We have,
2q + 6 = 12
\(\Rightarrow\) 2q + 6-6 = 12 - 6
[Subtraction 6 from both sides]
\(\Rightarrow\)2q = 6
\(\Rightarrow \quad \frac { 2q }{ 2 } =\frac { 6 }{ 2 } \) 
Dividing both sides by 2]
\(\Rightarrow\) q = 3
So, q = 3 is the solution of the given equation.

In the given figure, two sides of the triangle are equal.
We know that, the base angles opposite to equal sides are equal.
So, x = 60°
Hence, the value of x is 60°.

\(\frac { 1 }{ 2 } \)

In ΔPQS and ΔPRS, ㄥPSQ = ㄥPSR = 90°
Hypotenuse PQ = Hypotenuse PR = 3 cm
PS = PS                  [common]
Therefore, by RHS congruence rule, two triangles are congruent. The correspondence is P ↔️ P, Q ↔️ R,
S ↔️ S.
In symbolic form, ΔPQS ≌ ΔPRS

In ΔPQR and ΔMNL,
ㄥR = ㄥL = 60° [given]
RQ= LN = 6 cm [given]
ㄥQ = ㄥN = 30° [given]
Therefore, by ASA congruence rule, two triangles are congruent.
The correspondence is P ↔️ M, Q ↔️ N and R ↔️ L.
In symbolic form, ΔPQR ≅ ΔMNL

(i) We have, \(\frac{3}{4}\)of 16=\(\frac{3}{4}\times 16\)=\(\frac{3\times 16}{4}=\frac{48}{4}=12\)
(ii) We have, \(\frac{3}{4}\)of 27=\(\frac{3}{4}\times 36\)=\(\frac{3\times 36}{3}=\frac{108}{4}=27\)

Isosceles triangle

\(\angle \)y + 40° = 180°             [by linear pair]
\(\Rightarrow\) \(\angle \)y = 180° - 40°\(\Rightarrow\) \(\angle \)y = 140°
\(\angle \).z = 40° [verticallyopposite angles]
\(\because\) \(\angle \)x + 25° + \(\angle \)z = 180°    [by linear pair]
\(\angle \)x + 25° + 40° = 180°
\(\Rightarrow\)\(\angle \)x=180°-65°=1l5°
Hence, \(\angle \)x = 115°, \(\angle \)y = 140° and \(\angle \)z = 40°

(a) In the given figure,\(\bar { AC } =6cm,\bar { BC } =6cm\bar { ,AB } =4cm,\) \(\quad \because \bar { AC } =\bar { BC } \)
Hence, \(\triangle ABC\) is an isosceles triangle.
(b) In the given figure, \(\bar { AC } =4cm,\bar { AB } =4cm,\bar { BC } =4cm\)  \(\because \bar { AC } =\bar { AB } =\bar { BC } \)
Hence, \(\triangle ABC\) is an equilateral triangle.

5, 0, -3, -7, -10

Given, a = 21 and b = 18
\(\therefore\) LHS = a-(-b) = 21-(-18) = 21+18 = 39
and RHS = a + b = 21 + 18 = 39
Hence, LHS = RHS

In this pair, measures of the given angles are 110° and 50°.
\(\therefore\) Sum of the given angles = 110° + 50° = 160°,
which is less than 180°.
So, this pair of angles is not supplementary.

In this pair, sum of two angles = 70° + 20° = 90° So, this pair of angles is complementary.

According to the question,
Sum of x and 4 = x + 4 and the sum = 9
Hence, the required equation is x + 4 = 9.

We know that, whole numbers are those, which start from zero (0).
So, first five whole numbers are 0, 1,2,3 and 4.
\(\therefore\) \(mean=\frac { sum\ of\ number }{ number\ of\ terms } =\frac { 0+1+2+3+4 }{ 5 } =\frac { 10 }{ 5 } =2\)
Hence, the mean of first five whole numbers is 2.

We have, \(\frac{7}{10}+\frac{2}{5}+\frac{3}{2}\)
\(\therefore \frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}\)
=\(\frac{26}{10}=\frac{13}{5}\)
[dividing numerator and denominator by 2]

x + 5 = 6, Add x and 5 to get 6.

∵ \(\angle \)COB = 50° [Given]
∴ \(\angle \)AOD = 50° [Vertically opposite angles]
Now, \(\angle \)AOC and \(\angle \)COB form a linear pair.
Thus, \(\angle \)AOC + \(\angle \)COB = 180°
=> \(\angle \)AOC + 50° = 180°
=> \(\angle \)AOC = 180° - 50° = 130°
Also, \(\angle \)AOC and \(\angle \)BOD are vertically opposite angles.
∴ \(\angle \)BOD = \(\angle \)AOC = 130°
Thus, the required measures of the remaining
three angles are:​​​​​​​\(\begin{matrix} \angle AOD={ 50 }^{ o } \\ \angle AOC=130^{ o } \\ \angle BOD=130^{ o } \end{matrix} \)}

Let the smallest integer be x.
\(\therefore\)The three consecutive integers are x, (x + 1) and (x + 2).
Sum of the integers = x + (x + 1) + (x + 2) = 3x + 3
According to the condition,
[Sum of the consecutive integers]
= 2 x [The smallest integer] + 12
\(\Rightarrow \)(3x + 3) = 2(x) + 12
\(\Rightarrow \)3x + 3 = 2x + 12
Transposing 3 to R.H.S. and 2x to L.H.S., we have
3x - 2x = 12 - 3
\(\Rightarrow \)x=9
Thus, the three consecutive numbers are 9, 9 + 1 and 9 + 2 or 9, 10 and 11.
Hence, the required numbers are 9, 10 and 11.

For rectangle,
Area = Length \(\times\) Breadth
Since, Length = 0.5 m
Breadth = 1.5 m
∴ Area = (1.5\(\times\)0.5)m²
= \(\left( \frac { 15 }{ 10 } \times \frac { 5 }{ 10 } \right) \)m²
= \(\frac { 75 }{ 100 } \)m²
=0.75 m²

(i) \(\frac{2}{3}\times \Box=\frac{10}{30}\)
Here, 2 X5 = 10 and 3 X10 = 30
Hence, the required number in the box is \(\frac{5}{10}\).
(ii) Simplest form of \(\frac{5}{10}=\frac{1}{2}\)

Given, ABC is a triangle in which AB = AC and D is the mid-point of \(\bar { BC } \)
(i) The three pairs of equal parts in ΔADB and ΔADC are
AD = AD [common]
AB = AC [given]
DB = DC     [since, D is the mid-point of BC]
(ii) Yes, from part (i), we have
ΔADB ≅ ΔADC    [by SSS congruence rule]
(iii) Yes, because ΔADB ≅ ΔADC
⇒ ㄥB = ㄥC
[since, corresponding part of congruent triangles are equal]

\(\angle A=90^0\) [Hint Use the concept of is isosceles triangle]

Given, \(\angle \)1= 30°
Since, \(\angle \)1 and  \(\angle \)2 form a linear pair.
\(\therefore\) \(\angle \)1+ \(\angle \)2 = 180° \(\Rightarrow\) 30° + \(\angle \)2 = 180°
[transposing 30° to RHS]
\(\Rightarrow\)  \(\angle \)2 = 180° - 30°\(\Rightarrow\) \(\angle \) 2 = 150°
Now, \(\angle \)3 = \(\angle \)1= 30° [vertically opposite angles]
Hence, \(\angle \)2 = 150° and \(\angle \)3 = 30°

From the given figure, it is clear that the directions East and West are opposite to each other.
According to the question,

If moving towards East is a positive integer, then moving towards West is represented by a negative integer.
Now, distance moved towards West = - 30 km 
and distance moved towards East = 20 km
\(\therefore\) Rita's final position from A = 20 + (- 30) = 20 - 30
= -10 km (i.e. West)

Given, the enrolment of a school during six consecutive years
1555,1670,1750,2013,2540,2820
Now, sum of enrolments
= 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 12348
Number of years = 6
\(\therefore\)  Mean enrolment of the school for this period
 = \(\frac { sum\quad of\quad enrolments }{ number\quad of\quad years } =\frac { 12348 }{ 6 } =2058\)
Hence, the mean enrolment of the school for this period is 2058.

Since, arithmetic mean of 3,9,4,x 9,7,6 is 6
Arithmetic mean = \(\frac { sum\ of\ observations }{ Number\ of\ observations } \)
6 = \(\frac { 3+9+4+x+9+7+6 }{ 7 } \)
6 x 7 = 38 + x \(\Rightarrow\) 42 = 38 + x
x = 42-38 = 4
x = 4