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M01-36

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M01-36 [#permalink]

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Re M01-36 [#permalink]

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New post 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.


Answer: E
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Re: M01-36 [#permalink]

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New post 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..
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Re: M01-36 [#permalink]

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New post 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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Re: M01-36 [#permalink]

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New post 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.


Answer: E
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Re: M01-36 [#permalink]

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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.


Answer: E


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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Re: M01-36 [#permalink]

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New post 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.


Answer: E


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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Re: M01-36 [#permalink]

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New post 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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Re: M01-36 [#permalink]

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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M01-36 [#permalink]

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New post 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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