NCERT Solutions For Class 10 Math Chapter – 3 Exercise – 3.5

NCERT Solutions For Class 10 Math Chapter – 3 Exercise – 3.5

Q1. Which of the following pairs of linear equation has unique solution , no solution or infinitely many solutions ? In case there is a unique solution find it by using cross multiplication method.

X – 3y – 3 = 0 ; 3x – 9y – 2 = 0
2x + y = 5 ; 3x + 2y = 8
3x – 5y 20 ; 6x – 10y = 40
X – 3y – 7 = 0 ; 3x – 3y – 15 = 0

Solution:-

X – 3y – 3 = 0

3x – 9y – 2 = 0

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

2x + y = 5

3x+ 2y = 8

A₁/a₂ ≠ b₁/b₂

By cross multiplication method.

X/b₁c₂ – b₂c₁ = y/c₁a₂ – c₂a₁ = 1/a₁b₂-a₂b₁

X/-8-(-10) = y/-15 + 16 = 1/4 – 3

X = 2 y = 1

3x – 5y =20

6x – 10 = 40

A₁/a₂ = b₁/b₂ = c₁/c₂

X – 3y – 7 = 0

3x – 3y – 15 = 0

A₁/a₂ ≠ b₁/b₂

By cross – multiplication

X/45 – (21) = y/21 – (-15) = 1/-3 – (-9)

X = 4 and y = 1

Q2. (I) For which values of a and b does the following pair of linear equation have an infinite number of solutions ?

2x + 3y = 7

Solution:-

2x + 3y – 7 = 0

(a – b)x + (a + b)y – (3a + b – 2) = 0

A₁/a₂ = 2/a – b = 1/2

B₁/b₂ = -7/a + b and

C₁/c₂ = -7/-(3a + b – 2)

= 7/(3a + b – 2)

2/a-b = 7/3a + b – 26a + 2b- 4 = 7a – 7b

A – 9b = 4(I)

2/a-b = 3/a + b

2a + 2b = 3a – 3

A – 5b = 0 (ii)

Subtracting equation (I) from (ii) we get

4b = 4

B = 1

Substituting b = 1 in equation (2) we obtain

A – 5 x 1 = 0

A = 5

Hence a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

(ii) 3x + y – 1 = 0

(2k – 1)x + (k – 1)y – 2k – 1 = 0

3/2k – 1 = 1/k – 1 ≠ 1/2k + 1

3/2k – 1 = 2k – 1

3k – 3 = 2k – 1

K = 2

Q3. Solve the following pair of linear equations by the substitution and cross – multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:-

8x + 5y = 9 (I)

3x + 2y = 4 (ii)

From equation (2) we obtain

3x + 2y = 4

3x = 4 – 2y

X = 4 – 2y/3 (iii)

Put value of x in equation (I)

8(4 – 2y/3) + 5y = 9

32 – y = 27

Y = 5

Subsituting y = 5 in equation (iii) we obtain

X = 4 – 2 x 5/3

X = -2

Again by cross – multiplication method

8x + 5y = 9

3x + 3y = 4

8x + 5y – 9 = 0

3x + 2y – 4 =0

X/-20-(-18) = y/-27-(-32) = 1/16 – 15

X = -2 and y = 5

Q4. From the pair of linear equation in the following problems and find their solutions (if they exist ) by any algebraic method:

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 11000 as hostel charges whereas a student B , who takes food for 1000 days pays Rs 1180 as hostel charges Find the fixed charges and the cost of food per day.
A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its denominator Find the fraction.
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 ?
Places A and B are 100 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 5 hours. If they travel towards each other , they meet in 1 hour. What are the speeds of the two cars ?
The area of a rectangle gets reduced by 9 square units, if 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:-

X + 20y = 100 (I)

X + 26y = 1180 (ii)

Subtracting equation (i) from equation (ii)

6y = 180

Y = 30

Substituting y = 30 in equation (I) we obtain

X x 20 x 30 = 1000

X = 1000 – 600

X = 400

3x – y = 3(I)

4x – y = 8 (ii)

Subtracting equation (I) from equation(ii) we obtain

X = 5

Putting x = 5 in equation (I) we obtain

3 x 5 – y = 3

Y = 12

3x – y = 40 (I)

2x – y = 25 (ii)

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

x = 15

Substituting this in equation (2) we obtain

2 x 15 – y = 25

Y = 5

U – v = 20 (I)

U + v = 100 (ii)

Adding both the equations we obtain

2u = 120

U = 60

Substituting this value in equation (ii) we obtain

60 + v = 100

V = 40

3x – 5y – 6 = 0

2x + 3y – 61 = 0

X/305 – (-18) = y/-12-(-193) = 1/9 – (-19)

x/323 = y/171 = 1/19

X = 17 , y = 9