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If a and b are integers and a*b=2, is a>b?

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If a and b are integers and a*b=2, is a>b?  [#permalink]

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New post 15 Jul 2017, 21:54
1
2
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

58% (01:32) correct 42% (01:28) wrong based on 67 sessions

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If a and b are integers and a*b=2, is a>b?

(1) \(\frac{a}{b}>1\)
(2) \(a^2>b^2\)

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Re: If a and b are integers and a*b=2, is a>b?  [#permalink]

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New post 16 Jul 2017, 00:18
IMO the answer is
E.


Stem : a and b and integers and their product is 2. Possible values of a,b = (2,1) or (1,2) or (-2,-1) or (-1, -2)
Needed: Is a > b?

I found it easier to keep track of the possibilities by using a table and Working through S1, S2 and S1+S2. Table is attached.

From the Table we can see

S1 : a and b could be (2,1) or (-2,-1). So a > b or a < b. Insuff

S2: \(a^2 > b^2 => a^2-b^2 > 0\).

a and b could be (2,1) or (-2,-1). So a > b or a < b. Insuff

S1+S2: a and b could still be (2,1) or (-2,-1). So a > b or a < b. Insuff

Even with both statements its not suff.

I hope I am correct
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If a and b are integers and a*b=2, is a>b?  [#permalink]

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New post 14 Aug 2017, 03:52
chetan2u wrote:
If a and b are integers and a*b=2, is a>b?

(1) \(\frac{a}{b}>1\)
(2) \(a^2>b^2\)

self made


By using numbers we can arrive at E but I have a doubt in using algebra.

Since a*b=2, both a and b has to be same sign.

St.1: \(\frac{a}{b}>1\)

If a and b are positive, we can conclude a>b.
If a and b are negative, \(\frac{-a}{-b}\), will those signs not cancel out and become just a/b ? Should we invert > sign even when we cancel sign in numerator and denominator?

Please advise.
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Re: If a and b are integers and a*b=2, is a>b?  [#permalink]

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New post 14 Aug 2017, 04:23
1
Bang2919 wrote:
chetan2u wrote:
If a and b are integers and a*b=2, is a>b?

(1) \(\frac{a}{b}>1\)
(2) \(a^2>b^2\)

self made


By using numbers we can arrive at E but I have a doubt in using algebra.

Since a*b=2, both a and b has to be same sign.

St.1: \(\frac{a}{b}>1\)

If a and b are positive, we can conclude a>b.
If a and b are negative, \(\frac{-a}{-b}\), will those signs not cancel out and become just a/b ? Should we invert > sign even when we cancel sign in numerator and denominator?

Please advise.



Hi..
When a and b are NEGATIVE, it does not become -a/-b, it remains a/b because NEGATIVE sign is within a and b..

So a =-x, b =-y...
Now -x/-y>1....X>y.... Substitute in terms of a and B..
-a>-b.....Multiply by -1 and invert sign..
a<b..

Or a/b>1... \(1-\frac{a}{b}<0......\frac{b-a}{b}<0.....\)
So b-a and b will be of different sign..
b is NEGATIVE, so b-a>0...b>a

Follow any of above..
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Re: If a and b are integers and a*b=2, is a>b?  [#permalink]

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New post 14 Aug 2017, 04:54
Great question.
Here is what I did on this one -->


Since a and b are integers => a*b=2 => The possible cases are ==>
1,2
2,1
-1,-2
-2,-1


Lets see the 2 statements =>
a/b>1
Hence the two possible pairs are =>
(2,1)=> YES a>b is true
(-2,-1)=> NO a>b is false.

Hence not sufficient.

Statement 2 =>
a^2>b^2

Again => the two possible pairs are =>
(2,1)=> YES a>b is true
(-2,-1)=> NO a>b is false.


Not sufficient.

Combining the two statements => Two possible pairs are =>
(2,1)=> YES a>b is true
(-2,-1)=> NO a>b is false.

Hence not sufficient.

Hence E.

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Re: If a and b are integers and a*b=2, is a>b?   [#permalink] 14 Aug 2017, 04:54
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