Q20

[andreperez7]

Answer B.

Let’s follow the cascade of inferences and scenarios that occur if you put O in 3.

O in 3 means that, because of rule #3 and #2, M cannot go into 2, 4, and 7.
Now we look at rule #4 and decide where our HM chunk can go. It can go into 4 and 7, but not 5 because of rule #5. (Remember Q19 pointed this out as well.)
Now we get two scenarios:
__1_2_3_4_5_ 6 _7 _ 8
1) M J O H M G/L J G/L
2) J H O G/L J G/L H M

1) You get the first scenario because HM in 4 leads to M in 5, and then M/1(Rule#1), which lead to J in 2 (Rule #5). The two open spaces are 6 and 8, with only G and L left.

2) You get the second scenario because HM in 7 leads to H in 2 (Rule #2), which leads to J in 1 (rule #5), which then leads to J in 1 (Rule #1). The only spaces left are 4 and 6, with G and L floating between them.

Hope this helps someone.