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Mathematics

A shopkeeper buys x books for ₹ 720.

(i) Write the cost of 1 book in terms of x.

(ii) If the cost per book be ₹ 5 less, the number of books that could be bought for ₹ 720 would be 2 more.

Write down the equation in x and solve it to find x.

Quadratic Equations

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Answer

(i) Given,

Cost of x books = ₹ 720

The cost of one book = ₹ 720x\dfrac{720}{x}

Hence, cost of one book = ₹ 720x\dfrac{720}{x}.

(ii) If cost of per book is ₹5 less,

Then new cost per book = 720x5\dfrac{720}{x} - 5

According to question,

720720x5=x+27207205xx=x+2720x7205x=x+2720x=(x+2)(720x5x)720x=720x5x2+144010x720x720x+5x21440+10x=05x2+10x1440=05(x2+2x288)=0x2+2x288=0x2+18x16x288=0x(x+18)16(x+18)=0(x16)(x+18)=0(x16)=0 or (x+18)=0 ….[Using zero-product rule]x=16 or x=18\Rightarrow \dfrac{720}{\dfrac{720}{x} - 5} = x + 2 \\[1em] \Rightarrow \dfrac{720}{\dfrac{720 - 5x}{x}} = x + 2 \\[1em] \Rightarrow \dfrac{720x}{720 - 5x} = x + 2 \\[1em] \Rightarrow 720x = (x + 2)(720x - 5x) \\[1em] \Rightarrow 720x = 720x - 5x^2 + 1440 - 10x \\[1em] \Rightarrow 720x - 720x + 5x^2 - 1440 + 10x = 0 \\[1em] \Rightarrow 5x^2 + 10x - 1440 = 0 \\[1em] \Rightarrow 5(x^2 + 2x - 288 )= 0 \\[1em] \Rightarrow x^2 + 2x - 288 = 0 \\[1em] \Rightarrow x^2 + 18x - 16x - 288 = 0 \\[1em] \Rightarrow x(x + 18) - 16(x + 18) = 0 \\[1em] \Rightarrow (x - 16)(x + 18) = 0 \\[1em] \Rightarrow (x - 16) = 0 \text{ or } (x + 18) = 0 \text{ ….[Using zero-product rule]} \\[1em] \Rightarrow x = 16 \text{ or } x = -18

Since, number of books cannot be negative.

∴ x = 16

Hence, equation obtained is x2 + 2x - 288 = 0 and the value of x = 16.

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