 QUESTION

# Express the given expression in the form of $\dfrac{p}{q}:0.\overline{38}+1.\overline{27}$.

Hint: First convert the repeating decimals into $\dfrac{p}{q}$ form, check periodicity at which it repeats. Then according to that multiply it by powers of 10. Then subtract the obtained number from the original number, then represent it in the form of $\dfrac{p}{q}$ fraction, then simply add them to get the desired result.

In the question we are asked to write the $0.\overline{38}+1.\overline{27}$ in the form of $\dfrac{p}{q}$.
Rational numbers are numbers that can be expressed in the form of $\dfrac{p}{q}$. And the decimal expansions of a rational number either terminate after a finite number of digits or keep repeating in the same finite sequence of digits over and over.
So, the two given numbers are rational numbers, i.e., $0.\overline{38},1.\overline{27}$ are rational numbers with repeating digits.
Let $x=0.\overline{38},y=1.\overline{27}$.
So, they can be written as,
$x=0.38383838....,y=1.27272727....---(i)$
Now we can see that two digits are repeating, so we will multiply each with 100, we get
$100x=38.38383838....,100y=127.27272727....---(ii)$
Subtracting ‘x’ and ‘y’ from equation (i) and (ii) simultaneously, we get
\begin{align} & 99x=38,99y=126 \\ & \Rightarrow x=\dfrac{38}{99},y=\dfrac{126}{99} \\ \end{align}
So, we have written the given numbers in terms of $\dfrac{p}{q}$ form.
Now we will add ‘x’ and ‘y’, we get
$x+y=\dfrac{38}{99}+\dfrac{126}{99}=\dfrac{38+126}{99}=\dfrac{164}{99}$
Substituting the value of ‘x’ and ‘y’, we get
$0.\overline{38}+1.\overline{27}=\dfrac{164}{99}$
This of $\dfrac{p}{q}$ form, so the required answer

Note: If the digits are repeating after every two digits, then the repeating decimal number should be multiplied by 100.
Students often make mistakes in multiplying with the correct multiple of 10. They also make mistakes by directly adding the numbers then convert it in $\dfrac{p}{q}$ form, which will lead to wrong answers.