Question

A grocer wants to mix two kinds of candy. One kind sells for $1.30 per pound, and the other sells for $2.40 per pound. He wants to mix a total of 19 pounds and sell it for $1.55 per pound. How many pounds of each kind should he use in the new mix? (Round off the answers to the nearest hundredth.)

Answer 1

Step-by-step explanation:

We are given the following information;

A grocer wants to mix two types of candy, which we shall call x and y.

He wants to mix a total of 19 pounds which means, he would have the following;

`x+y=19---(1)`

He also intends to sell the total mix for $1.55 per pound. At that rate, his total sales would be;

`\begin{gathered} \text{Total mix}=1.55*19 \\ \text{Total mix}=29.45 \end{gathered}`

Note that one kind of candy sells for $1.30 per pound, that is;

`1.30x`

The other kind sells for $2.40 per pound, that is;

`2.40y`

The total mix would now sell for;

`1.30x+2.40y=29.45---(2)`

We can now solve the system of equations and determine the values of x and y as follows;

`\begin{gathered} x+y=19---(1) \\ 1.30x+2.40y=29.45---(2) \end{gathered}`

From equation (1), make x the subject of the equation and we'll have;

`x=19-y`

Substitute for the value of x into equation (2)

`\begin{gathered} 1.30(19-y)+2.40y=29.45 \\ 24.70-1.30y+2.40y=29.45 \end{gathered}`

We can now combine like terms;

`\begin{gathered} 2.40y-1.30y=29.45-24.70 \\ 1.10y=4.75 \end{gathered}`

Divide both sides by 1.10;

`\begin{gathered} (1.10y)/(1.10)=(4.75)/(1.10) \\ y=5.225 \\ \text{Rounded to the nearest hundredth;} \\ y=5.26 \end{gathered}`

We can now substitute for the value of y into equation (1);

`\begin{gathered} x+y=19 \\ x+5.26=19 \\ x=19-5.26 \\ x=13.74 \end{gathered}`

Therefore, he should use the following mix;

ANSWER:

For the $1.30 candy = 13.74 pounds

For the $2.40 candy = 5.26 pounds