Question

A piece of cloth costs Rs 200. If the piece was 5 m longer and each metre of cloth costs Rs 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre?

Answer 1

Step 1:

Form a quadratic equation:

Let the length of the piece be $x\mathrm{meters}$.

The rate of the piece per meter will be $\mathrm{Rs}.\frac{200}{x}$.

If the piece was 5 m longer, then, the new length will be $\left(x+5\right)\mathrm{meters}$.

The new rate of piece per meter will be $\mathrm{Rs}.\frac{200}{x+5}$

According to the question, the difference between the rate of piece per meter is $\mathrm{Rs}.2$, then,

$\begin{array}{rcl}\frac{200}{x}-\frac{200}{x+5}& =& 2\\ & \Rightarrow & \frac{200\left(x+5\right)-200x}{x\left(x+5\right)}=2\\ & \Rightarrow & \frac{200x+1000-200x}{x\left(x+5\right)}=2\\ & \Rightarrow & \frac{1000}{x\left(x+5\right)}=2\\ & \Rightarrow & 1000=2x\left(x+5\right)\\ & \Rightarrow & 500=x^2+5x\\ & \Rightarrow & x^2+5x-500=0\end{array}$

Step 2:

Determine the length of the piece:

Use the factorization method to solve the equation:

$$\begin{gathered} x^2+5x-500=0 \\ \begin{array}{rcl}& \Rightarrow & x^2+25x-20x-500=0\\ & \Rightarrow & x\left(x+25\right)-20\left(x+25\right)=0\\ & \Rightarrow & \left(x-20\right)\left(x+25\right)=0\\ & \Rightarrow & x-20=0\mathrm{or}x+25=0\\ & \Rightarrow & x=20\mathrm{or}x=-25\end{array} \end{gathered}$$

But the length cannot be negative, rejecting $x=-25$

$\therefore x=20$

Therefore, the length of the piece is $20\mathrm{meters}$.

Step 3:

Find the original rate per meter:

$\begin{array}{rcl}\mathrm{Rate}\mathrm{per}\mathrm{metre}& =& \frac{\mathrm{Original}\mathrm{price}}{\mathrm{Original}\mathrm{length}\mathrm{of}\mathrm{piece}}\\ & =& \frac{200}{x}\\ & =& \frac{200}{20}\\ & =& \mathrm{Rs}.10\mathrm{per}\mathrm{meter}\end{array}$

Therefore, the original rate of piece per meter is $\mathrm{Rs}.10$.

Final Answer:

Therefore, the length of the piece is 20 meters and the original rate of piece per meter is $\mathrm{Rs}.10$.