Q17

[QIAOH648]

For Q17, I could not understand why the answer is D, instead of E.

Thank you very much.

[Laura Damone]

There are two rules in play here: There must be at least two Topazes, and W and Z can't both be selected.

In order to satisfy the requirement that there are 2 or more Topazes, we'll have to select at least 2 from among these 4 - W, X, Y, and Z. Since you can't select both W and Z, at least one of them must be out.

If W is out, you'll need to select at least two of the remaining three - X, Y, and Z. No matter which two you choose, you'll always have X, Y, or both.

If Z is out, you'll need to select at least two of the remaining three - W, X, and Y No matter which two you choose, you'll still always have X, Y, or both.

Therefore, it must be true that the selection of stones includes X, Y, or both.

This is a pretty common occurrence in In/Out games with subgroups! If two elements in a single subgroup can't both be selected, I always ask myself who must be selected in order to fulfill the numeric requirement of that subgroup. That will often result in a dual option situation just like this, where at least one of the remaining elements must be In.

Hope this helps!

[AnnaT620]

Really struggling with these questions! Why is the answer here not A? My understanding is that the numerical distribution would be 3 - 2 - 1?

Thank so much!

[Laura Damone]

The game never tells us that each subgroup needs to be represented, and we only need 6 stones total. It's therefore possible that we have all three sapphires and 3 of the 4 topazes.

The other rule that would require at least one ruby is conditional: If we have exactly 2 topazes, we get exactly 1 ruby. Since we could have three sapphires, that rule isn't necessarily triggered either.