ so \: (a + b) \: \: s \: 4 \: days \: work = \frac{1}{12} \times 4 = \frac{1}{3} \: part \\ remaining \: work = 1 - \frac{1}{3} = \frac{3 - 1}{3} = \frac{2}{3} \: part \\ so \: time \: taken \: to \: finish \: the \: remaining \: work \: by \: a = \frac{ \frac{2}{3} }{ \frac{1}{30} } \\ = \frac{2}{3} \times \frac{30}{1} = 2 \times 10 = 20 \: days$

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Answer 2

Given:—

• A can do a piece of work in 30 days and B can do it in 20 days. They work together for 4 days and then B leaves the job.

To Find:—

• They work together for 4 days and then B leaves the job. ?

Solution:—

→ LCM of 30 and 20 = 60 units = Let total work .

So,

→ Efficiency of A = Total work / Time taken = 60/30 = 2 units / day .

and,

→ Efficiency of B = Total work / Time taken = 60/20 = 3 units / day .

then,

→ Work done by (A + B) in one day

• = (2 + 3)
• = 5 units

so,

→ Work done by (A + B) in first 4 days

• = 5×4
• = 20 units

Now,

→ Work left to be done = 60 - 20

= 40 units

since B left, only A has to complete left work .

therefore,

→ Time taken by A to complete the remaining work = Lefy work / Efficiency of A = 40/2 = 20 days (Ans.)

Hence,

A will finish the remaining work in 20 days .