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**Ques 1: The ages of two friends Ani and Biju differ by 3 years. Aniâ€™s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.****Sol:** Let the age of Ani = x years

And the age of Biju = y years

Case I:

y > x**According to 1st condition:**

y - x = 3 ...(1)

âˆµ [Age of Aniâ€™s father] = 2 [Age of Ani]

âˆ´ = 2x years

Also, [Age of Bijuâ€™s sister] **According to IInd condition:**

â‡’ 4x âˆ’ y = 60 ...(2)

Adding (1) and (2),

â‡’ x = 63/3 = 21

From (1), y - 21 = 3 â‡’ y = 3 + 21 = 24

âˆ´ Age of Ani = 21 years

Age of Biju = 24 years

Case II:

x > y

âˆ´ x - y = 3 ...(1)

According to the condition:

â‡’ 4x âˆ’ y = 60 ...(2)

Subtracting (1) from (2),

From (1), 19 - y = 3

â‡’ - y = 3 - 19

â‡’ - y = - 16 â‡’ y = 16

â‡’ Age of Ani = 19 years

Age of Biju = 16 years.**Ques ****2: One says, â€˜â€˜Give me a hundred, friend! I shall then become twice as rich as youâ€™â€™. The other replies, â€˜â€˜If you give me ten, I shall be six times as rich as youâ€™â€™. Tell me what is the amount of their (respective) capital ? [From the Bijaganita of Bhaskara II]****[Hint: x + 100 = 2 (y - 100), y + 10 = 6 (x - 10)].****Sol: **Let the capital of 1st friend = â‚¹ x

And capital of 2nd friend = â‚¹ y

According to the condition,

x + 100 = 2 (y - 100)

â‡’ x + 100 - 2y + 200 = 0

â‡’ x - 2y + 300 = 0 ...(1)

And 6 (x - 10) = y + 10

â‡’ 6x - 60 - y - 10 = 0

â‡’ 6x - y - 70 = 0 ...(2)

From (1), x = - 300 + 2y

From (2), 6x - y - 70 = 0

â‡’ 6 [- 300 + 2y] - y - 70 = 0

â‡’ - 1800 + 12y - y - 70 = 0

â‡’ - 1870 + 11y = 0

â‡’ y = 1870/11= 170

â‡’ y = 1870/11 = 170

Now, x = - 300 + 2y

= - 300 + 2 (170)

= - 300 + 340

= 40

Thus, 1st friend has â‚¹ 40 and the 2nd friend has â‚¹ 170.**Ques ****3: A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.****Sol: **Let the actual speed of the train = x km/hr

And the actual time taken = y hours

âˆµ Distance = speed Ã— time**1st Condition:**

(x + 10) Ã— (y - 2) = xy

â‡’ xy - 2x + 10y - 20 = xy

â‡’ 2x - 10y + 20 = 0 ...(1)**2nd Condition:**

(x - 10) Ã— (y + 3) = xy

â‡’ xy + 3x - 10y - 30 = xy

â‡’ 3x - 10y - 30 = 0 ...(2)

Using cross multiplication for solving (1) and (2):

a_{1} = 2, b_{1} = - 10, c_{1} = 20

a_{2} = 3, b_{2} = - 10, c_{2} = - 30

âˆ´

â‡’

â‡’

â‡’

Thus, the distance covered by the train = 50 Ã— 12 km = 600 km.**Ques ****4: The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.****Sol:** Let the number of students = x

And the number of rows = y

âˆ´ Number of students in each row = **1st Condition:**

=

[Number of students in a row Ã— Number of rows = Number of students]

â‡’

â‡’ ...(1)**2nd Condition:****â‡’ ****â‡’ **...(2)

Let x/y = p

âˆ´ Equations (1) and (2) can be expressed as

p - 3y + 3 = 0 ...(3)

2p - 3y - 6 = 0 ...(4)

Subtracting (3) from (4), we get

From (3), we get

9 âˆ’ 3y + 3 = 0 â‡’âˆ’ 3y = âˆ’ 12

â‡’ y = -12/-3 = 4

Since, x /y = 9 [âˆµ p = 9]

x/4 = 9 â‡’ x = 4 Ã— 9 = 36

Thus, number of students in the class

= xy

= 36 Ã— 4 = 144.**Ques ****5: In a Î” ABC, âˆ C = 3 âˆ B = 2 (âˆ A + âˆ B). Find the three angles.****Sol:** âˆµ Sum of angles of a triangle = 180Â°

âˆ´ âˆ A + âˆ B + âˆ C = 180Â° ...(1)

âˆµ âˆ C = 3 âˆ B = 2 (âˆ A + âˆ B) ...(2)

From (1) and (2), we have:

âˆ A + âˆ B + 2 (âˆ A + âˆ B) = 180Â°

â‡’ âˆ A + âˆ B + 2 âˆ A + 2 âˆ B = 180Â°

â‡’ 3 âˆ A + 3 âˆ B = 180Â°

â‡’ âˆ A + âˆ B = 60 ...(3)

Also,

âˆ A + âˆ B + 3 âˆ B = 180Â°

â‡’ âˆ A + 4 âˆ B = 180Â° ...(4)

Subtracting (3) from (4),

â‡’

From (4),

âˆ A + 4 (40) = 180Â° â‡’ âˆ A = 180Â° - (4 Ã— 40)

â‡’ âˆ A = 180Â° - 160 = 20Â°

Again âˆ C = 3 âˆ B = 3 Ã— 40Â° = 120Â°

Thus, âˆ A = 20Â°, âˆ B = 40Â° and âˆ C = 120Â°.**Ques ****6: Draw the graphs of the equations 5x - y = 5 and 3x - y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.Sol:** To draw the graph of 5x - y = 5, we get

5x - y = 5

and for equation 3x - y = 3, we get

Plotting the points (1, 0), (2, 5) and (0, - 5), we get a straight line l_{1}. Plotting the points (2, 3), (3, 6) and (0, - 3) we get a straight line l_{2}.

From the figure, obviously, the vertices of the triangle formed are :

A (1, 0), B (0, - 5) and C (0, - 3).

**Ques ****7: Solve the following pair of linear equations:****(i) **px + qy = p - q

qx - py = p + q

**(ii) **ax + by = c

bx + ay = 1 + c

**(iii) **

ax + by = a^{2} + b^{2}

**(iv)** (a - b) x + (a + b) y = a^{2} - 2ab - b^{2}

(a + b) (x + y) = a^{2} + b^{2}

**(v)** 152x - 378y = - 74

- 378x + 152y = - 604**Sol:** (i) We have:

px + qy = p - q ...(1)

qx - py = p + q ...(2)

Multiply (1) by p and (2) by q, and adding:

â‡’ (p^{2} + q^{2}) x = p^{2} + q^{2}

â‡’

From (1),

p (1) + q y = p âˆ’ q

â‡’ p + q y = p âˆ’ q

â‡’ q y = p âˆ’ q âˆ’ p = q

â‡’

Thus, the required solution is:

(ii) We have:

ax + by = c ...(1)

bx + ay = (1 + c) ...(2)

By cross multiplication, we have:

A_{1} = a, B_{1} = b, C_{1} = - c

A_{2} = b, B_{2} = a, C_{2} = - (1 + c)

âˆ´

â‡’

â‡’

âˆ´

or

(iii) We have:

= ...(1)

ax + by = a^{2} + b^{2} ...(2)

From (1), we have:

Substituting in (2), we have

â‡’

â‡’

â‡’ x = a

Substituting x = a in

â‡’ x = b

Thus, the required solution is

x = a

y = b

(iv) We have:

(a - b) x + (a + b) y = a^{2} - 2ab - b^{2} ...(1)

(a + b) (x + y) = a^{2}_{ }+ b^{2} ...(2)

From (2),

(a + b) x + (a + b) y = a^{2} + b^{2} ...(3)

Subtracting (3) from (1), we get

â‡’ x [a âˆ’ b âˆ’ a âˆ’ b]= âˆ’ 2ab âˆ’ 2b^{2}

â‡’ x (âˆ’ 2b)= âˆ’ 2b (a + b)

â‡’

â‡’ x =(a + b

Substituting x = (a + b) in (1),

(a - b) (a + b) + (a + b) y = a^{2} - 2ab - b^{2}

â‡’ a^{2} - b^{2} + (a + b) y = a^{2} - 2ab - b^{2}

â‡’ (a + b) y = a^{2} - 2ab - b^{2} - a^{2} + b^{2}

â‡’ (a + b) y = - 2ab

â‡’

Thus, x = a + b

Thus,

(v) We have:

152x - 378y = - 74 ...(1)

- 378x + 152y = - 604 ...(2)

Adding (1) and (2), we have:

- 226x - 226y = - 678

â‡’ x + y = 3 ...(3)

[Dividing throughout by - 226]

Subtracting (1) from (2),

â‡’âˆ’ x + y = âˆ’ 1

â‡’ x âˆ’ y = 1 ...(4)

Adding (3) and (4),

Subtracting (3) from (4),

Thus, the required solution is

x = 2 and y = 1**Ques 8: ABCD is a cyclic quadrilateral (see figure.). Find the angles of the cyclic quadrilateral.****Sol: **âˆµ ABCD is a cyclic quadrilateral.

âˆ´ âˆ A + âˆ C = 180Â° and âˆ B + âˆ D = 180Â°

â‡’ [4y + 20] + [- 4x] = 180Â°

â‡’ 4y - 4x + 20Â° - 180Â° = 0

â‡’ 4y - 4x - 160Â° = 0

â‡’ y - x - 40Â° = 0 ...(1)

[Dividing throughout by 4]

And

[3y - 5] + [- 7x + 5] = 180Â°

â‡’ 3y - 5 + 5 - 7x - 180Â° = 0

â‡’ 3y - 7x - 180Â° = 0 ...(2)

Multiplying (1) by 7 and subtracting from (2),

â‡’

Now, substituting y = 25Â° in (1)

y - x = 40 â‡’ - x = 40 - y

= 40Â° - 25Â° = 15Â°

â‡’ x = - 15Â°

âˆ´ âˆ A = 4y + 20Â° = 4 (25Â°) + 20Â°

= 100Â° + 20Â° = 120Â°

âˆ B = 3y - 5Â° = 3 (25Â°) - 5Â°

= 75Â° - 5Â° = 70Â°

âˆ C = - 4x = - (- 15Â°)

= 60Â°

âˆ D = - 7x + 5Â° = - 7 (- 15) + 5Â°

= 105Â° + 5Â° = 120Â°

Thus, âˆ A = 120Â°, âˆ B = 70Â°, âˆ C = 60Â°, âˆ D = 110Â°.

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