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# Is x|y| > y^2? (1) x > y (2) y > 0

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Re: Is x·|y| > y^2?  [#permalink]

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22 May 2016, 12:10
nishatfarhat87 wrote:
Is x·|y| > y^2?

(1) x > y
(2) y > 0

Note that y^2 will always be +ve. we have to see if x·|y| is +ve and >y^2

1) x>y
if x= 1/2 and y=1/3, then yes
if x and y are positive integers, then yes
but, if x= -1/3 and y= -1/2, then no

(2) y > 0
this does not tell us anything about x. insufficient.

combining both statements- y is positive and x>y, which means x is positive.

and, x*y> y^2

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Re: Is x·|y| > y^2?  [#permalink]

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23 May 2016, 03:48
1
nishatfarhat87 wrote:
Is x·|y| > y^2?

(1) x > y
(2) y > 0

Hi Bunuel,

Y^2 is always positive, and |Y| is always positive. Y^2 could be written as Y^2= |Y||Y|

So we can safely divide both sides by |Y| without problem, then the question becomes: is x>|Y|???

(1) x > y.

Put x=2 & Y=1 .. so it is yes.

Put x=2 & Y=-3.. so it is NO

Insuff

2) Y>0. clearly nkthing about X.
Insuff

Combining 1 & 2

X>Y>0. We can take x=2 & Y=1

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Updated on: 15 Nov 2016, 21:05
[quote="nishatfarhat87"]Is x·|y| > y^2?

(1) x > y
(2) y > 0

When you consider statement I- For x and y positive nos., it clearly holds true. Lets find out one case for which this statement doesn't hold true
Y is negative, say -4. And since x>y, lets assume x=-2. When we put these two values, it doesn't hold true for the question.

Similarly Statement II alone is not sufficient as x can have negative integer value or even 0, which doesn't satisfy question.

When you combine the 2 statements, its clear that y is positive and when x is greater than y, x also has to be positive.
thus, "xy>y^2" definetely holds true
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Originally posted by PKay on 07 Jun 2016, 05:28.
Last edited by PKay on 15 Nov 2016, 21:05, edited 1 time in total.
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Re: Is x·|y| > y^2?  [#permalink]

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07 Jun 2016, 08:01
nishatfarhat87 wrote:
Is x·|y| > y^2?

(1) x > y
(2) y > 0

Hi Bunuel,

1) Suppose x and y are -ve

then x·|y| < y^2 because |y| and y^2 are +ve

2) x and y are +ve
x·|y| > y^2

3) x and y are -ve
x·|y| < y^2

4) x is +ve and y is -ve
x·|y| > y^2

5) x is -ve and y is positive

x·|y| < y^2

6) x is 0 and y is -ve or +ve

x·|y| < y^2

7) y is 0 and x is +ve or -ve
x·|y| = y^2

Statement 1 x > y

x and y can be +ve or -ve or 0. not sufficient

Statement 2= y > 0

We are not told about value of x. Not sufficient

Combining statement 1 and 2
Y is +ve and x>y, which means x is +ve

a and y both are +ve in which x>y
hence x·|y| > y^2

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Re: Is x*|y| > y^2?  [#permalink]

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04 Jan 2017, 10:18
russ9 wrote:
Is x*|y| > y^2?

(1) x > y
(2) y > 0

I'll paste my question in the second post. Good Luck!

Great question, what its really asking is whether X > |y| since y^2 is always positive. X has to be greater than Y and Y has to be positive.

1) is an obvious trap if you follow the above^^ logic. X> y doesn't help us because y could be negative. If x is 2, and y is -5, x>y but (2) (-5) is not greater than (-5) ^2 which is 25.
2) by itself is not sufficient because doesn't tell us whether X > |x|, but if you add them together you get the answer!
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Re: Is x*|y| > y^2?  [#permalink]

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19 Jan 2017, 15:16
russ9 wrote:
Is x*|y| > y^2?

(1) x > y
(2) y > 0

I'll paste my question in the second post. Good Luck!

11 seconds to solve it.
1. i'll just provide the ways that don't work, as there are plenty of examples that do work - x can be 3, y can be -9. -9*-9 = 81. 3*9=27. AD out.
2. y>0. we know nothing about x. not sufficient. B is out.

1+2. y is not negative, therefore,it's sufficient to prove
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Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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03 Sep 2017, 16:18
devinawilliam83 wrote:
Is x|y|>y^2?

(1) x>y
(2) y>0

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There 2 variables and 0 equation. Thus we need 2 equations to solve for the variables; the conditions provide 2 equations, so there is high chance that (C) will be the answer.

The question $$x|y| > y^2$$ is equivalent to $$x|y| > |y|^2$$ or $$|y| ( x - |y| ) > 0$$.
The equivalent question is if $$|y| ( x - |y| ) > 0$$.

Condition 1)
$$x = 2$$, $$y = 1$$ : It is true.
$$x = 2$$, $$y = 0$$ : It is false.
This condtion is not true.

Condition 2)
This is not sufficient, since we don't know anything about $$x$$.

Condition 1) & 2)
Since $$y > 0$$, we have $$|y| > 0$$.
Since $$x > y = |y| > 0$$, $$x - |y| > 0$$.
Thus $$|y| ( x - |y| ) > 0$$.
Both conditions together are sufficient.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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12 Oct 2018, 23:36
Can any help me out with my doubts in this question?

Is x|y| > y^2?

here i understand y can have positive or negative value but my doubt is

can i put y as negative
x(-y)>y^2

is not question asking x * positive value of y > y^2?

2) if i put y as positive, eqsn will be xy>y^2
=>y(x-y)>0
which implies eithr y and x-y are both positive
so y>0 and x>y
or
both are negative
y<0 and x<y

So overall i get 4 conditions by solving the eqsn
1)y>0 and x>y
or
2)y<0 and x<y

so while proving it from statements given to us , do i need to prove both 1 and 2 or either of these.

Re: Is x|y| > y^2? (1) x > y (2) y > 0   [#permalink] 12 Oct 2018, 23:36

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