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# M01-36

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Math Expert
Joined: 02 Sep 2009
Posts: 43892

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15 Sep 2014, 23:16
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Difficulty:

65% (hard)

Question Stats:

44% (00:53) correct 56% (00:50) wrong based on 158 sessions

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Is the product of $$a$$ and $$b$$ equal to 1?

(1) $$a*b*a=a$$

(2) $$b*a*b=b$$
[Reveal] Spoiler: OA

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Joined: 02 Sep 2009
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15 Sep 2014, 23:16
Official Solution:

Question: is $$ab=1$$?

(1) $$a^2b=a$$

$$a^2b-a=0$$;

$$a(ab-1)=0$$: either $$a=0$$ (and $$b=\text{any value}$$, including zero) so in this case $$ab=0\neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(2) $$ab^2=b$$

$$ab^2-b=0$$;

$$b(ab-1)=0$$: either $$b=0$$ (and $$a=\text{any value}$$, including zero) so in this case $$ab=0 \neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(1)+(2) either $$a=b=0$$, so in this case $$ab=0 \neq 1$$ and the answer to the question is NO, OR $$ab=1$$ and the answer to the question is YES. Two different answers, not sufficient.

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20 Nov 2014, 19:46
why can't we conclude that:
a^2*b=a=>b=1/a
in this case a*1/a = 1
and statement 1 and 2 are both sufficient
hm..
Math Expert
Joined: 02 Sep 2009
Posts: 43892

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21 Nov 2014, 03:56
mvictor wrote:
why can't we conclude that:
a^2*b=a=>b=1/a
in this case a*1/a = 1
and statement 1 and 2 are both sufficient
hm..

Please re-read the solution above. Does not a = 0 satisfy the equation? You cannot write b=1/a from a^2*b=a because a could be 0, and we cannot reduce or divide by 0.
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19 Jun 2015, 06:45
Hi Bunuel,

I have a question based on the solution that you have mentioned below:

Let's say a question in DS is as follows?

If x is a positive integer, is x=4?

Stmt -1) X is a root of the quadratic equation - x^2 - 12x+12=0

Stmt -2) X is a root of the quadratic equation - x^2 - 9x+20=0

Now solving each statement alone,

Stmt 1 -> Allows x value to be either x= 3,4
Stmt 2 -> Allows x value to be either x= 4,5

Now for such a Data sufficiency question, don't we find that x=4 is a common solution in both statements together and hence mark the answer as Answer choice (C) - Both statements together are sufficient but neither alone.

Now based on this understanding, for the current question,

Both statement (1) and statement (2) when taken together show that ab = 1.
Can we not select option choice (c), based on this?

Please correct me if my understanding is wrong some where?

Vijay

Bunuel wrote:
Official Solution:

Question: is $$ab=1$$?

(1) $$a^2b=a$$

$$a^2b-a=0$$;

$$a(ab-1)=0$$: either $$a=0$$ (and $$b=\text{any value}$$, including zero) so in this case $$ab=0\neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(2) $$ab^2=b$$

$$ab^2-b=0$$;

$$b(ab-1)=0$$: either $$b=0$$ (and $$a=\text{any value}$$, including zero) so in this case $$ab=0 \neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(1)+(2) either $$a=b=0$$, so in this case $$ab=0 \neq 1$$ and the answer to the question is NO, OR $$ab=1$$ and the answer to the question is YES. Two different answers, not sufficient.

Math Expert
Joined: 02 Sep 2009
Posts: 43892

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19 Jun 2015, 06:55
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svijayaug12 wrote:
Hi Bunuel,

I have a question based on the solution that you have mentioned below:

Let's say a question in DS is as follows?

If x is a positive integer, is x=4?

Stmt -1) X is a root of the quadratic equation - x^2 - 12x+12=0

Stmt -2) X is a root of the quadratic equation - x^2 - 9x+20=0

Now solving each statement alone,

Stmt 1 -> Allows x value to be either x= 3,4
Stmt 2 -> Allows x value to be either x= 4,5

Now for such a Data sufficiency question, don't we find that x=4 is a common solution in both statements together and hence mark the answer as Answer choice (C) - Both statements together are sufficient but neither alone.

Now based on this understanding, for the current question,

Both statement (1) and statement (2) when taken together show that ab = 1.
Can we not select option choice (c), based on this?

Please correct me if my understanding is wrong some where?

Vijay

Bunuel wrote:
Official Solution:

Question: is $$ab=1$$?

(1) $$a^2b=a$$

$$a^2b-a=0$$;

$$a(ab-1)=0$$: either $$a=0$$ (and $$b=\text{any value}$$, including zero) so in this case $$ab=0\neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(2) $$ab^2=b$$

$$ab^2-b=0$$;

$$b(ab-1)=0$$: either $$b=0$$ (and $$a=\text{any value}$$, including zero) so in this case $$ab=0 \neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(1)+(2) either $$a=b=0$$, so in this case $$ab=0 \neq 1$$ and the answer to the question is NO, OR $$ab=1$$ and the answer to the question is YES. Two different answers, not sufficient.

When we consider two statements together we should take the values which satisfy both statements. For this question $$ab=1$$ satisfies both statement, but $$a=b=0$$ also satisfies both statements. So what you call "common solution" for this question is: $$ab=1$$ OR $$ab=0\neq{1}$$.
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15 Aug 2016, 20:45
Sorry Bunuel, but if b=0, it will not satisfy the statement 1, will it?

Bunuel wrote:
svijayaug12 wrote:
Hi Bunuel,

I have a question based on the solution that you have mentioned below:

Let's say a question in DS is as follows?

If x is a positive integer, is x=4?

Stmt -1) X is a root of the quadratic equation - x^2 - 12x+12=0

Stmt -2) X is a root of the quadratic equation - x^2 - 9x+20=0

Now solving each statement alone,

Stmt 1 -> Allows x value to be either x= 3,4
Stmt 2 -> Allows x value to be either x= 4,5

Now for such a Data sufficiency question, don't we find that x=4 is a common solution in both statements together and hence mark the answer as Answer choice (C) - Both statements together are sufficient but neither alone.

Now based on this understanding, for the current question,

Both statement (1) and statement (2) when taken together show that ab = 1.
Can we not select option choice (c), based on this?

Please correct me if my understanding is wrong some where?

Vijay

Bunuel wrote:
Official Solution:

Question: is $$ab=1$$?

(1) $$a^2b=a$$

$$a^2b-a=0$$;

$$a(ab-1)=0$$: either $$a=0$$ (and $$b=\text{any value}$$, including zero) so in this case $$ab=0\neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(2) $$ab^2=b$$

$$ab^2-b=0$$;

$$b(ab-1)=0$$: either $$b=0$$ (and $$a=\text{any value}$$, including zero) so in this case $$ab=0 \neq 1$$ OR $$ab=1$$. Two different answers, not sufficient.

(1)+(2) either $$a=b=0$$, so in this case $$ab=0 \neq 1$$ and the answer to the question is NO, OR $$ab=1$$ and the answer to the question is YES. Two different answers, not sufficient.

When we consider two statements together we should take the values which satisfy both statements. For this question $$ab=1$$ satisfies both statement, but $$a=b=0$$ also satisfies both statements. So what you call "common solution" for this question is: $$ab=1$$ OR $$ab=0\neq{1}$$.
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05 Sep 2017, 18:42
I am confused,

Why cant we say
a*b*a = a
a*b = a/a
a*b = 1

What am i missing Bunuel
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Math Expert
Joined: 02 Sep 2009
Posts: 43892

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05 Sep 2017, 19:40
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mbsingh wrote:
I am confused,

Why cant we say
a*b*a = a
a*b = a/a
a*b = 1

What am i missing Bunuel

You cannot reduce a*b*a = a by a, because a can be 0, and we cannot divide by 0. By doing so you are loosing a root, namely a = 0.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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Joined: 03 Aug 2016
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GMAT 1: 660 Q44 V38
GMAT 2: 690 Q46 V40
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05 Sep 2017, 19:42
Bunuel wrote:
mbsingh wrote:
I am confused,

Why cant we say
a*b*a = a
a*b = a/a
a*b = 1

What am i missing Bunuel

You cannot reduce a*b*a = a by a, because a can be 0, and we cannot divide by 0. By doing so you are loosing a root, namely a = 0.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.

Interesting , didnt think of that. Thanks Bunuel
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Re: M01-36   [#permalink] 05 Sep 2017, 19:42
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# M01-36

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