Angela bought apples and bananas at the fruit stand. She bought 20 pieces of fruit and spent $11.50. Apples cost $.50 and bananas cost $.75 each. How many of each did she buy?

2 Answers

1414 apples and 66 bananas.

Explanation:

So let's say the amount of apples is xx, this makes the amount of bananas 20-x20x. With this and the prices, you can create an equation.

11.50 = .5*(x)+ .75*(20-x)11.50=.5(x)+.75(20x)

Which simplifies to:

11.50 = .5x + 15 - .75x11.50=.5x+15.75x

Combine similar terms:

-3.50 = -.25x 3.50=.25x

Multiply -44 on both sides:

x = 14x=14

So the amount of apples is 1414 and the amount of bananas is (20-14)(2014), which equals 66.

Jan 31, 2018

14 apples and 6 bananas

Every step shown so you can see where things comes from.

You would normally expect the solution to look like that of Llama's

Explanation:

color(blue)("Setting up the known relationships")Setting up the known relationships

Let the total count of apples be A_cAc
Let the total cost (value) of apples be A_vAv at $0.50$0.50 each

Let the total total count of bananas be B_cBc
Let the total cost (value) of bananas be B_vBv at $0.75$0.75 each

color(brown)("Relating counts")Relating counts
It is given that A_c+B_c=20" "...................Equation(1)

color(brown)("Relating costs")
"[Apple count x cost each]"+"[Banana count x cost each]=Total cost"

[color(white)("ddd")A_c color(white)("d.dd")xxcolor(white)("d")$0.5color(white)("ddd")]+[color(white)("ddddd")B_c color(white)("dd.d")xxcolor(white)("d")$0.75]=$11.50

$0.5A_c+$0.75B_c=$11.50" ".........................Equation(2)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Answering the question")

From Eqn(1) we have color(red)(A_c=20-B_c)" "......Equation(1_a)

Using Eqn(1_a) substitute for color(red)(A_c) is Eqn(2)

color(green)($0.5color(red)((A_c)) color(white)("dddd")+$0.75B_c=$11.50)" ".........Equation(2)

color(green)($0.5color(red)((20-B_c))+$0.75B_c=$11.50)

$10.0-$0.5B_c+$0.75B_c=$11.50

Subtract $10 from both sides

$0.75B_c-$0.5B_c=$1.50

$0.25B_c=$1.50

To 'get rid' of the decimals multiply both sides by 100

$25B_c=$150

Divide both sides by $25

B_c=6
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Eqn(1) substitute for color(red)(B_c)

color(green)(A_c+color(red)(B_c)=20)" "......................Equation(1)

color(green)(A_c+color(red)(6)=20)

Subtract 6 from both sides 'A_c=14

So the answer is:

14 apples and 6 bananas