December 2000 LSAT Section 4 Question 13

# Which one of the following could be the selection of stones?

8 Replies

Mehran on September 4, 2013

Sorry about that. Would you like a video explanation for this game?Leila on September 4, 2013

I got them right just wanted to see if there was a faster way because it took me a bit longer than expected :)Victor on April 14, 2020

Yes please, a video explanation will help.Victor on April 14, 2020

or preferably the set up!!Michelle on April 25, 2020

I would also greatly appreciate a video explanation for the set up of this game!Ben on May 7, 2020

Hi @mfujii, thanks for reaching out.Unfortunately, as of now we do not have a video explanation for this game yet. We will be sure to let the content team know that it has been requested!

If you have any other questions, please let us know!

on August 13, 2020

Yes to the video explanation! In the meantime, would you be able to explain this game and the questions/answers here? Thanks so much! Not really sure how to approach this.Victoria on August 17, 2020

Hi @Anna2020,Happy to help!

We know that a jeweller will select six out of ten stones for six rings. We can conceptualize this as an "in/out" game in that six of the stones will be in rings and four of the stones will be "out" i.e. not selected.

_ _ _ _ _ _ | _ _ _ _

We learn from the stimulus that there are three different types of stones: rubies, sapphires, and topazes.

Rubies: F, G, H

Sapphires: J, K, M

Topazes: W, X, Y, Z

Now let's go through the rules.

Rule 1 - at least two of the topazes are selected.

Therefore, we cannot have a selection which has one topaz or no topazes.

Rule 2 - if exactly two of the sapphires are selected, then exactly one of the rubies is selected.

2S --> 1R

Not 1R --> Not 2S

Therefore, if exactly two sapphires are selected, then there must be three topazes selected to make six total stones.

Rule 3 - if W is selected, neither H nor Z is selected.

W --> Not H and Not Z

H or Z --> Not W

Rule 4 - if M is selected, W is also selected.

M --> W

Not W --> Not M

Notice that we can combine Rules 3 and 4 to make a transitive chain.

M --> W --> Not H and Not Z

H or Z --> Not W --> Not M

This is a tough one to have a full diagram for, but at least there aren't too many rules. Basically, what we have is:

_ _ _ _ _ _ | _ _ _ _

T T

It's important to remember that, while we must have at least the two Ts on the diagram, we can have more.

We also need to remember:

2S --> 1R --> 3T

and

M --> W --> Not H and Not Z

Now let's address the question stem.

We are looking for the answer choice which represents a possible selection of stones.

We can start by eliminating answer choice (B) because it only has one topaz, breaking Rule 1.

We can then eliminate answer choice (C) because it has exactly two sapphires, but also has two rubies, breaking Rule 2.

We should now be able to eliminate two other answer choices based on our transitive chain.

We can eliminate (A) because M is selected, but W is not, breaking Rule 4.

We can also eliminate answer choice (E) because W is selected, but so are both H and Z, breaking Rule 3.

Therefore, answer choice (D) should be our correct answer, but let's double check against our rules.

Rule 1 is satisfied because we have three topazes.

Rule 2 is irrelevant because we do not have exactly two sapphires.

Rule 3 is satisfied because both H and Z are selected while W is not.

Rule 4 is irrelevant because M has not been selected.

Hope this helps! Please let us know if you have any further questions.