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mmgmat2008
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What is the value of a?
(1) a^2-a=12
(2) b^2-b=2
(1) a^2-a=12
(2) b^2-b=2
I am getting D).
Whats the official answer?
For integers a and b if square root of (a^3-a^2-b)=7,
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dmateer25
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(1)
a^2-a-12=0
(a-4)(a+3)=0
a= 4 | a= -3
We can't determine from the information given whether a will be -4 or 3 because depending on what b is either of these values are possible.
(2)
b^2-b-2=0
(b-2)(b+1)=0
b= 2 | b=-1
If b is 2 then no integer for a will make the sqrt come out to 7.
therefore b must be -1 and knowing b is -1 the only possible integer for a that will make the sqrt=7 is 4
I pick B
a^2-a-12=0
(a-4)(a+3)=0
a= 4 | a= -3
We can't determine from the information given whether a will be -4 or 3 because depending on what b is either of these values are possible.
(2)
b^2-b-2=0
(b-2)(b+1)=0
b= 2 | b=-1
If b is 2 then no integer for a will make the sqrt come out to 7.
therefore b must be -1 and knowing b is -1 the only possible integer for a that will make the sqrt=7 is 4
I pick B
Last edited by dmateer25 on Fri Dec 05, 2008 11:59 am, edited 1 time in total.
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cramya
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Dmateer,a^2-a-12=0
(a-4)(a+3)=0
a= -4 | a= 3
(a-4)(a+3)=0
a= -4 | a= 3 should this not be a=4 and a=-3
Only 4 works for a^2-a = 12
If a=3 3^2-3 = 6 and not 12
Let me know ur thoughts
Hence I choose D)
Actually my approach was
a^2-a = 12
a(a-1) = 12
only possible when a=4 a-1=3 since its given a and b are integers
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logitech
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For integers a and b if square root of (a^3-a^2-b)=7,
What is the value of a?
(1) a^2-a=12
(2) b^2-b=2
In DS questions, both statement always give us TRUE facts, either we need them or not!
Statement 1 tells us: a=4 or a=-3
Statement 2 tells us: b=2 or b=-1
So actually we have 4 possible pairs for (a,b)
(4,2) (4,-1) (-3,2) and (-3,-1)
Let's put these values in our equation:
square root of (a^3-a^2-b)=7
Only (4,-1) gives us the solution.
This solution can be obtained from each statements. So do NOT choose (D)
Read the posts below:
What is the value of a?
(1) a^2-a=12
(2) b^2-b=2
In DS questions, both statement always give us TRUE facts, either we need them or not!
Statement 1 tells us: a=4 or a=-3
Statement 2 tells us: b=2 or b=-1
So actually we have 4 possible pairs for (a,b)
(4,2) (4,-1) (-3,2) and (-3,-1)
Let's put these values in our equation:
square root of (a^3-a^2-b)=7
Only (4,-1) gives us the solution.
This solution can be obtained from each statements. So do NOT choose (D)
Read the posts below:
Last edited by logitech on Fri Dec 05, 2008 12:33 pm, edited 1 time in total.
LGTCH
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dmateer25
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cramya wrote:Dmateer,a^2-a-12=0
(a-4)(a+3)=0
a= -4 | a= 3
(a-4)(a+3)=0
a= -4 | a= 3 should this not be a=4 and a=-3
Only 4 works for a^2-a = 12
If a=3 3^2-3 = 6 and not 12
Let me know ur thoughts
Hence I choose D)
Actually my approach was
a^2-a = 12
a(a-1) = 12
only possible when a=4 a-1=3 since its given a and b are integers
Cramya, yes you were correct a = 4 and a = -3. I did this correct on my scratch paper but typed it incorrect.
-3 will also work for a^2-a=12
-3^2-(-3)=12
9+3=12
12=12
However, I still feel that B is correct.
Stmt 1:
a = 4 and a = -3
When a=4
sqrt(4^3-4^2-b)=7
sqrt(64-16-b)=7
sqrt(48-b)=7
b must be -1
when a=-3
sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be 67
so we don't know if a is -3 or 4. INSUFF
From stmt 2 we know that b=2 or b=-1
from the original statment we know that a and b are both integers.
so if b=2 there is no integer that solves the solution for a.
sqrt(a^3-a^2-2)=7
No integer solves this.
Now trying -1 as the solution.
sqrt(a^3-a^2-(-1))=7,
a=4 is the only integer that will make this correct.
So we know that a must be 4.
Maybe I am missing something on this.
Last edited by dmateer25 on Fri Dec 05, 2008 12:12 pm, edited 1 time in total.
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logitech
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On the same token, statement two gives you two values for a and -3 can not be the answer because we can not have negative number inside the root. So IMO , D is the correct answer here.dmateer25 wrote:
Maybe I am missing something on this.
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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dmateer25
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Logitech,logitech wrote:On the same token, statement two gives you two values for a and -3 can not be the answer because we can not have negative number inside the root. So IMO , D is the correct answer here.dmateer25 wrote:
Maybe I am missing something on this.
Could you explain where I am going wrong with statement 1:
Stmt 1:
a = 4 and a = -3
When a=4
sqrt(4^3-4^2-b)=7
sqrt(64-16-b)=7
sqrt(48-b)=7
b must be -1
when a=-3
sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be -67
so we don't know if a is -3 or 4. INSUFF
Last edited by dmateer25 on Fri Dec 05, 2008 12:32 pm, edited 1 time in total.
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logitech
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-3 ( -3-1) = 12 Cramya both values for a works in this case.cramya wrote:
Actually my approach was
a^2-a = 12
a(a-1) = 12
only possible when a=4 a-1=3 since its given a and b are integers
LGTCH
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logitech
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My friend, there is nothing wrong with your logic. ( Oh by the way b=-67).dmateer25 wrote:
when a=-3
sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be 67
so we don't know if a is -3 or 4. INSUFF
In fact Statement 1 is INSUF since both a values satisfy the first question.
So B 'should' be the OA.
LGTCH
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dmateer25
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oops. Thanks for correcting me on the -67! I corrected it in my post.logitech wrote:My friend, there is nothing wrong with your logic. ( Oh by the way b=-67).dmateer25 wrote:
when a=-3
sqrt(-3^3-(-3^2)-b)=7
sqrt(-27+9-b)=7
b must be 67
so we don't know if a is -3 or 4. INSUFF
In fact Statement 1 is INSUF since both a values satisfy the first question.
So B 'should' be the OA.
This was a very tricky question!
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cramya
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Logitech,
I agree both of us missed that!
Dmateer,
You are absolutely correct. I made a mistake in not considering -3 as a possibility in stmt I (should have set it up as a quadratic like u).
The answer should be B) like u originally said!
Hope I am not messing up again
Should b not be -85?
sqrt(a^3-a^2-b) = 7
a^3-a^2-b=49
If a=-3
-27-((-3)^2) - b = 49
-27-9-b=49
-36-b=49
-b=85
b=-85
Regards,
Cramya
I agree both of us missed that!
Dmateer,
You are absolutely correct. I made a mistake in not considering -3 as a possibility in stmt I (should have set it up as a quadratic like u).
The answer should be B) like u originally said!
Hope I am not messing up again
Should b not be -85?
sqrt(a^3-a^2-b) = 7
a^3-a^2-b=49
If a=-3
-27-((-3)^2) - b = 49
-27-9-b=49
-36-b=49
-b=85
b=-85
Regards,
Cramya
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dmateer25
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Cramya,cramya wrote:Logitech,
I agree both of us missed that!
Dmateer,
You are absolutely correct. I made a mistake in not considering -3 as a possibility in stmt I (should have set it up as a quadratic like u).
The answer should be B) like u originally said!
Hope I am not messing up again
Should b not be -85?
sqrt(a^3-a^2-b) = 7
a^3-a^2-b=49
If a=-3
-27-((-3)^2) - b = 49
-27-9-b=49
-36-b=49
-b=85
b=-85
Regards,
Cramya
Yes, you are correct. I distributed the negative sign before squaring -3.
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mmgmat2008
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The answer is B. I think because the first one can come out with a^3-a^2=48. a= 4 or -3.cramya wrote:I am getting D).
Whats the official answer?
the second one is b=-1, or 2.
B=-1 a=4
B=2, no values for a.
thus answer is B.
