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- DanielleL473
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Vinny Gambini
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- Joined: October 05th, 2021
Re: Q15
Hi, glad to help:)
First lets start with the conditions:
1:/xx
2./GP
3.S→HS
4./I1→H3orH4 so H1orH2→I1 which leads to H2→I1
From con.3 and 4 its obvious that S and H may determine the schedule once settled, so I firstly took a glance at answer E, cuz con.3 told me S determines H, and con.4 told me H determines I, so S may be the most important to determine the schedule.
Luckily, it did save my time.
So let's take a look at the correct answer E
If S is shown twice, then HS happens twice.
Because we have con.1, HS has to be shown 1st and 3rd or 2nd and 4th
But if HS are shown 1st, which means H is shown first, then it contradicts with con.4
So HS has to be shown 2nd and 4th, again con.4 helps here, we know I has to be the first.Then we know G could only be shown 3rd.
Then con.2 helps us to know P had to be first, so R has to be 3rd.
Hope this helps
First lets start with the conditions:
1:/xx
2./GP
3.S→HS
4./I1→H3orH4 so H1orH2→I1 which leads to H2→I1
From con.3 and 4 its obvious that S and H may determine the schedule once settled, so I firstly took a glance at answer E, cuz con.3 told me S determines H, and con.4 told me H determines I, so S may be the most important to determine the schedule.
Luckily, it did save my time.
So let's take a look at the correct answer E
If S is shown twice, then HS happens twice.
Because we have con.1, HS has to be shown 1st and 3rd or 2nd and 4th
But if HS are shown 1st, which means H is shown first, then it contradicts with con.4
So HS has to be shown 2nd and 4th, again con.4 helps here, we know I has to be the first.Then we know G could only be shown 3rd.
Then con.2 helps us to know P had to be first, so R has to be 3rd.
Hope this helps
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