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Mathematics

A trader buys x identical articles for ₹ 600. If the cost per article were ₹ 5 more, the number of articles that can be bought for ₹ 600 would be four less. Find the value of x.

Linear Equations

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Answer

Given,

A trader buys x identical articles for ₹ 600.

Cost of each article = ₹ 600x\dfrac{600}{x}

If cost is ₹ 5 more then,

New cost of each article = ₹ 600x+5\dfrac{600}{x} + 5

Given,

With the new cost the number of articles that can be bought for ₹ 600 would be four less.

600600x+5=x4600600+5xx=x4600x600+5x=(x4)600x=(600+5x)(x4)600x=600x2400+5x220x600x600x=5x220x24005(x24x480)=0x24x480=0x224x+20x480=0x(x24)+20(x24)=0(x+20)(x24)=0x+20=0 or x24=0x=20 or x=24.\therefore \dfrac{600}{\dfrac{600}{x} + 5} = x - 4 \\[1em] \Rightarrow \dfrac{600}{\dfrac{600 + 5x}{x}} = x - 4 \\[1em] \Rightarrow \dfrac{600x}{600 + 5x} = (x - 4) \\[1em] \Rightarrow 600x = (600 + 5x)(x - 4) \\[1em] \Rightarrow 600x = 600x - 2400 + 5x^2 - 20x \\[1em] \Rightarrow 600x - 600x = 5x^2 - 20x - 2400 \\[1em] \Rightarrow 5(x^2 - 4x - 480) = 0 \\[1em] \Rightarrow x^2 - 4x - 480 = 0 \\[1em] \Rightarrow x^2 - 24x + 20x - 480 = 0 \\[1em] \Rightarrow x(x - 24) + 20(x - 24) = 0 \\[1em] \Rightarrow (x + 20)(x - 24) = 0 \\[1em] \Rightarrow x + 20 = 0 \text{ or } x - 24 = 0 \\[1em] \Rightarrow x = -20 \text{ or } x = 24.

Since, no. of articles cannot be negative.

∴ x = 24.

Hence, x = 24.

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