any easier way to solve this without numbers?
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
-
beeparoo
- Master | Next Rank: 500 Posts
- Posts: 200
- Joined: Sun Jun 17, 2007 10:46 am
- Location: Canada
- Thanked: 9 times
I don't think questions like this can be solved algebraically. Besides, such a method would just slow you down... Here's my strategy.ildude02 wrote:Is X > Y?
1. SQRT(X) > Y;
2. X ^3 > Y;
Is there any algebraic way to solve this?
Statement 1: Notice that X = [SQRT(X)]^2. If SQRT(X) is greater than or equal to 1, then surely its squared value will be even greater. Then indeed, X > Y, no matter what the value of Y is.
However, if SQRT(X) < 1, like say, SQRT(X) = 1/2, then it's squared value will be reduced. Try to work this out with y = 1/3.
INSUFFICIENT
Statement 2: Regardless of whether X is positive or negative, X^3 > Y does not immediately clarify whether X > Y. You can plug in values to test if you need, but most can just tell innately.
INSUFFICIENT
Combined 1 & 2: From Statement 2, we know that SQRT(X) should be equal to or greater than 1 (because X^3 means that SQRT(X) has been raised to power of 6 and it certainly cannot be not reduced).
SUFFICIENT
I choose C. I hope it's the OA.
-
ildude02
- Master | Next Rank: 500 Posts
- Posts: 320
- Joined: Sun Jan 13, 2008 10:00 pm
- Thanked: 10 times
Yes, the answer is C. But can you please clarify a bit about how you deduced that SQRT(X) should be equal to or greater than 1. If we can come up with that conclusion without having to plu and play with numjbers, that will really spped up the process of solving the question whne conbining both the statements. Appreciaet your response.beeparoo wrote: Combined 1 & 2: From Statement 2, we know that SQRT(X) should be equal to or greater than 1 (because X^3 means that SQRT(X) has been raised to power of 6 and it certainly cannot be not reduced).
SUFFICIENT
-
moliver
- Senior | Next Rank: 100 Posts
- Posts: 63
- Joined: Wed Jul 16, 2008 6:43 am
- Thanked: 3 times
- GMAT Score:680
I think the difficulty of this problem is to find if it is a C or E.ildude02 wrote:Is X > Y?
1. SQRT(X) > Y;
2. X ^3 > Y;
Is there any algebric way to solve this?
If you put the two statements together you can write something like this:
if x>1 then
x^3>x>x^(1/2)>y
and if x <1 then
y<x^3<x<x>x>x^3>y
you you can answered in both cases, so the answer is C
-
beeparoo
- Master | Next Rank: 500 Posts
- Posts: 200
- Joined: Sun Jun 17, 2007 10:46 am
- Location: Canada
- Thanked: 9 times
Ew man, check your spelling!ildude02 wrote:Yes, the answer is C. But can you please clarify a bit about how you deduced that SQRT(X) should be equal to or greater than 1. If we can come up with that conclusion without having to plu and play with numjbers, that will really spped up the process of solving the question whne conbining both the statements. Appreciaet your response. You better, muchacho!beeparoo wrote: Combined 1 & 2: From Statement 2, we know that SQRT(X) should be equal to or greater than 1 (because X^3 means that SQRT(X) has been raised to power of 6 and it certainly cannot be not reduced).
SUFFICIENT
I don't think there's anything wrong with plugging in numbers. GMAT students shy away from this method because they think it is too random and prone to error. But that is a myth! If you do it systematically, it can be a VERY FAST method compared to algebraic solving - especially for questions involving Inequalities and Number Properties. Anyway, to answer your question about how I deduced SQRT(X) =/> 1 (equal to or greater than):
Let SQRT(X) = 1/2 and Y = 1/3. So, SQRT(X) > Y.
X = [SQRT(X)]^2 = 1/4. X is NOT > Y.
X^3 = [SQRT(X)]^6 = 1/64 Again, X is NOT > Y.
In order to satisfy 2nd Statement, make SQRT(X) =/> 1.
Let SQRT(X) = 2 and Y = 1.
X = [SQRT(X)]^2 = 4. So, X > Y.
X^3 = [SQRT(X)]^6 = 64. Again, X > Y.
Sandra
