SOLUTION: Sam buys 3 kinds of fruit for 40 dollars, 10 dollars, and 1 dollar each. He pays 259 dollars for 100 pieces of fruit. How many of the cheapest did he buy?
Question 173499: Sam buys 3 kinds of fruit for 40 dollars, 10 dollars, and 1 dollar each. He pays 259 dollars for 100 pieces of fruit. How many of the cheapest did he buy? Found 2 solutions by ankor@dixie-net.com, josmiceli:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! : Sam buys 3 kinds of fruit for 40 dollars, 10 dollars, and 1 dollar each. He pays 259 dollars for 100 pieces of fruit. How many of the cheapest did he buy?
:
Three kinds of fruit x, y, z
:
x + y + z = 100
and
40x + 10y + z = 259
:
Subtract the 1st equation from the 2nd
40x + 10y + z = 259
x + y + z = 100
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39x + 9y = 159
Simplify, divide by 3
13x + 3y = 53
3y = 53-13x
y =
We know there can't be many $40 fruits (x) and y has to be an integer
try x = 2
y =
y =
y = 9 ea $10 fruit, this is the only positive integer solution
:
Find z
2 + 9 + z = 100
z = 100 - 11
z = 89 ea $1 animals, (the cheapest fruit)
:
Check in the Cost equation
40(2) + 10(27) + 1(89)
80 + 90 + 89 = 259; confirms our solutions
You can put this solution on YOUR website! Call the quantities of the 3 kinds of fruit ,, and
Given:
(1)
(2)
---------------------------
Just observing, I see I have 3 unknowns and only 2
equations. It looks like I need another equation,
but I'll try to solve anyway
Subtract (1) from (2)
divide both sides by
I'll make a chart and see if I can get
----------------------------------------------
-- a -- b -- c -- 13a+3b -- a+b+c
----------------------------------------------
-- 4 --- 0 --- 1 --- 53 ----- 5
-- 3 -- 14/3 (no good, is not a whole number)
-- 1 -- 40/3 (no good, is not a whole number)
-- 2 --- 9 --- 47 --- 53 --- 100
These are the only values for a,b, and c that work out
He bought 47 of the cheapest fruit
