LG Game 4 Question 23 Explanation, by LSATHacks
This question says that balls 2, 3 and 4 are the same color as each other.
That can only be green or red. We can’t have three whites, because there must be more red than white (rule 1).
I made both diagrams (red and green):
Let’s apply the rules. There must be a green before the reds. And there are at least two reds. In the first diagram, that means we need to put green first. In the second diagram, that means we need to put two reds in 5-6.
Next, we to place a WG block (rule 3). In the first diagram, only 5-6 has space. In the second diagram, we can put a white in spot 1:
D is CORRECT. In both diagrams it must be true that there’s only one white.
Thank you very much for the explanations.
I got all other Games in this section correct, but this one game killed me. It was easier than I thought.
Question 20 still confuses me, I don’t know how to read it. But your explanations are very helpful.
Thank you again!
I believe this question is flawed (Prep Test 63, sec. 2, #23). Both answers A and D could be the correct answer – am I wrong? Answer “A” states that “exactly two of the boxes contain a green ball”. This is true in both example of “3 reds” and “3 greens”. In no place does the question state that we are accounting for all boxes EXCEPT 2,3, and 4. If you would be willing to send me an e-mail and explain this to me, I would greatly appreciate it.
Thank you,
Dallin Littlefield
I think you may have misunderstood what answer A meant. A is saying “In ALL scenarios there are EXACTLY two green balls.” That refers to all six places. 2,3,4 are only for the question setup, but the answers haven’t nothing to do with 2,3,4 only.
This isn’t true. One of the scenarios I made above has THREE greens. So this proves A wrong.
Thank you so much! I understand now. The word “exactly” confused me earlier today as well when I was studying. Thanks again.