Answer
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Hint:
Consider the time taken by B as ‘x’. A takes 6 days less than B, so time taken by A is (x - 6). So work done by \[B=\dfrac{1}{x}\]and \[A=\dfrac{1}{x-6}\]. Total work done by (A + B) is \[\dfrac{1}{4}\]. Solve the equation formed to get the time taken by B to finish the work.
Complete step-by-step answer:
Suppose B alone takes x days to finish the work. Then A alone can finish the work in (x - 6) days.
Now we know A’s one day work\[=\dfrac{1}{x-6}\].
B’s work per day\[=\dfrac{1}{x}\].
The total work done by A and B in one day\[=\dfrac{1}{x}+\dfrac{1}{x-6}-(1)\]
\[A+B=\dfrac{1}{x}+\dfrac{1}{x-6}\].
It’s said that A and B can finish a work together in 4 days.
\[\therefore A+B=\dfrac{1}{4}-(2)\]
Now, Equating equation (1) and equation (2),
\[\dfrac{1}{x}+\dfrac{1}{x-6}=\dfrac{1}{4}\]
Simplifying the LHS, \[\dfrac{\left( x-6 \right)+x}{x\left( x-6 \right)}=\dfrac{1}{4}\]
\[\Rightarrow \dfrac{2x-6}{x\left( x-6 \right)}=\dfrac{1}{4}\]
Now cross multiplying them, then the equation becomes,
\[4\left( 2x-6 \right)=x\left( x-6 \right)\]
Opening the brackets and simplifying them,
\[\begin{align}
& 8x-24={{x}^{2}}-6x \\
& \Rightarrow {{x}^{2}}-6x-8x+24=0 \\
& {{x}^{2}}-x\left( 6+8 \right)+24=0 \\
& {{x}^{2}}-14x+24=0-(3) \\
\end{align}\]
Equation (3) is similar to the general quadratic equation, \[a{{x}^{2}}+bx+c=0\].
Comparing both general equation and equation (3), we get the values of constants a = 1, b = -14, c = 24.
Now, applying the above value in the quadratic formula,
\[\begin{align}
& \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-\left( -14 \right)\pm \sqrt{{{\left( -14 \right)}^{2}}-4\times 1\times 24}}{2\times 1} \\
& =\dfrac{14\pm \sqrt{196-96}}{2}=\dfrac{14\pm \sqrt{100}}{2}=\dfrac{14\pm 10}{2} \\
\end{align}\]
The roots are \[\left( \dfrac{14+10}{2} \right)\]and \[\left( \dfrac{14-10}{2} \right)\]= 12 and 2.
The value of x cannot be less than 6.
If B takes the time of 2 days, then the time taken by A becomes, \[x-6=2-6=-4\]days.
i.e. time taken cannot be negative.
\[\therefore x=12\].
i.e. It would take B a period of 12 days to finish the job.
Note:
If we were asked to find the time taken by A to finish the work alone then we got x=12 i.e. the time taken by B alone to do a work. So, time taken by A is 12 – 6 = 6 days. As A takes 6 days less than the time taken by B. If we are adding time taken by (A + B) together\[=\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{12+6}{12\times 6}=\dfrac{1}{4}\].
Consider the time taken by B as ‘x’. A takes 6 days less than B, so time taken by A is (x - 6). So work done by \[B=\dfrac{1}{x}\]and \[A=\dfrac{1}{x-6}\]. Total work done by (A + B) is \[\dfrac{1}{4}\]. Solve the equation formed to get the time taken by B to finish the work.
Complete step-by-step answer:
Suppose B alone takes x days to finish the work. Then A alone can finish the work in (x - 6) days.
Now we know A’s one day work\[=\dfrac{1}{x-6}\].
B’s work per day\[=\dfrac{1}{x}\].
The total work done by A and B in one day\[=\dfrac{1}{x}+\dfrac{1}{x-6}-(1)\]
\[A+B=\dfrac{1}{x}+\dfrac{1}{x-6}\].
It’s said that A and B can finish a work together in 4 days.
\[\therefore A+B=\dfrac{1}{4}-(2)\]
Now, Equating equation (1) and equation (2),
\[\dfrac{1}{x}+\dfrac{1}{x-6}=\dfrac{1}{4}\]
Simplifying the LHS, \[\dfrac{\left( x-6 \right)+x}{x\left( x-6 \right)}=\dfrac{1}{4}\]
\[\Rightarrow \dfrac{2x-6}{x\left( x-6 \right)}=\dfrac{1}{4}\]
Now cross multiplying them, then the equation becomes,
\[4\left( 2x-6 \right)=x\left( x-6 \right)\]
Opening the brackets and simplifying them,
\[\begin{align}
& 8x-24={{x}^{2}}-6x \\
& \Rightarrow {{x}^{2}}-6x-8x+24=0 \\
& {{x}^{2}}-x\left( 6+8 \right)+24=0 \\
& {{x}^{2}}-14x+24=0-(3) \\
\end{align}\]
Equation (3) is similar to the general quadratic equation, \[a{{x}^{2}}+bx+c=0\].
Comparing both general equation and equation (3), we get the values of constants a = 1, b = -14, c = 24.
Now, applying the above value in the quadratic formula,
\[\begin{align}
& \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-\left( -14 \right)\pm \sqrt{{{\left( -14 \right)}^{2}}-4\times 1\times 24}}{2\times 1} \\
& =\dfrac{14\pm \sqrt{196-96}}{2}=\dfrac{14\pm \sqrt{100}}{2}=\dfrac{14\pm 10}{2} \\
\end{align}\]
The roots are \[\left( \dfrac{14+10}{2} \right)\]and \[\left( \dfrac{14-10}{2} \right)\]= 12 and 2.
The value of x cannot be less than 6.
If B takes the time of 2 days, then the time taken by A becomes, \[x-6=2-6=-4\]days.
i.e. time taken cannot be negative.
\[\therefore x=12\].
i.e. It would take B a period of 12 days to finish the job.
Note:
If we were asked to find the time taken by A to finish the work alone then we got x=12 i.e. the time taken by B alone to do a work. So, time taken by A is 12 – 6 = 6 days. As A takes 6 days less than the time taken by B. If we are adding time taken by (A + B) together\[=\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{12+6}{12\times 6}=\dfrac{1}{4}\].
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