SOLUTION: A shopkeeper buys a certain number of books for Rs.720.if the cost per book was Rs.5 less, thenumber of books that could be bobought for Rs.720 could be 2 more.taking the original

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A shopkeeper buys a certain number of books for Rs.720.if the cost per book was Rs.5 less, thenumber of books that could be bobought for Rs.720 could be 2 more.taking the original       Log On


   



Question 867179: A shopkeeper buys a certain number of books for Rs.720.if the cost per book was Rs.5 less, thenumber of books that could be bobought for Rs.720 could be 2 more.taking the original cost of each book to be Rs.x, write an equation and solve it.
Found 2 solutions by richwmiller, dkppathak:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
n*c=720,
(n+2)*(c-5)=720
c = 45, n = 16
new cost 40 new number 18


Answer by dkppathak(439) About Me  (Show Source):
You can put this solution on YOUR website!
A shopkeeper buys a certain number of books for Rs.720.if the cost per book was Rs.5 less, then umber of books that could be bought for Rs.720 could be 2 more.taking the original cost of each book to be Rs.x, write an equation and solve it.
Answer :
let cost of 1 book purchased is Rs X
number of books purchased for Rs 720 will be = 720/x
as per given conditions
if cost of one book is reduced by RS 5 each
than new cost of 1 book will be = Rs (X-5)
number of books will be =720/(x-5)
if cost reduced RS 5 for each books than 2 books will purchased more
720/(x-5)-720/x=2
720x-720(x-5)/x(x-5) =2
720(x-x+5)/x^2-5x=2
720x5 =2(x^2-5x)
1800=X^2-5x
X^2-5x-1800=0
by splitting middle term
X^2-45x+40x-1800 =0
x(x-45)+40(x-45)=0
(x-45)(x+40)=0
x=45 or x=-40
neglecting -40 as cost of book can not be negative so cost of book will be Rs 45 each
therefor cost of 1 book will be Rs 45 each