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.
Answers
total profit = Rs 80
Number of articles sold = 80/2 = 40
Total number of articles bought = 40+ 5 = 45
we can also find the buying price of the articles = Rs 900/45
= Rs 20 each
Answer:
75
Step-by-step explanation:
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.
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Syeda , SubjectMatterExpert
Member since Jan 25 2017
Answer.
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