# Q1.A shopkeeper buys a number of pens for Rs80.If he had bought 4 more pens for the same amount , each pen would have cost him Rs 1 less. How many pens did he buys?

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by shab

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by shab

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No. of pens bought=16

A shopkeeper buys a number of pens 4 rs. 80.

If he had bought 4 more pens for the same amount,

each pen would have cost him 1 less.

How many pens did he buy?

Let the total number of pens bought by a man be x

Total cost of pen = Rs 80 / x

Cost of each pen = Rs 80 / (x + 4)

According to question 80 / (x - 80) / (x + 4) = 1

80 (1/(x - 1)(x + 4) = 1

(x + 4 - x) / x(x + 4) = 1/80

4 / x^2 + 4x = 1/80

80 (4) = 1(x^2 + 4x)... (by cross multiplication)

320 = x^2 + 4x x^2 + 4x - 320 = 0 ...(ta

If he had bought 4 more pens for the same amount,

each pen would have cost him 1 less.

How many pens did he buy?

Let the total number of pens bought by a man be x

Total cost of pen = Rs 80 / x

Cost of each pen = Rs 80 / (x + 4)

According to question 80 / (x - 80) / (x + 4) = 1

80 (1/(x - 1)(x + 4) = 1

(x + 4 - x) / x(x + 4) = 1/80

4 / x^2 + 4x = 1/80

80 (4) = 1(x^2 + 4x)... (by cross multiplication)

320 = x^2 + 4x x^2 + 4x - 320 = 0 ...(ta

320 = x^2 + 4x x^2 + 4x - 320 = 0 ...(taking R.H.S. to L.H.S)

x^2 + 20x - 16x - 320 = 0

(x^2 + 20x ) - (16x - 320) = 0

x(x + 20) - 16(x + 20) = 0

(x - 16)(x+ 20) = 0 either x - 16 = 0 or x + 20 = 0

x = 16 or x = -20

Since we had assumed the total number of pen to be x

here x =16 (x does not equal to -20 as the number of pens is never negative)

So the total number pens bought by a man is 16

x^2 + 20x - 16x - 320 = 0

(x^2 + 20x ) - (16x - 320) = 0

x(x + 20) - 16(x + 20) = 0

(x - 16)(x+ 20) = 0 either x - 16 = 0 or x + 20 = 0

x = 16 or x = -20

Since we had assumed the total number of pen to be x

here x =16 (x does not equal to -20 as the number of pens is never negative)

So the total number pens bought by a man is 16