Data Sufficiency

Question




Answer


Even after looking at the answer. I still don't understand. Can someone help?

If n is an integer, is n positive?

1) (2n+1)/(n+1) is an integer
2) n = -n

Statement 1: (2n+1)/(n+1) is an integer
In other words, 2n+1 is a multiple of n+1.
If n>0, then the smallest possible multiple of n+1 -- after n+1 itself -- is as follows:
2(n+1) = 2n+2.
Thus, if n>0, it is not possible that 2n+1 -- a value LESS than 2n+2 -- is a multiple of n+1.
Implication:
n cannot be positive.
SUFFICIENT.

Statement 2: n=-n
Adding n to both sides, we get:
2n = 0
n = 0.
Thus, n is not positive.
SUFFICIENT.

The correct answer is D.

dnakarting wrote:If n is an integer, is n positive?

1) (2n+1)/(n+1) is an integer
2) n = -n

Statement 1:

This tells us that 2n + 1 is a multiple of n + 1. That does not seem possible with a positive n. Try some values.

If n = 0, (2n+1)/(n+1) = 1, an integer.

If n = 1, (2n+1)/(n+1) = 3/2, which is not an integer.

If n = 2, (2n+1)/(n+1) = 5/3.

If n = 3, (2n+1)/(n+1) = 7/4

I can see what's going on here now. If n is positive, the higher n gets, the closer (2n+1)/(n+1) gets to 2, but (2n+1)/(n+1) is never going to equal 2. So if (2n+1)/(n+1) is an integer, then n is not positive.

Sufficient.

Statement 2:

A positive number cannot equal a negative number. So n = -n only works if n = 0.

Sufficient.

So the correct answer is D.

An easier way, I think:

S1::

(2n + 1)/(n + 1) =

(n + (n+1))/(n + 1) =

n/(n+1) + (n+1)/(n+1) =

n/(n+1) + 1

Since this entire expression is an integer, n/(n+1) must be an integer. But this is only possible if n = 0 or n = -2. So n is NOT positive, and S1 is sufficient.

S2::

2n = 0, n = 0, sufficient as well.

After reading Several times, Now i get it. Thanks Guys. I have another question, here. Hope you guys can help me again?

Question:


Answer


Questions that i Have:

1. In the solutions, there is a 30% + 30% - 9%. I'm cracking my head to answer, Where did the 9% come out from?

9% is the probability of Matt beating Larry in both the first game and the last game.

30% of 30% is 9%.

I don't really like that explanation though, as it seems to imply that he can't win by winning all three games.

Here's a better explanation, one that better matches how the situation actually works.

There are three ways via which he could make the team.

He could W W L.

The probability of doing that is .3 x 1 x .7 = .21

or

He could L W W.

The probability of doing that is .7 x 1 x .3 = .21

or

He could win W W W.

The probability of doing that is .3 x 1 x .3 = .09

Add those probabilities to get the total probability of making the team and you get .21 + .21 + .09 = .51.

Bingo.

So even if in order to maximize the probability that Matt makes the team you make Matt sure to beat Steve, you get .51.

So the answer to the question is no, and Statement 2 is sufficient.

Marty Murray wrote:9% is the probability of Matt beating Larry in both the first game and the last game.

30% of 30% is 9%.

I don't really like that explanation though, as it seems to imply that he can't win by winning all three games.

Here's a better explanation, one that better matches how the situation actually works.

There are three ways via which he could make the team.

He could W W L.

The probability of doing that is .3 x 1 x .7 = .21

or

He could L W W.

The probability of doing that is .7 x 1 x .3 = .21

or

He could win W W W.

The probability of doing that is .3 x 1 x .3 = .09

Add those probabilities to get the total probability of making the team and you get .21 + .21 + .09 = .51.

Bingo.

So even if in order to maximize the probability that Matt makes the team you make Matt sure to beat Steve, you get .51.

So the answer to the question is no, and Statement 2 is sufficient.

Thanks Marty!!! However i'm still a little stuck. if Matt can choose the 3 game combination of his choices, how come we would have to add up the probability of all the scenarios of him winning? i thought he could only use one combination?

I get how you got the 51%, but i don't understand why it needs to be added up.

I'm sorry if i'm slow to catch this.

The three game combination he is choosing is the order in which he plays against Steve and Larry. We want to see whether we can get above 51%. So we give him the greatest probability of winning by having him play Larry twice and play Steve in the middle game.

Once that is set, Matt is not choosing any more, at least for the purposes of this question. He is taking his chances, rolling the dice in a sense. He plays three games and we want to know, given the probability of his winning each, what the probability of his winning two in a row is.

There are multiple ways he could do that. He could win the first two, he could win the last two, or he could win all three, and in any of those cases he would make the team. On the other hand if he were to win only the middle game he would not make the team.

When you have multiple ways to win, your probability of winning is greater than it would be were you to have only one way to win, right? Consider a six sided die. If you only win if you roll a 6, you don't have as high a probability of winning as you would were the rule to be that you win if you roll a 6 or a 1.

Look at that example. The probability of rolling a 6 is 1/6, and the probability of rolling a 1 is 1/6. The probability of rolling a 1 or a 6 is 1/3 or 2/6. So to get the probability of doing one thing OR another, you add the probabilities of each of those mutually exclusive outcomes occurring. If they were not mutually exclusive, meaning only one of them can occur at a time, then the method would be a little different.

Similarly Matt has multiple mutually exclusive ways via which he can make the team. So to figure out the total probability of his making the team, you add up the probabilities of the three ways.

Thank you so much Marty. I Think i get it now. Because they are mutually exclusive occurrences, we must add them up to give him the highest probability he can succeed.

Marty Murray wrote:The three game combination he is choosing is the order in which he plays against Steve and Larry. We want to see whether we can get above 51%. So we give him the greatest probability of winning by having him play Steve twice and play Larry in the middle game.

This makes intuitive sense but is actually wrong in a crucial way. Given S1, Matt should play Larry twice.

Logically speaking, we'd justify this by saying that Matt needs only to win two games in a row. This means that the middle game is the most important: he MUST win that game to have any chance of being admitted to the snobby, twisted tennis club. So he wants the critical game to be against the weaker opponent.

Algebraically, we'd sort it out as follows. Suppose Matt's chances of beating Larry are x and Matt's chances of beating Steve are y, where y > x > 0.

Playing Steve twice, Matt MUST win the middle game and AT LEAST ONE of the other games.

p = x * (1 - (1 - y)²)

Playing Larry twice, Matt MUST win the middle game and AT LEAST ONE of the other games.

p = y * (1 - (1 - x)²)

Let's pretend that playing Steve twice is better. That gives us the inequality

x * (1 - (1 - y)²) > y * (1 - (1 - x)²)

or

2xy - xy² > 2xy - x²y

or

x²y > xy²

or

x > y

But this contradicts y > x.

Matt@VeritasPrep wrote:

Marty Murray wrote:The three game combination he is choosing is the order in which he plays against Steve and Larry. We want to see whether we can get above 51%. So we give him the greatest probability of winning by having him play Steve twice and play Larry in the middle game.

This makes intuitive sense but is actually wrong in a crucial way. Given S1, Matt should play Larry twice.

OK cool. That was a typo/word reversal. I just edited it to make it accurate.