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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