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If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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Updated on: 12 Mar 2015, 05:39
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If y = 2 + 2K and \(y\neq{0}\), then 1/y + 1/y + 1/y + 1/y = ? A. 1/(8+8k) B. 2/(1+k) C. 1/(8+k) D. 4/(8+k) E. 4/(1+k)
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Originally posted by zatspeed on 11 Mar 2015, 19:39.
Last edited by Bunuel on 12 Mar 2015, 05:39, edited 3 times in total.
Renamed the topic, edited the question and added the OA.




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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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11 Mar 2015, 19:51
zatspeed wrote: If Y = 2 + 2K and Y(not equal to) 0 , Then 1/Y + 1/Y + 1/Y + 1/Y = ? You should give the options. 1/Y + 1/Y + 1/Y + 1/Y = 4/Y = 4/(2 + 2K) = 2/(1 + K)
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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11 Mar 2015, 20:18
VeritasPrepKarishma wrote: zatspeed wrote: If Y = 2 + 2K and Y(not equal to) 0 , Then 1/Y + 1/Y + 1/Y + 1/Y = ? You should give the options. 1/Y + 1/Y + 1/Y + 1/Y = 4/Y = 4/(2 + 2K) = 2/(1 + K) Sorry, I have updated it now.



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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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11 Mar 2015, 21:58
Hi zatspeed, This prompt can be solved in a couple of different ways. While most Test Takers would probably take an Algebraic approach, you can solve it by TESTing VALUES. We're told that Y = 2 + 2K. We're asked for the value of 1/Y + 1/Y + 1/Y + 1/Y. While I normally would NOT TEST 0 or 1, the answer choices are sufficiently different from one another that using those numbers would not be a problem here.... IF.... K = 0 Y = 2+0 = 2 The answer to the question is 1/2 + 1/2 + 1/2 + 1/2 = 2 So we're looking for an answer that = 2 when K = 0. Answer A: 1/(8+0) = 1/8 This is NOT a match Answer B: 2/(1+0) = 2 This IS a match Answer C: 1/(8+0) = 1/8 This is NOT a match Answer D: 4/(8+0) = 1/2 This is NOT a match Answer E: 4/(1+0) = 4 This is NOT a match Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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31 Mar 2017, 22:48
VeritasPrepKarishma wrote: zatspeed wrote: If Y = 2 + 2K and Y(not equal to) 0 , Then 1/Y + 1/Y + 1/Y + 1/Y = ? You should give the options. 1/Y + 1/Y + 1/Y + 1/Y = 4/Y = 4/(2 + 2K) = 2/(1 + K) Responding to a pm: Quote: For you last two steps, this involves using the conjugate of the denominator and then recognizing that 1k2 can be expressed as a difference of squares, correct?
The question gives you that \(Y = 2 + 2K\) So you simply substitute that in place of Y in the denominator. \(\frac{4}{Y} = \frac{4}{(2 + 2K)}\) Then just take 2 common from the numerator and denominator to get: \(\frac{4}{(2 + 2K)} = \frac{2*2}{2(1 + K)}\) Cancel off the 2 of the numerator with the 2 of the denominator to get: \(\frac{2*2}{2(1 + K)} = \frac{2}{(1 + K)}\)
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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02 Apr 2017, 22:16
y=2(1+k) 4(1/y)= 4(1/2(1+k))= 2/(1+k)
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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22 Jul 2017, 17:04
zatspeed wrote: If y = 2 + 2K and \(y\neq{0}\), then 1/y + 1/y + 1/y + 1/y = ?
A. 1/(8+8k) B. 2/(1+k) C. 1/(8+k) D. 4/(8+k) E. 4/(1+k) y = 2+2k 1/y +1/y + 1/y+1/y = 4/y 4/(2+2k) = 2/(1+k)



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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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11 Dec 2017, 14:54
zatspeed wrote: If y = 2 + 2K and \(y\neq{0}\), then 1/y + 1/y + 1/y + 1/y = ?
A. 1/(8+8k) B. 2/(1+k) C. 1/(8+k) D. 4/(8+k) E. 4/(1+k) y= 2+ 2k 1/y+1/y+1/y+1/y= 4/y plug 2+2k in for y 4/2+2k> simplify to 2/1+K



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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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01 Feb 2019, 18:13
zatspeed wrote: If y = 2 + 2K and \(y\neq{0}\), then 1/y + 1/y + 1/y + 1/y = ?
A. 1/(8+8k) B. 2/(1+k) C. 1/(8+k) D. 4/(8+k) E. 4/(1+k) Adding the given fractions we have: 4/y Since y = 2 + 2k, we have: 4/(2 + 2k) = 4/[2(1 + k)] = 2/(1 + k) Answer: B
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?
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01 Feb 2019, 18:13






