Question

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?

Answer 1

`14` apples and `6` bananas.

Explanation:

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

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

Which simplifies to:

`11.50 = .5x + 15 - .75x`

Combine similar terms:

`-3.50 = -.25x`

Multiply `-4` on both sides:

`x = 14`

So the amount of apples is `14` and the amount of bananas is `(20-14)`, which equals `6`.

Answer 2

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")`

Let the total count of apples be `A_c`
Let the total cost (value) of apples be `A_v` at `$0.50` each

Let the total total count of bananas be `B_c`
Let the total cost (value) of bananas be `B_v` at `$0.75` each

`color(brown)("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