A trader brought a number of articles for Rs. 900,five articles were found damaged. He sold each of the remaining articles at Rs.2 more than what he paid for it. He got a profit of Rs.80 on the whole transactions. Find the number of articles he bought.
A trader brought a number of articles for Rs. 900,five articles were found damaged. He sold each of the remaining articles at Rs.2 more than what he paid for it. He got a profit of Rs.80 on the whole transactions. Find the number of articles he bought.
let the total no. of articles be n
let the initial value of 1 article be x
the question says that he bought n articles for Rs 900
so n*x = 900 ;
so x = 900/n
It also says that since five of them were damaged, he sold the remaining for Rs. 2 extra
and he also got a Rs.80 profit
number of items=n-5
New price. =x+2
Profit +cost =900+80
so the equation formed would be (n-5)(x+2) = 900+ 80
nx -5x +2n -10 = 980
900 - 5(900/n) + 2n = 990 [as nx = 900 and x = 900/n]
2n - (4500/n) = 90
n - (2250/n) = 45
n2 - 2250 = 45n
n2 -45n -2250 = 0
n2 - 75n + 30n -2250 = 0
n (n-75) +30 (n-75) = 0
(n-75) (n+30) = 0
n = 75 or -30
since n is the no. of goods, it is always positive
so the number of articles he bought was 75
let the total no. of articles be n
let the initial value of 1 article be x
the question says that he bought n articles for Rs 900
so n*x = 900 ;
so x = 900/n
It also says that since five of them were damaged, he sold the remaining for Rs. 2 extra
and he also got a Rs.80 profit
number of items=n-5
New price. =x+2
Profit +cost =900+80
so the equation formed would be (n-5)(x+2) = 900+ 80
nx -5x +2n -10 = 980
900 - 5(900/n) + 2n = 990 [as nx = 900 and x = 900/n]
2n - (4500/n) = 90
n - (2250/n) = 45
n2 - 2250 = 45n
n2 -45n -2250 = 0
n2 - 75n + 30n -2250 = 0
n (n-75) +30 (n-75) = 0
(n-75) (n+30) = 0
n = 75 or -30
since n is the no. of goods, it is always positive
so the number of articles he bought was 75
