#4 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=4541)

The correct answer choice is (A)

Answer choice (A) is proven possible by the following hypothetical:

O93_Game_#1_#4_diagram 1.png

[ccampise]

For question number 5 I understand that A can be a correct answer, but I don't understand why the others can't be correct too. For instance, couldn't there also be three gardenias?

[Jon Denning]

Hey ccampise,

Thanks for the question. You mention A is correct for "question 5," but I think you mean question 4 (as your title indicates). So that's the one I'll explain.

For corsages 2 and 3 to be exactly the same, we initially have three options: R R G, G G R, or R G V for both. Let's consider each:

RRG: that gives us 4 R so far, and 2 G. So corsage 1 would have to have O and V (all must be used), as either OOV, or VVO. Regardless, answer choice A is proven possible with exactly 2 O in corsage 1.Technically you could stop here since you've proven A works, but let's look at the other two to be sure.

GGR: that gives us 4 G and 2 R so far. Corsage 1 again has O and V, but must be VVO so that R is at least double O. This scenario does not fit any answer choice.

RGV: that gives us 2 R, 2 G, and 2 V. Corsage 1 must have O, and it must be a single O. The double spot in 1 could be R, G, or V, but that would give a total of 4 R, G, or V, which again fails to satisfy any of the answer choices.

Hence only A is possible from the choices given.

I hope that helps!

[rameday]

Hello,

I was just curious about question 4. I am not quite sure how powerscore got to their hypothetical to get the correct answer of A. This game was interesting for me because i misread one of the rules I thought their had to be exactly twice as many roses as orchids but I was still able to get 4 out of the 7 questions right and in time as well.

[David Boyle]

Hello,

I was just curious about question 4. I am not quite sure how powerscore got to their hypothetical to get the correct answer of A. This game was interesting for me because i misread one of the rules I thought their had to be exactly twice as many roses as orchids but I was still able to get 4 out of the 7 questions right and in time as well.

Hello rameday,

No, there have to be at least twice as many roses, not exactly twice as many, as orchids.
If you had two orchids in corsage 1, along with one violet; and in both of the other corsages, two roses and one gardenia, answer A could be correct.
Other answers don't work as well, of course: having three roses or gardenias is difficult in that corsage 2 and 3 have to mirror each other, so that each must have at least one rose and one gardenia. This helps keep an even number of roses or gardenias, and three is an odd number. (Also, corsage 1 can have only two kinds of flower, so this will tend to reserve it for orchids and violets, which may be underrepresented in the other two corsages.)
And five violets wouldn't leave much room for everything else; finally, five roses is an uneven number, and like five violets, would leave little room for other things, especially since corsage 1 can have only two kinds of flower.

Hope this helps,
David

[dimi.wassef@yahoo.com]

I am still confused as to why there can't be 3 gardenias? (GRR in both 2 and 3?)

[Rachael Wilkenfeld]

Hi dimi,

GRR in both 2 and 3 is only 2 gardenias. We couldn't put another gardenia on 1 in that scenario, because both O and V must be used, and thus go in the first group. The first group only has two types of flowers, so once we've placed O and V there, we cannot place any other type. Is there some other way that you would place the three Gs?

Rachael

[dimi.wassef@yahoo.com]

I was thinking GGG
RRR
ROV

[Adam Tyson]

Looks to me like that solution does not have corsages 2 and 3 being identical, which is what the question required, Dimi. If 2 and 3 re both GRR, then 1 would have to have O and V, and the third flower could be one or the other of those and not another G or R (because corsage 1 must use exactly two types per the first rule). So you could have GRR in 2 and 3, but answer A would still be the only one possible out of the choices presented.