Question

The total cost of a certain length of cloth is $200$. If the piece was $5m$ longer and each meter of cloth cost $2$ less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per meter?

Answer 1

Hint: In order to find the length of the piece and original cost, we need two equations: the first equation will be for the original length and cost per meter and the second equation will be when the length of the piece of cloth is increased and the cost is decreased. Solve both the equations and get the values needed.

Complete step by step answer:
Considering the original length of the cloth to be $x{\text{ m}}$ and the original cost per meter to be $y$.
The first equation:
Since, the total cost of the cloth given is $200$.
So, the first original length and cost equation numerically becomes:
$xy = 200$ --------- (1)
Dividing both sides by $x{\text{ meter}}$ and we get:
$ \Rightarrow \dfrac{{xy}}{x} = \dfrac{{200}}{x}$
$ \Rightarrow y = \dfrac{{200}}{x}$ ----------(2)

The second equation:
The length is increased $5m$ longer, so the length becomes $\left( {x + 5} \right)m$ and each meter of cloth costs $2$ less, so the cost becomes $\left( {y - 2} \right)$.
Since, there was no effect on the total cost, so we are substituting the length and the cost in the same equation that is equation 1, and we get:
$\left( {x + 5} \right)\left( {y - 2} \right) = 200$ ----------(3)
Which is our second equation for the condition when changes are made.
Substituting the value of $y$ from equation 2 in equation 3, and we get:
$\left( {x + 5} \right)\left( {\dfrac{{200}}{x} - 2} \right) = 200$
Opening the parenthesis:
$ \Rightarrow x.\dfrac{{200}}{x} - 2x + 5.\dfrac{{200}}{x} - 5 \times 2 = 200$
$ \Rightarrow 200 - 2x + \dfrac{{1000}}{x} - 10 = 200$
$ \Rightarrow 190 - 2x + \dfrac{{1000}}{x} = 200$
Taking $x$ as a common denominator on the left side, and we get:
$ \Rightarrow \dfrac{{190x}}{x} - \dfrac{{2{x^2}}}{x} + \dfrac{{1000}}{x} = 200$
$ \Rightarrow \dfrac{{190x - 2{x^2} + 1000}}{x} = 200$
Multiplying both sides by $x$:
$ \Rightarrow \dfrac{{190x - 2{x^2} + 1000}}{x}.x = 200x$
$ \Rightarrow 190x - 2{x^2} + 1000 = 200x$
Subtracting both sides by $200x$:
$ \Rightarrow 190x - 2{x^2} + 1000 - 200x = 200x - 200x$
$ \Rightarrow - 10x - 2{x^2} + 1000 = 0$
Taking $ - 2$ common, we get:
$ \Rightarrow - 2\left( {5x + {x^2} - 500} \right) = 0$
Dividing both the sides by $ - 2$:
$ \Rightarrow \dfrac{{ - 2\left( {5x + {x^2} - 500} \right)}}{{ - 2}} = \dfrac{0}{{ - 2}}$
$ \Rightarrow {x^2} + 5x - 500 = 0$
Since we obtained a quadratic equation, we will use mid-term factorization to solve it.
For that, we can split the mid-term $5x$ as $25x - 20x$, which on multiplying gives $500{x^2}$ and on adding or subtracting gives the mid-term, which is needed for mid-term factorization.
So, we write:
$ \Rightarrow {x^2} + 25x - 20x - 500 = 0$
Taking the first pair in one bracket and the second pair in another.
$ \Rightarrow \left( {{x^2} + 25x} \right) - \left( {20x + 500} \right) = 0$
Taking $x$ common from the first pair and $20$ common from the second equation:
$ \Rightarrow x\left( {x + 25} \right) - 20\left( {x + 25} \right) = 0$
Taking $\left( {x + 25} \right)$ common on the left side and we get:
$ \Rightarrow \left( {x + 25} \right)\left( {x - 20} \right) = 0$
Equating the two brackets with $0$, we get:
$x = 20,\left( { - 25} \right)$
But, since $x$ was the length and length cannot be negative, so we neglected $ - 25$.
So, the value of $x = 20$.
Therefore, the original length of cloth is $20{\text{ m}}$.
Substituting the value of $x = 20$ in equation (2), and we get:
$ \Rightarrow y = \dfrac{{200}}{{20}} = 10$
Therefore, the original length of cloth is $20{\text{ m}}$.
Therefore, the original cost of the piece is $Rs.{\text{ }}10.$

Note:
1. Remember you can check that the values obtained are correct or not by substituting the values that are obtained in the two equations that are equation 1 and equation 3, if they verify the values on the LHS and RHS, then the results are correct.
2. The quadratic equation can also be solved using the quadratic formula either and will give the same results.