<p>How many pounds of a $ 1.2 per pound nut mixture must be mixed with two pounds of a 90 cent-per-pound mixture to produce a mixture that sells for $1.00 per pound?
A) 0.5
B) 1
C) 1.5
D) 2
E)2.5</p>
<p>(2<em>90+x</em>120)/(2+x)=100. Solve it. Or just realize that B) is a pretty easy solution to it. (120<em>1+90</em>2)/3 = 300/3 = 100.</p>
<p>so, that equation is completely right, but here’s a way (at least for me), that makes these problems easier to do:</p>
<p>Type of Mixture Cost Per Pound Amount of Pounds
Mixture A 1.2 x
Mixture B .90 2
Mixture C 1 2+x</p>
<p>You simply multiply across:</p>
<p>1.2x+(.90*2)=2+x
1.2x+1.8=2+x
1.2x=.2+x
.2x=.2
x=1</p>
<p>Then, all you need to do is plug x in to find the answer!
Hope this helps!</p>
<p>why is mixture C = 1 2+x?</p>
<p>zslavitz’s post is supposed to be a table. The middle column is “Cost per pound”, and going down the rows you have 1.2, .90, and 1. Then the last column is “amount of pounds” with x, 2, and 2+x.</p>
<p>now I understand, but phuriku, how can you come up with the equation: (2<em>90+x</em>120)/(2+x)=100? Is the original equation the same as zlatvit’s?</p>
<p>Sorry about that, the formatting changed for my table.
(2<em>90+x</em>120)/(2+x)=100 is just another way of writing it. He uses 90, while I use .9. I use .9 because this theoretically refers to only 90 cents, and spits out a number that does not require any formatting. So, if you move everything over, you get:
instead of (2<em>90), you get (2</em>.9).
Instead of (120<em>x), you get (1.2</em>x).</p>