Quadratic Equation word problem...A trader bought a number of articles for Rs.1200. Ten were damaged and he sold each of the rest at Rs. 2 more than what he paid for it, thus clearing a profit of...

Quadratic Equation word problem...

A trader bought a number of articles for Rs.1200. Ten were damaged and he sold each of the rest at Rs. 2 more than what he paid for it, thus clearing a profit of Rs.60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it.

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krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The number of articles purchased = x

Total cost of buying x number of articles = Rs.1200 (given)

Purchase price = (Total cost)/(Number of articles) = 1200/x

Number of articles spoiled = 10 (Given)

Therefore: Number of articles sold = x - 10

Selling price = Purchase price + 2 (Given)

= 1200/x + 2

Profit made = Rs.60 (Given)

Total receipts from sale = (Cost of buying) +Profit = 1200 + 60 = 1260.   ...

We can represent total receipts from sales also as:

(Number of articles sold)*(selling price) = (x - 10)*(1200/x + 2)

= 1200 - 12000/x + 2x - 20 = 2x + 1180 - 12000/x

Therefore:

1260 = 2x + 1180 - 12000/x

Taking all the terms of this equation on left hand side of equation and multiplying each term by -x/2 we get:

x^2 - 40x - 6000 = 0

This equation can be represented also as:

(x- 100)*(x + 60) = 0

Therefore x = 100 or - 60

As number of articles purchased cannot be negative, the number of articles purchase is 100.

Unit purchase price = 1200/x = 1200/100 = Rs.12

Unit sale price = (Unit purchase price) + 2 = 12 + 2 = Rs.14

 

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let the number of articles purchased  be x.

Then the  price  of the article= Rs1200/x.

Excluding the 10 damaged number of the rest of the articles in good condition = x-10.

The selling price of him  is 2 more than the price he paid implies Rs(1200/x +2) is algebraically the selling price. So the amount he gets by selling the x-10 articles in good condition = Rs(1200/x +2)(x-10). But this Rs(1200/x +2)(x-10) is equal to a profit of Rs 60 which  means he gets Rs(1200+60) by sale of x-10 articles in good condition. So the required equation in x   to get the number of articles he purchased is :

Rs(1200/x +2)(x-10) = Rs1260. Multiplying by x  we solve for x:

(1200+2x)(x-10) = 1260x Or

1200x-12000 +2x^2-20x =1260x. Or

2x^2-80x-12000 = 0 Or dividing by 2 we get:

x^2-40x-6000 = 0 is equivalent modified equation , which is a quadratic equation, which could be solved by factorising the left side and equating the factors to zero to  get a practical solution. Or else the left side could be expresed as  a difference of  two squares  and find a solution.

x^2-100x+60x-6000=0 Or

x(x-100)+60(x-100) = 0 Or

(x-100)(x+40) = 0 .

x-100 = 0 or x+40 = 0 Or

x=100 is the practical solution.

Tally:

Rs1200/100 = Rs12 is the price of the article purchased.

100-10 = 90  are the articles in good condition.

90 artiles in good condition he sells at the rate Rs(12+2). So the amount he gets Rs90*12 = Rs1260 by which he gets a profit of  Rs 60  which is in tune with the data.

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