Divide 77 into two parts such that the sum of their reciprocals is $\frac{7}{12}$ . From which of the following equations, we can get the required numbers? (Take one required number as x).

$\frac{1}{x} + \frac{1}{77 - x} = \frac{7}{12}$

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$\frac{1}{x} + \frac{1}{77 - x} = \frac{12}{7}$

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$\frac{- 1}{x} + \frac{1}{77 - x} = \frac{7}{12}$

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$\frac{1}{x} + \frac{1}{77 + x} = \frac{7}{12}$

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Solution

The correct option is A
$\frac{1}{x} + \frac{1}{77 - x} = \frac{7}{12}$

Let one number be x. Then, the other number will be 77-x.

Reciprocals of the given numbers are :

$\frac{1}{x}$ and $\frac{1}{77 - x}$

It is given that sum of their reciprocals is $\frac{7}{12}$

$\Rightarrow \frac{1}{x} + \frac{1}{77 - x} = \frac{7}{12}$

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Similar questions

30 (x^{2} + \frac{1}{x^{2}}) - 77 (x - \frac{1}{x}) - 12 = 0

Divide 77 into two parts such that the sum of their reciprocals is $\frac{7}{12}$ . From which of the following equations, we can get the required numbers? (Take one required number as x).

Divide 77 into two parts such that the sum of their reciprocals is $\frac{7}{12} $. From which of the following equations, we can get the required numbers? (Take one required number as x).

Q. If

30 (x^{2} + \frac{1}{x^{2}}) - 77 (x - \frac{1}{x}) - 12 = 0

, then

x = 1, \frac{- 1}{2}, \frac{2}{3}, \frac{- 7}{5}

Total number of real values of x such that $\frac{(12 + x)^{\frac{1}{7}}}{x} + \frac{(12 + x)^{\frac{1}{7}}}{12} = \frac{64}{3} x^{\frac{1}{7}}$ is/are

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Divide 77 into two parts such that the sum of their reciprocals is 712. From which of the following equations, we can get the required numbers? (Take one required number as x).

The correct option is A 1x+177−x=712 Let one number be x. Then, the other number will be 77-x. Reciprocals of the given numbers are : 1x and 177−x It is given that sum of their reciprocals is 712 ⇒1x+177−x=712

Divide 77 into two parts such that the sum of their reciprocals is $\frac{7}{12}$ . From which of the following equations, we can get the required numbers? (Take one required number as x).