Math, asked by Chaitanya6677, 1 year ago

A trader bought a number of articles for rs.900 . five articles were found to be damaged . he sold each of the remaining articles for rs.2 more than what he paid for it.he got a profit of rs.80 on the whole transaction . find the number of articles he had bought.

Answers

Answered by amitnrw
19

Answer:

75 Articles

Step-by-step explanation:

A trader bought a number of articles for rs.900 . five articles were found to be damaged . he sold each of the remaining articles for rs.2 more than what he paid for it.he got a profit of rs.80 on the whole transaction . find the number of articles he had bought.

Let Say Trader bought = X articles

He Paid Rs 900 in Total

Cost Price of one article = 900/X  Rs

Damaged Articles = 5

Sell-able articles = X - 5

Selling Price = 900/X  + 2   = (900 + 2X)/X  Rs

Total Selling Amount = (X-5)((900 + 2X)/X)  Rs

He got a profit of Rs 800

so total Selling Price = 900 + 80 = Rs 980

(X-5)((900 + 2X)/X) = 980

=> (X-5)(900 + 2X) = 980X

=> 2X² + 890X - 4500 = 980X

=> 2X² - 90X - 4500 = 0

=> X² -45X - 2250 = 0

=> X² -75X + 30X - 2250 = 0

=> X (X -75) + 30(X -75) = 0

=> (X+30)(X-75) = 0

=> X = 75  as Numbers of articles can be negative

So Trader Bought 75 Articles

Answered by gokulumenon008
10

Answer:

Step-by-step explanation:

Let the inner of articles be x

Let the cost of one article be a

Cost of article = x×a=900

a=900/x

According to the information

{x-5}{a+2}=900+80

xa-5a+2x-10=980

900-5×900/x+2x=990

2x-4500/x=90

X-2250/x=45

x^-45x-2250=0

Using splitting the middle term formula

a+b =-45. ab=-2250

Let a =-75 and b =30

-75+30=-45 and -75×30=-2250

{x^-75x}+{30x-2250}=0

x{x-75}+30{x-75}=0

{x-75}{x+30}

So x=75. OR. x=-30

Cost of one article = 900/x

=900/75

=Rs12

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