Word problem

[Isucnmath]

A candy company needs to mix a 30% fat content chocolate with a 12%
fat content chocolate to obatain 50 lbs of 20% fat content chocolate. How many pounds of each kind of chocolate must be used ?

How would I set this up and work it out ?

I was thinking

.30x + .12y = 50 lbs of .20 fat chocolate

but not sure .

 

[Guest]

0.30x is the pounds of fat in x pounds of 30% fat content chocolate. So 0.30x + 0.12y = 50 means there's 50 pounds of fat.

First, you should really define your variables:

Let x = amount of pounds of 30% fat content chocolate
Let y = amount of pounds of 12% fat content chocolate

You need the total amount of pounds to be 50, so x + y = 50.

The pounds of fat in the 30% fat content chocolate is 0.30x. The pounds of fat in the 12% fat content chocolate is 0.12y. The amount of pounds of fat desired in the final mix is 0.20 * 50. So 0.30x + 0.12y = 0.20 * 50.

 

[Isucnmath]

Thanks , but how would I solve that with the two variables ...everytime I have an equation with 2 variables I have trouble with where in the world to start ..lol ...

 

[soroban]

Hello, Isucnmath!

Alex already set up the equations for us:

[1] . 0.30x + 0.12y .= .10
[2] . . . . x .+ . . .y . .= .50

First, I'd get rid of those annoying decimals.
Multiply through [1] by 100:

[3] . 30x + 12y .= .1000
[2] . . . x + . . y .= . .50

We want to eliminate one of the variables.

Here's one way to do: multiply through [2] by -12:

[3] . 30x + 12y .= .1000
[4] .-12x - .12y .= . -600

Now add the equations:
. . . . 18x .= .400 . . ---> . . x .= .200/9

Substitute into [2] and get: . y .= .250/9

. . (Unusual answers, but they check out!)