Question

You have one type of chocolate that sells for $2.10/lb and another type of chocolate that sells for $9.90/lb. You would like to have 62.4 lbs of a chocolate mixture that sells for $2.40/lb. How much of each chocolate will you need to obtain the desired mixture?

You will need __ lbs of the cheaper chocolate and __ lbs of the expensive chocolate.

Answer 1

This is a problem of weighted averages and can be solved with a system of equations.

Let's define:

- `x` as the amount (in lbs) of the cheaper chocolate,

- `y` as the amount (in lbs) of the expensive chocolate.

We can set up the following equations:

1. `x + y = 62.4` (the total amount of chocolate we want)

2. `$2.10x + $9.90y = $2.40 * 62.4` (the total cost of the chocolate mixture)

The second equation represents the total value of the chocolate, while the first represents the total weight.

We can solve this system of equations using substitution or elimination. In this case, it may be simpler to use elimination.

Let's first transform the second equation to get rid of the dollar signs:

2.10x + 9.90y = 2.40 * 62.4

2.10x + 9.90y = 149.76

Let's now multiply the first equation by 2.10 (the price per lb of the cheaper chocolate):

2.10x + 2.10y = 2.10 * 62.4

2.10x + 2.10y = 131.04

Subtracting the second equation (multiplied) from the original second equation, we get:

9.90y - 2.10y = 149.76 - 131.04

7.80y = 18.72

Solving for y, we get:

y = 18.72 / 7.80 = 2.4 lbs

Substituting y = 2.4 into the first equation:

x + 2.4 = 62.4

x = 62.4 - 2.4 = 60 lbs

So, you will need 60 lbs of the cheaper chocolate and 2.4 lbs of the expensive chocolate to obtain the desired mixture.