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by ChessWriter » Fri Jan 20, 2012 6:53 am
If x is not equal to zero,
is 1/x greater than one?

(1)y/x > y
(2)x*x*x > x*x

PS : I have used the asterix * for the multiply sign in statement (2). So Statement (2) is the same as "x raised to the power of three is greater than the square of x".
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Source: — Data Sufficiency |
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by rijul007 » Fri Jan 20, 2012 7:12 am
ChessWriter wrote:If x is not equal to zero,
is 1/x greater than one?

(1)y/x > y
(2)x*x*x > x*x

PS : I have used the asterix * for the multiply sign in statement (2). So Statement (2) is the same as "x raised to the power of three is greater than the square of x".

Ques can be rephrased as
Is 0<x<1?

(1)y/x > y

If y is +ve
0<x<1

If y is -ve
x<0 or x>1

Is 0<x<1?
We dont know

Not Sufficient

(2)x*x*x > x*x
x>1
Is 0<x<1?
No
Sufficient

Option B
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by neelgandham » Fri Jan 20, 2012 7:25 am
Nice explanation Rijul!

ChessWriter: Please find attached the graph of 1/x which will help you understand better.
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by [email protected] » Sat Jan 21, 2012 4:51 am
Rijul! could u plss show me how did u solve the first inequality...

(1)y/x > y

If y is +ve
0<x<1

If y is -ve
x<0 or x>1

Is 0<x<1?
We dont know


I got ur explanation but i m not getting the inequality solved
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by rijul007 » Sat Jan 21, 2012 11:38 am
[email protected] wrote:Rijul! could u plss show me how did u solve the first inequality...

(1)y/x > y

If y is +ve
0<x<1

If y is -ve
x<0 or x>1

Is 0<x<1?
We dont know


I got ur explanation but i m not getting the inequality solved
y/x > y
y/x - y > 0
y(1-x)/x > 0

here we need to find the range of values of x
Lets say y is +ve

then, we can move y to the RHS without changing the inequality sign
(1-x)/x > 0

0<x<1

Now lets say y is negative

y(1-x)/x > 0
Multiply both sides by -1
|y| (1-x)/x < 0

Now you can shift |y| to RHS
(1-x)/x < 0

x<0 and x>1


Heres how the graph would look like
the shaded region contains all the possible values of (x,y)




Image
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