MCQ
In Fig. the values of $x$ and $y$ are:


- A$x = 20, y = 130$
- B$x = 20, y = 140$
- ✓$x = 40, y = 140$
- D$x = 15, y = 140$

$\angle \text{ACB}+\angle \text{ACD}=180^\circ$
$\Rightarrow 40^\circ + \text{y}^\circ = 180^\circ$
$\Rightarrow \text{y}^\circ = 140^\circ$
$\Rightarrow \text{y} = 140$
Now, $\angle \text{ACD}=\angle \text{ABC}+\angle \text{BAC}$ [Exterior angle property]
$\Rightarrow 3\text{x}^\circ + 4\text{x}^\circ = \text{y}^\circ$
$\Rightarrow 7\text{x}^\circ = 140^\circ$
$\Rightarrow \text{x} = 20$
Hence, the correct answer is option $(c).$
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