Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 29 Jun 2017
Posts: 411
GPA: 4
WE: Engineering (Transportation)

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
04 Sep 2017, 02:35
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



VP
Joined: 09 Mar 2016
Posts: 1252

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
07 Jan 2018, 05:05
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 !



Math Expert
Joined: 02 Sep 2009
Posts: 64999

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
07 Jan 2018, 05:10
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)
_________________



VP
Joined: 09 Mar 2016
Posts: 1252

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
07 Jan 2018, 05:30
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



Math Expert
Joined: 02 Sep 2009
Posts: 64999

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
07 Jan 2018, 05:33
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.
_________________



Manager
Joined: 23 Oct 2017
Posts: 62

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
07 Jan 2018, 06:41
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)



Intern
Joined: 20 Feb 2018
Posts: 2

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
27 Mar 2018, 08:43
Hi all,
Can anyone advise why its not good logic to flip the reciprocals as the first step?



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 17034
Location: United States (CA)

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
27 Mar 2018, 09:52
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. GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



VP
Joined: 09 Mar 2016
Posts: 1252

In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
13 Apr 2018, 08:11
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) \(r = \frac{xy}{(y+x)}\)



Intern
Joined: 04 Sep 2018
Posts: 25
GPA: 3.33

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
31 Dec 2018, 09:35
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.



CEO
Joined: 03 Jun 2019
Posts: 3182
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)

Re: In an electric circuit, two resistors with resistances x and
[#permalink]
Show Tags
15 Sep 2019, 04:14
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 1/r = 1/x+1/y 1/r = (x+y)/xy r = xy / (x+y) IMO D Posted from my mobile device
_________________
Kinshook Chaturvedi Email: kinshook.chaturvedi@gmail.com




Re: In an electric circuit, two resistors with resistances x and
[#permalink]
15 Sep 2019, 04:14



Go to page
Previous
1 2
[ 31 posts ]

