Quadratic word problem....A trader bought a number of articles for Rs 900. Five articles were damaged and he sold each of the rest at Rs 3 more than what he paid for it, thus getting a profit of Rs...

Quadratic word problem....

A trader bought a number of articles for Rs 900. Five articles were damaged and he sold each of the rest at Rs 3 more than what he paid for it, thus getting a profit of Rs 150 on the whole transaction. Find the number of articles he bought.

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Top Answer

krishna-agrawala's profile pic

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

Posted on

Let the number of article bought be = x

As the total amount paid for x numbers is Rs. 900

Purchase cost paid per unit = 900/x

Selling price per unit is Rs, 3 more than purchase cost per unit,Therefore selling price = 900/x + 3

Units sold = (Units purchased) - (Units damaged) = x - 5

Total value of sales = (Unit selling price)*(Units sold) = (900/x + 3)*(x - 5)

= 900 - 4500/x + 3x - 15 = 3x + 885 - 4500/x

However as it is given that total profit is Rs. 150. The total value of sales is also

= Total purchase cost + profit = 900 + 150 = 1050

Therefore:

3x + 885 - 4500/x = 1050

Taking all the terms on left hand side the equation and multiplying both sides by x/3 the equation becomes:

x^2 - 55x  - 1500 = 0

We then factorise the left hand side of equation in steps given below:

x^2 - 75x + 20x - 1500 = 0

x(x - 75) + 20(x - 75) = 0

(x + 20)(x - 75) = 0

Therefor x = - 20, or x = 75.

As the number of articles bought cannot be negative, 75 is the correct value.

Answer:

Number of articles bought = 75

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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 mount of by sale of 70 articles = 70*15 = 1050

Profit = 1050 - 900 = 150

 

 

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