Problema Solution
A bookstore bought 100 books for $ 100. the manager paid three different prices $ 10, $3, and $0.50. If she bought at least one of each type of book, how many books were purchased at each price?
Answer provided by our tutors
let
x = the number of books purchased at $10, x >= 1
y = the number of books purchased at $3, y >= 1
z = the number of books purchased at $0.50, z >= 1
x + y + z = 100
z = 100 - (x + y)
10x + 3y + 0.50z = 100
plug z = 100 - (x + y) into the last equation:
10x + 3y + 0.50(100 - (x + y)) = 100
by simplifying we get:
x = 5(20 - y)/19
click here to see the step by step simplification:
since x needs to be integer and x>0 follows 20 - y must be dividable by 19 thus y = 1
for y = 1
x = 5(20 - 1)/19
x = 5
z = 100 - (5 + 1)
z = 94
there were 5 books purchased at $10, 1 book purchased at $1 and 94 purchased at $0.50.