To convert the repeating decimal $0.\overline{13}$ into a fraction $\frac{p}{q}$, follow these steps:
Let the given number be $x$. So, $x = 0.\overline{13}$.
This means $x = 0.131313...$
Since there are two repeating digits ('13'), multiply the equation by $100$ (which is $10^2$):
$100x = 100 \times 0.131313...$
$100x = 13.131313...$
Subtract the original equation ($x = 0.131313...$) from the new equation ($100x = 13.131313...$):
$100x - x = (13.131313...) - (0.131313...)$
$99x = 13$
Solve for $x$ by dividing both sides by $99$:
$x = \frac{13}{99}$
Therefore, the number $0.\overline{13}$ in the form of $\frac{p}{q}$ is $\frac{13}{99}$. This corresponds to Option 3.
Which fraction among the following is the least ?
\(\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}\)
Find the value of the following expression:
\(\frac{{3 \div 1 \times 2 + 5 - 2}}{{3 \times 3 - 2}}\)
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:
The value of \(9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3})\;of\;\frac{2}{9}]\) is: