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

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*(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#.

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:

#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