In an electric circuit, two resistors with resistance 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 OA is D.
$$\frac{1}{r}=\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}\ \Rightarrow r=\frac{xy}{x+y}$$
Option D.
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In an electric circuit, two resistors with resistances x and
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BTGmoderatorLU
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Brent@GMATPrepNow
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The reciprocal of r is equal to the sum of the reciprocals of x and yBTGmoderatorLU wrote:In an electric circuit, two resistors with resistance 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
So, 1/r = 1/x + 1/y
We must solve this equation for r.
Take: 1/r = 1/x + 1/y
Rewrite the right side with a COMMON DENOMINATOR of xy: 1/r = y/xy + x/xy
Combine the two terms to get: 1/r = (y + x)/xy
Since we have two equivalent fractions, their reciprocals will also be equal.
That is: r/1 = xy/(y + x)
Or, r = xy/(y + x)
Answer: D
Cheers,
Brent
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Jeff@TargetTestPrep
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We are given the reciprocal of r is equal to the sum of the reciprocals of x and y. Thus, we can say:BTGmoderatorLU wrote:In an electric circuit, two resistors with resistance 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
Getting the common denominator of xy 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)
Answer: D
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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: D
GMAT assassins aren't born, they're made,
Rich
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: D
GMAT assassins aren't born, they're made,
Rich
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deloitte247
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r as the combined resistor = Total resistance of a parallel circuit (Rt).
To find total Resistance in parallel circuit we have
$$\frac{1}{Rt}=\ \frac{1}{R_1}+\frac{1}{R_2}$$
Rt= r
R1 = x
R2 = y
relationship of r, x, and y is such=
$$\frac{1}{r}\ =\ \frac{1}{x}\ +\ \frac{1}{y}$$
$$\frac{1}{r}=\ \frac{\left(y+x\right)}{xy}$$
$$xy=r\left(x+y\right)$$
$$divide\ both\ sides\ by\ \left(x+y\right)$$
$$we\ have\ \frac{xy}{\left(x+y\right)}\ =\ r\ \frac{\left(x+y\right)}{\left(x+y\right)};\ r=\frac{xy}{\left(x+y\right)}$$
Therefore option D is the correct answer
To find total Resistance in parallel circuit we have
$$\frac{1}{Rt}=\ \frac{1}{R_1}+\frac{1}{R_2}$$
Rt= r
R1 = x
R2 = y
relationship of r, x, and y is such=
$$\frac{1}{r}\ =\ \frac{1}{x}\ +\ \frac{1}{y}$$
$$\frac{1}{r}=\ \frac{\left(y+x\right)}{xy}$$
$$xy=r\left(x+y\right)$$
$$divide\ both\ sides\ by\ \left(x+y\right)$$
$$we\ have\ \frac{xy}{\left(x+y\right)}\ =\ r\ \frac{\left(x+y\right)}{\left(x+y\right)};\ r=\frac{xy}{\left(x+y\right)}$$
Therefore option D is the correct answer
