
A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.
Answer
521.4k+ views
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\[=\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{12+6}{12\times 6}=\dfrac{1}{4}\].
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Trending doubts
Truly whole mankind is one was declared by the Kannada class 10 social science CBSE

Explain the three major features of the shiwaliks class 10 social science CBSE

Find the area of the minor segment of a circle of radius class 10 maths CBSE

Distinguish between the reserved forests and protected class 10 biology CBSE

A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

A gulab jamun contains sugar syrup up to about 30 of class 10 maths CBSE
