Question

Express the following in the form p/q where p and q are integers and q is not equal to 0 0.4 par bar +0.18 par bar

Answer 1

Given.

The repeating decimal numbers $\mathrm{0.\overline{4}}$ and $\mathrm{0.\overline{18}}$

To find. their sum

Solution.

First number. $\mathrm{0.\overline{4}}$

Let $\mathrm{x=0.\overline{4}=0.44444.....}$

Then $\mathrm{10x=4.4444.....}$

Now $\mathrm{10x-x=4.4444.....-0.4444.....}$

$\Rightarrow \mathrm{9x=4}$

$\Rightarrow \mathrm{x=\frac{4}{9}}$

$\Rightarrow \mathrm{0.\overline{4}=\frac{4}{9}}$ which is of the form $\mathrm{\frac{p}{q}}$ where $\mathrm{p}$ and $\mathrm{q}$ are integers and $\mathrm{q\neq 0}$.

Second number. $\mathrm{0.\overline{18}}$

Let $\mathrm{x=0.\overline{18}=0.1818.....}$

Then $\mathrm{100x=18.1818.....}$

Now $\mathrm{100x-x=18.1818.....-0.1818.....}$

$\Rightarrow \mathrm{99x=18}$

$\Rightarrow \mathrm{x=\frac{18}{99}}$

$\Rightarrow \mathrm{0.\overline{18}=\frac{18}{99}}$ which is of the form $\mathrm{\frac{p}{q}}$ where $\mathrm{p}$ and $\mathrm{q}$ are integers and $\mathrm{q\neq 0}$.

Finding the required sum.

Now $\mathrm{0.\overline{4}+0.\overline{18}}$

$\mathrm{=\frac{4}{9}+\frac{18}{99}}$

$\mathrm{=\frac{44+18}{99}}$ where LCM$\mathrm{(9,99)=99}$

$\mathrm{=\frac{62}{99}}$

This sum can also be written as $\mathrm{0.\overline{62}}$.

Answer. $\boxed{\mathrm{0.\overline{4}+0.\overline{18}=0.\overline{62}=\frac{62}{99}}}$

Answer 2

Answer:

0.4 + 0.18 =0.62

here this is ur answer..

this is too easy!!