Question

A tradesman bought a number of meters of cloth for Rs. $50$. He kept $5$ meters and sold the rest at Re. $1$ per meter more than he paid for and received Rs. $10$ more than he originally spent. How many meters of cloth did he buy?

Answer 1

Step 1: Determining the quadratic equation in $x$

Let the quantity of the cloth bought for Rs. $50$ be $x$ meters.

The cost price of $1$ meter of cloth will be Rs. $\frac{50}{x}$.

It is given that, the tradesman retains $5$ meters of cloth and sells the rest for Rs. $1$ profit per meter.

So, the new quantity of cloth becomes $(x-5)$ meters and the new cost price per meter becomes Rs. $\frac{50}{x-5}$.

As per the given condition and using the formula $\mathrm{Selling}\mathrm{Price}=\mathrm{Cost}\mathrm{Price}+\mathrm{Profit},$ we get

$$\begin{gathered} (\frac{50}{x}+1)(x-5)=50+10 \\ \Rightarrow \left(\frac{50+x}{x}\right)(x-5)=60 \\ \Rightarrow (x+50)(x-5)=60x \\ \Rightarrow x(x-5)+50(x-5)-60x=0 \\ \Rightarrow x^2-5x+50x-250-60x=0 \\ \Rightarrow x^2-15x-250=0 \end{gathered}$$

Step 2: Determining the number of meters of cloth the tradesman brought

By factorization method,

$x^2-15x-250=0$

For splitting the middle term, i.e. $-15$ we need to find two numbers such that their sum is $-15$ and product is $-250$.

Such numbers can be $-25$ and $10$.

$$\begin{gathered} \Rightarrow x^2-25x+10x-250=0 \\ \Rightarrow x(x-25)+10(x-25)=0 \\ \Rightarrow (x-25)(x+10)=0 \\ \therefore x=-10,25 \end{gathered}$$

Since the quantity of the cloth cannot be negative,

$\therefore x=25$

Therefore, the tradesman bought $25$ meters of cloth.