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In an electric circuit, two resistors with resistances x and

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In an electric circuit, two resistors with resistances x and [#permalink] New post 29 Dec 2012, 04:39
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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
[Reveal] Spoiler: OA
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Re: In an electric circuit, two resistors with resistances x and [#permalink] New post 28 Dec 2013, 02:42
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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

Where'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 [#permalink] New post 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).

Answer: D.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Manager
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Re: In an electric circuit, two resistors with resistances x and [#permalink] New post 29 Dec 2012, 04:54
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
Manager
Manager
Joined: 03 Nov 2009
Posts: 63
Followers: 1

Kudos [?]: 7 [0], given: 16

Re: In an electric circuit, two resistors with resistances x and [#permalink] New post 29 Dec 2012, 04:54
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
Manager
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Re: In an electric circuit, two resistors with resistances x and [#permalink] New post 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

Where's the flaw in my calculation?
Manager
Manager
User avatar
Joined: 12 Jan 2013
Posts: 247
Followers: 0

Kudos [?]: 7 [0], given: 47

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

By the way, thank you for all of your help!
Re: In an electric circuit, two resistors with resistances x and   [#permalink] 28 Dec 2013, 04:15
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