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for integers a & b , if sqt (a^3a^2b) = 7 ...wat is [#permalink]
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27 Dec 2005, 13:51
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This topic is locked. If you want to discuss this question please repost it in the respective forum. for integers a & b , if sqt (a^3a^2b) = 7 ...wat is the value of a ???
1. a^2  a = 12
2. b^2  b = 2
sorry guys but I don't know the OA ...I am curious to know the solution for this....



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Joined: 27 Jun 2005
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jinesh wrote: for integers a & b , if sqt (a^3a^2b) = 7 ...wat is the value of a ??? 1. a^2  a = 12 2. b^2  b = 2 sorry guys but I don't know the OA ...I am curious to know the solution for this....
E
1. a^2  a = 12
No info about b so Insuff
2 b^2  b = 2
solving this b = 1 , 2 ..with these two values of b will get two values of a so Insuff
taking together
12 a = 7+b ( but from stm2 "b" can have two values which will lead to 2 different values of a hence INsuff



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Joined: 16 Oct 2003
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I think it is C
From I we get a = 7 + b/12
From II we get b = 2



Manager
Joined: 15 Aug 2005
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Bhai wrote: I think it is C
From I we get a = 7 + b/12 From II we get b = 2
Agree with C
From Stmt I & II, only value of a & b can be
a=4 & b=1



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Joined: 15 Apr 2005
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jinesh wrote: for integers a & b , if sqt (a^3a^2b) = 7 ...wat is the value of a ??? 1. a^2  a = 12 2. b^2  b = 2 sorry guys but I don't know the OA ...I am curious to know the solution for this....
From stmt1 we get a^2a12=0 solving for a we get 4,3. We can substitute in the original qun to see which of this is value of A. but we do not know the value of b. Hence insuff.
From stmt2: we get b=2 or b=1. We still do not know the value of A. so insuff.
Combinging if a = 4 and b = 1 then the original equn is 7. so my answer is C.



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Thanx a lot guys...... got it!!!!!
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Joined: 28 Dec 2005
Posts: 1557

guys, i have a question about these type of questions.
When they ask for 'value', do they mean a numerical value, or are they just looking to get the variable a in terms of the variable b ? Because if its just a in terms of b, then I would answer A.
You can take statement 1 , and break the stem question like such:
sqt (a^3a^2b) = 7
sqt (a(a^2a)b) = 7
sqt (a(12)b) = 7
square both sides to get : 12ab=49 , and a=(49+b)/12
Correct ?



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Joined: 30 Dec 2005
Posts: 5
Location: Columbus, OH

pmenon,
In such questions, they are looking for an absolute single numeric value.
C looks right. A tricky one.



Director
Joined: 04 Oct 2005
Posts: 582
Location: Chicago

Bhai wrote: I think it is C
From I we get a = 7 + b/12 From II we get b = 2
Bhai How did you got b=2 from the second part
infact b can have 2 values 1 and 2 ..Isnt it?
Am i missing anything?
So my guess would be E



Director
Joined: 26 Sep 2005
Posts: 572
Location: Munich,Germany

agree it is C
stmt1,
a= 3, 4  insuff
stmt2,
b=1,2  insuff
consider C, only a=4, b=1 satisfies a^3a^2b=49, Hence C.



Intern
Joined: 09 Feb 2006
Posts: 20
Location: Raleigh, NC

Im a little confused on this one.
From the original statement,
SQRT [ a^3  a^2  b ] = 7
Hence a^3  a^2  b = 49 [ IS this step flawed ???]
=> a * ( a^2  a)  b = 49
Consider (i)
No matter what a is, the value of a^2a = 12
Substitute in the main equation, a * 12  b = 49 [ Not suff]
from (ii)
b = 2, 1 Again not sufficient. [Eliminate B]
(i) and (ii) together, again consider eq 12ab = 49,
a can be 4, 3 & B could be 2,1.
a=4 and b=1 satisfies 12ab=49, hence C
Dunno if I have done it right, will wait on the OA for sure.



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Joined: 03 Jan 2005
Posts: 2233

andy_gr8 wrote: Bhai wrote: I think it is C
From I we get a = 7 + b/12 From II we get b = 2 Bhai How did you got b=2 from the second part infact b can have 2 values 1 and 2 ..Isnt it? Am i missing anything? So my guess would be E
Actually from II we get b=1 or 2. When we combine them we know that b has to be odd for 12ab=49. You finally confirm that a=4 and b=1 is the only solution that satisfies all conditions.
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Manager
Joined: 17 Dec 2005
Posts: 162

HongHu wrote: andy_gr8 wrote: Bhai wrote: I think it is C
From I we get a = 7 + b/12 From II we get b = 2 Bhai How did you got b=2 from the second part infact b can have 2 values 1 and 2 ..Isnt it? Am i missing anything? So my guess would be E Actually from II we get b=1 or 2. When we combine them we know that b has to be odd for 12ab=49. You finally confirm that a=4 and b=1 is the only solution that satisfies all conditions.
I got C and that was my rationale. i think i'm starting to understand this concept. thanks all!!!










