Question

22-(9-(6-(10-bar4-bar3)))

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Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\sf{0. \overline{3} \: and \: 0. \overline{4}}$

$\bf \underline{To\: find-}$

$\textsf{their sun = ?}$

$\bf \underline{Solution-}$

$\sf{Let \: x = 0. \overline{3} = 0.333.... \: \: \: \: \: - - - - (i)}$

$\sf{\therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 10x = 3. \overline{3} \: \: \: \: \: - - - (ii)}$

$\textsf{Subtracting Eqⁿ (i) from Eqⁿ (ii), we get}$

$\sf{9x = 3}$

$\sf{ \Rightarrow \: x = \frac{3}{9} }$

$\sf{ \Rightarrow \: x = \frac{1}{3} }$

$\sf{Again, let \: y = 0. \overline{4} = 0.444.... \: \: \: \: \: \: - - - - (iii) }$

$\sf{\therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 10y = 4. \overline{4} \: \: \: \: \: - - - (vi)}$

$\textsf{Subtracting Eqⁿ (iii) from Eqⁿ (iv), we get}$

$\sf{9y = 4}$

$\sf{ \Rightarrow \: y = \frac{4}{9} }$

$\sf{Now,adding \: both \: the \: value \: of \: 0. \overline{3} \: and \: 0. \overline{4},we \: get}$

$\sf{ \therefore \: \: 0. \overline{3} \: and \: 0. \overline{4}}$

$\sf {= \frac{1}{3} + \frac{4}{9} }$

$\textsf{Take the LCM of 3 and 9 is 9.}$

$\sf { = \frac{3 + 4}{9} }$

$\sf {= \frac{7}{9} }$

$\bf \underline{Answer-}$

$\underline{\bf{Hence,the \: sum \: of \: 0. \overline{3} \: and \: 0. \overline{4}} \: is \: \frac{7}{9} .}$