Question

Samuel can complete a work in 20 days while Richard can complete the same work in 30 days. If both of them work on it together, in how many days can the work be completed?
10
12
25
50

Answer 1

Given : ↴

$\sf \: Samuel \: can \: complete \: a \: work \: in \: 20 \: days.$

$\sf \: While \: Richard \: can \: complete \: the \: same \: work \: in \: 30 \: days.$

To Find : ↴

We have to find that how many days they will take to complete the work, if they will work together.

So, Let's Start : ↴

$\sf \: Time \: taken \: by \: Samuel \: to \: complete \: the \: work \: = \: 20 \: days.$

$\sf \: Work \: done \: by \: Samuel \: in \: 1 \: day \: = \: \cfrac{1}{20}$

$\sf \: Time \: taken \: by \: Richard \: to \: complete \: the \: work \: = \: 30 \: days.$

$\sf \: Work \: done \: by \: Richard \: in \: 1 \: day \: = \: \cfrac{1}{30}$

$\sf \: Work \: done \: by \: Samuel \: and \: Richard \: in \: 1 \: day \: = \: \cfrac{1}{20} \: + \cfrac{1}{30}$

$\sf \longrightarrow\: \cfrac{1}{20} \: + \cfrac{1}{30} \:$

$\sf \: First \: of \: We \: have \: to \: make \: the \: denominator \: same.$

$\sf \longrightarrow\: \cfrac{1}{20} \: \times \: \cfrac{3}{3} \: = \: \cfrac{3}{60}$

$\sf \longrightarrow\: \cfrac{1}{30} \: \times \: \cfrac{2}{2} \: = \: \cfrac{2}{60}$

$\sf \longrightarrow \: \cfrac{3}{60} \: + \cfrac{2}{60} \: = \: \cfrac{3 \: + 2}{60}$

$\sf \longrightarrow \: \cfrac{3 \: + 2}{60} \: = \: \cfrac{5}{60}$

$\sf \longrightarrow \: \cfrac{5}{60} \: = \: \cfrac{1}{12}$

$\sf \: Thus \: time \: take \: by \: both \: to \: finish \: work \: :$ ↴

$\sf \: \longrightarrow \: 1 \: \div \: \cfrac{1}{12} \: = \: 12 \: days.$

$\sf \: So, \: they \: will \: take \: 12 \: days, \: if \: they \: will \: work \: together.$

$\boxed { \bf \longrightarrow Answer \: = \: 12 \: days}$

Answer 2

• the answer is 12. multiply the each numbers