LG Game 2 Question 12 Explanation, by LSATHacks
This question places M first. That triggers the third rule: If M is before P, then H is before G.
Further, since only H and G can be last, then if H is before G, J must be last. C is CORRECT.
Here’s the diagram for reference:
You could also redraw the deductions above into a new diagram, but I found it would have been too much work on this question.
Hi Graeme,
Could you possibly explain why answer (a) is wrong? Because if the third rule dictates that H appears earlier than G, then the last three can only be in the order of H-G-J, leaving both (a) and (c) correct… or am I missing something?
Thanks,
Michelle
P could be after H – G, no? I just glanced at it, but nothing in the diagram forces P to be earlier than them.
I feel as if the last three can only be G-H-J. Is there any other sequence that cannot have G as fifth?
G-H-J actually wouldn’t work. If M goes first, then we know for sure that H comes before G (Rule 3). Rule 1 then tells us that L is before H (but is after M for this case since M is first).
So the actual sequence we know is: M – L- H – G
J just has to be after L, and P has to be before J.
So if we don’t want to have G as fifth, it would have to be:
M – L – H – G – P – J