Question

A can do a piece of work in 25 days which B alone can finish in 20 days. Both together work for 5 days and then A leaves off. How many days will B take to finish
the remaining work?

Answer 1

$\huge\mathbb\colorbox{black}{\color{red}{Question}}$

• $\sf\red{How \: many \: days \: will \: B \: take \: to \: finish}$
• $\sf\red{the \: remaining \: work?}$

$\huge\mathbb\colorbox{black}{\color{red}{Answer:-}}$

• $\sf\red{11 \: days}$

$\small\mathbb\colorbox{black}{\color{orange}{Solution \: For \: The \: Correct \: Answer-}}$

Work done by A in 1 day =

$\frac{1}{25}$

Work done by B in 1 day =

$\frac{1}{20}$

• =>
• $\frac{1}{25} + \frac{1}{20} \\ \\ = > \frac{5}{100} + \frac{4}{100} \\ \\ = > \frac{5 + 4}{100} = \frac{9}{100}$

Both worked for 5 days together =

• $\small\bold\orange{5×1 \: day \: work=5}$

• => 9 × 5 = 45

Remaining work is done by B

• =>
• $1 - \frac{45}{100} \\ \\ = > \frac{100}{100} - \frac{45}{100} \\ \\ = > \frac{100 -4 5}{100} \\ \\ = > \frac{55}{100}$
• Now let B take x days to complete the work.

=>

$\frac{1}{20} \times x = \frac{55}{100} \\ \\ = > x = \frac{55 \times 20}{100} \\ \\ = > x = \frac{1100}{100} = > x = 11$

:. B took 11 days to complete the work alone.