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**Short Answer Type Questions****Ques 1: For what value of p, the pair of linear equations****px = 2y****2x - y + 5 = 0 has unique solution?****Sol:** We have:

px = 2y

â‡’ p - 2y = 0

2x = y + 5

â‡’ 2x - y = - 5

Here, a_{1} = p, b_{1} = - 2,

c_{1} = 0

a_{2 }= 2, b_{2} = - 1,

c_{2} = - 5

For a unique solution,

â‡’

â‡’ p â‰ 2 Ã— 2

â‡’ p â‰ 4**Ques ****2: In a cyclic quadrilateral PQRS, âˆ P = (2x + 4)Â°, âˆ Q = (y + 3)Â°, âˆ R = (2y + 10)Â° and âˆ S = (4x - 5)Â°. Find its four angles. Sol:** In a cyclic quadrilateral, the opposite angles are supplementary.

âˆ´âˆ P + âˆ R = 180Â°

â‡’ (2x + 4)Â° + (2y + 10)Â° = 180Â°

â‡’ 2x + 2y + 14 - 180 = 0

â‡’ 2x + 2y - 166 = 0

â‡’ x + y - 83 = 0 ...(1)

Also âˆ Q + âˆ S = 180Â°

âˆ´ (y + 3)Â° + (4x - 5)Â° = 180Â°

â‡’ y + 4x - 2 - 180 = 0

â‡’ y + 4x - 182 = 0 ...(2)

From (1) and (2),

a

a

â‡’

â‡’

âˆ´ âˆ P = (2x + 4)Â° = [(2 Ã— 33) + 4]Â° = 70Â°

âˆ Q = (y + 3)Â° = [50 + 3]Â° = 53Â°

âˆ R = (2y + 10)Â° = [2 Ã— 50 + 10]Â° = 110Â°

âˆ S = (4x - 5)Â° = [4 Ã— 33 - 5Â°] = 127Â°

23x + 35y = 209 ...(1)

35x + 23y = 197 ...(2)

88x + 88y = 406 [Adding (1) and (2)]

â‡’ x + y = 7 ...(3) [Dividing by 88]

Subtracting (1) from (2),

35x + 23y = 197

23x + 35y = 209

(-) (-) (-)

12x - 12y = - 12

â‡’ x - y = 1 ...(4) [Dividing by 12]

Adding (3) and (4),

2x = 8 â‡’ x = 4

From (3) x + y = 7 â‡’ 4 + y = 7

â‡’ y = 3

So, x = 4 and y = 3.

a

a

â‡’

âˆ´

Thus, the required solution is:

.

6x - 3y + 10 = 0 ...(1)

2x - y + 9 = 0 ...(2)

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

a

a

Now,

âˆ´We have

=

This condition represents parallel lines. Hence, the given pair represents parallel lines.

Also = 0 â‡’ y = ...(2)

Plotting the points (0, 1), (2, 4), (- 2, - 2) and (0, 3), (2, 6), (â€“2, 0) we get two straight lines l

âˆ´ The given equations are inconsistent.

From the graph we observe that line l

From 1st condition,

â‡’ 3x + 6 = y + 2

â‡’ 3x âˆ’ y + 4 = 0 ...(1)

From 2nd conditon,

â‡’ 5x + 15 = 2y + 6

â‡’ 5x âˆ’ 2y + 9 = 0 ...(2)

From (1) and (2), we have:**Ques ****8: Check graphically whether the pair of equations****3x + 5y = 15****x - y = 5****is consistent. Also find the coordinates of the points where the graphs of equations meet the y-axis.****Sol:** We have

3x + 5y = 15

â‡’

âˆ´

And from x - y = 5

â‡’ y = x - 5

Plotting the above two sets of points we get two straight lines l1 and l2 which intersect at the point (5, 0).

Thus, the given system is consistent.

Obviously the line l_{1} meets the y-axis at (0, 3) and line l_{2} meets the y-axis at (0, - 5).**Ques ****9: Places A and B are 160 km apart on highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 8 hours, but if they travel towards each other, they meet in 2 hours. What are the speeds of the two cars? ****Sol: **Let the car-I and car-II starts from A and B at x km/hr and y km/hr respectively.**Case-I:** [Cars are moving in the same direction]

Let the two cars meet at C after 8 hours.

Distance covered:

by car-I = AC = 8x km

by car-II = BC = 8y km

âˆ´ AB = AC - BC

â‡’ 160 = 8x - 8y

â‡’ x - y = 20 ...(1)**Case-II:** [Cars are moving in opposite directions]

Let, after 2 hours, the cars meet at D.

âˆ´ Distance cover after 2 hours,

by car-I = AD = 2x km

by car-II = BD = 2y km

â‡’ AB = AD + BD

â‡’ 160 = 2x + 2y

â‡’ 80 = x + y

â‡’ x + y = 80 ...(2)

Adding (1) and (2), we get

x + y = 80

x - y = 20

2x = 100

â‡’ x = 100/2 = 50

â‡’ Substituting x = 50 in (1), we get

x - y = 20 â‡’ 50 - y = 20

â‡’ y = 50 - 20 = 30

â‡’ Speed of car-I = 50 km/hr

Speed of car-II = 30 km/hr.**Ques ****10: Solve for x and y:****Sol: **We have:

= ...(1)

ax âˆ’ by = 2ab ...(2)

Dividing (2) by a, we have:

â‡’ ...(3)

From (1) and (3), we have

â‡’

From (2),

ab - by = 2ab

â‡’ - by = 2ab - ab = ab

â‡’

y = - a

Thus, x = b and y = - a.**Ques ****11: The sum of two numbers is 8. Determine the numbers if the sum of their reciprocals is 8/15.****Sol:** Let the two numbers be x and y.

According to the conditions:

x + y = 8 ...(1)

= ..(2)

From (1), x = (8 - y)

Substituting x = (8 - y) in (2),

â‡’

â‡’ 8 Ã— 15 = 8 Ã— y (8 - y)

â‡’ 64y - 8y^{2} - 120 = 0

â‡’- y^{2} + 8y - 15 = 0

â‡’ y^{2} - 8y + 15 = 0

â‡’ y^{2} - 5y - 3y + 15 = 0

â‡’ y (y - 5) - 3 (y - 5) = 0

â‡’ (y - 5) (y - 3) = 0

â‡’ If y - 5 = 0 then y = 5

or if y - 3 = 0, then y = 3

Since x = 8 - y

â‡’ when y = 5,

then x = 8 - 5 = 3

when y = 3, then

x = 8 - 3 = 5

â‡’ The required numbers are (3, 5) or

(5, 3).**Ques ****12: Solve the following pair of equations:****Sol:** Let

â‡’The given system of equations becomes:

5p + q = 2 ...(1)

6p - 3q = 1 ...(2)

Multiplying (1) by 3 and adding to (2),

and 5p + q = 2 â‡’

â‡’

â‡’

Since,

âˆ´

= 3 â‡’ x = 4

Also,

â‡’ y - 2 = 3 â‡’ y = 5

Thus, x = 4 and y = 4

â‡’ 3y - 6 = 1

â‡’ 3y = 1 + 6 = 7

â‡’ y = 7/3**Ques ****13: Solve the following pair of equations:****Sol:** Let

â‡’ The given pair of equation is expressed as

10p + 2q = 4 â‡’ 5p + q = 2 ...(1)

15p - 5q = - 2 ...(2)

Multiplying (1) by 5 and adding to (2)

From (1),

â‡’1 + q = 2 â‡’ q = 2 - 1 = 1

Since,

â‡’

â‡’ x + y = 5 ...(3)

And

â‡’ x âˆ’ y = 1 ...(4)

Adding (3) and (4),

From (3), 3 + y = 5 â‡’ y = 2

Thus, x = 3 and y = 2.**Ques ****14: Solve for x and y:****37x + 43y = 123****43x + 37y = 117****Sol:** We have:

37x + 43y = 123 ...(1)

43x + 37y = 117 ...(2)

Adding (1) and (2)

Dividing both sides by 80, we get

x + y = 3 ...(3)

Subtracting (2) from (1),

- 6x + 6y = 6 â‡’ - x + y = 1 ...(4)

Adding:

â‡’ y = 4/2 = 2

Putting y = 2 in x + y = 3, we get

x + 2 = 3 â‡’ x = 3 - 2 = 1

Thus, x = 1 and y = 2.**Ques ****15: Solve for â€˜xâ€™ and â€˜yâ€™:****(a - b) x + (a + b) y = a ^{2} - 2ab - b^{2}**

(a - b) x + (a + b) y = a

(a + b) x + (a + b) y = a

âˆ’ 2b x = âˆ’ 2ab âˆ’ b

â‡’ (âˆ’ 2b) x = âˆ’ 2b (a + b)

â‡’

From (2),

(a + b) (a + b) + (a + b) y = a

â‡’(a + b)2 + (a + b) y = (a

â‡’ (a + b) y = (a

â‡’ (a + b) y = a

â‡’ (a + b) y = a

â‡’ (a + b) y = - 2ab

â‡’

Thus, x = (a + b) and

x + 3y = 6

and 2x - 3y = 12

Plotting the above points, we get two straight lines l_{1} and l_{2} such that they intersect at (6, 0) as shown below:

Obviously,

The line l_{1} meets the y-axis at (0, 2).

The line l_{2} meets the y-axis at (0, - 4).**Ques ****17: Solve for x and y:****Sol:** Let

âˆ´ We have:

5p + q = 2 ...(1)

6p - 3q = 1 ...(2)

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

âˆ´

â‡’

âˆ´

And

Now,

â‡’ x âˆ’ 1 = 3 â‡’ x = 4

And

â‡’ y âˆ’ 2 = 3 â‡’ y = 5

Thus x = 4 and y = 5**Ques ****18: For what values of â€˜aâ€™ and â€˜bâ€™ does the following pair of equations have an infinite number of solutions?****2x + 3y = 7****a (x + y) - b (x - y) = 3a + b - 2****Sol:** We have:

2x + 3y = 7 ...(1)

a (x + y) - b (x - y) = 3a + b - 2 ...(2)

From (2), we have:

a (x + y) - b (x - y) = 3a + b - 2

â‡’ ax + ay - bx + by = 3a + b - 2

â‡’ ax - bx + ay + by = 3a + b - 2

â‡’ (a - b) x + (a + b) y = 3a + b - 2

Now, A_{1} = 2, B_{1}= 3,

C_{1} = - 7

A_{2} = (a - b), B_{2}

= (a + b),

C_{2} = - [3a + b - 2]

For infinite number of solutions,

i.e.,

âˆ´

â‡’ 2 (a + b) = 3 (a âˆ’ b)

â‡’ 2a + 2b âˆ’ 3a + 3b = 0

â‡’ âˆ’ a + 5b = 0

â‡’ a = 5b ...(3)

Also =

â‡’ 3 (3a + b - 2) = 7 (a + b)

â‡’ 9a + 3b - 6 = 7a + 7b

â‡’ 9a - 7a + 3b - 7b = 6

â‡’ 2a - 4b = 6

â‡’ a - 2b = 3 ...(4)

From (3) and (4),

5b - 2b = 3

â‡’ 3b = 3 â‡’ b = 1

Thus, a = 5 Ã— b

â‡’ a = 5 Ã— 1 = 5

i.e., a = 5 and b = 1.**Ques ****19: Solve the following pairs of equations for x and y:****Sol: **Let

âˆ´ We have:

15p + 22q = 5 ...(1)

40p + 55q = 13 ...(2)

From (1) and (2), we get

Adding (3) and (4), we have

2x = 16 â‡’ x = 16/2 =8

From (4), 8 + y = 11 â‡’ y = 11 - 8 = 3

Thus, x = 8 and y = 3.**Ques ****20: Draw the graph of the pair of linear equations x â€“ y + 2 = 0 and 4x â€“ y â€“ 4 = 0. Calculate the area of triangle formed by the lines so drawn and x-axis. Sol: **To draw the graph of the given pair of equations, we have the table of ordered pairs

Plot the points A(0, 2), B(â€“2, 0); C(0, â€“4) and D(1, 0) on the graph paper and join the points to form the lines AB and CD :

From the graph, we find that the points P(2, 4) is common to both the lines AB and CD.

These lines meet x-axis at B(â€“2, 0) and D(1, 0).

Thus, the triangle BDP is formed by the lines and the x-axis.

The vertices of this Î” are

B(â€“2, 0), D(1, 0) and P(2, 4)

Now, base of Î”BDP = BD

= (BO + OD)

= (2 + 1) units

= 3 units

Altitude of the Î”BDP = PQ

= 4 units

âˆ´ Area of Î”BDP =

= 6 sq. units

The time taken by the pipe of smaller diameter to fill the pool separately = y hours.

âˆ´ Part of the pool fill by the pipe of larger diameter in 1 hour = 1/x.

Part of fool filled by the pipe of larger diameter in 4 hours = 4/x.

Similarly, Part of the pool filled by the pipe of smaller diameter in 9 hours = 9/y.

âˆ´We have = ...(1)

Since, the pool is fill by both the pipes together in 12 hours.

âˆ´ ...(2)

To solve (1) and (2), multiplying (1) by 3 and subtracting (2) from it, we have

Substituting, y = 30 in (2), we have

â‡’

âˆ´

â‡’ x = 20

â‡’ Required time taken by pipe of larger diameter = 20 hours

Required time taken by pipe of smaller diameter = 30 hours

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