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Posts: 26
Updated on: 12 Jun 2015, 15:22
Originally posted by tjerkrintjema on 10 Jun 2015, 02:26.
Last edited by Harley1980 on 12 Jun 2015, 15:22, edited 1 time in total.
Last edited by Harley1980 on 12 Jun 2015, 15:22, edited 1 time in total.
Changed OA
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If a and b are positive integers larger than 1, is a^b > b^a?
(1) a = b^2
(2) b > 2
Source: Mindful Gmat
(1) a = b^2
(2) b > 2
Source: Mindful Gmat
a and b positive integers > 1
Looking at statement 1 you can conclude that a > b, since for example 4 = 2^2, or 16 = 4^2
However 4^2 = 2^4, and looking at 16^4 < 4^16
Statement 1 is insufficient
Statement 2 tells us b > 2, but does not give any information about a, therefore statement 2 alone can't be sufficient.
Hence answer E
Looking at statement 1 you can conclude that a > b, since for example 4 = 2^2, or 16 = 4^2
However 4^2 = 2^4, and looking at 16^4 < 4^16
Statement 1 is insufficient
Statement 2 tells us b > 2, but does not give any information about a, therefore statement 2 alone can't be sufficient.
Hence answer E
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
tjerkrintjema
This is a reworded OG question: if-x-and-y-are-nonzero-integers-is-x-y-y-x-168279.html
Similar question: if-x-and-y-are-integers-is-x-y-y-x-114980.html
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EMPOWERgmatRichC
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Expert reply
Updated on: 12 Jun 2015, 16:31
Originally posted by EMPOWERgmatRichC on 10 Jun 2015, 19:41.
Last edited by EMPOWERgmatRichC on 12 Jun 2015, 16:31, edited 1 time in total.
Last edited by EMPOWERgmatRichC on 12 Jun 2015, 16:31, edited 1 time in total.
Hi tjerkrintjema,
This DS question is perfect for TESTing VALUES.
We're told that A and B are POSITIVE INTEGERS that are GREATER than 1. We're asked if A^B > B^A. This is a YES/NO question.
Fact 1: A = B^2
IF....
B = 2
A = 4
4^2 is NOT > 2^4 ....and the answer to the question is NO
IF....
B = 3
A = 9
9^3 = (3^2)^3 = 3^6
9^3 IS NOT > 3^9 ....and the answer to the question is NO
IF....
B = 4
A = 16
16^4 = (4^2)^4 = 4^8
16^4 IS NOT > 4^16 ....and the answer to the question is NO
The answer to the question will ALWAYS BE NO.
Fact 1 is SUFFICIENT
Fact 2: B > 2
With this Fact, we don't know ANYTHING about the value of A....
IF....
B = 3
A = 9
9^3 IS NOT > 3^9 ....and the answer to the question is NO
IF....
B = 4
A = 3
3^4 IS > 4^3 ....and the answer to the question is YES
Fact 2 is INSUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This DS question is perfect for TESTing VALUES.
We're told that A and B are POSITIVE INTEGERS that are GREATER than 1. We're asked if A^B > B^A. This is a YES/NO question.
Fact 1: A = B^2
IF....
B = 2
A = 4
4^2 is NOT > 2^4 ....and the answer to the question is NO
IF....
B = 3
A = 9
9^3 = (3^2)^3 = 3^6
9^3 IS NOT > 3^9 ....and the answer to the question is NO
IF....
B = 4
A = 16
16^4 = (4^2)^4 = 4^8
16^4 IS NOT > 4^16 ....and the answer to the question is NO
The answer to the question will ALWAYS BE NO.
Fact 1 is SUFFICIENT
Fact 2: B > 2
With this Fact, we don't know ANYTHING about the value of A....
IF....
B = 3
A = 9
9^3 IS NOT > 3^9 ....and the answer to the question is NO
IF....
B = 4
A = 3
3^4 IS > 4^3 ....and the answer to the question is YES
Fact 2 is INSUFFICIENT
Final Answer:
GMAT assassins aren't born, they're made,
Rich
Fact 1 : if B=2 A=4 then as per que we check is a^b>b^a ans is 16>16 'NO'
if B=3 A=9 then 9^3>3^9 we get 'NO'
in both cases i get ans as 'NO' & when i put a=2 b=2 i get ans as a^b>b^a 'NO'
now put B=4 A=2 ans is 'NO'
put B=9 A=3 ans is 'YES'
So even if i i take B>2 i do not any single answer so clearly ans is 'E'
Jimmy
if B=3 A=9 then 9^3>3^9 we get 'NO'
in both cases i get ans as 'NO' & when i put a=2 b=2 i get ans as a^b>b^a 'NO'
now put B=4 A=2 ans is 'NO'
put B=9 A=3 ans is 'YES'
So even if i i take B>2 i do not any single answer so clearly ans is 'E'
Jimmy
EMPOWERgmatRichC
EMPOWERgmatRichC
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Expert reply
Hi jimmy02,
By TESTing VALUES in Fact 1 and Fact 2, you were able to gather PROOF that each Fact was Insufficient on its own. However, you have NOT provided any proof when you combined Facts....so how do you know what the correct answer is?
GMAT assassins aren't born, they're made,
Rich
By TESTing VALUES in Fact 1 and Fact 2, you were able to gather PROOF that each Fact was Insufficient on its own. However, you have NOT provided any proof when you combined Facts....so how do you know what the correct answer is?
GMAT assassins aren't born, they're made,
Rich
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Bunuel
I think this question is little different as it says a and b are 'positive' integers. The OG questions say non zero integers.
Also I am getting 'A' as the answer. Can anybody else check this again?
My solution is:
Given: a, b are integers and greater than 1, i.e. 2, 3, 4, 5...
Question: Is \(a^b > b^a\)?
Statement 1: \(a=b^2\)
b=2, a=4 : \(a^b = 4^2 = 16; b^a = 2^4 = 16;\) Is \(a^b > b^a =>\) Is 16>16. Answer is NO.
b=3, a=9 : \(a^b = 9^3; b^a = 3^9 = 3^6 . 3^3 = 9^3 . 3^3;\) Is \(a^b > b^a =>\) Is \(9^3 > 9^3 . 3^3\). Answer is NO.
We can do the same for b=4, 5... and will notice that answer will always be NO.
So, statement 1 is sufficient to answer.
Statement 2: b>2
No information about a. So, statement 2 is insufficient to answer.
So, final answer shall be A.
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soumyajit_nayak
Hello soumyajit_nayak
I think you are right.
Here is misplaced parts of inequality in explanations:
EMPOWERgmatRichC
should be \(a^b = 9^3\) and \(b^a = 3^9\) so the answer is no.
So final answer is A
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EMPOWERgmatRichC
Hi EMPOWERgmatRichC
In your solution, you are trying to answer 'Is B^A > A^B?'. Question is 'Is A^B > B^A?'.
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Algebraically we can see why the answer is A. If we know that \(a = b^2\), we can substitute for a in the question. Bear in mind what we're doing - we're rephrasing the question, or in other words, finding out under what conditions the answer to the question will be 'yes'.
Is \(a^b > b^a\) ?
Is \((b^2)^b > b^{(b^2)}\) ?
Is \(b^{2b} > b^{(b^2)}\) ?
and since our base is greater than 1, this inequality will only be true if the exponent on the left is larger than the exponent on the right, so our question becomes:
Is \(2b > b^2\) ?
and since b is positive, we can divide by b without worrying about reversing the inequality:
Is \(2 > b\) ?
So the answer to the question is only 'yes' when \(b < 2\) is true. Since the question tells us this is absolutely not true (it tells us b is an integer greater than 1, so we know b > 2), we know the answer to the question must be 'no'.
One final comment:
One needs to be careful using logic like this in GMAT DS questions. It's very possible for a statement that doesn't mention a here to be sufficient, and the trickiest real GMAT DS questions are often ones with statements that, at first glance, don't appear to have a chance to be sufficient, but which turn out to be. I can quickly make up a simple example, but far more devious examples are possible:
If k and z are positive integers, is \(k^z \geq z^k\)?
1. \(z = 1\)
2. \(k > z\)
Here if you thought "statement 1 doesn't mention k, so it cannot be sufficient" you'd be dismissing Statement 1 too quickly. If we know z=1 from Statement 1, then substituting, the question becomes "is \(k^1 \geq 1^k\)?", so the question becomes "Is \(k \geq 1\)?" which we know is true. So Statement 1 is sufficient. Statement 2 is not (you can get a 'yes' answer whenever z=1, but a 'no' answer when k=4 and z=3, say), so A would be the answer here.
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Is \(a^b > b^a\) ?
Is \((b^2)^b > b^{(b^2)}\) ?
Is \(b^{2b} > b^{(b^2)}\) ?
and since our base is greater than 1, this inequality will only be true if the exponent on the left is larger than the exponent on the right, so our question becomes:
Is \(2b > b^2\) ?
and since b is positive, we can divide by b without worrying about reversing the inequality:
Is \(2 > b\) ?
So the answer to the question is only 'yes' when \(b < 2\) is true. Since the question tells us this is absolutely not true (it tells us b is an integer greater than 1, so we know b > 2), we know the answer to the question must be 'no'.
One final comment:
soumyajit_nayak
One needs to be careful using logic like this in GMAT DS questions. It's very possible for a statement that doesn't mention a here to be sufficient, and the trickiest real GMAT DS questions are often ones with statements that, at first glance, don't appear to have a chance to be sufficient, but which turn out to be. I can quickly make up a simple example, but far more devious examples are possible:
If k and z are positive integers, is \(k^z \geq z^k\)?
1. \(z = 1\)
2. \(k > z\)
Here if you thought "statement 1 doesn't mention k, so it cannot be sufficient" you'd be dismissing Statement 1 too quickly. If we know z=1 from Statement 1, then substituting, the question becomes "is \(k^1 \geq 1^k\)?", so the question becomes "Is \(k \geq 1\)?" which we know is true. So Statement 1 is sufficient. Statement 2 is not (you can get a 'yes' answer whenever z=1, but a 'no' answer when k=4 and z=3, say), so A would be the answer here.
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EMPOWERgmatRichC
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Expert reply
Updated on: 13 Jun 2015, 14:04
Originally posted by EMPOWERgmatRichC on 12 Jun 2015, 16:33.
Last edited by EMPOWERgmatRichC on 13 Jun 2015, 14:04, edited 1 time in total.
Last edited by EMPOWERgmatRichC on 13 Jun 2015, 14:04, edited 1 time in total.
Quote:
Hi soumyajit_nayak,
Good eye! I've edited my solution.
GMAT assassins aren't born, they're made,
Rich
Posts: 3
I got this one wrong also. But after some thoughts I found the solution.
For (1), if a=b^2, then we can use this condition to translate the problem
a^b=(b^2)^b=b^2b
b^a=b^(b^2)
Knowing that the base b>1, we can just compare the powers 2b and b^2, 2b-b^2=b(2-b)<=0 (because b>=2), so we have 2b<=b^2, thus a^b<=b^a, the "=" only applies when b=2 and a=4, otherwise it is always true a^b<b^a. (1) alone is sufficient. If gives us the generalization that when both the base and power are greater than 2, the power is more important than the base.
For 2, not much to be said, we can easily find examples that lead to different answers, so, insufficient.
For (1), if a=b^2, then we can use this condition to translate the problem
a^b=(b^2)^b=b^2b
b^a=b^(b^2)
Knowing that the base b>1, we can just compare the powers 2b and b^2, 2b-b^2=b(2-b)<=0 (because b>=2), so we have 2b<=b^2, thus a^b<=b^a, the "=" only applies when b=2 and a=4, otherwise it is always true a^b<b^a. (1) alone is sufficient. If gives us the generalization that when both the base and power are greater than 2, the power is more important than the base.
For 2, not much to be said, we can easily find examples that lead to different answers, so, insufficient.
Posts: 2
[quote="tjerkrintjema"]If a and b are positive integers larger than 1, is a^b > b^a?
(1) a = b^2
(2) b > 2
Source: Mindful Gmat
[spoiler=]a and b positive integers > 1
Looking at statement 1 you can conclude that a > b, since
for example 4 = 2^2, or 16 = 4^2
However 4^2 = 2^4, and looking at 16^4 < 4^16
Statement 1 is insufficient
Statement 2 tells us b > 2, but does not give any information about a, therefore statement 2 alone can't be sufficient.
So , A,D and B are not answers , now, taking both statements together
a=b^2 and b > 2
so minimum value of b has to be 3 ( both a and b are greater than 1 )
that makes a = b^2 = 9
now putting values of a and b in equation
a^b = 9^3 = 729
b^a = 3^9 = 729x27
so a^b < b^a
Next value of b = 4
a = 16
a^b = 16^4 = 2^16
b^a = 4^16 = 2^32
again a^b < b^a
so for every value of b > 2 and a=b^2
we will have a^b < b^a
so answer is (C)
(1) a = b^2
(2) b > 2
Source: Mindful Gmat
[spoiler=]a and b positive integers > 1
Looking at statement 1 you can conclude that a > b, since
for example 4 = 2^2, or 16 = 4^2
However 4^2 = 2^4, and looking at 16^4 < 4^16
Statement 1 is insufficient
Statement 2 tells us b > 2, but does not give any information about a, therefore statement 2 alone can't be sufficient.
So , A,D and B are not answers , now, taking both statements together
a=b^2 and b > 2
so minimum value of b has to be 3 ( both a and b are greater than 1 )
that makes a = b^2 = 9
now putting values of a and b in equation
a^b = 9^3 = 729
b^a = 3^9 = 729x27
so a^b < b^a
Next value of b = 4
a = 16
a^b = 16^4 = 2^16
b^a = 4^16 = 2^32
again a^b < b^a
so for every value of b > 2 and a=b^2
we will have a^b < b^a
so answer is (C)
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IanStewart
Hi Ian,
I concluded my transformation of the inequality in the question with relation to statement 1 at the highlighted bit shown, and then began to test values.
Upon testing b=2, I get 2^4 > 2^4, which of course is not true. The inequality holds for integers above 2 and this led me to disregard A as the answer choice as no uniform solution was obtained.
Could you please highlight why this approach is faulty? It seems perfectly fine to me and I'm unable to wrap my head around why it doesn't work.
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ShreyasJavahar
It's the part I've highlighted that is not correct. If you plug in b = 3, for example, the inequality becomes 3^6 > 3^9, which again is not true, and you'll find the inequality is never true if you plug in larger values. Because the inequality is not true for any valid values of b, we can be certain the answer to the question is 'no', and that means Statement 1 is sufficient alone.
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