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Senior Manager  P
Joined: 29 Jun 2017
Posts: 417
GPA: 4
WE: Engineering (Transportation)
Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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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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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).

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 !
Math Expert V
Joined: 02 Sep 2009
Posts: 59725
Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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dave13 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).

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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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
Math Expert V
Joined: 02 Sep 2009
Posts: 59725
Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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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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Joined: 23 Oct 2017
Posts: 60
Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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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)
Intern  B
Joined: 20 Feb 2018
Posts: 2
Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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Hi all,

Can anyone advise why its not good logic to flip the reciprocals as the first step?
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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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.

GMAT assassins aren't born, they're made,
Rich
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Joined: 09 Mar 2016
Posts: 1228
In an electric circuit, two resistors with resistances x and  [#permalink]

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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

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)

$$r = \frac{xy}{(y+x)}$$ Intern  B
Joined: 04 Sep 2018
Posts: 27
GPA: 3.33
Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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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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Posts: 1885
Location: India
Re: In an electric circuit, two resistors with resistances x and  [#permalink]

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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

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

IMO D

Posted from my mobile device Re: In an electric circuit, two resistors with resistances x and   [#permalink] 15 Sep 2019, 05:14

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