Out of two numbers, the first number is three-fourth of the second number. If the average of the reciprocal of the two numbers is $\frac{7}{48}$ , then the first number is :
6
Let's solve the problem step-by-step:
Let the two numbers be \( x \) and \( y \), where the first number \( x \) is three-fourths of the second number \( y \). Therefore, we have:
\(x = \frac{3}{4} \times y\)
According to the problem, the average of the reciprocals of the two numbers is given as \(\frac{7}{48}\). This can be expressed as:
\(\frac{1}{2} \left( \frac{1}{x} + \frac{1}{y} \right) = \frac{7}{48}\)
Let's eliminate the fractions by multiplying the entire equation by 2 to simplify:
\(\frac{1}{x} + \frac{1}{y} = \frac{7}{24}\)
Substitute the value of \( x \) in terms of \( y \) from the earlier equation:
\(\frac{1}{\frac{3}{4}y} + \frac{1}{y} = \frac{7}{24}\)
Which simplifies to:
\(\frac{4}{3y} + \frac{1}{y} = \frac{7}{24}\)
To add the fractions, find a common denominator:
\(\frac{4 + 3}{3y} = \frac{7}{24}\)
\(\frac{7}{3y} = \frac{7}{24}\)
Cancelling 7 on both sides gives:
\(\frac{1}{3y} = \frac{1}{24}\)
Cross-multiplying, we find:
\(3y = 24 \Rightarrow y = 8\)
Now, substitute \( y = 8 \) back into the expression for \( x \):
\(x = \frac{3}{4} \times 8 = 6\)
Therefore, the first number is:
6
Hence, the correct answer is 6, which corresponds to option 3.
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