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In an electric circuit, two resistors with resistances x and
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29 Dec 2012, 05:39
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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
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Re: In an electric circuit, two resistors with resistances x and
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29 Dec 2012, 05:42



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Re: In an electric circuit, two resistors with resistances x and
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29 Dec 2012, 05:54
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
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29 Dec 2012, 05:54
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
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27 Dec 2013, 10:31
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 Where's the flaw in my calculation?



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Re: In an electric circuit, two resistors with resistances x and
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28 Dec 2013, 03:42
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 + yWhere'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
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28 Dec 2013, 05:15
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
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22 Oct 2014, 07:56
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 + yWhere'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
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20 May 2015, 13:10
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
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20 May 2015, 20:17
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 Final Answer: GMAT assassins aren't born, they're made, Rich
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In an electric circuit, two resistors with resistances x and
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Updated on: 26 Jan 2017, 08:12
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 reciprocal 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
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Originally posted by AsadAbu on 15 Oct 2015, 23:33.
Last edited by AsadAbu on 26 Jan 2017, 08:12, edited 3 times in total.



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Re: In an electric circuit, two resistors with resistances x and y are con
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16 Oct 2015, 03:08
1/r = 1/x + 1/y => r = xy/(x+y) Please correct the OA provided Answer D . This question has already been discussed  inanelectriccircuittworesistorswithresistancesxandyarecon28295.html
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Re: In an electric circuit, two resistors with resistances x and y are con
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16 Oct 2015, 03:37
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  inanelectriccircuittworesistorswithresistancesxandyarecon28295.htmlMy given answer is correct (B)....So, try for the second time!
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Re: In an electric circuit, two resistors with resistances x and
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17 Oct 2015, 10:47
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 mistranscribed. With this transcription error, the "correct" answer would change though  it would become . GMAT assassins aren't born, they're made, Rich
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Re: In an electric circuit, two resistors with resistances x and
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19 Oct 2015, 11:02
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 mistranscribed. With this transcription error, the "correct" answer would change though  it would become . 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
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28 Apr 2016, 18:54
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
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02 May 2016, 07:20
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 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) The answer is D.
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Re: In an electric circuit, two resistors with resistances x and
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22 Aug 2017, 11:17
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
1/r=(1/x)+(1/y)=xy/(x+y)



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Re: In an electric circuit, two resistors with resistances x and
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04 Sep 2017, 03:23
BunuelWhy are we not considering r= x+y as it is mentioned that r is the combined resistance? Is it because we don't know how they are connected.
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Re: In an electric circuit, two resistors with resistances x and
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