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# Variable in Exponent DS problem

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Current Student
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Variable in Exponent DS problem [#permalink]  07 Dec 2010, 17:19
00:00

Difficulty:

35% (medium)

Question Stats:

61% (02:53) correct 38% (01:39) wrong based on 34 sessions
I would appreciate it if someone could walk me through this problem. Thanks

Is 5^k less than 1000?

1. 5^k+1 > 3000
2. 5^k-1 = 5^k -500
[Reveal] Spoiler: OA
Ms. Big Fat Panda
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Re: Variable in Exponent DS problem [#permalink]  07 Dec 2010, 17:25
Did you mean to say that 5^k^-^1 = 5^k - 500?
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Re: Variable in Exponent DS problem [#permalink]  07 Dec 2010, 17:50
Expert's post
tonebeeze wrote:
I would appreciate it if someone could walk me through this problem. Thanks

Is 5^k less than 1000?

1. 5^k+1 > 3000
2. 5^k-1 = 5^k -500

I am assuming the question is:

Is 5^k less than 1000?

1. 5^{k+1} > 3000
2. 5^{k-1} = 5^k -500

5^4 = 625 and 5^5 = 3125 (even if you do not know this, it is fine. You don't need to calculate. Just observe that 625*5 will be greater than 3000)

Statement 1: 5^{k+1} > 3000
This means k + 1 is greater than 4 so k is greater than 3 (It doesnt mean that k + 1 is at least 5 because the question doesn't say that k is an integer. k + 1 could be 4.999 making k = 3.999) Since k can take values less than 4 and more than 4, 5^k could be less than 1000 or more than 1000. Not sufficient.

Statement 2: 5^{k-1} = 5^k -500
Re-arrange: 500 = 5^k -5^{k-1}
5^3 *4 = 5^{k-1}(5 - 1)
Hence k - 1 = 3 and k = 4
So 5^k = 625 which is less than 1000. Answer is 'Yes'. Sufficient.

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Karishma
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Manager Joined: 13 Jul 2010 Posts: 172 Followers: 1 Kudos [?]: 13 [0], given: 7 Re: Variable in Exponent DS problem [#permalink] 07 Dec 2010, 18:18 Great point that (K+1) in stmt 1 does not mean more than 4 or at least 5, we forget non-integers when considering DS questions. Thanks. Intern Joined: 19 Dec 2010 Posts: 30 Followers: 0 Kudos [?]: 3 [0], given: 4 Re: Variable in Exponent DS problem [#permalink] 02 Jan 2011, 19:03 Can you go over how you went from 500 = 5^k -5^{k-1} to 5^3 *4 = 5^{k-1}(5 - 1)? Thanks so much in advance! Current Student Status: First-Year MBA Student Joined: 19 Nov 2009 Posts: 127 GMAT 1: 720 Q49 V39 Followers: 6 Kudos [?]: 43 [0], given: 209 Re: Variable in Exponent DS problem [#permalink] 02 Jan 2011, 19:30 m990540 wrote: Can you go over how you went from 500 = 5^k -5^{k-1} to 5^3 *4 = 5^{k-1}(5 - 1)? Thanks so much in advance! Please refer to this link. It will help you easily post mathematic symbols in your posts. writing-mathematical-symbols-in-posts-72468.html Regarding the problem: 500 = 5^k - 5^{k-1} 500 can be simplified into 125 x 4, which is equal to 5^3 x 4 Step 1: Recognize that 5^k is a common factor in both 5^k and 5^{k -1}. Proceed to factor out 5^k Step 2: 5^k (1 - 5^{-1}) =500 ---> 5^k (1 - \frac {1}{5}) = 500 ---> 5k (\frac {4}{5}) = 500 --->5^k = \frac {5}{4} (500) ---> 5^k = 625 Hope this helps Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4037 Location: Pune, India Followers: 862 Kudos [?]: 3623 [0], given: 144 Re: Variable in Exponent DS problem [#permalink] 03 Jan 2011, 07:29 Expert's post m990540 wrote: Can you go over how you went from 500 = 5^k -5^{k-1} to 5^3 *4 = 5^{k-1}(5 - 1)? Thanks so much in advance! Left hand side: 500 = 5*100 = 5*25*4 = 5^3*4 (Since your concern is the power of 5, separate 5s from the rest) Right hand side: 5^k -5^{k-1} = 5^{k - 1} ( 5 - 1)(Take 5^{k - 1} common. e.g. if you have 5^4 - 5^3, you can take 5^3 common and you will be left with (5 - 1)) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Variable in Exponent DS problem [#permalink]  03 Jan 2011, 08:14
5^k+1 > 3000
5. 5^k > 3000
5^k > 600 (This is not sufficient to tell that it is less than 1000)
5^k-1 = 5^k -500
5^k / 5 = 5^k – 500
5^k = 5^k+1 – 2500
5^k (5 -1) = 2500
5^k = 2500 / 4 = 625 (this is sufficient)
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Re: Variable in Exponent DS problem [#permalink]  23 Nov 2011, 10:51
St1: 5^k >600 ----insufficient as we don't know if 5^k <1000

St2: 5 ^ {k-1} = 5^K - 500
500 = 5^k (1- \frac{1}{5})
500 (\frac{5}{4})= 5^k ----sufficient as we can determine if 5^k <1000

hence B!
Re: Variable in Exponent DS problem   [#permalink] 23 Nov 2011, 10:51
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