SOLUTION: Okay, I am taking an online Math class and have come across all of the types of work problems, Mystery #s, Money, Mixture & Motion problems. I am so stuck on this Mixture Problem.

Question 87916: Okay, I am taking an online Math class and have come across all of the types of work problems, Mystery #s, Money, Mixture & Motion problems. I am so stuck on this Mixture Problem. :
Ms. Post wants to mix 2 kinds of cereal. Type A sells at $2.20/lb and Type B sells at $4.20/lb. How many lbs of Type A should she use if she wants a 20 lb mixture selling at $3.00/lb.?
Well, I tried using problems in my lesson as an example for the chart.
This is what I came up with and am pretty sure I got it wrong...
2.20x+4.20(20-x)=3.00(20)
2.20x+84-4.20x=60
+4.20x -84=
6.4x divided by itself, cancelling the 6.4, then dividing -24 by the 6.4, I ended up w/ x=-3.75
** I'm pretty sure I got this wrong. SO could you please help me so I can finish my work? Thanks very much for anything you can give me.!!!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Ms. Post wants to mix 2 kinds of cereal. Type A sells at $2.20/lb and Type B sells at $4.20/lb. How many lbs of Type A should she use if she wants a 20 lb mixture selling at $3.00/lb.?
-----------------
Let amt of Type A be "x" ; value of this is 220x cents
-----------------------------------
Amt of Type B is "20-x" ; value of this is 420(20-x) = 8400-420x cents
-----------------------
Amt of mixture is 20 lbs. ; value of this is 300*20= 6000 cents
========================
EQUATION:
value + value = value
220x + 8400-420x = 6000
-200x = -2400
x = 12 lbs (amount of Type A cereal)
20-x = 8 lbs (amount of Type B cereal)
======================
Cheers,
Stan H.
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