# 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 Jan 30, 2018 #### Answer: $14$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. $11.50 = .5 \cdot \left(x\right) + .75 \cdot \left(20 - x\right)$Which simplifies to: $11.50 = .5 x + 15 - .75 x$Combine similar terms: $- 3.50 = - .25 x$Multiply $- 4$on both sides: $x = 14$So the amount of apples is $14$and the amount of bananas is $\left(20 - 14\right)$, which equals $6$. Jan 31, 2018 #### Answer: 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: $\textcolor{b l u e}{\text{Setting up the known relationships}}$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 $\textcolor{b r o w n}{\text{Relating counts}}$It is given that ${A}_{c} + {B}_{c} = 20 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$$\textcolor{b r o w n}{\text{Relating costs}}$$\text{[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) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Answering the question}}$From $E q n \left(1\right)$we have $\textcolor{red}{{A}_{c} = 20 - {B}_{c}} \text{ } \ldots \ldots E q u a t i o n \left({1}_{a}\right)$Using $E q n \left({1}_{a}\right)$substitute for $\textcolor{red}{{A}_{c}}$is $E q n \left(2\right)$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 $E q n \left(1\right)$ substitute for $\textcolor{red}{{B}_{c}}$

$\textcolor{g r e e n}{{A}_{c} + \textcolor{red}{{B}_{c}} = 20} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

$\textcolor{g r e e n}{{A}_{c} + \textcolor{red}{6} = 20}$

Subtract 6 from both sides $' {A}_{c} = 14$

So the answer is:

14 apples and 6 bananas