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A trader brought a number of articles for ₹ 900, five were damaged and he sold each of the rest at ₹ 2 more than what he paid for it. If on the whole he gains ₹ 80, find the number of articles brought.

Linear Equations

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Answer

Let no. of articles brought be x.

Cost of each article = ₹ 900x\dfrac{900}{x}

5 articles were damaged.

No. of articles left = (x - 5).

Given,

No. of articles left were sold at ₹ 2 more than what he paid for it.

S.P. of each article = ₹ 900x+2\dfrac{900}{x} + 2

Given,

⇒ Gain = ₹ 80.

⇒ Total S.P. - Total C.P. = ₹ 80

(x5)(900x+2)900=80x×900x+2x5×900x5×2=80+900900+2x4500x10=980900x+2x2450010xx=980900x+2x2450010x=980x2x2+900x980x10x4500=02x290x4500=02(x245x2250)=0x245x2250=0x275x+30x2250=0x(x75)+30(x75)=0(x+30)(x75)=0x=30 or x=75.\Rightarrow (x - 5)\Big(\dfrac{900}{x} + 2\Big) - 900 = 80 \\[1em] \Rightarrow x \times \dfrac{900}{x} + 2x - 5 \times \dfrac{900}{x} - 5 \times 2 = 80 + 900 \\[1em] \Rightarrow 900 + 2x - \dfrac{4500}{x} - 10 = 980 \\[1em] \Rightarrow \dfrac{900x + 2x^2 - 4500 - 10x}{x} = 980 \\[1em] \Rightarrow 900x + 2x^2 - 4500 - 10x = 980x \\[1em] \Rightarrow 2x^2 + 900x - 980x - 10x - 4500 = 0 \\[1em] \Rightarrow 2x^2 - 90x - 4500 = 0 \\[1em] \Rightarrow 2(x^2 - 45x - 2250) = 0 \\[1em] \Rightarrow x^2 - 45x - 2250 = 0 \\[1em] \Rightarrow x^2 - 75x + 30x - 2250 = 0 \\[1em] \Rightarrow x(x - 75) + 30(x - 75) = 0 \\[1em] \Rightarrow (x + 30)(x - 75) = 0 \\[1em] \Rightarrow x = -30 \text{ or } x = 75.

Since, no. of articles cannot be negative.

Hence, no. of articles = 75.

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