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# If f(w) = (1/w) + 1$$w+8) and the function f(w) equals 1/(w-1) then a  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 08 Jun 2017 Posts: 12 If f(w) = (1/w) + 1\(w+8) and the function f(w) equals 1/(w-1) then a [#permalink] ### Show Tags Updated on: 15 Oct 2017, 00:41 3 00:00 Difficulty: 15% (low) Question Stats: 81% (01:55) correct 19% (03:11) wrong based on 54 sessions ### HideShow timer Statistics If \(f(w) =\frac{1}{w} + \frac{1}{w+8}$$ and the function f(w) equals $$\frac{1}{w-1}$$, then a possible value for w could be

A) -8
B) -4
C) -2
D) -1
E) 3

(Source: Princeton Review GMAT Practice Test 6)

Originally posted by aashaybaindurgmat on 14 Oct 2017, 21:54.
Last edited by Bunuel on 15 Oct 2017, 00:41, edited 2 times in total.
Renamed the topic and edited the question.
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
Re: If f(w) = (1/w) + 1$$w+8) and the function f(w) equals 1/(w-1) then a [#permalink] ### Show Tags 14 Oct 2017, 22:01 1 aashaybaindurgmat wrote: If f(w) =\(\frac{1}{w} + \frac{1}{w+8}$$and the function f(w) equals $$\frac{1}{w-1}$$ , then a possible value for w could be

A) -8
B) -4
C) -2
D) -1
E) 3

(Source: Princeton Review GMAT Practice Test 6)

given $$\frac{1}{w}+\frac{1}{(w+8)} = \frac{1}{(w-1)}$$

solve this to get: $$w^2-2w-8=0$$

or $$(w-4)(w+2)=0$$

so $$w=4$$ or $$-2$$

From the given options only possible value is $$-2$$

Option C

Hi aashaybaindurgmat

kindly rename the subject or ask any moderator to do it, if you are facing any difficulty.
Manager
Joined: 03 May 2014
Posts: 161
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: If f(w) = (1/w) + 1$$w+8) and the function f(w) equals 1/(w-1) then a [#permalink] ### Show Tags 17 Oct 2017, 23:03 or plugin answer choices. given 1/w+1/(w+8)=1/(w−1) let us start with c ie w=-2 -1/2+1/(-2+8)=1/(-2-1) -1/2+1/6=-1/3 -1/3=-1/3 answer c Manager Joined: 03 May 2014 Posts: 161 Location: India WE: Sales (Mutual Funds and Brokerage) Re: If f(w) = (1/w) + 1\(w+8) and the function f(w) equals 1/(w-1) then a [#permalink] ### Show Tags 17 Oct 2017, 23:09 or plugin answer choices. given \(\frac{1}{w}$$+$$\frac{1}{(w+8)}$$=$$\frac{1}{(w−1)}$$

$$\frac{-1}{2}$$+$$\frac{1}{{-2+8}}$$=$$\frac{1}{(-2-1)}$$

$$\frac{-1}{2}$$+$$\frac{1}{6}$$=$$\frac{-1}{3}$$

$$\frac{-1}{3}$$=$$\frac{-1}{3}$$ answer c
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Joined: 22 May 2016
Posts: 2091
If f(w) = (1/w) + 1$$w+8) and the function f(w) equals 1/(w-1) then a [#permalink] ### Show Tags 20 Oct 2017, 07:26 aashaybaindurgmat wrote: If \(f(w) =\frac{1}{w} + \frac{1}{w+8}$$ and the function f(w) equals $$\frac{1}{w-1}$$, then a possible value for w could be

A) -8
B) -4
C) -2
D) -1
E) 3

(Source: Princeton Review GMAT Practice Test 6)

This math just looks time-consuming. It is not. Numbers are easy.*

$$f(w) =\frac{1}{w} + \frac{1}{w+8}$$

And $$f(w) = \frac{1}{w-1}$$. Set them equal.

$$\frac{1}{w-1} = \frac{1}{w} + \frac{1}{w+8}$$*

$$\frac{1}{w-1} = \frac{(w+8)+w}{w(w+8)}$$

$$\frac{1}{(w-1)} = \frac{2w+8}{w^2+8w}$$

$$(2w+8)(w-1) = w^2 + 8w$$

$$2w^2 - 2w + 8w - 8 = w^2 + 8w$$

$$w^2 - 2w - 8 = 0$$

$$(w - 4)(w + 2) = 0$$

$$w = 4$$ or $$w = -2$$

First option is not a choice. w = -2

*Use $$\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$$
If f(w) = (1/w) + 1\(w+8) and the function f(w) equals 1/(w-1) then a &nbs [#permalink] 20 Oct 2017, 07:26
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