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In an electric circuit, two resistors with resistances x and

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

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New post 04 Sep 2017, 03:35
Ans is simply D

kind of physics problem
R1 is first resistance R2 is another
if they are in series then R equivalent is R1+R2

if they are in parallel R equivalent is R1R2/(R1+R2)
so r1=x and r2=y therefore , xy/(x+y) is the answer... Ans is D



Also by what is given in ques

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

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

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New post 07 Jan 2018, 06:05
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.


Bunuel as per this wording "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" i did this 1/r = 1/x + 1/y and chose C

can you explan how you got this ? --> r=xy/(x + y)

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

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New post 07 Jan 2018, 06:10
dave13 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.


Bunuel as per this wording "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" i did this 1/r = 1/x + 1/y and chose C

can you explan how you got this ? --> r=xy/(x + y)

thanks !


It's very simple algebraic manipulation.

1/r = 1/x + 1/y

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

xy = r*(x + y)

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

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New post 07 Jan 2018, 06:30
It's very simple algebraic manipulation.

1/r = 1/x + 1/y

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

xy = r*(x + y)

r = xy/(x + y)

Bunuel thanks all clear - just one question after this xy = r*(x + y) we need to express r in terms of x and y - correct ?

xy = r*(x + y) this equation has two parts "xy" and "r*(x + y)"

hence shouldnt we divide both parts of equation by r*(x + y) to express r in terms of x and y

xy /r*(x + y)= r*(x + y)/ r*(x + y) ? :? please explan
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Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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New post 07 Jan 2018, 06:33
dave13 wrote:
It's very simple algebraic manipulation.

1/r = 1/x + 1/y

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

xy = r*(x + y)

r = xy/(x + y)

Bunuel thanks all clear - just one question after this xy = r*(x + y) we need to express r in terms of x and y - correct ?

xy = r*(x + y) this equation has two parts "xy" and "r*(x + y)"

hence shouldnt we divide both parts of equation by r*(x + y) to express r in terms of x and y

xy /r*(x + y)= r*(x + y)/ r*(x + y) ? :? please explan


Divide xy = r*(x + y) by x + y to get xy/(x + y) = r.
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Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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New post 07 Jan 2018, 07:41
The only relevant part is "the reciprocal of r is equal to the sum of the reciprocals of x and y" which indicates:
1/r = 1/x + 1/ y
Now rearrange it a bit, to get x = f(x,y)
r = xy/(x+y)
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Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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New post 27 Mar 2018, 09:43
Hi all,

Can anyone advise why its not good logic to flip the reciprocals as the first step?
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Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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New post 27 Mar 2018, 10:52
HiItsAlif wrote:
Hi all,

Can anyone advise why its not good logic to flip the reciprocals as the first step?


Hi HiItsAlif,

The 'math step' that you're thinking about is NOT mathematically correct. For example:

1/2 = 1/3 + 1/6

However, if you 'flip' the reciprocals, you end up with...

2 = 3 + 6

...which is not correct.

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

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New post 13 Apr 2018, 09:11
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


finally I drummed into my head what it means r in terms of x and y :) after ten years of study lol

\(\frac{1}{r}= \frac{1}{x}+\frac{1}{y}\)

\(xy = r(y+x)\)

no we need to ISOLATE \(r\) :) r in terms of x and y ---> means R should be on the RIGHT SIDE and X and Y on the LEFT SIDE

\(r(y+x) = xy\) now divide by (y+x)

\(r = \frac{xy}{(y+x)}\) :)
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Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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New post 31 Dec 2018, 10:35
It's very simple algebraic manipulation.

1/r = 1/x + 1/y

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

xy = r*(x + y)

r = xy/(x + y)[/quote]



Bunnel, all clear except the second step which you did. "1/r = (x + y)/(xy)"

How did you get this "(x + y)/(xy)" from the 1st step which is 1/r = 1/x + 1/y. Is this some algebra rule?

Thanks in advance. Sorry for asking such a basic thing, but actually just a novice in GMAT Quant.
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Re: In an electric circuit, two resistors with resistances x and y are con  [#permalink]

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Re: In an electric circuit, two resistors with resistances x and y are con   [#permalink] 08 Feb 2019, 03:05

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