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A shopkeeper buys a certain number of books for ₹960. If the cost per book was ₹8 less, the number of books that could be bought for ₹960 would be 4 more. Taking the original cost of each book to be ₹x, write an equation in x and solve it to find the original cost of each book.

Quadratic Equations

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Answer

Let the cost of each book be ₹x

Number of books that can be bought for ₹960 = 960x\dfrac{960}{x}

If the cost of book is ₹8 less then number of books that can be bought for ₹960 = 960x−8\dfrac{960}{x - 8}

Given, number of books would be 4 more if price would be ₹8 less

∴960x−8−960x=4⇒960x−960(x−8)x(x−8)=4⇒960x−960x+7680x2−8x=4⇒7680=4(x2−8x)⇒4x2−32x−7680=0⇒4(x2−8x−1920)=0⇒x2−8x−1920=0⇒x2−48x+40x−1920=0⇒x(x−48)+40(x−48)=0⇒(x−48)(x+40)=0⇒x−48=0 or x+40=0x=48 or x=−40\therefore \dfrac{960}{x - 8} - \dfrac{960}{x} = 4 \\[1em] \Rightarrow \dfrac{960x - 960(x - 8)}{x(x - 8)} = 4 \\[1em] \Rightarrow \dfrac{960x - 960x + 7680}{x^2 - 8x} = 4 \\[1em] \Rightarrow 7680 = 4(x^2 - 8x) \\[1em] \Rightarrow 4x^2 - 32x - 7680 = 0 \\[1em] \Rightarrow 4(x^2 - 8x - 1920) = 0 \\[1em] \Rightarrow x^2 - 8x - 1920 = 0 \\[1em] \Rightarrow x^2 - 48x + 40x - 1920 = 0 \\[1em] \Rightarrow x(x - 48) + 40(x - 48) = 0 \\[1em] \Rightarrow (x - 48)(x + 40) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x + 40 = 0 \\[1em] x = 48 \text{ or } x = -40

Since, cost of book cannot be negative hence, x ≠ -40

∴ x = 48

Hence, the cost of each book is ₹48.

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