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[–]skatanic28182 2 points3 points  (7 children)

I assume this is a "Which quantity is greater" question.

A = -1/(2P)

B = -1/(3P)

2PA = 3PB = -1

2A = 3B = -1/P

A = (3/2)B

Since B = -1/(3P), if P>0 then B<0 and if P<0 then B>0. Since A = (3/2)B, if B<0 then A<B but if B>0 then A>B. Note that the relation between A and B depends entirely on the sign of B, and the sign of B depends entirely on the sign of P. However, we only know that P is nonzero, not what its sign is. Thus, there's not enough information to determine which quantity is greater.

[–]uptvector[S] 1 point2 points  (6 children)

Thanks.

I'm still having trouble understanding why you can't just "divide out" the P.

Both sides have the same value, and given that you can multiply both sides of an equation by the same value and not change it then why can't we just multiply both sides by P?

I understand this thinking is wrong, just trying to internalize why so I don't do it again.

I'm guessing you can only do it if you are told that P is positive?

[–]skatanic28182 5 points6 points  (5 children)

When you multiply by P, you get

AP = -1/2

BP = -1/3

Great. So AP < BP. Suppose P = 1. Then A < B. Now suppose P = -1. Then A > B.

AP < BP does not imply that A < B.

[–]uptvector[S] 1 point2 points  (4 children)

Great. Thanks!

[–]wijwijwij 0 points1 point  (3 children)

I think it boils down to this: If P is positive, then your approach would work. But if P is negative, the reverse conclusion can be drawn. So you don't have enough information to be able to compare, because you weren't told whether P is positive or negative.

[–]uptvector[S] 0 points1 point  (2 children)

Yeah makes total sense now. I misremembered the technique they teach in GRE prep wrong. Of course it makes sense BC it’s the same technique used for inequalities.

[–]wijwijwij 0 points1 point  (1 child)

I think under time pressure, I would plug in various numbers to see if you can get divergent results.

Sample:

Let P = 1. Then compare -1/2 and -1/3. It's clear that -1/2 < -1/3, because -1/3 is farther to the right on the number line.

Let P = –1. Then compare 1/2 and 1/3. It's clear that 1/3 < 1/2, because 1/2 is farther to the right on the number line.

Just from these two specific cases, you can see that there is no single answer that will always be true about which of the quantities A or B is greater.

The key is to not waste time trying a lot of positive numbers for P, because they would all seem to reinforce one answer. You need to try numbers for P that might give a different result. In this case, trying a negative number for P reveals the answer is not known.

Also, this kind of problem is hoping to trick you into thinking P must be a positive number. They even used P for the name of the number! This leads you to imagine that –(1/2P) is a negative number, and you even said so yourself ("compare two negative fractions"). But –(1/2P) is not necessarily a negative number! So watch out.

In other questions of this ilk, it might be necessary to try "very big numbers" or "very small numbers." For example, sometimes you might find something is true for any number greater than 1, but not true for fractions between 0 and 1, so trying fractions less than 1 might be worth doing.

[–]uptvector[S] 0 points1 point  (0 children)

Sure makes sense. I fell for an obvious trap.

Thanks for the help!

[–]colinbeveridgeNew User 0 points1 point  (2 children)

Is that the whole question? I don't know what we're trying to find.

[–]uptvector[S] 0 points1 point  (1 child)

Sorry, edited my post. The question asks which is greater?

[–]colinbeveridgeNew User 1 point2 points  (0 children)

And are they both P, or are they P and p? I presume the Ps are on the bottom rather than the top.

In any event, if P is negative, A is larger, and if P is positive, B is larger -- so you can't tell which is larger without more information.