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How many roots does this equation have? A*x + A = B1) 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
Joined: 25 Aug 2007
Posts: 959
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Followers: 38
Kudos [?]:
555
[0], given: 40
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CEO
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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 .
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Director
Joined: 25 Aug 2007
Posts: 959
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Followers: 38
Kudos [?]:
555
[0], given: 40
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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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GMAT Club team member
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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. Answer: A. Hope it's clear.
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CEO
Joined: 15 Aug 2003
Posts: 3550
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626
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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. Answer: A. Hope it's clear. Agree about the question. thanks.
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