Chapter 6.2, Problem 18P

Suppose you got 8 mangoes and 3 apples for $18 and 3 mangoes and 5 apples for $14.50. How much does each mango and each apple cost?

To determine

To find: The price of each mango and each apple by given conditions.

Answer to Problem 18P

The cost of mango is $1.5 and the cost of apple is $2.

Explanation of Solution

Given information:

The price of each mango and each apple when you got 8 mangoes and 3 apples is

$18 and 3 mangoes and 5 apples is $14.50

Calculation:

Suppose you got 8 mangoes and 3 apples for $18 and 3 mangoes and 5 apples for

$14.50. Let us assume the cost of mango is x and the cost of apple is y .

Hence we can write the above statement in equation form as follows:

  $\begin{array}{l}8x+3y=18\\ 3x+5y=14.50\end{array}$

In order to solve two systems by substitution, we solve one of the equations for one

of the variables, and plug this into other equation.

  $\begin{array}{l}8x+3y=18\\ 3x+5y=14.50\end{array}$

As it is given that

  $\begin{array}{l}8x+3y=18\\ \Rightarrow x=\frac{18}{8}-\frac{3}{8}y\end{array}$

Now we substitute $x=\frac{18}{8}-\frac{3}{8}y$ into $3x+5y=14.50$ , we get,

  $\begin{array}{l}3x+5y=14.50\\ \Rightarrow 3\left(\frac{18}{8}-\frac{3}{8}y\right)+5y=14.50\\ \Rightarrow \frac{54}{8}-\frac{9}{8}y+5y=14.50\\ \Rightarrow \left(\frac{-9+40}{8}\right)y=\frac{116-54}{8}\\ \Rightarrow \frac{31}{8}y=\frac{62}{8}\\ \Rightarrow y=\frac{62}{31}=2\end{array}$

By plugging $y=2$ in equation $x=\frac{18}{8}-\frac{3}{8}y$ we get

  $\begin{array}{l}x=\frac{18}{8}-\frac{3}{8}y\\ \Rightarrow x=\frac{18}{8}-\frac{6}{8}\\ \Rightarrow x=\frac{12}{8}=1.5\end{array}$

Therefore the solution is

  $\left(x,y\right)=(1.5,2)$

Therefore the cost of mango is $1.5 and the cost of apple is $2.