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?

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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?

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Answer 1 · 2018-01-07T17:39:10

10 lbs of the more expensive chocolate was used.

Explanation:

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

Based on total pounds we know: $x + y = 30$ $x + y = 30$
Based on cost we know $\frac{1.50 \cdot x + 0.90 \cdot y}{x + y} = 1.10$ $(1.50*x+0.90*y)/(x+y) = 1.10$

We can simplify the cost equation:

$\frac{1.50 \cdot x + 0.90 \cdot y}{x + y} = 1.10$ $(1.50*x+0.90*y)/(x+y) = 1.10$

cross multiply:

$1.50 \cdot x + 0.90 \cdot y = 1.10 (x + y)$ $1.50*x+0.90*y = 1.10(x+y)$

distribute:

$1.50 \cdot x + 0.90 \cdot y = 1.10 x + 1.10 y$ $1.50*x+0.90*y = 1.10x+1.10y$

move everything to one side:

$0.4 x - 0.20 y = 0$ $0.4x -0.20y=0$

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

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

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

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

expanding:

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

collecting:

$12 - 0.60 y = 0$ $12-0.60y=0$

subtract 12 from both sides:

$- 0.60 y = - 12$ $-0.60y=-12$

divide through by $- 0.60$ $-0.60$ :

$y = 20$ $y=20$

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

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

Answer 2 · 2018-01-07T17:55:32

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

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 $0.6$

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

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

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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?

2 Answers

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