A trader bought a number of articles for rupees 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
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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.
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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Answer:
let the total no of aarticles be n
let the value of one article ne a
given:
na=900
a=900/n
(n-5)*(a+2)=980
na-5a+2n=990
900-5(900/n)+2n=990
2n-4500/n=90
n2-2250=45n
n2-45-2250=0
n(n-75)30(n-75)
(n+30) (n-75)
n=-30 n=75
sine the no of arrticles cannot be in negative
the answer is 75
Step-by-step explanation:
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