<p>Coffee A normally cost 75c per pound. It is mixed with Coffee B, which normally costs 80c per pound, to form a mixture that costs 78c per pound. If there are 10 pounds of the mix, how many pounds of Coffee A were used in the mix?</p>
<p>a 3
b 4
c 4.5
d 5
e 6</p>
<p>answer is 4.</p>
<p>Let x= pounds of Coffee A,</p>
<p>.8(10-X) + .75x = (.78)(10)
8 - .8x + .75x = 7.8
8 - .05x = 7.8
.05x = .2
x=4</p>
<p>^ in order to get that equation</p>
<p>.75¢/lb * (x)lb = .75x</p>
<p>.80¢/lb * (y)lb = .8y</p>
<p>.78¢/lb * (10)lb = 7.8</p>
<p>from that you find that </p>
<p>x + y = 10
.75x + .8y = 7.8</p>
<p>from first equation find that y = (10 - x)
and plug in in second equation to get 4.</p>
<p>^ Yeah. I was making the substitutions logically.</p>
<p>Snorlax, why is .75x+ .8y= 7.8?</p>
<p>Basically, the equation says:</p>
<p>(Cost of Coffee A per pound)(Pounds of Cofee A) + (Cost of Coffee B)(Pounds of Coffee B) = (Cost of mixture per pound)(Pounds of mixture)</p>
<p>The x stands for pounds of Coffee A and y for pounds of Coffee B. In my equation, I knew that whatever x was, y would be x less than 10 because 10 is the total number of pounds. I therefore substituted (10-x) for y, so</p>
<p>.8(10-X) + .75x = (.78)(10)</p>
<p>thank you silverturtle, but I’m also interested in Snorlax’s answer</p>
<p>I was explaining the logic behind his equation as well.</p>