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