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# If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?

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Intern
Joined: 14 Oct 2014
Posts: 19
If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?  [#permalink]

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Updated on: 12 Mar 2015, 05:39
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Question Stats:

91% (00:48) correct 9% (01:19) wrong based on 313 sessions

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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)

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 = ?  [#permalink]

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11 Mar 2015, 19:51
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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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Karishma
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##### General Discussion
Intern
Joined: 14 Oct 2014
Posts: 19
Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?  [#permalink]

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11 Mar 2015, 20:18
1
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 = ?  [#permalink]

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11 Mar 2015, 21:58
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1
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

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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?  [#permalink]

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31 Mar 2017, 22:48
1
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 1-k2 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 = ?  [#permalink]

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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 = ?  [#permalink]

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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)
Intern
Joined: 18 Jan 2017
Posts: 26
Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?  [#permalink]

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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 = ?  [#permalink]

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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)

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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?  [#permalink]

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01 Jul 2020, 04:39
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?   [#permalink] 01 Jul 2020, 04:39