<p>Blue book, page 718, #8</p>
<p>The price of ground coffee beans is d dollars for 8 ounces and each ounce makes c cups of brewed coffee. In terms of c and d, what is the dollar cost of the ground coffee beans required to make 1 cup of brewed coffee?</p>
<p>A) d/8c
B) cd/8
C) 8c/d
D) 8d/c
E) 8cd</p>
<p>Any help would be appreciated, thanks!</p>
<p>Whenever you come across these problems, ALWAYS pick #s. Let's say it costs $8(d) to make 8 ounces. Each ounce can make 1(C) cup of coffee. So you need $1 to get the one ounce to get the 1 cup. So A) works-- 8/8*1</p>
<p>But aren't there other ones that work as well using the numbers you picked? I always try picking numbers, and I tried picking numbers for this one too, but I couldn't come up with one single answer. </p>
<p>Look at C, 8*1/8 = 1 also...</p>
<p>Well they do always say that numbers like 1 and 0 are dangerous numbers. I can try some new ones then. $4 for 8lb, and 1lb=2 cups. So each ounce is $.50, so if .50=2 cups, then .25=1 cup. A) comes out again.</p>
<p>Yeah, that works, thanks.
I tried 2 sets of numbers and they both worked with 2 answers so I gave up. I guess I should have kept picking. </p>
<p>Also, I was kinda hoping that someone would show the algebraic way of working that one out.</p>
<p>
[quote]
The price of ground coffee beans is d dollars for 8 ounces and each ounce makes c cups of brewed coffee. In terms of c and d, what is the dollar cost of the ground coffee beans required to make 1 cup of brewed coffee?
[/quote]
</p>
<p>Dimensional analysis:</p>
<p>(d dollars)/(8 oz ) * (1 oz)/(c cups brew) = (d dollars) / (8c cups brew) = d/(8c) dollars/cups brew</p>
<p>d/(8c) is (A)</p>
<p>I use dimensinoal analysis all the time in Chemistry, and while I might use it to check over my work, usually picking #s works best for me. In this case the #s didn't come out to work everytime, but 1 is a dangerous number. I'd rather focus on other questions though because most of the time, the #s you pick work to one example(I would rather use this way to check than use dimensional analysis).</p>
<p>Wow amber, for some reason, I didnt realize how simple the problem was. Thanks.</p>