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Quadratic Equations
RD Sharma
CBSE
Grade 10

Question:

A piece of cloth costs $₹200$. If the piece were $5\text{ m}$ longer and each metre 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 metre?

Step-by-Step Solution

Key Concept: Length $= x\text{ m}$, rate $= 200/x$. $(x + 5)\left(\dfrac{200}{x} - 2\right) = 200 \Rightarrow 200 - 2x + \dfrac{1000}{x} - 10 = 200 \Rightarrow -2x - 10 + \dfrac{1000}{x} = 0 \Rightarrow 2x^2 + 10x - 1000 = 0 \Rightarrow x^2 + 5x - 500 = 0 \Rightarrow (x + 25)(x - 20) = 0 \Rightarrow x = 20\text{ m}$. Rate $= 200/20 = ₹10$ per metre.
$(x + 5)\left(\dfrac{200}{x} - 2\right) = 200 \Rightarrow -2x - 10 + \dfrac{1000}{x} = 0$. [1.5 Marks]
$2x^2 + 10x - 1000 = 0 \Rightarrow x^2 + 5x - 500 = 0$. [1.5 Marks]
$(x + 25)(x - 20) = 0 \Rightarrow x = 20\text{ m}$. [1.0 Mark]
Original rate $= 200/20 = ₹10$ per metre. Length $= 20\text{ m}$, Rate $= ₹10$/m. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Forming cost equation: 1.5 Marks
Simplifying to $x^2 + 5x - 500 = 0$: 1.5 Marks
Solving length $x = 20\text{ m}$: 1.0 Mark
Finding rate $= ₹10$/m: 1.0 Mark

Correct Answer:
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