If X^2=Y^2 is it true that X>0?

[ghong14]

X^2=Y^2 is it true that X>0??

1) x=2Y+1
2)y<=-1

Hi this is a GMAT prep problem that popped up. I need some help with picking numbers on this one? What are two numbers that I can pick to show that statement one is insufficient. I have picked x=-1 and Y=-1 to show that X>0 is not true. But can't seem to find a pair that will prove X>0 and X=2Y+1. In addition why is that we need the second statement for this to be sufficient?

[ghong14]

Sorry don't know how to post the image

[tim]

Sorry, when your original post contains obvious errors we're going to need to see a screenshot so we can see EXACTLY what the actual problem looks like.

[ghong14]

[RonPurewal]

X^2=Y^2 is it true that X>0??

1) x=2Y+1
2)y<=-1

Hi this is a GMAT prep problem that popped up. I need some help with picking numbers on this one? What are two numbers that I can pick to show that statement one is insufficient. I have picked x=-1 and Y=-1 to show that X>0 is not true. But can't seem to find a pair that will prove X>0 and X=2Y+1. In addition why is that we need the second statement for this to be sufficient?

you're not going to be readily able to pick numbers for statement 1, because statement 1 is two simultaneous equations: x^2 = y^2 (from the prompt) and x = 2y + 1.

the thing that you should know here, though, is that "x^2 = y^2" implies either that x and y are the same (x = y) or else that they are opposites (x = -y, or y = -x).
if you don't see why this is the case, just fish around for two numbers that you can square to get the same value, and you should see what's going on pretty quickly.

so, just write each of these cases, substitute into x = 2y + 1, and solve.

if y = x:
x = 2y + 1
x = 2x + 1
x = -1
in this case, x = y, so y = -1.

if y = -x:
x = 2y + 1
x = -2x + 1
3x = 1
x = 1/3
in this case, y = -x, so y = -1/3.

so, there are two cases that satisfy statement 1 (and, of course, the prompt): (x, y) = (-1, -1) and (x, y) = (1/3, -1/3).
so, statement 1 is not sufficient, since one of these x-values is positive and the other is negative.

--

that brings us to trying the two statements together. if we bring statement 2 into the mix, then only the pair (x, y) = (-1, -1) survives. in that case, you have concrete values for both x and y, so definitely sufficient.
answer should be (c).

[FarukhA694]

Hi Ron,

1- How can we see this question and decide whether we need to pick numbers or not?

2- In other words, combining the two statements here is fairly easy, but how do we get it to a 50-50 between C and E in 5 to 10 seconds?

3 - In other words, how do we ensure that we don't waste time plugging in numbers in statement 1?

[RonPurewal]

5-10 seconds? huh?

to show that statement 1 is not sufficient, you have to try solving y = 2x + 1 for x = y, and then try solving it again for x = –y.
these things don't take a huge amount of time, but, they're also not things you can do in your head.

more importantly, you should NEVER have to think about "minutes" and/or "seconds" WHILE DOING PROBLEMS!
...remember, time management really just boils down to (1) acknowledging when you're stuck on a problem, and (2) STOPPING WHAT YOU'RE DOING if you're genuinely stuck.

[RonPurewal]

...and in statement 1, you have to solve y = 2x + 1 subject to the constraint that x^2 = y^2 (...which is the same as saying x = ±y).

it's not possible to "pick numbers" that will satisfy BOTH the equation AND the constraint, so, it will become clear pretty quickly that you're going to have to use algebra for that statement.