(i) x + y = 1200                                  ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 180⁰
\(\Rightarrow\)x + 2y = 1800                                  ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°

Let ABCD be the quadrilateral in which AC is a diagonal, such that AC = 40 cm

Suppose DE \(\bot \) AC and BF \(\bot \) AC, such that
DE = 8.4 cm and BF = 6.6 cm.
Now, area of the quadrilateral ABCD
= [Area of \(\Delta\)ABC] + [Area of \(\Delta\)ADC]
=\(\left[ \frac { 1 }{ 2 } \times AC\times BF \right] \)+\(\left[ \frac { 1 }{ 2 } \times AC\times DE \right] \)
=\(\left[ \frac { 1 }{ 2 } \times 40cm\times 66cm\right] \)+\(\left[ \frac { 1 }{ 2 } \times 40cm\times 8.4cm\right] \)
=\(\left[ \frac { 1 }{ 2 } \times 40\times \frac { 66 }{ 10 } { cm }^{ 2 } \right] \)+​​​​​​​​​​​​​​\(\left[ \frac { 1 }{ 2 } \times 40\times \frac { 84 }{ 10 } { cm }^{ 2 } \right] \)
= [2 x 66 cm²] + [2 x 84 cm²]
= 132 cm² + 168 cm² = 300 cm²
Thus, the area of the quadrilateral ABCD = 300 cm².

Let the cost price of each pen be Rs.1.
∴ CP of 6 pens = Rs.6
CP of 4 pens =Rs.4
∴ SP of 4 pens = [CP of 6 pens] = Rs.6
⇒Gain = Rs.6 - Rs.4 = Rs.2
Gain per cent = \({2\over 4}\times100\%\)
[∵ Gain% = \({gain\over CP}\)x 100%]]
\(={2\times25\over1}\%=50\%\)

Here, we observe that the parallelogram having same perimeters ean have different areas.

(a) \(\frac{12^4\times 9^3\times 4}{6^3\times 8^2\times 27} =\frac{(3\times 2^2)^4\times (3^2)^3\times 2^2}{(2\times 3)^3\times (2^3)^2\times 3^3}\)
\(=\frac{3^4\times 2^8\times 3^6\times 2^2}{2^3\times3^3\times2^6\times3^3}\)
\(=\frac{2^{8+2}\times 3^{6+4}}{2^{6+3}\times 3^{3+3}}\)
\(=\frac{2^{10}\times 3^{10}}{2^9\times 3^6}=2^{10-9}\times 3^{10-6}\)
= 2 x 3⁴ = 2 x 81 = 162.
(b) 2³ x a³ x 5a⁴
= 8 x a³ x 5 x a⁴
= 8 x 5 x a³ x a⁴
= 40 x a³ + 4
= 40 x a⁷
= 40a⁷.

(25)^n-1 + 100 = 5^(2n-1)
⇒ (5²)^n-1 + 100 = 5^(2n-1)
⇒ 5^2n-2 + 100 = 5^2n-1
⇒ 5^2n-2 - 5^2n-1 = - 100
⇒ 5^{2n - 1} - 5^{2n- 2} = 100
⇒ 5^2n-2 x (5 -1) = 100
⇒ 5^2n-2 x 4 = 100
\(⇒\ 5^{2n-2}={100\over 4}=25\)
Thus, 5^{2n- 2} = 5²
As base is same on both the sides
∴ 2n-2 = 2
⇒ 2n=2+2
⇒ 2n = 4
\(⇒\ n={4\over 2}=2\)

Quantity of pure alcohol in 400 ml. of 15% = 400 \(\times \frac { 15 }{ 100 } \)
= 60 ml
Now, we add x ml of pure alcohol to the sample.
So, total pure alcohol = (60 + x) ml.
But volume of new sample = (400 + x) ml.
\(\therefore\) Percentage of pure alcohol in new sample \(=\frac { (60+x) }{ (400+x) } \times 100\)
which is equal to 32%
\(\Rightarrow \quad \frac { 60+x }{ 400+x } \times 100=32\)
\(\Rightarrow \quad \frac { 60+x }{ 400+x } =\frac { 32 }{ 100 } \)
\(\Rightarrow\) 100(60 + x) = 32 (400 + x)
\(\Rightarrow\) 100x + 6000 = 32x + 12800
\(\Rightarrow\) 100x-32x = 12800 - 6000
\(\Rightarrow\)  68x = 6800
\(\Rightarrow \ x=\frac { 6800 }{ 68 } =100\ ml\)

Given, \(\frac { 2x-3 }{ 5 } +\frac { x+3 }{ 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 4(2x-3) }{ 5\times 4 } +\frac { 5(x+3) }{ 5\times 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12 }{ 20 } +\frac { 5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12+5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 13x+3 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow\) 7(13x + 3) = 20(4x + 1)
\(\Rightarrow\) 91x + 21 = 80x + 20
\(\Rightarrow\) 91x - 80x = 20 - 21
\(\Rightarrow\) 11x = -1
Thus, \(x=-\frac { 1 }{ 11 } \)

Let the number of students = 100
Passed students = 12
Hence, failed students = 100 - 12 = 28
\(\because \) When failed are 28, then total no. of students
=100
 \(\because \)When failed are 392, then total no. of students
=\(\frac { 100 }{ 28 } \times \)392
=1400

Given pattern is
\(-\frac { 1 }{ 3 } ,\frac { 2 }{ 6 } ,\frac { 3 }{ 9 },-\frac { 4 }{ 12 } ...\)
Here, \(-\frac { 1 }{ 3 } =\frac { (-1)\times 1 }{ 3\times 1 } \)
\(-\frac { 2 }{ 6 } =\frac { (-1)\times 2 }{ 3\times 2 } \)
\(-\frac { 3 }{ 9 } =\frac { (-1)\times 3 }{ 3\times 3 } \)
and \(-\frac { 4 }{ 12 } =\frac { (-1)\times 4 }{ 3\times 4 } \)
Hence, next four numbers are
\(\frac { (-1)\times 5 }{ 3\times 5 } =-\frac { 5 }{ 15 } \)
\(\frac { (-1)\times 6 }{ 3\times 6 } =-\frac { 6 }{ 18 } \)
\(\frac { (-1)\times 7 }{ 3\times 7 } =-\frac { 7 }{ 21 } \)
\(\frac { (-1)\times 8 }{ 3\times 8 } =-\frac { 8 }{ 24 } \).