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M01-36

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M01-36  [#permalink]

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New post 15 Sep 2014, 23:16
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A
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D
E

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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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New post 19 Jun 2015, 06:55
4
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
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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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]

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New post 06 Jul 2018, 09:18
We have seen numerous DS questions where statement (1) gives two values of x and statement (2) gives another two values but when only one of the values is common in the solutions from both the statements then we conclude with option "C" that is "both statements taken together are sufficient to answer the question, but neither statement alone is sufficient" because we get one definite value (which is common) of x from both the statements.

Say for example if we have a question like:-
What is the value of x?
1) x^2 = x (we find that x could be 0 or 1 therefore not sufficient)
2) (x-1)(x-2) = 0 ( we find that x could be 1 or 2)

Now answer me what would be the answer of this question, will it be "option C" that x = 1, i.e. common solution; or
"Option E" that x could either be 0 or 2 and x could be 1. Clearly when x will be 0 or 2 then it cannot be 1 and vice versa because x can assume only one value at a given point of time.

Applying the same logic for this question I am unable to understand the logic why "E" is the right answer? ab = 1 is common to both the statements.

Peripheral issue is: - by definition if a & b can assume only one value, i.e. that is "zero" then no matter what you do, you cannot have a solution of ab = 1 which is a contradiction in the question & statements itself?
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M01-36  [#permalink]

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New post 21 Jul 2018, 07:43
Bunuel or chetan2u, I was able to establish the equations from statements 1 and 2.
1) a = 0 or ab = 1 (not suff.)
2) b = 0 or ab = 1 (not suff.)

I couldn't understand how to synthesize these equations together. I guess this is a very fundamental doubt. That is when we move to combine these equations, how exactly do we do it?
I combined these equations and thought ab = 1 is the only COMMON solution, hence C is the answer. Can you help me with this?
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New post 21 Jul 2018, 07:54
sandman13 wrote:
Bunuel or chetan2u, I was able to establish the equations from statements 1 and 2.
1) a = 0 or ab = 1 (not suff.)
2) b = 0 or ab = 1 (not suff.)

I couldn't understand how to synthesize these equations together. I guess this is a very fundamental doubt. That is when we move to combine these equations, how exactly do we do it?
I combined these equations and thought ab = 1 is the only COMMON solution, hence C is the answer. Can you help me with this?



Hi...
If the two statements had given two different values of a,
I) say a=0 or ab=1
2) say a=2 or ab=1
Them a cannot be both values and hence only possibility is ab=1

But here we have two solutions..
1) ab=1
Or
2) a=b=0

Both the above will satisfy the equation..
So ab=0 or ab=1
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New post 21 Jul 2018, 08:58
chetan2u wrote:
Hi...
If the two statements had given two different values of a,
I) say a=0 or ab=1
2) say a=2 or ab=1
Them a cannot be both values and hence only possibility is ab=1

But here we have two solutions..
1) ab=1
Or
2) a=b=0

Both the above will satisfy the equation..
So ab=0 or ab=1


Hi chetan2u, thanks for the prompt reply.
Just a clarification: why did you pick that COMBINATION of possible solutions?
Let me make this a bit abstract.
Say,
S1: X or Y
(where X and Y are some sets of solution(s))
S2: Z or Y

Now, combining the two: (X or Y) and (Z or Y) ... is what? I think I studied this back in school, but I have no clear recollection of how those compliments and unions work.
I was able to DRAW the event (X or Y) and (Z or Y) on a Venn diagram and get the result Y or (X and Z)

In our problem, this would basically translate as follows:

S1: a = 0 OR ab = 1
S2: b = 0 OR ab = 1

Combining, we get EITHER
1. ab = 1
OR
2. a = 0 AND b = 0

But this seems like a REALLY long way to solve the problem. What am I missing?
>> !!!

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Re: M01-36  [#permalink]

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New post 16 Dec 2018, 02:53
Bunuel

"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."

If its given in problem a & b are not zero, then answer will be D. Right?

Pl. clarify!
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Re: M01-36  [#permalink]

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New post 16 Dec 2018, 06:42
16SK16 wrote:
Bunuel

"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."

If its given in problem a & b are not zero, then answer will be D. Right?

Pl. clarify!


Right. In this case we could reduce (1) by a and get ab = 1 and reduce (2) by b and get ab = 1.
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Re: M01-36 &nbs [#permalink] 16 Dec 2018, 06:42
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