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M31-53

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M31-53  [#permalink]

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New post 20 Jul 2015, 04:49
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

45% (01:06) correct 55% (00:53) wrong based on 73 sessions

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Re M31-53  [#permalink]

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New post 20 Jul 2015, 04:49
1
1
Official Solution:


If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

(1) \(6^a = 36^b\). Simplify: \(6^a = 6^{2b}\). Bases are equal, hence we can equate the powers: \(a=2b\). If \(a=b=0\), then \(a\) is NOT greater than \(b\) but if \(a=2\) and \(b=1\), then \(a\) IS greater than \(b\). Not sufficient.

(2) \(5^a = 35^b\). If both \(a\) and \(b\) are positive integers, then we'd have that \(5^a\) is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both \(a\) and \(b\) must be 0, giving a NO answer to the question whether \(a\) is greater than \(b\). Sufficient,


Answer: B
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Re M31-53  [#permalink]

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New post 17 Jul 2016, 02:20
I think this is a high-quality question and I agree with explanation. Great Trap on choice B! Excellent
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Re: M31-53  [#permalink]

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New post 04 Jun 2017, 07:31
Very good question. I made a mistake assuming that a = 2b => a> b without considering b =0
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Re: M31-53  [#permalink]

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New post 09 Jul 2017, 02:23
Hi Bunuel,

Since it is given a & b are positive, how can we assume a=0 from statement 1? (0 is neither -ve nor +ve)

Thanks
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Re: M31-53  [#permalink]

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New post 09 Jul 2017, 02:29
petal123 wrote:
Hi Bunuel,

Since it is given a & b are positive, how can we assume a=0 from statement 1? (0 is neither -ve nor +ve)

Thanks


We are not given that a and b are positive, we are given that a and b are non-negative. Non-negative, means 0 or positive (so those which are not negative).
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Re: M31-53  [#permalink]

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New post 26 Aug 2017, 03:15
Bunuel wrote:
Official Solution:


If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

(1) \(6^a = 36^b\). Simplify: \(6^a = 6^{2b}\). Bases are equal, hence we can equate the powers: \(a=2b\). If \(a=b=0\), then \(a\) is NOT greater than \(b\) but if \(a=2\) and \(b=1\), then \(a\) IS greater than \(b\). Not sufficient.

(2) \(5^a = 35^b\). If both \(a\) and \(b\) are positive integers, then we'd have that \(5^a\) is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both \(a\) and \(b\) must be 0, giving a NO answer to the question whether \(a\) is greater than \(b\). Sufficient,


Answer: B


Hi bunuel,

Not able to understand the statement 2 explaination. Could you explain in simpler words or with some example?

Thanks
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Re: M31-53  [#permalink]

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New post 26 Aug 2017, 04:06
1
ashikaverma13 wrote:
Bunuel wrote:
Official Solution:


If \(a\) and \(b\) are non-negative integers, is \(a > b\)?

(1) \(6^a = 36^b\). Simplify: \(6^a = 6^{2b}\). Bases are equal, hence we can equate the powers: \(a=2b\). If \(a=b=0\), then \(a\) is NOT greater than \(b\) but if \(a=2\) and \(b=1\), then \(a\) IS greater than \(b\). Not sufficient.

(2) \(5^a = 35^b\). If both \(a\) and \(b\) are positive integers, then we'd have that \(5^a\) is equal to some multiple of 7 (because 35=5*7), which is not possible since 5 in any positive integer power has only 5's in it. Therefore, both \(a\) and \(b\) must be 0, giving a NO answer to the question whether \(a\) is greater than \(b\). Sufficient,


Answer: B


Hi bunuel,

Not able to understand the statement 2 explaination. Could you explain in simpler words or with some example?

Thanks


If a and b are not both 0, then we'd have that {not a multiple of 7} = {a multiple of 7}, which cannot be true, thus both \(a\) and \(b\) must be 0.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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M31-53  [#permalink]

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New post 26 Feb 2018, 04:00
Hi Bunuel,

I worked the second statement out in the following way:

5^a=5^b*7^b

Since exponents with the same bases can be equated...I then got a=b (by equating the exponents of the two 5s)
If a=b, a cannot be greater than b...hence sufficient.

I was still uncomfortable with my working out because I wasn't sure whether ignoring the 7^b was correct...was my approach correct?

Thanks.
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Re M31-53  [#permalink]

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New post 31 Oct 2018, 14:44
I think this the explanation isn't clear enough, please elaborate.
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Re: M31-53  [#permalink]

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New post 31 Oct 2018, 21:42
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Re: M31-53  [#permalink]

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New post 19 Nov 2018, 08:44
A quick and important tip for such questions!

ALWAYS REMEMBER IN DS questions statement 1 can never contradict statement 2!
So even in case u made mistake of taking a > b from statement 1, while doing statement 2 u realize\(a = b = 0\).
You should immediately stop and check to see why statement 1 and 2 are contradicting.
Probably because u made some mistake.
This simple tip will help u avoid in a lot of DS questions.
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Re: M31-53 &nbs [#permalink] 19 Nov 2018, 08:44
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