MCQ
The value of $2.\overline{45}+0.\overline{36}$ is:
- A$\frac{67}{33}$
- B$\frac{24}{11}$
- ✓$\frac{31}{11}$
- D$\frac{167}{110}$
✓
Answer
Correct option: C.
$\frac{31}{11}$
Let $\text{x}=2.\overline{45}$
i.e., $\text{x}=2.4545 \ ...(\text{i})$
$\Rightarrow100\text{x}=245.4545 \ ...(\text{ii})$
On subtracting $(i)$ and $(ii),$ we get
$99\text{x}=243$
$\Rightarrow\text{x}=\frac{243}{99}$
Let $\text{y}=0.\overline{36}$
i.e., $\text{y}=0.3636 \ ...(\text{iii})$
$\Rightarrow100\text{y}=36.3636 \ ...(\text{iv})$
On subtracting $(iii)$ and $(iv),$ we get
$99\text{y}=36$
$\Rightarrow\text{y}=\frac{36}{99}$
$\therefore2.\overline{45}+0.\overline{36}=\text{x}+\text{y}=\frac{\text{243}}{\text{99}}+\frac{\text{36}}{\text{99}}=\frac{\text{279}}{\text{99}}=\frac{\text{31}}{\text{11}}$
Hence, the correct answer is option $(c).$
i.e., $\text{x}=2.4545 \ ...(\text{i})$
$\Rightarrow100\text{x}=245.4545 \ ...(\text{ii})$
On subtracting $(i)$ and $(ii),$ we get
$99\text{x}=243$
$\Rightarrow\text{x}=\frac{243}{99}$
Let $\text{y}=0.\overline{36}$
i.e., $\text{y}=0.3636 \ ...(\text{iii})$
$\Rightarrow100\text{y}=36.3636 \ ...(\text{iv})$
On subtracting $(iii)$ and $(iv),$ we get
$99\text{y}=36$
$\Rightarrow\text{y}=\frac{36}{99}$
$\therefore2.\overline{45}+0.\overline{36}=\text{x}+\text{y}=\frac{\text{243}}{\text{99}}+\frac{\text{36}}{\text{99}}=\frac{\text{279}}{\text{99}}=\frac{\text{31}}{\text{11}}$
Hence, the correct answer is option $(c).$
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