Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,

x =d  / t

Or, d = xt (i)

Using the information given in the question, we obtain

By using equation (i), we obtain

3x − 10t = 30 (iii)

Adding equations (ii) and (iii), we obtain

x = 50

Using equation (ii), we obtain

(−2) × (50) + 10t = 20

−100 + 10t = 20

10t = 120

t = 12 hours

From equation (i), we obtain

Distance to travel = d = xt

= 50 × 12

= 600 km

Hence, the distance covered by the train is 600 km.

Let the number of rows be x and number of students in a row be y.

Total students of the class

= Number of rows × Number of students in a row

= xy

Using the information given in the question,

Condition 1

Total number of students = (x − 1) (y + 3)

xy = (x − 1) (y + 3) = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3 (i)

Condition 2

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6 (ii)

Subtracting equation (ii) from (i),

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 3 + 6

y = 9

By using equation (i), we obtain

3x − 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Number of total students in a class = xy = 4 × 9 = 36

Let the number of rows be x and number of students in a row be y.

Total students of the class

= Number of rows × Number of students in a row

= xy

Using the information given in the question,

Condition 1

Total number of students = (x − 1) (y + 3)

xy = (x − 1) (y + 3) = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3 (i)

Condition 2

Total number of students = (x + 2) (y − 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6 (ii)

Subtracting equation (ii) from (i),

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 3 + 6

y = 9

By using equation (i), we obtain

3x − 9 = 3

3x = 9 + 3 = 12

x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Number of total students in a class = xy = 4 × 9 = 36

Given that,

∠C = 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B)

3∠B = 2∠A + 2∠B

∠B = 2∠A

2 ∠A − ∠B = 0 … (i)

We know that the sum of the measures of all angles of a triangle is 180°. Therefore,

∠A + ∠B + ∠C = 180°

∠A + ∠B + 3 ∠B = 180°

∠A + 4 ∠B = 180° … (ii)

Multiplying equation (i) by 4, we obtain

8 ∠A − 4 ∠B = 0 … (iii)

Adding equations (ii) and (iii), we obtain

9 ∠A = 180°

∠A = 20°

From equation (ii), we obtain

20° + 4 ∠B = 180°

4 ∠B = 160°

∠B = 40°

∠C = 3 ∠B

= 3 × 40° = 120°

Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

5x − y = 5

Or, y = 5x − 5

The solution table will be as follows.

x	0	1	2
y	−5	0	5

3x − y = 3

Or, y = 3x − 3

The solution table will be as follows.

x	0	1	2
y	− 3	0	3

The graphical representation of these lines will be as follows.

 

It can be observed that the required triangle is ΔABC formed by these lines and y-axis.

The coordinates of vertices are A (1, 0), B (0, − 3), C (0, − 5).

(i)px + qy = p − q … (1)

qx − py = p + q … (2)

Multiplying equation (1) by p and equation (2) by q, we obtain

p²x + pqy = p² − pq … (3)

q²x − pqy = pq + q² … (4)

Adding equations (3) and (4), we obtain

p²x + q² x = p² + q²

(p² + q²) x = p² + q²

From equation (1), we obtain

p (1) + qy = p − q

qy = − q

y = − 1

 

(ii)ax + by = c … (1)

bx + ay = 1 + c … (2)

Multiplying equation (1) by a and equation (2) by b, we obtain

a²x + aby = ac … (3)

b²x + aby = b + bc … (4)

Subtracting equation (4) from equation (3),

(a² − b²) x = ac − bc − b

From equation (1), we obtain

ax + by = c

Or, bx − ay = 0 … (1)

ax + by = a² + b² … (2)

Multiplying equation (1) and (2) by b and a respectively, we obtain

b²x − aby = 0 … (3)

a²x + aby = a³ + ab² … (4)

Adding equations (3) and (4), we obtain

b²x + a²x = a³ + ab²

x (b² + a²) = a (a² + b²)

x = a

By using (1), we obtain

b (a) − ay = 0

ab − ay = 0

ay = ab

y = b

(iv) (a − b) x + (a + b) y = a²− 2ab − b² … (1)

(a + b) (x + y) = a² + b²

(a + b) x + (a + b) y = a² + b² … (2)

Subtracting equation (2) from (1), we obtain

(a − b) x − (a + b) x = (a² − 2ab − b²) − (a² + b²)

(a − b − a − b) x = − 2ab − 2b²

− 2bx = − 2b (a + b)

x = a + b

Using equation (1), we obtain

(a − b) (a + b) + (a + b) y = a² − 2ab − b²

a² − b² + (a + b) y = a²− 2ab − b²

(a + b) y = − 2ab

(v) 152x − 378y = − 74

76x − 189y = − 37

 … (1)

− 378x + 152y = − 604

− 189x + 76y = − 302 … (2)

Substituting the value of x in equation (2), we obtain

− (189)² y + 189 × 37 + (76)² y = − 302 × 76

189 × 37 + 302 × 76 = (189)² y − (76)² y

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

From equation (1), we obtain

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

Therefore, ∠A + ∠C = 180

4y + 20 − 4x = 180

− 4x + 4y = 160

x − y = − 40 (i)

Also, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 (ii)

Multiplying equation (i) by 3, we obtain

3x − 3y = − 120 (iii)

Adding equations (ii) and (iii), we obtain

− 7x + 3x = 180 − 120

− 4x = 60

x = −15

By using equation (i), we obtain

x − y = − 40

−15 − y = − 40

y = −15 + 40 = 25

∠A = 4y + 20 = 4(25) + 20 = 120°

∠B = 3y − 5 = 3(25) − 5 = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = − 7x + 5 = − 7(−15) + 5 = 110°