|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 20 Jul 2004
Posts: 601
Followers: 1
Kudos [?]:
3
[0], given: 0
|
For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is [#permalink]
 05 Oct 2004, 18:52
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
For integers, a and b, if sqrt(a^3 - a^2 - b) = 7, what is the value of a?
I) a^2 - a = 12
II) b^2 - b = 2
|
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
1.a=4 or-3. Two values for a and hence insuff.
2. b can be 2 or -1. Applying the values of b in a^3-a^2-b=49, we have a as either some fraction or a=4. Since a need to be an integer, a is 4.
hence B.
|
|
|
|
|
|
Manager
Joined: 30 Apr 2004
Posts: 51
Location: NYC
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Quote: Applying the values of b in a^3-a^2-b=49, we have a as either some fraction or a=4. Since a need to be an integer, a is 4.
Please explain your method in solving this. My calc with polynomial equations is hazy. Thanks.
|
|
|
|
|
|
Director
Joined: 31 Aug 2004
Posts: 619
Followers: 1
Kudos [?]:
8
[0], given: 0
|
I think it is 49 = a.(a^2-a)-b so a = (49+b)/12
knowing that (stat. 2) b = 2 or -1, a being an integer, b = -1 and a = 4
|
|
|
|
|
|
Director
Joined: 08 Jul 2004
Posts: 613
Followers: 1
Kudos [?]:
4
[0], given: 0
|
I think C
1 gives 12-b=49
2 gives quadratic b=-1 and 2
Only b=-1, gives a as integer=4
S
|
|
|
|
|
|
Director
Joined: 31 Aug 2004
Posts: 619
Followers: 1
Kudos [?]:
8
[0], given: 0
|
I think you just prove statement 2 is suff !
|
|
|
|
|
|
Director
Joined: 08 Jul 2004
Posts: 613
Followers: 1
Kudos [?]:
4
[0], given: 0
|
how do you solve
a^3-a^2-48=0 to find the value of a to conclude B as answer?
S
|
|
|
|
|
|
Manager
Joined: 21 Aug 2004
Posts: 145
Followers: 1
Kudos [?]:
3
[0], given: 0
|
venksune wrote: 1.a=4 or-3. Two values for a and hence insuff.
2. b can be 2 or -1. Applying the values of b in a^3-a^2-b=49, we have a as either some fraction or a=4. Since a need to be an integer, a is 4.
hence B.
Hi Venk, I see that your solution for a in statement 1 neglects the original statement which states that if sqrt(a^3 - a^2 - b) = 7, what is the value of a, You solve statement 1 independ of the original equation and that does not seem right.
At best, from the original equation we can say that
a^3 - a^2 - b = 49
Now given 1, a2-a=12 implies that 12a -b =49 from (a(a^2-a) -b =49, making statement 1 insufficient.
Statement 2 now gives you two values of b (2, -1) which when substitude into the original equation yield 2 values, one an integer of 4 and the other a non integer. Thus for a to be an integer it has to be 4. B is sufficient
|
|
|
|
|
|
Manager
Joined: 21 Aug 2004
Posts: 145
Followers: 1
Kudos [?]:
3
[0], given: 0
|
saurya_s wrote: how do you solve
a^3-a^2-48=0 to find the value of a to conclude B as answer?
S
It is posible, try picking numbers adnd you will land at a = 4 or a = 3.25.
It is easy to fall into the trap that
statement 1 combine with the original statement yields
12a -b = 49 and thus the moment one gets b, one can find a making C the dangerous answer. But if you substitue the value of B calculated and substitude into the original equation you might get only one integer for a and that is 4.
|
|
|
|
|
|
Director
Joined: 16 Jun 2004
Posts: 929
Followers: 1
Kudos [?]:
4
[0], given: 0
|
sdanquah,
I am also saying only B is sufficient. In stem 1, we have two values of 'a' - question is what is the value of 'a'. We cant conclusively say the value of 'a'. Also, when applied to the original equation, we still need 'b'. I didnt neglect it, I thought the added data is not necessary to prove insufficiency of stem 1.
|
|
|
|
|
|
Manager
Joined: 07 Sep 2004
Posts: 62
Followers: 1
Kudos [?]:
0
[0], given: 0
|
for the second we have
b=-1,2
for b=-1 , sqrt(a^3-a^2+1)=7. implies a^3-a^2+1=49
or a^3 - a^2 -48 = 0
a = 4 is a solution
divide a^3-a^2-48 by a-4 gives the polynomial a^2+3a+12
so (a^2+3a+12)(a-4)=0
a^2+3a+12 is never zero since discriminant is negative (no roots for a) 
giving a =4
for b = 2
we get a^3-a^2=51
a^2(a-1)=51
from this we cannot find any value of A
hope this helps.
|
|
|
|
|
|
Director
Joined: 20 Jul 2004
Posts: 601
Followers: 1
Kudos [?]:
3
[0], given: 0
|
OA is B. From Princeton.
OE = "Statement 2 gives b = 2, -1. Since a cant be an interger if b = 2, it must be true that b = -1 and a = 4."
|
|
|
|
|
|
Manager
Joined: 21 Aug 2004
Posts: 145
Followers: 1
Kudos [?]:
3
[0], given: 0
|
venksune wrote: sdanquah,
I am also saying only B is sufficient. In stem 1, we have two values of 'a' - question is what is the value of 'a'. We cant conclusively say the value of 'a'. Also, when applied to the original equation, we still need 'b'. I didnt neglect it, I thought the added data is not necessary to prove insufficiency of stem 1.
Agreed with your answer venksume, Just that I found that you neglected orignial statement in your earlier solution.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
For integers a and b, if sqrt(a^3 - a^2 - b) = 7, what is
|
tyagel |
3 |
22 Oct 2004, 13:23 |
|
|
|
|
For Integers A and B, if SQRT(A^3 - A^2 -B) = 7, what is the
|
Dookie |
7 |
15 Nov 2004, 13:10 |
 |
|
|
For integers a and b, if square root of (a^3 - a^2 - b) = 7,
|
rahulraao |
16 |
22 Oct 2005, 22:22 |
|
|
|
|
For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is
|
anandsebastin |
6 |
24 Oct 2006, 13:53 |
|
|
1
|
 |
For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the
|
aalriy |
3 |
08 Dec 2010, 09:39 |
|
|
|
|
|
|
close
