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 is the value of $\frac{(0.5)^{3}-(0.1)^{3}}{(0.5)^{2}+0.5 \times 0.1+(0.1)^{2}}?$

The value of $\frac{1.073 \times 1.073 - 0.927 \times 0.927}{1.073 - 0.927} + \frac{(3^4)^4 \times 9^6}{(27)^7 \times 3^9}$ is

If $x^{2} + \frac{1}{x^{2}} = 1$,then what is the value of $x^{48} + x^{42} + x^{38} + x^{30} + x^{24} + x^{18} + x^{12} + x^{6} + 1$?

If $a^{3} b = a b c = 180$,and $a, b, c$ are positive integers,then the value of $c$ is

If $2 = x + \frac{1}{1 + \frac{1}{3 + \frac{1}{4}}}$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module