SOLUTION: A man bought a certain number of chairs for Rs 10,000.He kept one for his own use and sold the rest at the rate Rs 50 more than he gave for one chair.Besides getting his own chair

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Question 550377: A man bought a certain number of chairs for Rs 10,000.He kept one for his own use and sold the rest at the rate Rs 50 more than he gave for one chair.Besides getting his own chair for nothing,he made a profit of Rs 450.How many chairs did he buy?pls...help me.
Found 2 solutions by fcabanski, mananth:
Answer by fcabanski(1391) About Me  (Show Source):
You can put this solution on YOUR website!
n = number of chairs purchased
p = price paid per chair


np = 10,000 so p=10,000/n
(n-1)(p+50)=10450


np+50n-p-50=10450


10000+50n-(10000/n)-50=10450


Subtract 10,000 from both sides and add 50 to both sides. Add 50n and 10,000/n by using n as the common denominator.


(50n^2 - 10000)/n = 500


50n^2 - 10000 = 500n


50n^2 - 500n - 10000 =0


n^2 - 10n - 200 = 0


(n-20)(n+10)=0


n=20 and n=-10


Can't use -10 because there can't be a negative number of chairs. The answer is 20.

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Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
A man bought a certain number of chairs for Rs 10,000.He kept one for his own use and sold the rest at the rate Rs 50 more than he gave for one chair.Besides getting his own chair for nothing,he made a profit of Rs 450.How many chairs did he buy?pls...help me.
number of chairs bought = x
He sold x- 1 chairs
CP of one chair = 10000/x
SP= (10000/x) + 50
SP of (x-1) chairs =%28x-1%29%28%2810000%2Fx%29+%2B50%29
SP-CP= profit
%28x-1%29%28%2810000%2Fx%29+%2B50%29+-10000=+450
%28x-1%29%28%2810000%2B50x%29%2Fx%29%29-10000=450
multiply equation by x
%28x-1%29%2810000%2B50x%29-10000x=450x
10000x+50x^2-10000-50x-10000x=450x
50x^2-500x-10000=0
/50
x^2-10x-200=0
x^2-20x+10x-200=0
x(x-20)+10(x-20)=0
(x-20)(x+10)=0
x= 20 which is positive.
Number of chairs bought = 20