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Is ab = 1?

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04 Sep 2008, 23:28
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Is ab = 1?

(1) aba = a
(2) bab = b

OPEN DISCUSSION OF THIS QUESTION IS HERE: m01-71305.html
[Reveal] Spoiler: OA
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04 Sep 2008, 23:50
rao_1857 wrote:
IS ab = 1?

1) aba = a
2) bab = b

I assume that aba = a*b*a

then
1) a^2b - a = 0 => a(ab -1) = 0 => either a =0 or ab = 1,
Thus insuff

2) b^2a - b =0
b(ab-1) = 0 => b=0 or ab = 1
Thus insuff

Together,
We can have ab=0 or ab = 1 from 1) & 2)

thus insuff

I am not sure though
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04 Sep 2008, 23:58
I will go with C.

the reason is that the only common solution for both will be ab = 1. If, a is zero, b(ab-1) will be -b and nont zero. Similarly, if b = 9, a(ab-1) will be -a and not 0.

What is the OA?
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05 Sep 2008, 00:29
Agree with C. I solved it in a stupid way, though: ba=1/(ba) => ba=1.
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05 Sep 2008, 08:46
thanks guy .. OA is C.
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05 Sep 2008, 09:21
why can't a=b=0?
and we have ab=0 or ab=1??
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05 Sep 2008, 10:47
arjtryarjtry wrote:
why can't a=b=0?
and we have ab=0 or ab=1??

1st stm-- gives you two solutions.
a=0 or ab=1
that means a= 0 or a=1/b and both can't be true..
when a=0 ab=0 and not ab=1

2nd stm -- gives you two solutions.
b=0 or ab=1

combine.
{a=0 or ab=1} and {b=0 or ab=1}
-----> ab=1
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05 Sep 2008, 11:07
IS ab = 1?

1) aba = a
2) bab = b

Why cant we use the below? Not sure where i'm going wrong.
I assume that aba = a*b*a

1) a^2*b=a
divide both the sides by a
a*b=1

2) b^2*a=b
divide both the sides by b
a*b=1

D ??
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05 Sep 2008, 12:06
I don't think answer C is correct.

If a=b=0 then both statements are correct and if a=b=1 then both statements are correct

Graphically

Think of a and b as x and y. Then if you graphed both equations they would intersect at (0,0) and (1,1).
Without additional information there is no way to narrow it down.

Logically and algebraically

The statement "x or y" is true if x is true or y is true (including the possibility of both true)
The statement "x and y" is true only if both are true.

{ab=1 or a=0} and {ab=1 or b=} is true if ab=1 but it is also true if a=0 and b=0. Again, without additional information there is no way to narrow it down.

For those of you dividing by a or b, you cannot divide both sides of an equation by a quantity that might be zero. If you do this you will lose solutions.
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16 Feb 2013, 10:50
pm4553 wrote:
IS ab = 1?

1) aba = a
2) bab = b

Why cant we use the below? Not sure where i'm going wrong.
I assume that aba = a*b*a

1) a^2*b=a
divide both the sides by a
a*b=1

2) b^2*a=b
divide both the sides by b
a*b=1

D ??

Exactly what i thought ... what is wrong with this way?
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16 Feb 2013, 11:43
pm4553 wrote:
IS ab = 1?

1) aba = a
2) bab = b

Why cant we use the below? Not sure where i'm going wrong.
I assume that aba = a*b*a

1) a^2*b=a
divide both the sides by a
a*b=1

2) b^2*a=b
divide both the sides by b
a*b=1

D ??

i think of it in a different way.
if ab=1 there can be two cases a=b=1 or a=1/b.

S1) aba=a -------> with a=b=1 ------> 1x1x1=1 true. -------> with a=1/b---------> (1/b)b(1/b)=1/b -------> (1/b)=(1/b) true. sufficient
S2) bab=b -------> with a=b=1 ------> 1x1x1=1 true. -------> with a=1/b---------> b(1/b)b=b --------------> b=b true sufficient
Ans should be D

Regards,

Abhijit
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17 Feb 2013, 08:58
Narenn wrote:

i think of it in a different way.
if ab=1 there can be two cases a=b=1 or a=1/b.

S1) aba=a -------> with a=b=1 ------> 1x1x1=1 true. -------> with a=1/b---------> (1/b)b(1/b)=1/b -------> (1/b)=(1/b) true. sufficient
S2) bab=b -------> with a=b=1 ------> 1x1x1=1 true. -------> with a=1/b---------> b(1/b)b=b --------------> b=b true sufficient
Ans should be D

What you've done above is assumed that the answer to the question is 'yes', and you have then tried to prove that the statements are true. That is backwards. The statements are facts; they cannot be wrong, so you should never be trying to prove that they're true. They are. The question, on the other hand, is a question; you don't know what the answer to the question is without more information, and that's the whole point of Data Sufficiency. You can't just assume the answer to the question is 'yes', because then you're assuming what you should be trying to prove. That's the logical fallacy known as 'begging the question'. It's crucially important to be clear about the approach to DS questions, because if you approach them backwards, you'll answer many DS questions incorrectly, including this one.

Here the answer is E, since even knowing both statements, we might have a=b=1 or a=b=0.
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17 Feb 2013, 09:41
IanStewart wrote:
Narenn wrote:

i think of it in a different way.
if ab=1 there can be two cases a=b=1 or a=1/b.

S1) aba=a -------> with a=b=1 ------> 1x1x1=1 true. -------> with a=1/b---------> (1/b)b(1/b)=1/b -------> (1/b)=(1/b) true. sufficient
S2) bab=b -------> with a=b=1 ------> 1x1x1=1 true. -------> with a=1/b---------> b(1/b)b=b --------------> b=b true sufficient
Ans should be D

What you've done above is assumed that the answer to the question is 'yes', and you have then tried to prove that the statements are true. That is backwards. The statements are facts; they cannot be wrong, so you should never be trying to prove that they're true. They are. The question, on the other hand, is a question; you don't know what the answer to the question is without more information, and that's the whole point of Data Sufficiency. You can't just assume the answer to the question is 'yes', because then you're assuming what you should be trying to prove. That's the logical fallacy known as 'begging the question'. It's crucially important to be clear about the approach to DS questions, because if you approach them backwards, you'll answer many DS questions incorrectly, including this one.

Here the answer is E, since even knowing both statements, we might have a=b=1 or a=b=0.

Yes Ian. I forgot to consider the possibility of a and b being zero. I also admit that the strategy which i applied to solve this problem is wrong and can be disastrous in the GMAT. Having said that i should also mention here that for the first time i solved any DS with this strategy.
Many thanks to you sir for making me alert on the right occasion.

Regards,

Abhijit.
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Re: Is ab = 1? [#permalink]

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18 Feb 2013, 05:07
OPEN DISCUSSION OF THIS QUESTION IS HERE: m01-71305.html
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Re: Is ab = 1?   [#permalink] 18 Feb 2013, 05:07
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