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Retired Moderator
Joined: 16 Apr 2020
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b v/s 3
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16 Aug 2021, 08:27
16 Aug 2021, 08:27
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Question Stats:
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\(\frac{3(ab)^3 + 9(ab)^2 − 54ab}{(a − 1)(a + 2)} = 0\), where \(a\) and \(b\) are both non-zero integers.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(b\) |
3 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: b v/s 3
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16 Aug 2021, 09:53
16 Aug 2021, 09:53
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a≠1, a≠-2
\((ab)^2 + 3(ab) − 18=0\)
substitute \(ab\) with \(u\)
\(u^2 + 3u − 18=0\), \((u+6)(u-3)=0\) and ab=-6 or ab=3
only, when a=1, b is 3 and a≠1
only, when a=-2, b is 3 and a≠-2. Clearly, b cannot be 3 and excluding choice C.
where \(a\) and \(b\) are both non-zero integers, we conclude the possible sets for ab=-6 condition such as a={-6,-3,-1,2,3,6}
and b={1,2,6 ....} don't even need to continue since the answer must be D
\((ab)^2 + 3(ab) − 18=0\)
substitute \(ab\) with \(u\)
\(u^2 + 3u − 18=0\), \((u+6)(u-3)=0\) and ab=-6 or ab=3
only, when a=1, b is 3 and a≠1
only, when a=-2, b is 3 and a≠-2. Clearly, b cannot be 3 and excluding choice C.
where \(a\) and \(b\) are both non-zero integers, we conclude the possible sets for ab=-6 condition such as a={-6,-3,-1,2,3,6}
and b={1,2,6 ....} don't even need to continue since the answer must be D
KarunMendiratta wrote:
\(\frac{3(ab)^3 + 9(ab)^2 − 54ab}{(a − 1)(a + 2)} = 0\), where \(a\) and \(b\) are both non-zero integers.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(b\) |
3 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
General Discussion
Re: b v/s 3
[#permalink]
27 Sep 2022, 22:05
27 Sep 2022, 22:05
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
