Question
Express in the form of $\frac{\text{p}}{\text{q}}:0.\overline{38}+1.\overline{27}.$
✓
Answer
Let $0.\overline{38}=\text{x}$
$1.\overline{27}=\text{y}$
$\text{x}=0.3838 \ ...(\text{i})$ Multiply with $100$ as there are $2$ repeating digits after decimals $100x = 38.3838 ...(ii)$
Subtracting $(i)$ from $(ii)$ we get $99\text{x}=38$
$\Rightarrow\text{x}=\frac{38}{99}$ Similarly,
we take $y = 1.2727 ...(iii)$ Multiply $y$ with $100$ as there are $2$ repeating digits after decimal.
$100y = 127.2727 ...(iv)$
Subtract $(iii)$ from $(iv)$ we get $99\text{y}=126$
$\Rightarrow\text{y}=\frac{126}{99}$ Now, $\text{x}+\text{y}=\frac{38}{99}+\frac{126}{99}=\frac{164}{99}$
$1.\overline{27}=\text{y}$
$\text{x}=0.3838 \ ...(\text{i})$ Multiply with $100$ as there are $2$ repeating digits after decimals $100x = 38.3838 ...(ii)$
Subtracting $(i)$ from $(ii)$ we get $99\text{x}=38$
$\Rightarrow\text{x}=\frac{38}{99}$ Similarly,
we take $y = 1.2727 ...(iii)$ Multiply $y$ with $100$ as there are $2$ repeating digits after decimal.
$100y = 127.2727 ...(iv)$
Subtract $(iii)$ from $(iv)$ we get $99\text{y}=126$
$\Rightarrow\text{y}=\frac{126}{99}$ Now, $\text{x}+\text{y}=\frac{38}{99}+\frac{126}{99}=\frac{164}{99}$
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