Question
Fri January 06, 2012 By: Mahesh Joshi

# Problem sum

Fri January 06, 2012

Le x be the number of articles purchased. The amount paid for the x articles is Rs 900.

So the price of the article = 900/x.

The number good articles = x-5.

The selling price is more than the purchase price by Rs 3 implies he sells at Rs(900/x +3) per article.

Therefore the amount he gets by selling (x-5) articles in good condition = number of artilcles sold * selling price = (x-5)(900/x +3) but this is equal to Rs150 profit. Or Rs(900+150) in cash. So,

(x-5)((900/x+3 ) = 1050. Multiply by x an we get:

(x-5)(900+3x) = 1050x Or

3x^2+(900-15)x-4500 = 1050x Or

x^2 +885x-1050x+4500 = 0 Or

3x^2 - 165x - 4500 = 0. This is a quadratic equation in x. This could be solved either by factorising the left side and equating each factor to zero. Or by the quadratic formula for roots: If ax^2+bx+c = 0, then x = {-b +or- (b^2-4ac)^0.5}/(2a). So,

x = [165 +or- sqrt(165^2- - 4*3*4500)^(1/2)]/(2*3)

= (165 +or- 285)/6

= 75.

So the number of articles he brought = 75

Tally:

No of articles : 900/75 =12.

The damaged =5. Good articles = 75 - 5 =70.

Selling price = 12+3 =15.

The amount of by sale of 70 articles = 70*15 = 1050

Profit = 1050 - 900 = 150

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