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# for integers a & b , if sqt (a^3-a^2-b) = 7 ...wat is

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Manager
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for integers a & b , if sqt (a^3-a^2-b) = 7 ...wat is [#permalink]

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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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27 Dec 2005, 14:27
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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SVP
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27 Dec 2005, 14:29
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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27 Dec 2005, 18:22
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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Senior Manager
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27 Dec 2005, 23:29
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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28 Dec 2005, 02:46
Thanx a lot guys...... got it!!!!!
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"Nothing is Impossible"

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03 Jan 2006, 13:09
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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Intern
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03 Jan 2006, 18:51
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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03 Jan 2006, 22:38
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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14 Feb 2006, 02:45
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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Intern
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14 Feb 2006, 07:28
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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16 Feb 2006, 20:19
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.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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Manager
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11 Jun 2006, 15:09
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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11 Jun 2006, 15:09
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