Question

Two teachers A and B can complete an academic work in 10 days and 15 days respectively. They started the work together, but A left after 5 days and another teacher C joined, who alone can complete the work in 60 days. In how many days the work got completed?​

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

As We know that: Two teachers A and B can complete an academic work in 10 days and 15 days respectively.

One day Work of both Teachers as:

${\sf{ \dfrac{1}{10} , \dfrac{1}{15} }}$ Respectively.

They started the work together and completed some amount of work of whole work.

$\colon\implies{\sf{ \dfrac{1}{10} + \dfrac{1}{15} }} \\ \\ \colon\implies{\sf{ \dfrac{3+2}{30} }} \\ \\ \colon\implies{\sf{ \cancel{ \dfrac{5}{30} } = \dfrac{1}{6} }}$

A and B's One day work is ${\sf{ \dfrac{1}{6} }}$

Teacher A and B worked for 5 days & then Teacher A left working but they together complete work for 5 days as.

$\colon\implies {\tt{ \dfrac{1}{6} \times 5 }} \\ \\ \colon\implies {\tt{ \dfrac{5}{6} }}$

So, They completed ${\tt{ \dfrac{5}{6} }}$ of whole work.

Now, We can find how much work is left after completing some amount of it.

$\colon\implies{\sf{ 1 - \dfrac{5}{6} }} \\ \\ \colon\implies{\sf{ \dfrac{6-5}{6} }} \\ \\ \colon\implies{\sf{ \dfrac{1}{6} }}$

So, We also Know that another teacher C joined, who alone can complete the work in 60 days.

Now, we have to find the work of Teacher B and C together as.

$\longrightarrow {\tt{ \dfrac{1}{15} + \dfrac{1}{60} = \dfrac{4+1}{60} }} \\ \\ \\ \longrightarrow {\tt{ \cancel{ \dfrac{5}{60} } = \dfrac{1}{12} }}$

Now,

• Total Work left = ${\sf{ \dfrac{1}{6} }}$

• Total Efficiency = ${\sf{ \dfrac{1}{12} }}$ (B & C)

According to Question:

$\colon\implies{\tt{ \dfrac{ \dfrac{1}{6} }{ \dfrac{1}{12} } = \dfrac{ 1 \times 12 }{6 \times 1} }} \\ \\ \colon\implies{\tt{ \cancel{ \dfrac{ 12 }{6} } }} \\ \\ \colon\implies{\tt\large\gray{ 2 \ days }}$

Hence,

The work got completed in 2 days after leftover of the Teacher A.

[PROVED]