<p>Yojo, the "official" solution to that problem is on CB's website. The answer is (C), 11/30, to see what they wrote. Their algebraic methods are very easy to follow/understand, but I tend to flub w/ the time crunch on the real SAT I when it comes to possible algebraic reasoning, unfortunately. I'm working on improving that pronto, lol. Anyway...</p>
<p>What I did was this...</p>
<p>So we know that for orange, you have a total of 5 parts. 3 of them are red, and 2 of them are yellow. IOW, 2/5 of the parts are yellow.</p>
<p>In the green sample, we have a total of 3 parts. 2 of them are blue, and 1 of them is yellow. IOW, 1/3 of the parts is yellow.</p>
<p>Now let's tack on some units to make it more realistic/less abstract/easier to think about. </p>
<p>Equal amounts of green and orange are mixed, so why not 1 L of orange and 1 L of green? In this case, we'd have 2/5 L of yellow from orange, and 1/3 L yellow from green. (2/5) + (1/3) = 11/15 L yellow total in the orange + green mixture. Now, we have to find the total volume to be able to find the PROPORTION of yellow in the combined sample. 1 L + 1 L = 2 L total.</p>
<p>[(11/15) L]/[2 L] = 11/30
*Note how the units cancel out to leave you with just the proportion.</p>