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

on August 14 at 04:46PM

What is the best strategy for this question?

What is the least time consuming strategy to solve this question?

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!