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In an electric circuit, two resistors with resistances x and [#permalink]
29 Dec 2012, 04:39

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This post was BOOKMARKED

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Difficulty:

15% (low)

Question Stats:

70% (01:54) correct
30% (01:02) wrong based on 504 sessions

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

Re: In an electric circuit, two resistors with resistances x and [#permalink]
29 Dec 2012, 04:42

Expert's post

Walkabout 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).

Re: In an electric circuit, two resistors with resistances x and [#permalink]
27 Dec 2013, 09:31

Bunuel wrote:

Walkabout 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).

Answer: D.

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

Re: In an electric circuit, two resistors with resistances x and [#permalink]
28 Dec 2013, 02:42

1

This post received KUDOS

Expert's post

aeglorre wrote:

Bunuel wrote:

Walkabout 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).

Answer: D.

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

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

In an electric circuit, two resistors with resistances x and [#permalink]
22 Oct 2014, 06:56

aeglorre wrote:

Bunuel wrote:

Walkabout 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).

Answer: D.

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)]

Re: In an electric circuit, two resistors with resistances x and [#permalink]
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]
20 May 2015, 19:17

1

This post received KUDOS

Expert's post

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

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

Last edited by iMyself on 18 Oct 2015, 09:20, edited 2 times in total.

When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford +1 Kudos if you find this post helpful

Re: In an electric circuit, two resistors with resistances x and [#permalink]
17 Oct 2015, 09:47

Expert's post

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

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

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