|
Author |
Message |
|
Intern
Joined: 05 Oct 2008
Posts: 14
Followers: 0
Kudos [?]:
2
[1] , given: 0
|
1
This post received
KUDOS
Question Stats:
30% (01:51) correct
69% (00:55) wrong based on 1 sessions
If x^3*y^4 = 2000, what is y? 1. x is an integer 2. y is an integer Source: GMAT Club Tests - hardest GMAT questions
|
|
|
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2530
Followers: 41
Kudos [?]:
358
[2] , given: 19
|
2
This post received
KUDOS
echizen wrote: x^3*y^4 = 2000 . What is y?
1. x is an integer
2. y is an integer Statement 1: x is an integer: if x is an integer, y could or could not be an integer. for ex: if x = 1, y = -/+(2000^(1/4)) if x = 2, y = 250^(1/4) if x = 5, y = 2 or -2 if x = 2000^(1/3), y = 1 or -1 Statement 2: y is an integer if y = 1 or -1, x = (2000^1/3) if y = 2 or -2, x = 125^(1/3) if y = 5, x = (16/5)^1/3 from 1 and 2: y = + or - 2. Therefore, E.
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Intern
Joined: 05 Oct 2008
Posts: 14
Followers: 0
Kudos [?]:
2
[0], given: 0
|
Thanks for your reply this helps 
|
|
|
|
|
|
Intern
Joined: 28 May 2010
Posts: 6
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Agree with E.
Statement 1: Insufficient.
Statement 2: Insuffiecient.
Combining 1 & 2: 5^3 x 2^4 =2000
Now, (-2)^4 = (2)^4= 16... thats a good catch here, + 2 & -2 raised to a +ve exponent yield the same positive result. Hence, y does not have definite value, it could be either +2 or -2.
Hence, E.
|
|
|
|
|
|
Manager
Joined: 15 Sep 2009
Posts: 148
Followers: 1
Kudos [?]:
8
[0], given: 2
|
I will go with option E as there are two set of values satisfying the above eqn..
x = 5 & y = 2
x = 5 & y = -2.
So option E
|
|
|
|
|
|
Director
Joined: 21 Dec 2009
Posts: 592
Concentration: Entrepreneurship, Finance
Followers: 13
Kudos [?]:
130
[0], given: 20
|
concurred, the answer should be E. integers doesn't mean only positive values.
_________________
KUDOS me if you feel my contribution has helped you.
|
|
|
|
|
|
Manager
Joined: 12 Jul 2010
Posts: 69
Followers: 1
Kudos [?]:
2
[0], given: 3
|
Its E.
For ex: 5 and 2 solved for X and Y resp.
But Y cannot be determined it can be +2 or -2!
|
|
|
|
|
|
Manager
Joined: 09 Sep 2009
Posts: 59
Followers: 2
Kudos [?]:
5
[0], given: 13
|
St1: x is an integer - insufficient, coz value of y can be anything, right from an integer to a fractional value...( the only fact that is derived here is x will be definitely positive)
St2: y is an integer - insufficient, x can take any value.
St1 & St2 - x should be positive, y can be positive or negative..
Therefore, E
|
|
|
|
|
|
Manager
Joined: 07 Aug 2010
Posts: 90
Followers: 1
Kudos [?]:
13
[0], given: 9
|
E came up with one solution using (1) 2000 = 1000 * 2 = 10^3 * (sqrt(sqrt(2)) ) ^ 4 similar solution can be obtained for (2) however, many questions remain unanswered. or simply put, if you plug in multiple values, you may or may not get 2000 as the product.
_________________
Click that thing - Give kudos if u like this
|
|
|
|
|
|
Manager
Joined: 04 Dec 2009
Posts: 76
Location: INDIA
Followers: 2
Kudos [?]:
6
[0], given: 4
|
Ans E: X must be +ve but Y can be +ve or -ve 2000=2*2*2*2*5*5*5 or 2000=-2*-2*-2*-2*5*5*5 so,E
_________________
MBA (Mind , Body and Attitude )
|
|
|
|
|
|
Manager
Joined: 15 Apr 2010
Posts: 200
Followers: 3
Kudos [?]:
14
[0], given: 29
|
Thanks all... this seemed like a slam-dunk E..... due to no "exclusions" or narrowing of scope with the statements 1 or 2. (eg: x>0, y>0)...
|
|
|
|
|
|
Intern
Joined: 25 Aug 2010
Posts: 42
Followers: 1
Kudos [?]:
10
[0], given: 0
|
my opinion as follows.
x^3*y^4=2000
(x*y)^3*y=2*10^3
thus, x*y must produce a multiple of 10.
(x*y)^3=k*10^3
k*y=2
if k is larger than 2, then y cannot be an integer.
if y was a fraction, then x might be a fraction. (8/3 and 30/8, the two produce 10)
|
|
|
|
|
|
Intern
Joined: 02 Apr 2010
Posts: 48
Location: Mumbai
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Answer is E. The catch point here is to pick numbers of different varieties in order to try to deny and support a statement.. Since both are integers, each can be +ve or -ve. Therefore, y has +2 or -2.
_________________
Consider kudos for good explanations.
|
|
|
|
|
|
Intern
Joined: 02 Nov 2010
Posts: 47
Followers: 0
Kudos [?]:
0
[0], given: 23
|
thanks everyone good explanations! I didn't think of 2 pos being neg. I should have, problem seemed too easy.
|
|
|
|
|
|
Manager
Status: He who asks is a fool for five minutes, but he who does not ask remains a fool forever
Joined: 20 Aug 2010
Posts: 104
Followers: 2
Kudos [?]:
2
[0], given: 2
|
concur with as Y has even power...
|
|
|
|
|
|
Intern
Joined: 17 Dec 2010
Posts: 10
Location: Lima, Peru
Followers: 0
Kudos [?]:
0
[0], given: 3
|
What is the fastest way to determine that x and y are 5 and +/- 2? Or would an easier way to answer this question be simply to note that because y is raised to an even exponent it will have to have an +/- answer and therefore the answer will be E?
_________________
To those who have struggled with them, the mountains reveal beauties that they will not disclose to those who make no effort. That is the reward the mountains give to effort. And it is because they have so much to give and give it so lavishly to those who will wrestle with them that men love the mountains and go back to them again and again. The mountains reserve their choice gifts for those who stand upon their summits. (Sir Francis Younghusband)
|
|
|
|
|
|
Manager
Joined: 28 Feb 2011
Posts: 95
Followers: 0
Kudos [?]:
29
[0], given: 2
|
Its E! Missed considering the negative part..Silly of me.. 
_________________
Fight till you succeed like a gladiator..or doom your life which can be invincible
|
|
|
|
|
|
Manager
Joined: 01 Jan 2011
Posts: 92
Schools: INSEAD,IIMA,IIMB
Followers: 0
Kudos [?]:
0
[0], given: 2
|
even and odd powers provide number of solutions....go ahead with E after a few quick substitutions
_________________
_________________________
Try and you will succeed !
|
|
|
|
|
|
Senior Manager
Joined: 20 Dec 2010
Posts: 261
Schools: UNC Duke Kellogg
Followers: 3
Kudos [?]:
19
[0], given: 4
|
In the DS section -- its useful to take a moment to just think about what concept the GMAT is likely trying to test. This makes it much easier to solve the problem than just using the brute force method.
|
|
|
|
|
|
Manager
Affiliations: NABE
Joined: 25 Apr 2010
Posts: 67
Location: United States
Concentration: Strategy
GPA: 3.1
WE: Marketing (Energy and Utilities)
Followers: 1
Kudos [?]:
14
[0], given: 8
|
I chose E, but for two diff reasons: 1) X and Y can be +ve or -ve 2) Is "star" (e.g. *) defined as multiplication? I thought multiplication was "x"? I figured since you don't know what function "*" is, you cannot solve the problem anyway. I encountered problems like these on Kaplan Advanced.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
