Question

A takes twice as much as B or thrice as many times as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:
A. 3 days
B. 7 days
C. 4 days
D. 6 days

Answer 1

Hint: Assume three variables for the number of days taken by A, B, and C to do the work alone. Using the unitary method find the amount of work done by each of them in 1 day. According to the question together they will do the work in 2 days. So, by unitary method together in one day, they will $\dfrac{1}{2}$ work. Obtain an equation by some of the work done by each of them in the day with $\dfrac{1}{2}$ and use the given relation in question to solve the equation.

Complete step-by-step solution:
Let us assume that A can alone do the work in x days and B can do the work alone in y days and C can do the work alone in z days.
According to the question, A takes twice as much time as B
So, $x=2y$ …………… (1)
And, A takes thrice as much as time as C,
So, $x=3z$ ……….(2)
And if they will work together, they can do the work in 2 days.
According to our question,
A can do the work in x days
By using the method, work done by A in 1 day $=\dfrac{1}{x}$ ………………. (3)
Similarly, work done by ‘B’ in 1 day $=\dfrac{1}{y}$ …………(4)
And work done by ‘C’ in 1 day $=\dfrac{1}{z}$ …………………(5)
According to the question,
Work is done by them together in 2 days.
So, by unitary method, work done by them in 1 day $=\dfrac{1}{2}$ i.e. work done by A,B and C together in 1 day $=\dfrac{1}{2}$ . i.e.
 $\begin{align}
  & \left( \text{work done by A in one day} \right)+\left( \text{work done by B in one day} \right)+ \\
 & \left( \text{work done by C in one day} \right)=\dfrac{1}{2} \\
\end{align}$ .
Using eq (3), (4) and (5), we will get,
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2}$ .
Using eq (1)
$\begin{align}
  & x=2y \\
 & \Rightarrow y=\dfrac{x}{2} \\
\end{align}$
On putting $y=\dfrac{x}{2}$ , we will get
$\dfrac{1}{x}+\dfrac{1}{\dfrac{x}{2}}+\dfrac{1}{z}=\dfrac{1}{2}$ .
Using eq (2)
$\begin{align}
  & x=3z \\
 & \Rightarrow z=\dfrac{x}{3} \\
\end{align}$
On putting $z=\dfrac{x}{3}$ , we get,
$\dfrac{1}{x}+\dfrac{1}{\dfrac{x}{2}}+\dfrac{1}{\dfrac{x}{3}}=\dfrac{1}{2}$ .
$\Rightarrow \dfrac{1}{x}+\dfrac{2}{x}+\dfrac{3}{x}=\dfrac{1}{2}$ .
Taking LCM and adding in LHS, we get
$\begin{align}
  & \Rightarrow \dfrac{1+2+3}{x}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{6}{x}=\dfrac{1}{2} \\
\end{align}$
On cross-multiplying, we get
$\begin{align}
  & x=2\times 6 \\
 & \Rightarrow x=12 \\
\end{align}$
On putting x=12 in eq (1) we will get
$12=2\left( y \right)$
On dividing both sides of equation by (2), we get,
$6=y$ .
We have assumed that B can do the work done in y days and we have obtained y=6.
So, B can do the work alone in 6 days, and option (D) is the correct answer.

Note: Students can also solve this question by this method. Using only one variable
Suppose A,B and C takes $x,\dfrac{x}{2},\dfrac{x}{3}$ days respectively to finish the work.
Then,
$\begin{align}
  & \left( \dfrac{1}{x}+\dfrac{2}{x}+\dfrac{3}{x} \right)=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{1+2+3}{x}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{6}{x}=\dfrac{1}{2} \\
 & \therefore x=12 \\
\end{align}$
So, B takes $\dfrac{x}{2}$ i.e. $\dfrac{12}{2}=6$ days to finish the work.