Answer
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Hint: First we have to write the given sentences to mathematical terms. We generate the equations and substitute them on one another. After that we have to solve the quadratic equation to get the final answer.
Complete step-by-step solution -
A and B working together can do a work in 6 days
Let x be the no of days taken by A.
Let y be the no of days taken by B.
The work done by A in one day = \[\dfrac{1}{x}\]
The work done by B in one day = \[\dfrac{1}{y}\]
So, The work done by both A and B together in one day is \[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\]. . . . . . . . . . . . . . . . . . . (a)
If A takes 5 days less than B to finish the work = \[x=y-5\]. . . . . . . . . . . . . . . . . . . . . (1)
By substituting (1) in (a) we get
\[\dfrac{1}{y-5}+\dfrac{1}{y}=\dfrac{1}{6}\]
By further solving we get
\[(2y-5)6={{y}^{2}}-5y\]
By further solving we get
\[12y-30={{y}^{2}}-5y\]
\[{{y}^{2}}-17y + 30=0\]
We have to find the roots of the quadratic equation.
\[y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of a, b, c. we get
\[y=\dfrac{-(-17)\pm \sqrt{{{\left( -17 \right)}^{2}}-4\left( 1 \right)\left( 30 \right)}}{2(1)}\]
By further solving we get the roots as \[\begin{align}
& y=15 \\
& y=2 \\
\end{align}\]
Put the values of y in (1) we get,
\[x=2-5=-3\]. This is not possible.
\[x=15-5=10\] This is possible.
The answer is B takes 15 days to complete the work.
Note: In the above y=2 is not considered because the number of days required to complete the work cannot be negative. So we have taken the positive value. \[\dfrac{1}{x}\] is the work done by x in one day. Similarly \[\dfrac{1}{y}\] is the work done by y in one day.
Complete step-by-step solution -
A and B working together can do a work in 6 days
Let x be the no of days taken by A.
Let y be the no of days taken by B.
The work done by A in one day = \[\dfrac{1}{x}\]
The work done by B in one day = \[\dfrac{1}{y}\]
So, The work done by both A and B together in one day is \[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\]. . . . . . . . . . . . . . . . . . . (a)
If A takes 5 days less than B to finish the work = \[x=y-5\]. . . . . . . . . . . . . . . . . . . . . (1)
By substituting (1) in (a) we get
\[\dfrac{1}{y-5}+\dfrac{1}{y}=\dfrac{1}{6}\]
By further solving we get
\[(2y-5)6={{y}^{2}}-5y\]
By further solving we get
\[12y-30={{y}^{2}}-5y\]
\[{{y}^{2}}-17y + 30=0\]
We have to find the roots of the quadratic equation.
\[y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of a, b, c. we get
\[y=\dfrac{-(-17)\pm \sqrt{{{\left( -17 \right)}^{2}}-4\left( 1 \right)\left( 30 \right)}}{2(1)}\]
By further solving we get the roots as \[\begin{align}
& y=15 \\
& y=2 \\
\end{align}\]
Put the values of y in (1) we get,
\[x=2-5=-3\]. This is not possible.
\[x=15-5=10\] This is possible.
The answer is B takes 15 days to complete the work.
Note: In the above y=2 is not considered because the number of days required to complete the work cannot be negative. So we have taken the positive value. \[\dfrac{1}{x}\] is the work done by x in one day. Similarly \[\dfrac{1}{y}\] is the work done by y in one day.
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