A trader bought 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 transaction. Find the number of articles he bought.
let the no. of articles be x then,
let the original price be y
(x-5)2 = 900+80
2x-10=980
2x = 970
x=970/2 = 485
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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
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
Hope it helps....Thumbs up plz!!
- 197
Let no. of articles = x and cost of each one be Rs. y.
According to the question,
xy = 900. Therefore, y = 900 / x. ------------ (1)
And, (x-5) (y+2) = 900 + 80
Implies, xy + 2x - 5y - 10 = 900 + 80 (since xy = 900 so they are the same thing)
Implies, 2x - 5y - 90 = 0
Implies, 2x - 5(900 / x) - 90 = 0
Implies, 2x2 - 4500 - 90x = 0
Implies, x2 - 45x - 2250 = 0
Implies, x2- 75n + 30n - 2250 = 0
Implies, x(x- 75) +30(x - 75) = 0
Implies,(x - 75) (x + 30) = 0
Therefore, x = 75 and x = -30.
Since no. of articles can never be -ve, x = 75.
Therefore, Number of articles = 75.
- 70
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
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
Hope it helps....Thumbs up plz!!
- 11
therefore c.pof per article= 900/x
profit on s.p= (x-5)2
real profit=( x-5)2- 5(900/x)
given profit = 80
therefore(x-5)2-5(900/x)= 80
2x- 10-4500/x= 80
2x^2 -10x- 4500= 80x
x^2 -5x-2250=40x
x^2-45x- 2250=0
x^2-75x+30x-2250=0
(x-75)(x+30)=0
therefore,
no. of articles =75
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