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A piece of cloth cost ₹300. If the piece was 5 metre longer and each metre of cloth cost ₹2 less, the cost of the piece would have remained unchanged. How long is the original piece of cloth and what is the rate per metre?

Quadratic Equations

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Answer

Let the length of original cloth be x metres

Since, cost of total piece = ₹300

So, cost of each metre = 300x\dfrac{300}{x}

Given, new length is 5 metre more i.e. (x + 5) metres and cost of each metre is ₹2 less i.e. 300x2\dfrac{300}{x} - 2

Since total cost remains unchanged

(x+5)×(300x2)=300(x+5)(3002xx)=300(x+5)(3002x)=300x (On cross multiplication) 300x2x2+150010x300x=02x2+150010x=02(x2+5x750)=0x2+5x750=0x2+30x25x750=0x(x+30)25(x+30)=0(x+30)(x25)=0x+30=0 or x25=0x=30 or x=25\therefore (x + 5) \times \big(\dfrac{300}{x} - 2 \big) = 300 \\[1em] \Rightarrow (x + 5)\big(\dfrac{300 - 2x}{x}\big) = 300 \\[1em] \Rightarrow (x + 5)(300 - 2x) = 300x \text{ (On cross multiplication) } \\[1em] \Rightarrow 300x - 2x^2 + 1500 - 10x - 300x = 0 \\[1em] \Rightarrow -2x^2 + 1500 - 10x = 0 \\[1em] \Rightarrow -2(x^2 + 5x - 750) = 0 \\[1em] \Rightarrow x^2 + 5x - 750 = 0 \\[1em] \Rightarrow x^2 + 30x - 25x - 750 = 0 \\[1em] \Rightarrow x(x + 30) - 25(x + 30) = 0 \\[1em] \Rightarrow (x + 30)(x - 25) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 25 = 0 \\[1em] x = -30 \text{ or } x = 25

Since, length cannot be negative hence, x ≠ -30

∴ x = 25 , 300x\dfrac{300}{x} = 12

Hence, the length of original cloth is 25 metre and cost per metre is ₹12.

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