SOLUTION: Mike bought some kilograms of apples and some kilograms of oranges and spent
$27.25. The apples cost $3.10 per kilogram and oranges cost $4.95 per kilogram.
If apples become $1
Question 211973: Mike bought some kilograms of apples and some kilograms of oranges and spent
$27.25. The apples cost $3.10 per kilogram and oranges cost $4.95 per kilogram.
If apples become $1.55 more expensive and oranges become $1.15 chaper he will
spend $30. How many ograms of apples and oranges did Mike buy? Answer by drj(1380) (Show Source): You can put this solution on YOUR website! Mike bought some kilograms of apples and some kilograms of oranges and spent
$27.25. The apples cost $3.10 per kilogram and oranges cost $4.95 per kilogram.
If apples become $1.55 more expensive and oranges become $1.15 cheaper he will
spend $30. How many kilograms of apples and oranges did Mike buy?
1. Let x=number of kilograms of apples and y=number of kilograms of oranges.
2. Cost of apples = x*$3.10 =3.10x
3. Cost of oranges= y*4.95=4.95y
4. Total cost = 27.25 = 3.10x+4.95y
5. Let's look at "If apples become $1.55 more expensive and oranges become $1.15 cheaper he will spend $30."
a. For apples 3.10+1.55=4.65
b. For oranges 4.95-1.15=3.80
c. New total cost = 30 = 4.65x+3.80y
d. Note we still have the same number of apples x and oranges y.
6. So now we have two equations and two unknowns
a. 27.25 = 3.10x+4.95y
b. 30 = 4.65x+3.80y
Need to solve for x and y. The steps below will show x=4 and y=3
Solve: We'll use substitution. After moving 4.95*y to the right, we get: , or . Substitute that
into another equation: and simplify: So, we know that y=3. Since , x=4.
Answer: .
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