Show that $0.2353535 \ldots=0.2 \overline{35}$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0$.
Show that $0.2353535 \ldots=0.2 \overline{35}$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0$.
Let $x=0.2 \overline{35}$. Over here, note that $2$ does not repeat, but the block $35$ repeats. since two digits are repeating, we multiply $x$ by $100$ to get
$100 x=23.53535 \ldots$
So, $100 x=23.3+0.23535 \ldots=23.3+x$
Therefore, $99 x=23.3$
i.e., $99 x=\frac{233}{10},$ which gives $x=\frac{233}{990}$
You can also check the reverse that $\frac{233}{990}=0.2 \overline{35}$.
Similar Questions
Find :
$(i)$ $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
$(ii)$ $\left(\frac{1}{3^{3}}\right)^{7}$
$(iii)$ $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
$(iv)$ $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$
Find :
$(i)$ $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
$(ii)$ $\left(\frac{1}{3^{3}}\right)^{7}$
$(iii)$ $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
$(iv)$ $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$
Visualise $3.765$ on the number line, using successive magnification.
Visualise $3.765$ on the number line, using successive magnification.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ?
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$ ?
Add $2 \sqrt{2}+5 \sqrt{3}$ and $\sqrt{2}-3 \sqrt{3}$
Add $2 \sqrt{2}+5 \sqrt{3}$ and $\sqrt{2}-3 \sqrt{3}$
Locate $\sqrt 3$ on the number line.
Locate $\sqrt 3$ on the number line.