Question

(a) In the given figure, POQ is a line, then the value of 'x'is [4 MARKS]

Figure


(b) In the figure below, x = ${30}^{\circ }$. The value of (6y – 3x) is ___.

Answer 1

(a) Steps : 1 Mark
Result : 1 Mark
(b) Steps : 1 Mark
Result : 1 Mark

(a) ${40}^{\circ }+4x+3x={180}^{\circ }$ [since POQ is straight line]

$7x={180}^{\circ }-{40}^{\circ }$

$x=\frac{140}{7}={20}^{\circ }$

Hence, the value of 'x' is

$x={20}^{\circ }$


(b) Given, x = ${30}^{\circ }$,

So, 2x = ${60}^{\circ }$

Also, 2x + 3y = ${180}^{\circ }$ (linear pair)

Replacing x in the above equation,
${60}^{\circ }$ + 3y = ${180}^{\circ }$

3y = ${120}^{\circ }$

y = ${40}^{\circ }$

The value of 6y - 3x = ${150}^{\circ }$