Question

A mixture of 30 pounds of candy sells for $1.10 a pound. The mixture consists of chocolates worth $1.50 a pound and chocolates worth 90 cents a pound. How many pounds of the $1.50 chocolate were used to make the mixture?

Answer 1

10 lbs of the more expensive chocolate was used.

Explanation:

Let `x` be the expensive chocolate ($1.50 per pound) and `y` be the cheap chocolate ($0.90 per pound).

Based on total pounds we know: `x + y = 30`
Based on cost we know `(1.50*x+0.90*y)/(x+y) = 1.10`

We can simplify the cost equation:

`(1.50*x+0.90*y)/(x+y) = 1.10`

cross multiply:

`1.50*x+0.90*y = 1.10(x+y)`

distribute:

`1.50*x+0.90*y = 1.10x+1.10y`

move everything to one side:

`0.4x -0.20y=0`

Now we can solve the system:
`x + y = 30` and `0.4x -0.20y=0`

From the first equation we know that `x=30-y`

Substituting into `0.4x -0.20y=0`, we have:

`0.4(30-y) -0.20y=0`

expanding:

`12-.4y-.2y=0`

collecting:

`12-0.60y=0`

subtract 12 from both sides:

`-0.60y=-12`

divide through by `-0.60`:

`y=20`

Taking `y=20` and substituting into `x+y=30` gives `x=10`.

Since `x` is the more expensive chocolate, we used 10 lbs of the more expensive.

Answer 2

This is a simultaneous equation problem.

Chocolate used at $1.50 per pound is 10 pounds

Chocolate used at $0.90 per pound is 20 pounds

Explanation:

Let the weight of chocolates worth $0.90 a pound be `c_(0.90)`
Let the weight of chocolates worth $1.50 a pound be `c_(1.50)`

Initial conditions:

`c_(0.90)+c_(1.50)=30^("lb")" ".....................Equation(1)`

The value of the blend is `($1.10)/("pound")` so the total value of the final blend is `($1.10)/cancel("pound")xx30 color(white)("d")cancel("pound") = $33.00`

Thus we have:

`$0.90c_(0.90)+$1.5c_(1.5)=$33.00" "................Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You may choose which one you so wish to substitute for. I choose `c_(0.90)`

Consider `Eqn(1)`

Write as `c_(0.90)=30-c_(1.50)" "......Equation(1_a)`

Using `Eqn(1_a)` substitute for `color(red)(c_(0.90))` in `Eqn(2)` giving:
Dropping the units of measerment for now.

`color(green)( 0.90color(red)(c_(0.90))color(white)("dddd.d")+1.5c_(1.50)=33.00)`

`color(green)(0.90( color(red)(30-c_(1.50)))+1.5c_(1.50)=33.00)`

`color(green)(color(white)("d")27-0.9c_(1.50)color(white)("d")+1.5c_(1.50)=33.00)`

`color(green)(color(white)("d")27color(white)("ddddd")+0.6c_(1.50)color(white)("dddd")=33.00)`

Subtract 26 from both sides

`color(green)(0.6c_(1.50)=6`

Divide both sides by `0.6`

`c_(1.50)=10 larr " pounds"`

`c_(0.90)= 30-10=20" pounds"`