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If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1

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Bunuel
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If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   22 Dec 2016, 02:35
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If y is not equal to zero, is ax > y?

(1) a = -y = |x|
(2) a < 1
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C
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vitaliyGMAT
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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   Updated on: 23 Dec 2016, 09:22
Bunuel wrote:
If y is not equal to zero, is ax > y?

(1) a = -y = |x|
(2) a < 1


Nobody wants to solve this one so I’ll give a shot. My first attempt in inequalities, hope I won’t screw it up.

(1) a = -y = |x|

From that we can make an inference that a > 0 and y < 0, but x can be either > 0 or < 0. We don’t know the sign nor the value of x. Insufficient.

(2) a < 1

Without value of x that gives us nothing. Ex. 1/2 *16 > 4 or 1/2 * -16 < 4. Not sufficient.

(1) & (2) together.

a = -y = |x| < 1

x > 0 we have: 1/2 * 1/2 > -1/2 ----> 1/4 > -1/2

x < 0 we have: 1/2 * - 1/2 > - 1/2 ----> -1/4 > - 1/2

Sufficient

Answer C.

Please correct me if i'm wrong

Originally posted by vitaliyGMAT on 23 Dec 2016, 08:43.
Last edited by vitaliyGMAT on 23 Dec 2016, 09:22, edited 1 time in total.
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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   23 Dec 2016, 09:15
Bunuel wrote:
If y is not equal to zero, is ax > y?

(1) a = -y = |x|
(2) a < 1


Answer is C

i will approach this sum by number substitution method.

i will tell some time saving technique

we can see in both statement "a", so we will assume a value for that.

s2: tells a<1 , so we will choose two values for a on both sides say a = -2 or 2


we will get yes and no , if you substitute.

when we combine we know a = -2

so C
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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   23 Dec 2016, 09:32
I don't think that \(a\) can be a negative integer.

a = |x| absolute value of x is always positive

a < 1 means tht a can only be a fraction < 1.

Please check again.
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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   23 Dec 2016, 09:47
vitaliyGMAT wrote:
I don't think that \(a\) can be a negative integer.

a = |x| absolute value of x is always positive

a < 1 means tht a can only be a fraction < 1.

Please check again.


ya , you are correct.

My bad :(
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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   23 Dec 2016, 16:43
From 1

A is +ve , y is -ve and x is non zero
Question becomes is
Ax-y> 0

Since we know nothing about x except it ain't zero

insuff

From 2

Clearly insuff

Both


C - edited from E.





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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   23 Dec 2016, 22:20
Expert Reply
Bunuel wrote:
If y is not equal to zero, is ax > y?

(1) a = -y = |x|
(2) a < 1



Hi
We are looking at ax>y..


(1) a = -y = |x|
This tells us the value of each discarding SIGN is same..
a is +, y is - and x can be any thing..

Substitute y=-a in equation ax>y....
So ax>-a.......ax+a>0.......a(1+x)>0...
We know a is positive, so if 1+x is positive ans is YES otherwise NO..
SIGN or value of a/y/x is not known ..
Insuff

(2) a < 1
Nothing about x and y ..
Clearly insuff

Let's see combined..
Is a(1+x)>0....
The answer depends on value of x..
Now a is between 0 and 1..
So x will be between -1 and 1, both exclusive and 0 not included..
Now 1+x will always be POSITIVE, when x is between -1 & 1..
So answer is YES
Suff

C
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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink] New post   05 Jul 2023, 08:07
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Re: If y is not equal to zero, is ax > y? (1) a = -y = |x| (2) a < 1 [#permalink]
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