June 2011 LSAT Section 2 Question 20

# The ball in which one of the following boxes must be the same color as at least one of the other balls?

1 Reply

Ravi on August 14 at 07:46PM

@ifuhrmanb,Great question. This is definitely a tough one, and it helps if you

have the four scenarios of ball distribution down first.

We know there are more red balls than white ones. This is huge, as we

also know that there is at least 1 ball of each color. This means

there are four possibilities of ball color distribution

1) 4 red, 1 white, and 1 green (with the white immediately below the

green to satisfy the final rule)

2) 3 red, 2 green, and 1 white

3) 2 red, 3 green, and 1 white

4) 3 red, 2 white, and 1 green

One way to answer this question is to simply try the answer choices

out, starting with (A).

The way you check these answer choices is to see if you can make a

scenario where the box stated in the answer choice is the only box

with a given color. If that scenario works, then you know that that

answer choice is out.

Looking at (A), we have box 2. If we put green in box 2, white in

boxes 1 and 3, and red in 4, 5, and 6, then we're good to go, so (A)

is out.

With (B), if we put green in 3, white in 1 and 2, and red in 4, 5, and

6, we're also good.

With (C), we could have green in 1, red in 2 and 3, and then white in

4, followed by green in 5 and 6, so (C) is out.

With (D), we could put green in 1, red in 2 and 3, green in 4, white

in 5, and green in 6, and we'd be good, so (D) is out.

By process of elimination, we're left with (E), which is correct.

Now, there is another way to solve this problem more quickly, but it

requires your ability to perceive why box 6 can't be a solo color.

Why can't 6 be solitary? It's because whatever the solitary thing is,

it's going to be green or white because there are at least 2 reds in

every scenario. It can't be green because there is a green that's

lower than any of the reds, and it can't be white because there is a

white that's lower than at least one green. Thus, we know for sure

that box 6 can't be solitary. This is the fastest way to solve the

problem, but it requires having a really strong grasp of how the rules

work together.

Does this make sense? Let us know if you have any other questions!