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# m11#10 - DS

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Director
Joined: 24 Aug 2007
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11 Jun 2010, 05:57
How many roots does this equation have? $$A*x + A = B$$

1) A does not equal 0
2) B does not equal 0

OA is A and OE is:
Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is $$(B-A)/A$$.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is:
For S1, x = (B-A)/A

If B=0 then x= -1
If A=B then x=0
If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

I dont know where I am missing.
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Director
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11 Jun 2010, 10:29
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11 Jun 2010, 13:18
ykaiim wrote:
How many roots does this equation have? $$A*x + A = B$$

1) A does not equal 0
2) B does not equal 0

OA is A and OE is:
Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is $$(B-A)/A$$.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is:
For S1, x = (B-A)/A

If B=0 then x= -1
If A=B then x=0
If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

I dont know where I am missing.

Perhaps we need to take a closer look at the problem statement. The question is : How many roots does the equation have?

With Statement 1, you can definitely answer, 3 roots. With Statement 2, we need to know something about A to answer the question .
Director
Joined: 24 Aug 2007
Posts: 954
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
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Kudos [?]: 1245 [0], given: 40

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11 Jun 2010, 20:35
Praetorian,

Thats correct. But, my question is untill we know both A and B we are not sure how many roots can be possible. If the question asked - What maximum number of roots this equation can have, then we could answer that it can be 3.
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12 Jun 2010, 08:16
ykaiim wrote:
How many roots does this equation have? $$A*x + A = B$$

1) A does not equal 0
2) B does not equal 0

OA is A and OE is:
Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is $$(B-A)/A$$.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is:
For S1, x = (B-A)/A

If B=0 then x= -1
If A=B then x=0
If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

I dont know where I am missing.

Not really the question GMAT would ask, but anyway:

I think the question asks about the # of roots for $$ax+a=b$$, where $$a$$ and $$b$$ are some particular real numbers and $$x$$ is the variable.

3 cases:
1. if $$a=0$$ and $$b\neq{0}$$, equation will have no root (0 roots), as no value of $$x$$ can satisfy the equation $$0*x+0=b$$, when $$b\neq{0}$$;

2. if $$a\neq{0}$$ and $$b=any \ number$$, equation will have have one root: $$x=\frac{b-a}{a}$$. This root can take any value depending on $$a$$ and $$b$$ (not only 3 values);

3. if $$a=0$$ and $$b=0$$, equation will have have infinite # of roots, as any value of $$x$$ will satisfy the equation $$0*x+0=0$$.

(1) $$a\neq{0}$$ --> we have the second case, hence equation has one root. Sufficient.
(2) $$b\neq{0}$$ --> we can have the first case, for which equation has no root OR the second case, for which equation has one root. Not sufficient.

Hope it's clear.
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12 Jun 2010, 08:50
Bunuel wrote:
ykaiim wrote:
How many roots does this equation have? $$A*x + A = B$$

1) A does not equal 0
2) B does not equal 0

OA is A and OE is:
Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is $$(B-A)/A$$.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is:
For S1, x = (B-A)/A

If B=0 then x= -1
If A=B then x=0
If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

I dont know where I am missing.

Not really the question GMAT would ask, but anyway:

I think the question asks about the # of roots for $$ax+a=b$$, where $$a$$ and $$b$$ are some particular real numbers and $$x$$ is the variable.

3 cases:
1. if $$a=0$$ and $$b\neq{0}$$, equation will have no root (0 roots), as no value of $$x$$ can satisfy the equation $$0*x+0=b$$, when $$b\neq{0}$$;

2. if $$a\neq{0}$$ and $$b=any \ number$$, equation will have have one root: $$x=\frac{b-a}{a}$$. This root can take any value depending on $$a$$ and $$b$$ (not only 3 values);

3. if $$a=0$$ and $$b=0$$, equation will have have infinite # of roots, as any value of $$x$$ will satisfy the equation $$0*x+0=0$$.

(1) $$a\neq{0}$$ --> we have the second case, hence equation has one root. Sufficient.
(2) $$b\neq{0}$$ --> we can have the first case, for which equation has no root OR the second case, for which equation has one root. Not sufficient.

Hope it's clear.

Re: m11#10 - DS   [#permalink] 12 Jun 2010, 08:50
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# m11#10 - DS

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