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If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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05 Nov 2014, 08:46
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Tough and Tricky questions: Algebra. If \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\), and a and b are both nonzero integers, which of the following could be the value of b? I. 2 II. 3 III. 4 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III Kudos for a correct solution.
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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09 Nov 2014, 20:45
\(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\)
\(3(ab)^3 + 9(ab)^2  54ab = 0\)
Divide the complete equation by 3ab
\((ab)^2 + 3(ab)  18 = 0\)
ab = 6 OR ab = 3
For ab = 6
if b = 2, then a = 3 (Satisfies denominator)
if b = 3, then a = 2 (Does not satisfy denominator)
For ab = 3
if b = 3, then a = 1 (Does not satisfy denominator)
Answer = A




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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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06 Nov 2014, 02:19
Bunuel wrote: Tough and Tricky questions: Algebra. If \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\), and a and b are both nonzero integers, which of the following could be the value of b? I. 2 II. 3 III. 4 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III Kudos for a correct solution.\(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\) we cannot have a=1, and a=2. because this will make denominator zero. now, consider the numerator. let ab=k thus we have k(3k^2 +9k54) =k(3k^2+18k9k54) =k(3k(k+6)9(k+6)) =k(3k9)(k+6) if 3k9=0 3k=9 k=3 i.e. ab=3 now if b=3, then a=1, which is not acceptable as a=1, will make the denominator equal to zero. if k+6=0 k=6 or ab=6 again, if b=3, then a=2, which is not acceptable as a=2, will make the denominator equal to zero. thus except b=3, all other values could be possible. hence answer should be D



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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06 Nov 2014, 11:43
manpreetsingh86 wrote: Bunuel wrote: Tough and Tricky questions: Algebra. If \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\), and a and b are both nonzero integers, which of the following could be the value of b? I. 2 II. 3 III. 4 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III Kudos for a correct solution.\(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\) we cannot have a=1, and a=2. because this will make denominator zero. now, consider the numerator. let ab=k thus we have k(3k^2 +9k54) =k(3k^2+18k9k54) =k(3k(k+6)9(k+6)) =k(3k9)(k+6) if 3k9=0 3k=9 k=3 i.e. ab=3 now if b=3, then a=1, which is not acceptable as a=1, will make the denominator equal to zero. if k+6=0 k=6 or ab=6 again, if b=3, then a=2, which is not acceptable as a=2, will make the denominator equal to zero. thus except b=3, all other values could be possible. hence answer should be D As ab=6 or ab=3, so if we consider value of b as 4, a becomes Fraction in both the cases, but a is a nonzero integer(Given in the question). Hence b cannot be 4, so the answer should be (A) I only



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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06 Nov 2014, 11:56
IMO ans shud be A
using numerator and taking ab= x we get 3x(x^2+3x18)=0 gives 2 equations 3x=0 .i.e 3 ab =0 i.e a=0 or b=0 # but questions has given both are non zero so it can be true therefore x^23x+18=0 , gives can be factorized as (x6)(x+3)=0 gives x=6 or ab=6 it has soloutions as (1,6),(6,1),(2,3),3,2) a=1 and a=2 is not possible as questions will become non mathematical. Therfore solutions possible are (6,1) and (3,2) Therfore ans is A b=2



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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06 Nov 2014, 23:37
manpreetsingh86 wrote: Bunuel wrote: Tough and Tricky questions: Algebra. If \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\), and a and b are both nonzero integers, which of the following could be the value of b? I. 2 II. 3 III. 4 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III Kudos for a correct solution.\(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\) we cannot have a=1, and a=2. because this will make denominator zero. now, consider the numerator. let ab=k thus we have k(3k^2 +9k54) =k(3k^2+18k9k54) =k(3k(k+6)9(k+6)) =k(3k9)(k+6) if 3k9=0 3k=9 k=3 i.e. ab=3 now if b=3, then a=1, which is not acceptable as a=1, will make the denominator equal to zero. if k+6=0 k=6 or ab=6 again, if b=3, then a=2, which is not acceptable as a=2, will make the denominator equal to zero. thus except b=3, all other values could be possible. hence answer should be D ab=6 consider b=4, then a=3/2 which is not a integer. But according to the question a,b are non zero integer. So, answer is A



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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10 Nov 2014, 23:46
I just have this question, why cant be a=1 or 2. Why cant the denominator equal zero..



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2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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10 Nov 2014, 23:54
sunaimshadmani wrote: I just have this question, why cant be a=1 or 2. Why cant the denominator equal zero.. If the denominator is zero, then the equation fails. Its value turn Infinite For example, \(\frac{5}{0} = Infinite\)



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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11 Nov 2014, 03:51
sunaimshadmani wrote: I just have this question, why cant be a=1 or 2. Why cant the denominator equal zero.. Division by 0 is not allowed: anything/0 is undefined.
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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14 Jan 2015, 04:06
PareshGmat wrote: \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\)
\(3(ab)^3 + 9(ab)^2  54ab = 0\)
Divide the complete equation by 3ab
\((ab)^2 + 3(ab)  18 = 0\)
ab = 6 OR ab = 3
For ab = 6
if b = 2, then a = 3 (Satisfies denominator)
if b = 3, then a = 2 (Does not satisfy denominator)
For ab = 3
if b = 3, then a = 1 (Does not satisfy denominator)
Answer = A One question here. For For ab = 3, why cant we have: ab=3 > ab=1(3), cannot be ab=3 > ab=3(1), couldn't this one also be? Or it is not included in the solution because there is not "1" as an answer option?



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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07 Jun 2016, 11:23
Hi Bunuel,
Can you please provide a proper solution to this question? I am not very clear from the solutions provided above.
Thanks.



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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08 Jun 2016, 01:37
Bunuel wrote: Tough and Tricky questions: Algebra. If \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\), and a and b are both nonzero integers, which of the following could be the value of b? I. 2 II. 3 III. 4 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III Kudos for a correct solution.\(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\) Take 3ab common in the numerator. \(\frac{3(ab) * [(ab)^2 + 3(ab)  18]}{(a1)(a + 2)} = 0\) Note that \((ab)^2 + 3ab  18\) is a quadratic which we can split into factors just like we do for \(x^2 + 3x  18\). \(x^2 + 3x  18 = (x + 6)(x  3)\) so \((ab)^2 + 3ab  18 = (ab + 6)*(ab  3)\) \(\frac{3(ab) * [(ab + 6)*(ab  3)]}{(a1)(a + 2)} = 0\) Now, when will this fraction be 0? When one of the factors in the numerator is 0. (Note that denominator cannot be 0. So a cannot the 1 or 2.) So either ab = 0 (a and b are non zero so not possible) or (ab + 6) = 0 i.e. ab = 6 or (ab  3) = 0 i.e. ab = 3 So, we get that one of these two must hold. Either ab = 6 or ab = 3.Now let's look at the possible values of b. Can b be 2? If b = 2, a = 3 (in this case, ab = 6. Satisfies) Can b be 3? If b = 3, a is 2 or 1. Both values are not possible so b cannot be 3. Can b be 4? If b = 4, a is not an integer in either case. So b cannot be 4. Answer (A)
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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08 Jun 2016, 03:20
Thanks for the quick reply Karishma! Now the question is crystal clear.



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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24 Jul 2017, 09:40
Given (a1) and (a+2) in the denominator, a is not equal to 1 or 2 Simplifying the numerator we get: [3(ab)^3 + 9(ab)^2 – 54ab] = 3ab[(ab)^2+3ab18] = 3ab[(ab)^2+6ab3ab18] = 3ab(ab+6)(ab3) Since the whole equation is equal to 0, we can say the numerator is also equal to 0. Therefore, (ignoring 3ab=0) ab+6=0 which implies b = 6/a say equ(1) ab3=0 which implies b = 3/a say equ(2) Since a and b are both nonzero integers (note, they could be negative) from equ (2) we can say a = {3,1,1,3} However, applying the constraint set by the denominator, a = {3,1,3} Substituting these values of 'a' in the equ(1) and equ(2), we get b = {3,2,1,1,2,6} Hence, the answer is option A.



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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05 Aug 2017, 18:14
PareshGmat wrote: \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\)
\(3(ab)^3 + 9(ab)^2  54ab = 0\)
Divide the complete equation by 3ab
\((ab)^2 + 3(ab)  18 = 0\)
ab = 6 OR ab = 3
For ab = 6
if b = 2, then a = 3 (Satisfies denominator)
if b = 3, then a = 2 (Does not satisfy denominator)
For ab = 3
if b = 3, then a = 1 (Does not satisfy denominator)
Answer = A which denominator are u talking about?



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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07 Aug 2017, 18:37
Start with initial analysis of the problem:
a and b are integers ab cannot = 0 We cannot have a = 1 or a = 2, since denominator cannot be 0 Since we want the solution = 0 , then the numerator must = 0
So, focusing on the numerator:
Factoring 3(ab)^3 + 9(ab)^2  54ab =3ab(ab)(ab) + 3ab(3)(ab)  3ab(18) =3ab(ab^2 + 3ab  18)
I know that 3ab cannot be 0, so focusing on:
(ab^2 + 3ab  18) = 0 =(ab + 6)(ab  3)
So, ab = 6 ab = 3 are my only solutions
I know that a cannot = 1 or 2, so b cannot = 6 or 3
b could be 3, 2, 1, 1, 2, 6
2 is only choice above provided in the answer. Answer is A



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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20 Aug 2018, 10:36
Why does the 3ab which is factored out not translate to ab=0 as a possible answer?



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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20 Aug 2018, 11:05
dwhitmer21 wrote: Why does the 3ab which is factored out not translate to ab=0 as a possible answer? The prompt indicates that a and b are NONZERO INTEGERS. Thus, ab=0 is not a valid case.
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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17 Nov 2018, 10:21
Bunuel wrote: Tough and Tricky questions: Algebra. If \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\), and a and b are both nonzero integers, which of the following could be the value of b? I. 2 II. 3 III. 4 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III Kudos for a correct solution.Please tell me where i am going wrong. Bunuel VeritasKarishma chetan2u3(ab)^3 + 9 (ab)^2=54 ab 3(ab)^2[ab+3]=54 ab div by 3 ab b/s ab[ab+3]=18 let ab =z z[z+3]=18 z[z+3]=3*6 hence, ab=3 so a cant b 1 so b cnt b 3 but m not getting rest of soltuins?i know i some how messed up by not making quadratic equation ,can u please guide me where i m going wrong?



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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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17 Nov 2018, 20:03
vanam52923 wrote: Bunuel wrote: Tough and Tricky questions: Algebra. If \(\frac{3(ab)^3 + 9(ab)^2  54ab}{(a1)(a + 2)} = 0\), and a and b are both nonzero integers, which of the following could be the value of b? I. 2 II. 3 III. 4 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III Kudos for a correct solution.Please tell me where i am going wrong. Bunuel VeritasKarishma chetan2u3(ab)^3 + 9 (ab)^2=54 ab 3(ab)^2[ab+3]=54 ab div by 3 ab b/s ab[ab+3]=18 let ab =z z[z+3]=18 z[z+3]=3*6 hence, ab=3 so a cant b 1 so b cnt b 3 but m not getting rest of soltuins?i know i some how messed up by not making quadratic equation ,can u please guide me where i m going wrong? You are missing out on other value of z... z(z+3)=18.. Here z can be 6 too. 6(6+3)=6*3=18 So ab=6.. when a is 3, b is 2.... So I is valid.. Now for b as 4, not possible, otherwise a will become fraction.
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a1)(a + 2)] = 0, and a and b are bo
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