
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
513.6k+ 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
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What constitutes the central nervous system How are class 10 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE
