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Is (y-10)^2 > (x+10)^2? [#permalink]
 15 Dec 2012, 11:31
Question Stats:
35% (03:08) correct
64% (02:38) wrong based on 4 sessions
Is (y-10)^2 > (x+10)^2? (1) -y > x+5 (2) x > y
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]
17 Dec 2012, 00:34
JJ2014 wrote: Is (y-10)^2 > (x+10)^2?
(1) -y > x+5
(2) x > y Hi JJ2014, 1. From St 1 , we get x+y<-5 X=10 Y= -4 Then St under consideration is not true i.e (y-10)^2> (x+10)^2 X=-11 y= 5 Then st under consideration is true So A and D ruled out. St 2 alone is not sufficient. Ex X=3, y=2, St not true X=-2, y=-3, St is true So B ruled out. Combining both the statements we get x+y<-5 and x>y We see that y<0 as otherwise if y>0 then the eqn X>y is not true. Hence ans C Thanks Mridul
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]
21 Dec 2012, 02:34
Karishma / Bunuel, Could you throw some light on the below query. I always have trouble in questions like this as to what numbers to pick. I end up going way beyond 3 minutes. Do tell me the certain set of numbers/range of numbers that one should always begin testing with P.S. I know it could vary from question to question but want a general set.
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]
22 Dec 2012, 06:23
we need to find if (y-10)^2 > (x+10)^2?
or, is y^2 - 20y + 100 > x^2 + 20x + 100?
or, is y^2 - 20y > x^2 + 20x?
or, is y^2 - x^2 > 20(x + y)?
or, is (y+x)(y-x) > 20(x+y)?
here, because of inequality, we cannot divide by (x+y) on both sides, as we don't know whether (x+y) is positive or negative. But we can break this in 2:
(1). if (x+y) is +ve, is (y-x) > 20?
(2). if (x+y) is -ve, is (y-x) < 20?
Let's see the 1st option:
-y > x+5
or, x+y < -5
or, x+y is -ve
So, eq (2) needs to be proved. is (y-x) < 20? Can't say. This does not prove eq (2).
Let's see the 2nd option:
x > y
or, x-y > 0
or y-x < 0
So, irrespective of whether (x+y) is +ve or -ve, (y-x) is less than 0. So, the inequality can be answered by this statement alone. The answer should be B. I am not sure why the spoiler says C as the answer.
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]
27 Dec 2012, 06:27
anandrajakrishnan wrote: we need to find if (y-10)^2 > (x+10)^2?
or, is y^2 - 20y + 100 > x^2 + 20x + 100?
or, is y^2 - 20y > x^2 + 20x?
or, is y^2 - x^2 > 20(x + y)?
or, is (y+x)(y-x) > 20(x+y)?
here, because of inequality, we cannot divide by (x+y) on both sides, as we don't know whether (x+y) is positive or negative. But we can break this in 2:
(1). if (x+y) is +ve, is (y-x) > 20?
(2). if (x+y) is -ve, is (y-x) < 20?
Let's see the 1st option:
-y > x+5
or, x+y < -5
or, x+y is -ve
So, eq (2) needs to be proved. is (y-x) < 20? Can't say. This does not prove eq (2).
Let's see the 2nd option:
x > y
or, x-y > 0
or y-x < 0
So, irrespective of whether (x+y) is +ve or -ve, (y-x) is less than 0. So, the inequality can be answered by this statement alone. The answer should be B. I am not sure why the spoiler says C as the answer. hey Anand Had a small question when x > y we get y - x < 0 How can y - x < 0 by itself be sufficient to solve (y+x)(y-x) > 20(x+y) ? I feel C is the correct answer .. we need both A and B to determine. Please clarify
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]
27 Dec 2012, 19:40
1
This post received
KUDOS
Adityam wrote: anandrajakrishnan wrote: we need to find if (y-10)^2 > (x+10)^2?
or, is y^2 - 20y + 100 > x^2 + 20x + 100?
or, is y^2 - 20y > x^2 + 20x?
or, is y^2 - x^2 > 20(x + y)?
or, is (y+x)(y-x) > 20(x+y)?
here, because of inequality, we cannot divide by (x+y) on both sides, as we don't know whether (x+y) is positive or negative. But we can break this in 2:
(1). if (x+y) is +ve, is (y-x) > 20?
(2). if (x+y) is -ve, is (y-x) < 20?
Let's see the 1st option:
-y > x+5
or, x+y < -5
or, x+y is -ve
So, eq (2) needs to be proved. is (y-x) < 20? Can't say. This does not prove eq (2).
Let's see the 2nd option:
x > y
or, x-y > 0
or y-x < 0
So, irrespective of whether (x+y) is +ve or -ve, (y-x) is less than 0. So, the inequality can be answered by this statement alone. The answer should be B. I am not sure why the spoiler says C as the answer. hey Anand Had a small question when x > y we get y - x < 0 How can y - x < 0 by itself be sufficient to solve (y+x)(y-x) > 20(x+y) ? I feel C is the correct answer .. we need both A and B to determine. Please clarify Hmm. I need some retrospection here: As I mentioned that the inequality can be divided in 2: (1). if (x+y) is +ve, is (y-x) > 20? (2). if (x+y) is -ve, is (y-x) < 20? With the 2nd option, we get y-x < 0. This doesn't resolve the inequality as we still need to prove which side of zero does (x+y) lies. So, 1st part is necessary. Option C is the right choice. Thanks for clarifying.
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Re: Is (y-10)^2 > (x+10)^2? [#permalink]
13 Jan 2013, 23:39
C. Individually each one is insuffient by plugging values. Now question stem: Is (y-10)^2 > (x+10)^2? = Is (y+x)(y-x-20) > 0 on simplification 1. gives (y+x)<-5 => (y+x) is negative. Using 1, therefore the stem becomes Is (y-x-20) < 0 ? or Is (y-x) < 20 ? On using 2. if x >y => (y-x) is negative and hence (y-x) < 20 and the answer to the stem is Yes. (combining 1. & 2.) Hence C. KUDOS... If YOU LIKE IT 
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Re: Is (y-10)^2 > (x+10)^2?
[#permalink]
13 Jan 2013, 23:39
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