Question

A takes 6 days less than the time taken by B to complete a piece of work. If both A and B together can complete the work in 4 days, find the time taken by B to finish the work.

Answer 1

Hint: We solve this problem by assuming the number of days taken by both A and B separately as some variables.
Then we use the given condition to form two equations of two variables so that we can solve them easily. In forming the equations we need to note that we cannot add or subtract the time taken when the work is done together by two persons but we can add or subtract the work done when they work together.
We consider a short time here 1 day and then we find work done in that 1 day to build the equations

Complete step by step answer:
We are given that A and B are working on some piece of work.
Let us assume that A takes \[x\] days to complete the work.
Now, let us assume a small amount of time that is 1 day.
So, we can say that A completes \[\dfrac{1}{x}\] part of work in 1 day
Now, let us assume that B takes \[y\] days to complete the same work
So, we can say that B completes \[\dfrac{1}{y}\] part of work in one day.
Let us consider the first statement given that A takes 6 days less than the time taken by B to complete a piece of work
By converting the above statement into mathematical equation then we get
\[\Rightarrow x=y-6.......equation(i)\]
We are given with the second statement that is both A and B together complete the work in 4 days
Let us rewrite the above statement in terms of work then we get that both A and B complete \[\dfrac{1}{4}\] part of work in 1 day
By converting the above statement into mathematical equation then we get
\[\Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{4}\]
By substituting the required value from equation (i) in above equation then we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{y-6}+\dfrac{1}{y}=\dfrac{1}{4} \\
 & \Rightarrow \dfrac{y+y-6}{y\left( y-6 \right)}=\dfrac{1}{4} \\
\end{align}\]
Now, by cross multiplying the terms in the above equation then we get
\[\begin{align}
  & \Rightarrow 8y-24={{y}^{2}}-6y \\
 & \Rightarrow {{y}^{2}}-14y+24=0 \\
\end{align}\]
Now, let us rewrite the middle term to get the factorisation method then we get
\[\begin{align}
  & \Rightarrow {{y}^{2}}-12y-2y+24=0 \\
 & \Rightarrow y\left( y-12 \right)-2\left( y-12 \right)=0 \\
 & \Rightarrow \left( y-12 \right)\left( y-2 \right)=0 \\
\end{align}\]
We know that if \[a\times b=0\] then either of \[a,b\] will be 0
Let us take the second term to 0 then we get
\[\begin{align}
  & \Rightarrow y-2=0 \\
 & \Rightarrow y=2 \\
\end{align}\]
Here, we get the value of \[x\] at \[y=2\] from equation (i) as
\[\begin{align}
  & \Rightarrow x=y-6 \\
 & \Rightarrow x=2-6=-4 \\
\end{align}\]
We know that the number of days can never be negative.
So, we can say that \[y=2\] is not correct.
By taking the other term then we get
\[\begin{align}
  & \Rightarrow y-12=0 \\
 & \Rightarrow y=12 \\
\end{align}\]
Here, we get the value of \[x\] at \[y=2\] from equation (i) as
\[\Rightarrow x=12-6=6\]
Therefore, we can conclude that A takes 6 days and B takes 12 days to complete the work.


Note:
Students may do mistakes in converting the first statement into a mathematical equation.
We are given that A takes 6 days less than B to complete the work.
Here, we can see that number of days taken by A should be less than that of B so that we get the equation as
\[x=y-6\]
But students may do mistake by taking the reverse order and assume the equation as
\[x-6=y\]
This gives the wrong answer because here the number of days taken by B is more in this condition.