# LSAT and Law School Admissions Forum

Get expert LSAT preparation and law school admissions advice from PowerScore Test Preparation.

## Logic game #8

hanvan
• Posts: 19
• Joined: Mar 17, 2012
#4452
Hi, I am working on the logic game page,( power score test preparation) about
pattern, identify the possibilities game. I really don't understand the question 22, and 23. I open the explanation, still don't get it...the pattern game is kind of tricky, huh? Please help!thanks.
Jon Denning
• PowerScore Staff
• Posts: 904
• Joined: Apr 11, 2011
#4453
Hey hanvan - thanks for the question. Pattern games can definitely be tricky, but fortunately not only are they extremely rare, but with some practice they become MUCH more manageable. I've addressed those two questions below:

22. For the first experiment to give us a single orange flask as the question says, we had to either combine 1 and 2 (red) and have 3 (green) and 4 (orange), or we could mix 1 and 3 (blue) and have 2 (blue) and 4 (orange). We can't mix 2 and 3 (orange) because we'd still have 4, a second orange. So experiment 2 could either mix 3 and 4 (if 1 and 2 got mixed in exp1) or 2 and 4 (if 1 and 3 got mixed in exp1). Answer choice E says 3 and 4 for that second experiment, so it's correct.

23. To perform an experiment and get no orange flasks, we have to mix 4 (orange) with something else. Otherwise it'll be unused and we'll still have orange. Note that mixing 4 with any of the other flasks doesn't make orange, so it doesn't matter what we mix it with. So that's answer choice E as well.

Thanks!
hanvan
• Posts: 19
• Joined: Mar 17, 2012
#4454
Jon Denning wrote:Hey hanvan - thanks for the question. Pattern games can definitely be tricky, but fortunately not only are they extremely rare, but with some practice they become MUCH more manageable. I've addressed those two questions below:

22. For the first experiment to give us a single orange flask as the question says, we had to either combine 1 and 2 (red) and have 3 (green) and 4 (orange), or we could mix 1 and 3 (blue) and have 2 (blue) and 4 (orange). We can't mix 2 and 3 (orange) because we'd still have 4, a second orange. So experiment 2 could either mix 3 and 4 (if 1 and 2 got mixed in exp1) or 2 and 4 (if 1 and 3 got mixed in exp1). Answer choice E says 3 and 4 for that second experiment, so it's correct.

23. To perform an experiment and get no orange flasks, we have to mix 4 (orange) with something else. Otherwise it'll be unused and we'll still have orange. Note that mixing 4 with any of the other flasks doesn't make orange, so it doesn't matter what we mix it with. So that's answer choice E as well.

Thanks!
Thanks so much, Jon....I got it now!
rima363
• Posts: 1
• Joined: Jan 31, 2013
#7481
This game is extremely difficult to me and I can't seem to comprehend the process of solving it. The answers and explanations in the back assume the student knows how to answer or the basics but I am lost. How can I answer this? Thanks!
Jon Denning
• PowerScore Staff
• Posts: 904
• Joined: Apr 11, 2011
#7489
Hey rima,

Thanks for the question. This is a tricky game for a lot of people, so you're certainly not alone. Fortunately Pattern games like this don't appear all that often, so you probably won't see one on the test (although, no guarantees!), and if you do they tend to behave pretty predictably, as this one does.

As the exaplanation notes, the key to this game is recognizing that the combinations of flasks can only happen in a limited number of ways: 6 options if only one mixing occurs, and 3 options if two mixings occur. I won't type all of these out since they're really well described in the answer key, but they're essentially the product of combinations, where you can only make 6 unique groups combining only two of the four flasks (1+2 3 4, 1+3 2 4, 1+4 2 3, 2+3 1 4, etc), and 3 unique groups combining all four flasks in groups of two (1+2 3+4, 1+3 2+4, 1+4 2+3).

Once you see that it's easy to determine what the resultant colors of each mixing combination would be by just following the rules--for instance, for the 1+2 3 4 scenario, 1+2 gives red, 3 is green (doesn't change), and 4 is orange (doesn't change)--and you've got a really nice template that describes every possible solution.

Take a question like 19 to illustrate that. From our template we know that if there are two experiments there are only three possible combinations: 1+2 3+4, 1+3 2+4, 1+4 2+3. So just determine what colors each of those would produce and find a matching answer. C is correct because 1+2 makes red, and 3+4 makes blue.

I hope that helps to clarify things!

Analyze and track your performance with our Testing and Analytics Package.