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# In an electric circuit, two resistors with resistances x and

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In an electric circuit, two resistors with resistances x and [#permalink]

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29 Dec 2012, 04:39
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In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y?

(A) xy
(B) x + y
(C) 1/(x + y)
(D) xy/(x + y)
(E) (x + y)/xy
[Reveal] Spoiler: OA
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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29 Dec 2012, 04:42
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In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y?

(A) xy
(B) x + y
(C) 1/(x + y)
(D) xy/(x + y)
(E) (x + y)/xy

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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29 Dec 2012, 04:54
From the statements:
x+y = r --> 1
1/r = 1/x+1/y --> 2

From 1 and 2

So 1/r = (x+y)/xy,

r = xy/(x+y)

Ans - D
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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29 Dec 2012, 04:54
From the statements:
x+y = r --> 1
1/r = 1/x+1/y --> 2

From 1 and 2

So 1/r = (x+y)/xy,

r = xy/(x+y)

Ans - D
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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27 Dec 2013, 09:31
Bunuel wrote:
In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y?

(A) xy
(B) x + y
(C) 1/(x + y)
(D) xy/(x + y)
(E) (x + y)/xy

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

Could you explain this more in depth?

I interpret it as:

" r is the combined of x and y" --> r = x + y

"the reciprocal of r is equal to the sum of the reciprocals of x and y" ---> 1/r = (1/x) + (1/y)

" What is r in terms of x and y?" ---> 1 = [(1/x) + (1/y)] * r ----> r = 1 * [(x/1) + (y/1)] ---> r = x + y

Where's the flaw in my calculation?
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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28 Dec 2013, 02:42
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aeglorre wrote:
Bunuel wrote:
In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y?

(A) xy
(B) x + y
(C) 1/(x + y)
(D) xy/(x + y)
(E) (x + y)/xy

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

Could you explain this more in depth?

I interpret it as:

" r is the combined of x and y" --> r = x + y

"the reciprocal of r is equal to the sum of the reciprocals of x and y" ---> 1/r = (1/x) + (1/y)

" What is r in terms of x and y?" ---> 1 = [(1/x) + (1/y)] * r ----> r = 1 * [(x/1) + (y/1)] ---> r = x + y

Where's the flaw in my calculation?

$$\frac{1}{r} = \frac{1}{x} + \frac{1}{y}$$;

$$\frac{1}{r} = \frac{y+x}{xy}$$;

$$r=\frac{xy}{x+y}$$.

Hope it's clear.
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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28 Dec 2013, 04:15
Bunuel wrote:

$$\frac{1}{r} = \frac{1}{x} + \frac{1}{y}$$;

$$\frac{1}{r} = \frac{y+x}{xy}$$;

$$r=\frac{xy}{x+y}$$.

Hope it's clear.

Bunuel, this is still not clear to me.

I remember asking this very same question (in another thread) about a week ago. Quite obviously there are flaws in my fundamental understanding of concepts in regards to addition of algebraic fractions. Can you direct me to some guide or tutorial that in a simple but efficient way explains these concepts?

Or rather, if you - as you read this - instantly understand where my flaws are and what rule/law/fundamental error I need to "fix", I would be very thankfull if you could explain to me what understanding I lack for algebraic addition of fractions.

By the way, thank you for all of your help!
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In an electric circuit, two resistors with resistances x and [#permalink]

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22 Oct 2014, 06:56
aeglorre wrote:
Bunuel wrote:
In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y?

(A) xy
(B) x + y
(C) 1/(x + y)
(D) xy/(x + y)
(E) (x + y)/xy

The wording is a bit confusing, though basically we are told that 1/r = 1/x + 1/y, from which it follows that r=xy/(x + y).

Could you explain this more in depth?

I interpret it as:

" r is the combined of x and y" --> r = x + y

"the reciprocal of r is equal to the sum of the reciprocals of x and y" ---> 1/r = (1/x) + (1/y)

" What is r in terms of x and y?" ---> 1 = [(1/x) + (1/y)] * r ----> r = 1 * [(x/1) + (y/1)] ---> r = x + y

Where's the flaw in my calculation?

The steps should be as follows:

-> 1 = [(1/x)+(1/y)] * r
-> r = 1 / [(1/x)+(1/y)]
-> r = 1/ [(y+x)/xy] ... addition in denominator
-> r = 1/ [(x+y)/xy]
-> r = 1/1 * [xy/(x+y)] ... dividing fraction using reciprocal (flipping)
-> r= [xy/(x+y)]
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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20 May 2015, 12:10
after some manipulations you get 1/R = x+y/xy you can just reverse this equation to get R --> R = xy/x+y
And yes, if a question is whether we can just reverse the equation -- 1/2 = 2/4 --> 2/1 = 4/2 (Both parts =2)
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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20 May 2015, 19:17
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Hi All,

This question can be solved with TESTing Values.

We're told that the reciprocal of R is equal to the SUM of the reciprocals of X and Y. This means….

1/R = 1/X + 1/Y

We're asked for the value of R in terms of X and Y

If X = 2 and Y = 3, then we have…

1/R = 1/2 + 1/3

1/R = 3/6 + 2/6 = 5/6

R = 6/5

So we need an answer that = 6/5 when X = 2 and Y = 3.

Answer A: XY = (2)(3) = 6 NOT a match
Answer B: X+Y = 2+3 = 5 NOT a match
Answer C: 1/(X+Y) = 1/5 NOT a match
Answer D: XY/(X+Y) = 6/5 This IS a match
Answer E: (X+Y)/XY = 5/6 NOT a match

[Reveal] Spoiler:
D

GMAT assassins aren't born, they're made,
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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Senior Manager Joined: 23 Feb 2015 Posts: 270 Followers: 3 Kudos [?]: 25 [0], given: 104 In an electric circuit, two resistors with resistances x and [#permalink] ### Show Tags 15 Oct 2015, 22:33 In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the reciprocals of the sum of x and y. What is r in terms of x and y ? (A) xy (B) x + y (C)1/(x + y) (D) xy/(x + y) (E) (x + y)/xy _________________ “The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow Last edited by iMyself on 18 Oct 2015, 09:20, edited 2 times in total. Math Forum Moderator Status: Greatness begins beyond your comfort zone Joined: 08 Dec 2013 Posts: 935 Location: India Concentration: General Management, Strategy GPA: 3.2 WE: Information Technology (Consulting) Followers: 39 Kudos [?]: 379 [0], given: 58 Re: In an electric circuit, two resistors with resistances x and y are con [#permalink] ### Show Tags 16 Oct 2015, 02:08 1/r = 1/x + 1/y => r = xy/(x+y) Please correct the OA provided Answer D . This question has already been discussed - in-an-electric-circuit-two-resistors-with-resistances-x-and-y-are-con-28295.html _________________ When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long +1 Kudos if you find this post helpful Senior Manager Joined: 23 Feb 2015 Posts: 270 Followers: 3 Kudos [?]: 25 [0], given: 104 Re: In an electric circuit, two resistors with resistances x and y are con [#permalink] ### Show Tags 16 Oct 2015, 02:37 skywalker18 wrote: 1/r = 1/x + 1/y => r = xy/(x+y) Please correct the OA provided Answer D . This question has already been discussed - in-an-electric-circuit-two-resistors-with-resistances-x-and-y-are-con-28295.html My given answer is correct (B)....So, try for the second time! _________________ “The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 8269 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 379 Kudos [?]: 2455 [0], given: 163 Re: In an electric circuit, two resistors with resistances x and [#permalink] ### Show Tags 17 Oct 2015, 09:47 Hi iMyself, Your "version" of the prompt differs from the one listed in the OGs (so it was either transcribed incorrectly or purposely changed). In the OGs, the prompt states that "the reciprocal of R is equal to the SUM of the RECIPROCALS of X and Y." In the post here, the prompt states that "the reciprocal of R is equal to the RECIPROCAL of the SUM of X and Y." EVERYTHING else (including the 5 answer choices) is exactly the same though, so I'm led to believe that this question was mis-transcribed. With this transcription error, the "correct" answer would change though - it would become [Reveal] Spoiler: B . GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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19 Oct 2015, 10:02
EMPOWERgmatRichC wrote:
Hi iMyself,

Your "version" of the prompt differs from the one listed in the OGs (so it was either transcribed incorrectly or purposely changed).

In the OGs, the prompt states that "the reciprocal of R is equal to the SUM of the RECIPROCALS of X and Y."

In the post here, the prompt states that "the reciprocal of R is equal to the RECIPROCAL of the SUM of X and Y."

EVERYTHING else (including the 5 answer choices) is exactly the same though, so I'm led to believe that this question was mis-transcribed. With this transcription error, the "correct" answer would change though - it would become
[Reveal] Spoiler:
B
.

GMAT assassins aren't born, they're made,
Rich

I've intentionally changed the question pattern. But, there is still something missing in your calculation or thinking.
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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28 Apr 2016, 17:54
1/r = 1/x + 1/y
1/r = y/y(1/x)+x/x(1/y)
1/r = (y/xy) + (x/xy)
1/r = (y+x/xy)
r=(xy/x+y)
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Re: In an electric circuit, two resistors with resistances x and [#permalink]

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02 May 2016, 06:20
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In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y?

(A) xy
(B) x + y
(C) 1/(x + y)
(D) xy/(x + y)
(E) (x + y)/xy

Solution:

We are given the reciprocal of r is equal to the sum of the reciprocals of x and y. Thus we can say:

1/r = 1/x + 1/y

Getting a common denominator for the right side of the equation we have:

1/r = y/xy + x/xy

1/r = (y + x)/xy

If we reciprocate both sides of the equation, we have:

r = xy/(y+x)

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Re: In an electric circuit, two resistors with resistances x and   [#permalink] 02 May 2016, 06:20
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