#17 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=15614)

The correct answer choice is (B)

If O joined the firm in 1965 and M joined the firm in 1967, then it must be true that I joined the firm in 1968 and that G joined the firm in 1966:

J91_Game_#3_#17_diagram 1.png

However, it is still not possible to determine the exact years in which H, N, K, and J each joined the firm. Therefore, if O joined the firm in 1965 and M joined the firm in 1967, one can determine the years in which exactly two of the other partners joined the firm. It follows that answer choice (B) is correct.

[salgado145]

Hi
I need help understanding why is G in 6 since I don't see a chain directly with M. I understand I is in place 8 because would be the only letter left over assuming that this could be a possibility

__K/J_. K/J__H/N___H/N____0___. __G__. ___M__. __I__
61. 62. 63. 64. 65. 66. 67. 68

. Thank you

[Robert Carroll]

salgado145,

Consider what's before G. N and J are, via the third rule. The first two rules put one thing before N and one before J. So G has four things before it, and thus can't be in the first four spots. As you point out, G can't be last either. Since, because of the local condition, there is only one place left, G must be there.

Robert Carroll

[kqw24]

Hello,

If we are told the years in which O and M joined, and it must be true that I and G joined at certain times, why is the answer B for 2 and not D for 4? Am I interpreting the question stem incorrectly?

[Robert Carroll]

kqw24,

The question stem asks about "other" partners. O and M shouldn't be included in the count.

Robert Carroll