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**Q.1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0(ii) 2x + y = 5 and 3x + 2y = 8(iii) 3x – 5y = 20 and 6x – 10y = 40(iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0Solution: **

a

(a

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

a

(a

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x / (b

x / (-8 - (-10)) = y / (15 + 16) = 1 / (4 - 3)

x / 2 = y / 1 = 1

(a

(b

(c

a

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

(a

(b

(c

a

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x / (45 - 21) = y / (-21 + 15) = 1 / (-3 + 9)

x / 24 = y / -6 = 1 / 6

x / 24 = 1 / 6 and y / -6 = 1 / 6

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Solution: (

(a + b)y + (a-b)y – (3a + b -2) = 0

a

For infinitely many solutions,

a

Thus 2 / (a - b) = 7 / (3a + b – 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4 ………………….(i)

2 / (a - b) = 3 / (a + b)

2a + 2b = 3a – 3b

a – 5b = 0 …………………(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a - 5 x 1= 0

a = 5

(2k - 1)x + (k - 1)y – 2k -1 = 0

a

For no solutions

a

3 / (2k - 1) = 1 / (k - 1) ≠ 1 / (2k + 1)

3 / (2k – 1) = 1 / (k - 1)

3k - 3 = 2k - 1

k = 2

8x + 5y = 9

3x + 2y = 4

Solution:

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = (4 – 2y ) / 3 ……………………. (3)

Using this value in equation 1, we get

8(4 - 2y) / 3 + 5y = 9

32 – 16y + 15y = 27

-y = -5

y = 5 ………………………(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x + 5y – 9 = 0

3x + 2y – 4 = 0

x / (-20 + 18) = y / (-27 + 32 ) = 1 / (16 - 15)

-x / 2 = y / 5 =1 / 1

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a 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 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from (ii) we get

6y = 180

y = Rs.30

Using this value in equation (ii) we get

x = 1180 - 26 x 30

x = Rs.400.

(ii) Let the fraction be x / y.

So, as per the question given,

(x - 1) / y = 1 / 3 => 3x – y = 3…………………(1)

x / (y + 8) = 1 / 4 => 4x – y = 8 ………………..(2)

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

x = 5 ……………………………(3)

Using this value in equation (2), we get,

(4 × 5) – y = 8

y = 12

According to the given question;

3x − y = 40……..(1)

4x − 2y = 50

⇒ 2x − y = 25…….(2)

Subtracting equation (2) from equation (1), we get;

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Or y = 5

Hence, total number of questions = 20

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………(i)

If the car travels in the opposite direction,

x + y = 100……………(ii)

Solving equation (i) and (ii), we get

x = 60 km/h………………………(iii)

Using this in equation (i), we get,

60 – y = 20

y = 40 km/h

Speed of car from point B = 40 km/h.

The length of rectangle = x unit

And breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy - 9

3x – 5y – 6 = 0……………………………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x / (305 + 18) = y / (-12 + 183) = 1 / (9 + 10)

x / 323 = y / 171 = 1 / 19

Therefore, x = 17 and y = 9.

And breadth of the rectangle = 9 units

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