# 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

• -38

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
mast !!
• -29
Answer is 24 all the best
• -41
answer is 75 not 24 ramacash\$
• -17
Ayushi is correct
• -25
• -14
• -17

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
• -23
Number of articles=x( say)

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
• 7
• -12
• -18
best of luck!!!
• -8
Fv
• -4
Sol: let the total number of articles be n. let the value of one article be a. Given that trader bought n articles for Rs 900. So that n×a = 900 and a = 900/n. Also given that five of them were damaged, he sold the remaining articles for Rs.2 extra and he also got a Rs.80 profit. So that the equation would be (n - 5)×(a + 2) = 900 + 80 ⇒ na - 5a +2n -10 = 980. ⇒ 900 - 5(900/n) + 2n = 990 [∴ na = 900 and a = 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 number of goods, it is always positive. Hence number of articles he bought was 75. Hence number of articles he bought was 75
• 0
/bf{That helps u}

• 0