Question

Need some help ASAP with some math. I'll post what I have done so far and I'll post the problem in the comments.

What you know is:
Percentage of A: 60% (60/100 as a fraction)
Percentage of B: 85% (85/100 as a fraction)
Final percentage: 80% (80/100 as a fraction)
Total Milliliters: 150 mL
(Percentage of dressing A) x (milliliters you need) + (Percentage of dressing B) x (milliliters you need) = (final percentage) x (Total milliliters)

Now you actually plug things in
(60/100) x A + (85/100) x B = (80/100) x 150
To solve:
First combine the things that do not involve a variable

Answer 1

The problem

Answer 2

To solve:
First combine the things that do not involve a variable (80/100) x 150
80x150 =12,000
Divided by 100 = 120
So now the equation reads (60/100) x A + (85/100) x B = 120

Answer 3

Now you are not able to solve for two variables at one time, so you need to eliminate one. To do this you use substitution.
Since you know that A+B= 150
Meaning, you will use A mL of dressing A and B mL of dressing B to get a total of 150 mL
So you get A by itself,
B= 150-A
Now plug this into your equation where B was
(60/100) x A + (85/100) x (150-A) = 120
Now you can distribute:
(60/100)A + (80/100 x 150) – (85/100 x A) = 120

Answer 4

And then I'm lost on what to do next

Answer 5

Amount of 60% solution to use: x
Amount of 85% solution to use: 150-x
.6x+.85(150-x)=.80(150)

Answer 6

.6x+127.5-,85x=120
-.25x=-7.5
x=30 pints of 60 % solution
150-30=120 pints of 85% solution

Answer 7

I'm sorry, I'm still totally lost. :( Math learning disabilities suck.

Answer 8

Somebody please help me out.

Answer 9

If you have 5 pounds of candy and 3 pounds of it is snickers and the rest is jelly beans wouldn't the jelly beans be 5-3?