
A piece of cloth costs Rs. 35. If the piece were 4 meter longer and each meter costs Rs. 1 less, the cost would remain unchanged. The length of the piece is _________ meter.
Answer
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Hint:Before solving this question, we must know about the cost of the total buying of the cloth can be calculated by multiplying the cost of cloth per meter and the total length of cloth which is bought (in meters).Now, in this question, it is particularly mentioned that the total money spent on buying the entire length of the cloth is remaining the same; hence, we can form two equations in two variable and then solve them to get the desired value.
Complete step-by-step answer:
As mentioned in the question, we have to find the length of cloth that is bought by using the information that is given in the question.
Now, let us take the length of the cloth be x meters and let the cost of 1 meter be Rs. y.
Now, as mentioned in the hint, we can write the following equations to get to the values of x and y as follows
\[\begin{align}
& (x)(y)=35\ \ \ \ \ ...(a) \\
& y=\dfrac{35}{x} \\
\end{align}\]
And they given if the piece of cloth were 4 meter long then cost of each meter will be less than 1 and total cost will remain unchanged then we get an equation as follows,
\[\begin{align}
& (x+4)(y-1)=35 \\
& xy+4y-x-4=35 \\
& xy+4y-x=39\ \ \ \ \ ...(b) \\
\end{align}\]
Now, on using equations (a) and (b) as follows
\[\begin{align}
& 35+4y-x=39 \\
& 4y-x=4\ \ \ \ \ ...(c) \\
\end{align}\]
Now, in equation (c), we can use equation (a) as follows
\[\begin{align}
& 4\times \dfrac{35}{x}-x=4 \\
& 140-{{x}^{2}}=4x \\
& {{x}^{2}}+4x-140=0 \\
\end{align}\]
Now, we will solve this equation as follows
\[\begin{align}
& {{x}^{2}}+4x-140=0 \\
& x=\dfrac{-4\pm \sqrt{{{(4)}^{2}}-4(-140)1}}{2\times 1} \\
& x=\dfrac{-4\pm \sqrt{16+560}}{2} \\
& x=\dfrac{-4\pm \sqrt{576}}{2} \\
& x=\dfrac{-4\pm 24}{2}=10,-14 \\
\end{align}\]
Here -14 have to neglect as length cannot be negative value
Hence, we get that the length of the cloth that is bought is 10m.
Note:-We have to analyse the question and form equations based on information given in the question.The students should know how to solve a system of linear equations in two variables.The methods of solving a system of linear equation in two variables using the following three methods which are
1. Elimination method
2. Substitution method
3. Cross multiplication method
Complete step-by-step answer:
As mentioned in the question, we have to find the length of cloth that is bought by using the information that is given in the question.
Now, let us take the length of the cloth be x meters and let the cost of 1 meter be Rs. y.
Now, as mentioned in the hint, we can write the following equations to get to the values of x and y as follows
\[\begin{align}
& (x)(y)=35\ \ \ \ \ ...(a) \\
& y=\dfrac{35}{x} \\
\end{align}\]
And they given if the piece of cloth were 4 meter long then cost of each meter will be less than 1 and total cost will remain unchanged then we get an equation as follows,
\[\begin{align}
& (x+4)(y-1)=35 \\
& xy+4y-x-4=35 \\
& xy+4y-x=39\ \ \ \ \ ...(b) \\
\end{align}\]
Now, on using equations (a) and (b) as follows
\[\begin{align}
& 35+4y-x=39 \\
& 4y-x=4\ \ \ \ \ ...(c) \\
\end{align}\]
Now, in equation (c), we can use equation (a) as follows
\[\begin{align}
& 4\times \dfrac{35}{x}-x=4 \\
& 140-{{x}^{2}}=4x \\
& {{x}^{2}}+4x-140=0 \\
\end{align}\]
Now, we will solve this equation as follows
\[\begin{align}
& {{x}^{2}}+4x-140=0 \\
& x=\dfrac{-4\pm \sqrt{{{(4)}^{2}}-4(-140)1}}{2\times 1} \\
& x=\dfrac{-4\pm \sqrt{16+560}}{2} \\
& x=\dfrac{-4\pm \sqrt{576}}{2} \\
& x=\dfrac{-4\pm 24}{2}=10,-14 \\
\end{align}\]
Here -14 have to neglect as length cannot be negative value
Hence, we get that the length of the cloth that is bought is 10m.
Note:-We have to analyse the question and form equations based on information given in the question.The students should know how to solve a system of linear equations in two variables.The methods of solving a system of linear equation in two variables using the following three methods which are
1. Elimination method
2. Substitution method
3. Cross multiplication method
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