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Updated on: 12 Mar 2015, 04:24
Originally posted by littlegirl on 09 Mar 2015, 18:01.
Last edited by littlegirl on 12 Mar 2015, 04:24, edited 1 time in total.
Last edited by littlegirl on 12 Mar 2015, 04:24, edited 1 time in total.
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C
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X and Y are positive numbers
Is X^4< y^3
(1) x^5< y^3
(2) x^4< y^5
Pls tell the best way to sovle this problem.Thanks!
Is X^4< y^3
(1) x^5< y^3
(2) x^4< y^5
Pls tell the best way to sovle this problem.Thanks!
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09 Mar 2015, 18:43
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Hi littlegirl,
Since the prompt restricts X and Y to POSITIVE numbers (NOT necessarily positive INTEGERS though), we're going to be dealing with Number Properties that interact with Exponents. You'll likely want to TEST VALUES and keep an open mind about the possible positive values that you can use...
We're asked if X^4 < Y^3. This is a YES/NO question.
Fact 1: X^5 < Y^3
IF....
X = 1
Y = 2
1 < 8
1^4 is < 2^3 and the answer to the question is YES.
IF....
X = 1/2
Y = 1/3
1/32 < 1/27
(1/2)^4 is NOT < (1/3)^3 and the answer to the questions is NO
Fact 1 is INSUFFICIENT
Fact 2: X^4 < Y^5
IF....
X = 1
Y = 2
1 < 32
1^4 is < 2^3 and the answer to the question is YES.
IF....
X = 3
Y = 4
81 < 1024
3^4 is NOT < 4^3 and the answer to the question is NO.
Fact 2 is INSUFFICIENT
Combined, we know..
X^5 < Y^3
X^4 < Y^5
IF....
X = 1
Y = 2
1 < 32
1^4 is < 2^3 and the answer to the question is YES.
To get a NO answer, we would have to think about using fractions (as we did in Fact 1). However, fractional values are not possible because the two given Facts, when combined, won't support a set of fractional answers.
To get a NO answer, we would need X^4 to be >= Y^3.
Using fractions, this would require that X >= Y.
As we raise X and Y to increasing 'powers', their respective values would DECREASE (since multiplying a fraction by another fraction makes the product SMALLER).
The two Facts give us one restriction is which X has the higher exponent and another restriction in which the Y has the higher exponent. There's no way for both of these restrictions to exist AND for X^4 to be > Y^3.
This is a very long way of saying that, when combined, the answer to the question is ALWAYS YES.
Combined, SUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
Since the prompt restricts X and Y to POSITIVE numbers (NOT necessarily positive INTEGERS though), we're going to be dealing with Number Properties that interact with Exponents. You'll likely want to TEST VALUES and keep an open mind about the possible positive values that you can use...
We're asked if X^4 < Y^3. This is a YES/NO question.
Fact 1: X^5 < Y^3
IF....
X = 1
Y = 2
1 < 8
1^4 is < 2^3 and the answer to the question is YES.
IF....
X = 1/2
Y = 1/3
1/32 < 1/27
(1/2)^4 is NOT < (1/3)^3 and the answer to the questions is NO
Fact 1 is INSUFFICIENT
Fact 2: X^4 < Y^5
IF....
X = 1
Y = 2
1 < 32
1^4 is < 2^3 and the answer to the question is YES.
IF....
X = 3
Y = 4
81 < 1024
3^4 is NOT < 4^3 and the answer to the question is NO.
Fact 2 is INSUFFICIENT
Combined, we know..
X^5 < Y^3
X^4 < Y^5
IF....
X = 1
Y = 2
1 < 32
1^4 is < 2^3 and the answer to the question is YES.
To get a NO answer, we would have to think about using fractions (as we did in Fact 1). However, fractional values are not possible because the two given Facts, when combined, won't support a set of fractional answers.
To get a NO answer, we would need X^4 to be >= Y^3.
Using fractions, this would require that X >= Y.
As we raise X and Y to increasing 'powers', their respective values would DECREASE (since multiplying a fraction by another fraction makes the product SMALLER).
The two Facts give us one restriction is which X has the higher exponent and another restriction in which the Y has the higher exponent. There's no way for both of these restrictions to exist AND for X^4 to be > Y^3.
This is a very long way of saying that, when combined, the answer to the question is ALWAYS YES.
Combined, SUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
littlegirl
hey littlegirl,
what is the source of this question?
OPTION A: \(X^5 < Y^3\)
at X=2 , Y=3 \(X^4<Y^3\)
but at \(X=1/2\) and \(Y=1/3\) \(X^4\) is not less than \(Y^3\)
OPTION B: \(X^4 < Y^5\)
at X=2 , Y=3 \(X^4<Y^3\)
but at \(X=4\) and \(Y=5\) \(X^4\) is not less than \(Y^3\)
combining A and B still insufficient
\(X^5 < Y^3\)
\(X^4 < Y^5\)
as we are told X and Y are positive numbers we can divide the inequalities safely to yield
\(( 1/X ) < Y^2\)
at X=2 and Y=1 this inequality holds but is \(2^4 < 1^3\) ? NO.
at X=\(1/3\) and Y=4 this inequality holds but is \((1/3)^4 < 4^3\) ? YES.
Hi Bunuel, VeritasPrepKarishma
I am not sure how the official answer is C . Kindly help , i am lost in this question .
Thanks
Lucky
) . 
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