Q21

 
ShehryarB30
Thanks Received: 2
Elle Woods
Elle Woods
 
Posts: 100
Joined: July 07th, 2018
 
 
 

Q21

by ShehryarB30 Thu Jan 17, 2019 1:13 pm

Could you pls explain this
User avatar
 
ohthatpatrick
Thanks Received: 3808
Atticus Finch
Atticus Finch
 
Posts: 4661
Joined: April 01st, 2011
 
 
 

Re: Q21

by ohthatpatrick Wed Jan 23, 2019 2:28 pm

Since presenting on M means that you automatically present on S,
W's two options could be M and S, or J and S.

j:
m: W
s: W

j: W
m:
s: W

The contrapositive of the 2nd rule says that
"if W is in S, then R cannot be in S"

Putting R into M would then send it to S as well, so we can't put R into M.
So, in both of these scenarios, we have to put R into J.

j: R
m: W
s: W

j: W, R
m:
s: W

At this point, G could go anywhere in the first scenario. But in the second one, G has to go in M, because SOMEONE has to go in M and we've already ruled out W and R. Once we put G in M, though, it will also have to appear in S.

j: R
m: W
s: W

j: W, R
m: G
s: W, G

I would go to the answers with these two partial scenarios. In the first scenario we NEED to add at least one G (to any of the groups), but we could have two or three. In the second scenario, we could still add a G to J if we wanted to.

(A) this isn't necessarily gonna happen (both scenarios)
(B) this isn't necessarily gonna happen (top scenario)
(C) this isn't necessarily gonna happen (top scenario)
(D) YES, this is happening in both scenarios.
(E) this won't happen in either scenario.

(D) is the answer.


(Just for what it's worth, which is very little .... If we had a good understanding of the "if you're in M, you're in S" rule, then we would know that (B) could not possibly be the correct answer ... if it were true that G had to be in M, then it would also be true that G has to be in S, as (C) says.)