#24 - Local, Must Be True, Suspension

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=303&t=7341)

The correct answer choice is (E)

The new condition in the question stem results in the following sequence:

Capture27.PNG

Because K still cannot attend the first class, this sequence forces a number of students into certain classes. Here is the scenario, with distribution information:

Capture28.PNG

Only H, N, and S are not accounted for on the above diagram. Because either N or S must still attend the first class (remember, H is a female and from the third rule cannot attend the first class), we can infer that H and the N/S remainder attend the sixth class. Accordingly, answer choice (E) is correct.

[SherryZ]

Oct 1999 LSAT, Sec 3 Game #4, Q24：

Why does H must attend 6th? I thought she can attend 5th too, as following:

K-IL-O-G-PHN-S,
or
K-O-IL-G-PHS-N.


Could you help me out? Thank you so much for your help!

---Sherry

[KelseyWoods]

Hi Sherry!

K can't be first because the 3rd rule says that she is not the first student. So if O and IL are between K and G now, that forces K into the 2nd slot, O and IL in 3/4 in some order, G into 5, and PH into 6:

N/S -- K -- O/IL -- IL/O -- G -- P H S/N

Hope that helps!

Best,
Kelsey