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Re: Data Sufficiency - Simultaneous Equations [#permalink]

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12 Apr 2011, 09:32

amp0201 wrote:

Is the product of A and B equal to 1?

1. A * B * A = A 2. B * A * B = B

1. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient 2. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient 3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient 4. EACH statement ALONE is sufficient 5. Statements (1) and (2) TOGETHER are NOT sufficient

Please justify your answer, as I am not sure if the OA is correct. Thanks.

1. \(A^2B-A=0\) \(A(AB-1)=0\)

Either A=0 OR AB=1 If A=0; AB=0 OR AB=1 AB can be 0 or 1. Not Sufficient.

2. \(B^2A-B=0\) \(B(AB-1)=0\)

Either B=0 OR AB=1 If B=0; AB=0 OR AB=1 AB can be 0 or 1. Not Sufficient.

Combining both statements:

Either A and B are both 0 OR AB=1 If A and B are both zeros: AB=0 or AB=1 Not Sufficient.

It's not a good practice to divide the parts of an equation by the unknown (A or B here) because this unknown can be 0. As we know, division by 0 can't be done.

jainanurag78 wrote:

I think it should be D. Why can we divide A*B*A=A by A on both the sides and it would give us AB =1 same with the S2.

For all who think we can divide both sides by A to have AB=1; Ok, we can try it, but just like absolute values, we have to consider two scenarios:

1) A is NOT equal to 0 : Well here AB=1

2) A is EQUAL to zero. Here since we can not divide, we have to find other ways to manipulate the statements. We arrive at A(AB-1)=0. Analyzing it, we can see AB can be both 1 and 0. Insuf.

I think D is irrelevant and if we are gonna become dubious b/w two choices, they are more likely to be C and E. As you know the general guideline is to choose C b/w C and E on tough questions when it comes to guessing. Solving it, I couldn't make my brain free of the thought that the statements together may be suf.
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What are wonderful question ! Reminds me of the Mona lisa.. Wonderfully simple in its presentation, yet on closer inspection, the painting reveals so many subtleities. Anyway, the key point to remember is that , on the GMAT, a) If the questions pertains to the "=" symbol, never ever cancel terms on both ends unless one is sure that the variable is not equal to 0. Remember that there is no such mathematical action as "Cancellation". Cancellation is division in disguise. b) If the question pertains to the ">" or "<" symbol, never ever divide, or cross multiply unless one is certain that the variable being acted upon is >0. Remember that there is no such mathematical operation as "Cross multiplying". It is multiplication in disguise

Apply the FOIN tests to ensure that you have covered all the values that a variable may assume :- F - Fraction o - ZERO i - Integer or Irrational N - Negative
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well, i first chose E but on a second thought, i chose A , cos, drug related is an entity and others' is another entity that must be recognised as well. so i go for A, i hope am right.

well, i first chose E but on a second thought, i chose A , cos, drug related is an entity and others' is another entity that must be recognised as well. so i go for A, i hope am right.

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 or its sign. We can not divide by zero.

BAB = B. =>BAB/B = 1 => BA = 1 => AB = 1 .... Suff.

Hence either of them is sufficient Hence D.

I thought of this in the beginning, but A or B might equal zero, which makes dividing by A or B undefined. There is nothing in the question that tells you that A and B are not zero. I think if the questions states that A and B are not zero, then your solution will be correct.

(1) \(a^2b=a\) --> \(a^2b-a=0\) --> \(a(ab-1)=0\) --> either \(a=0\) (and \(b=any \ value\), 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 \(a=any \ value\), including zero) so in this case \(ab=0\neq{1}\) OR \(ab=1\). Two different answers, not sufficient.

(1)+(2) As from (1) \(a(ab-1)=0\) and from (2) \(b(ab-1)=0\) then \(a(ab-1)=b(ab-1)=0\) --> 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.

Side note on dividing (1) by \(a\) and (2) by \(b\): 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 can not divide by zero.

Hope it helps.

Hi Bunnel, per my understanding when we combine the solution from 2 equations we look for common solution. from 2 equations we have a=0 OR b=0 OR ab=1 a=0, is not a solution since it doesn't satisfy II b=0, is not a solution since it doesn't satisfy I ab=1 satisfy both I and II so this is the solution or we have single solution for the problem.

(1) \(a^2b=a\) --> \(a^2b-a=0\) --> \(a(ab-1)=0\) --> either \(a=0\) (and \(b=any \ value\), 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 \(a=any \ value\), including zero) so in this case \(ab=0\neq{1}\) OR \(ab=1\). Two different answers, not sufficient.

(1)+(2) As from (1) \(a(ab-1)=0\) and from (2) \(b(ab-1)=0\) then \(a(ab-1)=b(ab-1)=0\) --> 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.

Side note on dividing (1) by \(a\) and (2) by \(b\): 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 can not divide by zero.

Hope it helps.

Hi Bunnel, per my understanding when we combine the solution from 2 equations we look for common solution. from 2 equations we have a=0 OR b=0 OR ab=1 a=0, is not a solution since it doesn't satisfy II b=0, is not a solution since it doesn't satisfy I ab=1 satisfy both I and II so this is the solution or we have single solution for the problem.

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 \(ab=0\neq{1}\).