The number 0.121212 ldots in the form frac{p} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $x = 0.121212 \ldots$ (Equation $1$)Since there are two repeating digits,multiply both sides by $100$:$100x = 12.121212 \ldots$ (Equation $2$)

Vedclass · Accepted Answer

$\frac{4}{33}$

Vedclass · Answer

(C) Let $x = 0.121212 \ldots$ (Equation $1$)Since there are two repeating digits,multiply both sides by $100$:$100x = 12.121212 \ldots$ (Equation $2$)Subtract Equation $1$ from Equation $2$:$100x - x = 12.121212 \ldots - 0.121212 \ldots$$99x = 12$$x = \frac{12}{99}$Dividing both numerator and denominator by their greatest common divisor,which is $3$:$x = \frac{12 \div 3}{99 \div 3} = \frac{4}{33}$

The number $0.121212 \ldots$ in the form $\frac{p}{q}$ is equal to

Similar Questions

What value will come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value)
$10^{3} \times 100^{3} + 999999999 = 10^{?} + 10^{?}$

If $P = \frac{96}{9597}$,$Q = \frac{97}{96 \times 98}$,and $R = \frac{1}{97}$,then which of the following is the smallest?

$11.6 + 9.28 + 0.464 - 0.2828 + 0.07 = ?$

If $x + \frac{1}{x} = \sqrt{13}$,then $\frac{3x}{x^2 - 1}$ is equal to:

$\left(\frac{24}{9}\right)^{2} \times \frac{399}{39} \div \frac{41}{899} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module