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Hello TLS,

I have a trouble understanding two conditional rules from Ace the LSAT- LG.

There are 6 sessions and 4 entities (Kiwi, Lime, Mud, Nutrient). The entities is used either once or twice.

1. "Kiwi may be used only once."
Does this mean kiwi "is" used only once, not that it can be used but it doesn't have to be used once? The former is how I have treated "can" rule in the LG. But I don't understand how "can" could become "is."

2. "If the kiwi or lime treatment is scheduled for session 5, the other treatment is scheduled for session 3, and if the kiwi or lime treatment is scheduled for session 3, the other treatment is scheduled for session 5.

I translated this as biconditional:
Kiwi 3 <-> lime 5 or lime 3 <-> kiwi 5

I understand biconditional as "always together" so I made two diagrams assigning kiwi 3 lemon5, and lemon 3 kiwi 5 in the other, assuming kiwi can't go to other sessions since it is used only once. (Basically treating both as locked enitities)
1. 2. 3.K/L 4. 5.K/L. 6.

But the explanation says kiwi and lime are not required to be on 3 and 5.

Aren't they required to be there if they are "always together" (=biconditional + or)????? Please help!!

Donna29 wrote:Hello TLS,

I have a trouble understanding two conditional rules from Ace the LSAT- LG.

There are 6 sessions and 4 entities (Kiwi, Lime, Mud, Nutrient). The entities is used either once or twice.

1. "Kiwi may be used only once."
Does this mean kiwi "is" used only once, not that it can be used but it doesn't have to be used once? The former is how I have treated "can" rule in the LG. But I don't understand how "can" could become "is."

2. "If the kiwi or lime treatment is scheduled for session 5, the other treatment is scheduled for session 3, and if the kiwi or lime treatment is scheduled for session 3, the other treatment is scheduled for session 5.

I translated this as biconditional:
Kiwi 3 <-> lime 5 or lime 3 <-> kiwi 5

I understand biconditional as "always together" so I made two diagrams assigning kiwi 3 lemon5, and lemon 3 kiwi 5 in the other, assuming kiwi can't go to other sessions since it is used only once. (Basically treating both as locked enitities)
1. 2. 3.K/L 4. 5.K/L. 6.

But the explanation says kiwi and lime are not required to be on 3 and 5.

Aren't they required to be there if they are "always together" (=biconditional + or)????? Please help!!

I'm basing this off of your description.

1) "K may be used only once" is not the same as "K muse be used exactly once." But the background rules state that every treatment needs to be used once or twice. The combination of the two constraints means that K will be used exactly once (to use your words, this is how the "can" rule turned into an "is" rule).

2) This tells you that if K or L are scheduled for 3, the other is scheduled on 5. Nothing says that you need to have them scheduled on 3 and 5; the condition merely tells you what happens if you try to schedule K/L on 3 or 5. If you think about it carefully, you should realize that the condition is still satisfied if you had M or N filling the 3 and 5 slots.