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Manhattan Advanced Quant pg 84 #8

This topic has 3 expert replies and 3 member replies
bpolley00 Master | Next Rank: 500 Posts
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Manhattan Advanced Quant pg 84 #8

Post Mon Jan 28, 2013 7:26 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If yx cannot equal 0, is 0<Y<1?

    1)y< 1/y
    2) y=z^2

    Can someone please explain why 1 is not sufficient? The Manhattan Explanation has two explanations; however, it says the same thing and says YES and NO. It starts If y<0 If y<0 and then gets yes and No. Can an expert please explain why 1 isn't sufficient here? I completely understand #2 but cannot wrap my mind around #1. Thank you.

    -BP

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    hemanthkumarmn Junior | Next Rank: 30 Posts Default Avatar
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    Post Tue Jan 29, 2013 4:30 am
    If yx cannot equal 0, is 0<Y<1?

    y< 1/y

    Sub Y=0.5

    0.5<1/0.5 - true

    Here we can conclude that Y lies between 0 and 1.

    Similarly Sub Y= -2

    -2< -1/2 - which is again true.

    Here we can conclude that Y doesn't lie between 0 and 1.

    Hence the statement contradicts each other.

    So it isn't sufficient.

    bpolley00 Master | Next Rank: 500 Posts
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    Post Tue Jan 29, 2013 5:38 am
    Wow, that was really relatively simple. Not quite sure what I was thinking last night. I could have just substituted -2 -1 -.5 0 .5 1 2 in and figured it out. Thank you for the response sir.

    -BP

    Post Tue Jan 29, 2013 12:28 pm
    Here's a good rule of thumb with DS questions - try to prove INSUFFICIENCY! That is, try to come up with an example that fits the statement and gives you a "yes" answer to the question, then try to come up with one that fits the statement but gives you a "no" answer. Hemanth came up with two good examples that prove this point.

    If you're trying to prove insufficiency with inequalities, there are several things you need to keep in mind: positives v. negatives, integers v. fractions, and 0, 1, and -1 (these are values that can sometimes make things equal, although 0 was ruled out in this particular problem). If you only tested integers v. fractions, you were forgetting about positives v. negatives.

    Statement 2 tells us that y must be positive, because it's a non-zero square. If it's positive, then it must be a positive fraction. The answer is C.

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    Post Tue Jan 29, 2013 12:35 pm
    hemanthkumarmn wrote:
    If yx cannot equal 0, is 0<Y<1?

    y< 1/y

    Sub Y=0.5

    0.5<1/0.5 - true

    Here we can conclude that Y lies between 0 and 1.

    Similarly Sub Y= -2

    -2< -1/2 - which is again true.

    Here we can conclude that Y doesn't lie between 0 and 1.

    Hence the statement contradicts each other.

    So it isn't sufficient.
    Your number testing strategy worked perfectly here! Just a couple of notes...

    Generally speaking, it's easier to deal with fractions than with decimals on the GMAT. They're much easier to manipulate when you don't have a calculator. I'd especially avoid mixing decimals and fractions together, like 1/0.5. In this case it didn't matter much, but on more complicated questions it could lead to confusion. (Again, this is not a rule, just a pro-tip).

    Another thing - I knew exactly what you meant, but be careful of saying "the statements contradict each other." That's not exactly true. The issue here was that we could take the same statement (which we know must be true), and come up with contradicting answers to the question. Basically, this gives us a "maybe" answer to the question. But it's not the statement itself that is contradictory! Statements are always true. Just wanted to clarify that for the sake of anyone else reading.

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    hemanthkumarmn Junior | Next Rank: 30 Posts Default Avatar
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    Post Tue Jan 29, 2013 11:21 pm
    ceilidh.erickson : Thanks for your valuable advice.. And it's indeed true that one got to be very precise while reasoning a GMAT question, not only for the sake of others but also for self.. Point taken.. Thanks again..

    Post Wed Jan 30, 2013 3:34 am
    Quote:
    If yz does not equal zero. Is 0<y<1?
    (1)y<1/y
    (2)y=z²
    The solution below offers a slightly different way to reason through statement 1.

    Statement 1: y < 1/y.
    The CRITICAL POINTS are where y = 1/y or where the inequality is undefined.
    y = 1/y when y=1 or y=-1.
    y < 1/y is undefined when y=0.
    When y is ANY OTHER VALUE, y < 1/y or y > 1/y.
    Thus, there are 4 ranges to consider: y < -1, -1<y<0, 0<y<1, and y>1.

    To determine the range of y, test one value to the left and right of each critical point.
    If y = -2, then the inequality becomes -2 < -1/2.
    This works. Thus, y<-1 is part of the range.

    If y = -1/2, then the inequality becomes -1/2 < -2.
    Doesn't work. Thus, -1<y<0 is not part of the range.

    If y = 1/2, then the inequality becomes 1/2 < 2.
    This works. Thus, 0<y<1 is part of the range.

    If y = 2, then the inequality becomes 2 < 1/2.
    Doesn't work. Thus, y>1 is not part of the range.

    Thus, it's possible that y<-1 or that 0<y<1.
    INSUFFICIENT.

    Statement 2: y=z².
    Since the square of a number cannot be negative, and it is given that yz≠0, we know that y>0.
    Thus, it's possible that y=1/2 (in which case 0<y<1) or that y=2 (in which case y>1).
    INSUFFICIENT.

    Statements 1 and 2 combined:
    The only range that satisfies both statements is 0<y<1.
    SUFFICIENT.

    The correct answer is C.

    An algebraic way to determine the CRITICAL POINTS of y<1/y is to multiply each side by y².
    We can safely multiply by y² for the following reasons:
    y≠0, since 1/y is undefined when y=0.
    y² cannot be negative, since the square of a value cannot be negative.
    Thus, y²>0.
    The result is that we can safely multiply each side by y², knowing that the direction of the inequality will not change:

    y²(y) < y²(1/y)
    y³ < y
    y³ - y < 0
    y(y² - 1) < 0
    y(y+1)(y-1) < 0.

    The CRITICAL POINTS are where y(y+1)((y-1) = 0:
    y=-1, y=0 and y=1.

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