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In the formula 1/R=1/R1

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Carcass
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In the formula 1/R=1/R1 [#permalink] New post   14 Jul 2021, 05:05
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In the formula \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\), if \(R=4 \) and \(R_1=10\), what is the value of \(R_2 \)?


Express your value as a fraction


Show: :: OA
\(\frac{20}{3}\)
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Re: In the formula 1/R=1/R1 [#permalink] New post   14 Jul 2021, 05:30
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Solution:

\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\)

\(\frac{1}{R}- \frac{1}{R_1}= \frac{1}{R_2}\)

\(\frac{1}{4 }- \frac{1}{10}= \frac{1}{R_2}\)

\(\frac{6}{40}= \frac{1}{R_2}\)

\(\frac{3}{20}= \frac{1}{R_2}\)

\(R_2=\frac{ 20}{3}\)


Answer: \(\frac{ 20}{3}\)

Hope this helps!
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In the formula 1/R=1/R1 [#permalink] New post   14 Jul 2021, 05:55
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\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\)

Substituting the value of \(R=4 \) and \(R_1=10\) we get

\(\frac{1}{4}=\frac{1}{10}+\frac{1}{R_2}\)
=> \(\frac{1}{R_2}\) = \(\frac{1}{4} - \frac{1}{10}\)
We know that LCM(4,10) = 20
2 | 4 , 10
-----------
1 | 2 , 5
=> LCM(4,10) = 2*2*5 = 20

=> \(\frac{1}{R_2}\) = \(\frac{5}{20} - \frac{2}{20}\) = \(\frac{5-2}{20}\) = \(\frac{3}{20}\)

=> \(R_2\) = \(\frac{20}{3}\)
Hope it helps!

To learn How to find LCM and GCD using multiple methods, watch the following video

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Re: In the formula 1/R=1/R1 [#permalink] New post   11 Nov 2021, 12:09
Carcass wrote:
In the formula \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\), if \(R=4 \) and \(R_1=10\), what is the value of \(R_2 \)?


Express your value as a fraction


Show: :: OA
\(\frac{20}{3}\)


Plug \(R=4 \) and \(R_1=10\) into the given formula to get: \(\frac{1}{4}=\frac{1}{10}+\frac{1}{R_2}\)

To eliminate two of the fractions, we can multiply both sides of the equation by \(20\) (the least common multiple of \(4\) and \(10\)).
When we do this we get: \(5=2+\frac{20}{R_2}\)

Subtract \(2\) from both sides of the equation to get: \(3=\frac{20}{R_2}\)

Multiply both sides of the equation by \(R_2\) to get: \((3)(R_2)=20\)

Divide both sides of the equation by \(3\) to get: \(R_2 = \frac{20}{3}\)

Answer: 20/3
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Re: In the formula 1/R=1/R1 [#permalink]
11 Nov 2021, 12:09
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