CBSE Class 10 Mathematics Linear Equations Assignment Set A. Students are advised to refer to the attached assignments and practise them regularly. This will help them to identify their weak areas and will help them to score better in examination. Parents should download and give the assignments to their children for practice.
Q1.Which of the following pairs of linear equations are consistent/inconsistent?
(i) x + y = 5, 2x + 2y = 10
(i) x + y = 5; 2x + 2y = 10
a1/a2 = 1/2
b1/b2 = 1/2 and
c1/c2 = 5/10 = 1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.
Q2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically.
Solution: Let cost of one bat = Rs x
Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
3x + 6y = 3900
Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x + 2y = 1300
Q3.Find the number of solutions of the following pair of linear equations:
x + 2y -8=0,2x+ 4y=16
Solution: a1/a2 =1/2
Here a1/a2 = b1/b2= c1/c2
Therefore, the equations have infinite number of solutions.
Q4. Two lines are given to be parallel. The equation of one of the lines is 4x+ 3y=14. Find the equation of the second line.
LEVEL-2 ( 2 MARKS EACH)
Q1. Solve the following pair of linear equations by the substitution method.
x + y = 14 ; x – y = 4
Solution: x + y = 14 ... (i)
x – y = 4 ... (ii)
From equation (i), we get
x = 14 - y ... (iii)
Putting this value in equation (ii), we get
(14 - y) - y = 4
14 - 2y = 4
10 = 2y
y = 5 ... (iv)
Putting this in equation (iii), we get
x = 9
∴ x = 9 and y = 5
Q2. 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
3x + y -1 = 0
(2k –1)x + (k –1)y - (2k + 1) = 0
a1/a2 = 3/2k-1
b1/b2 = 1/k-1 and
c1/c2 = -1/-2k-1 = 1/2k+1
For no solutions,
a1/a2 = b1/b2 ≠ c1/c2
3/2k-1 = 1/k-1 ≠ 1/2k+1
3/2k-1 = 1/k-1
3k - 3 = 2k - 1
k = 2
Hence, for k = 2, the given equation has no solution.
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