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Manager
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for integers a & b , if sqt (a^3-a^2-b) = 7 ...wat is [#permalink]
 27 Dec 2005, 13:51
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for integers a & b , if sqt (a^3-a^2-b) = 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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Director
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jinesh wrote: for integers a & b , if sqt (a^3-a^2-b) = 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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I think it is C
From I we get a = 7 + b/12
From II we get b = 2
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Manager
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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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jinesh wrote: for integers a & b , if sqt (a^3-a^2-b) = 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^2-a-12=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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Manager
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Thanx a lot guys...... got it!!!!! 
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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^3-a^2-b) = 7
sqt (a(a^2-a)-b) = 7
sqt (a(12)-b) = 7
square both sides to get : 12a-b=49 , and a=(49+b)/12
Correct ?
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pmenon,
In such questions, they are looking for an absolute single numeric value.
C looks right. A tricky one.
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Director
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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
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
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Director
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agree it is C
stmt1,
a= -3, 4 -- insuff
stmt2,
b=-1,2 -- insuff
consider C, only a=4, b=-1 satisfies a^3-a^2-b=49, Hence C.
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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^2-a = 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 12a-b = 49,
a can be 4, -3 & B could be 2,-1.
a=4 and b=-1 satisfies 12a-b=49, hence C
Dunno if I have done it right, will wait on the OA for sure.
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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 12a-b=49. You finally confirm that a=4 and b=-1 is the only solution that satisfies all conditions.
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keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
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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 12a-b=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!!! 
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