Unit Test (Solutions): Pair of Linear Equations in Two Variables | Mathematics (Maths) Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers  9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: Which of the following options represents a pair of linear equations in two variables?  (1 Mark) 
(a) 2x + 3y = 7
(b) x² + y² = 25
(c) 4x + 2y = 10
(d) 3x + 2y² = 8
Ans: (a)
A pair of linear equations in two variables must be of the form ax + by = c, where a, b, and c are constants. Option a) fits this form, while the other options involve either quadratic terms (b) or variable terms with exponents other than 1 (d), which do not represent linear equations.

Q2: The number of solutions for a pair of linear equations in two variables can be:  (1 Mark) 
(a) One
(b) Two
(c) Three
(d) None
Ans: (a)
A pair of linear equations in two variables can have one of the following types of solutions:

• Unique Solution (a single point of intersection)
• Infinitely Many Solutions (coincident lines)
• No Solution (parallel lines)

Q3: If two lines represented by a pair of linear equations are parallel, then:  (1 Mark) 
(a) They intersect at one point
(b) They do not intersect
(c) They intersect at two points
(d) None of the above
Ans: (b)
If two lines are parallel, they do not intersect at any point. For a pair of linear equations to represent parallel lines, their slopes must be equal, and their y-intercepts may differ.

Q4: Solve the pair of equations: 2x - 3y = 7 and 4x - 6y = 14.  (1 Mark) 
Ans: Dividing the second equation by 2, we get:
2x - 3y = 7
This is exactly the same as the first equation.
So, the given pair of equations are dependent (consistent) and have infinitely many solutions.

Q5: Check the consistency of the pair of linear equations and solve them graphically:  (1 Mark) 
4x - 3y = 9
8x - 6y = 18
Ans: To check the consistency of the equations, we can compare their slopes. Both equations are equivalent, which means they represent the same line. As a result, they have infinitely many solutions and are consistent.
To solve them graphically, we plot the line represented by 4x - 3y = 9 (or 8x - 6y = 18 since they are equivalent). The graph is a straight line passing through points (0, -3) and (9/2, 0) on the coordinate plane.

Q6: Solve the pair of linear equations:  (2 Marks) 
x - y = 4
2x + 3y = 10
Ans: To solve the pair of linear equations, we can use the method of substitution or elimination.
Substituting the value of x from the first equation into the second equation:
x = 4 + y
2(4 + y) + 3y = 10
Simplify the equation:
8 + 2y + 3y = 10
5y = 2
y = 2/5
Now, substitute the value of y into the first equation to find x:
x - (2/5) = 4
x = 4 + 2/5
x = 22/5
So, the solution to the pair of linear equations is x = 22/5 and y = 2/5.

Q7: Find the value of 'k' for which the following system of equations has no solution:  (2 Marks) 
2x + 3y = 5
4x + ky = 10
Ans: For a system of equations to have no solution, the lines represented by the equations must be parallel, i.e., they have the same slope. Therefore, we need to find the value of 'k' for which the slopes of the two equations are equal.
The slopes of the equations are given by the coefficients of y:
For the first equation: Slope_1 = 3/2
For the second equation: Slope_2 = k/4
For both lines to be parallel, Slope_1 must be equal to Slope_2:
3/2 = k/4
Now, cross-multiply to find the value of 'k':
3 * 4 = 2 * k
12 = 2k
k = 12/2
k = 6
So, the value of 'k' for which the system of equations has no solution is k = 6.

Q8: A sum of money amounts to Rs. 11,000 after 3 years and Rs. 13,750 after 5 years, when it is invested at a certain rate of simple interest. Calculate the rate of interest using a pair of linear equations in two variables.  (2 Marks) 
Ans: Let the principal amount be P and the rate of interest be R.
After 3 years, the amount is Rs. 11,000:
11,000 = P + (P * R * 3/100) ... (Equation 1)
After 5 years, the amount is Rs. 13,750:
13,750 = P + (P * R * 5/100) ... (Equation 2)
We have a pair of linear equations in two variables, P, and R.
To solve the equations, we can use the substitution or elimination method as mentioned earlier.

Q9: The sum of the digits of a two-digit number is 9. If we add 9 to the number, the digits get reversed. Find the number using a pair of linear equations in two variables.  (3 Marks) 
Ans: Let the tens digit of the two-digit number be x, and the units digit be y.
The number can be represented as 10x + y.
According to the first condition, the sum of the digits is 9:
x + y = 9 ... (Equation 1)
According to the second condition, if we add 9 to the number, the digits get reversed:
10x + y + 9 = 10y + x
Simplifying this equation:
10x - x = 10y - y - 9
9x = 9y - 9
x = y - 1 ... (Equation 2)
Now, we can solve the system of equations (Equation 1 and Equation 2) to find the values of x and y, which will give us the required two-digit number.

Q10: Solve the following system of equations using any appropriate method:  (3 Marks) 
3x - 2y = 8
6x - 4y = 16
Ans: Divide the second equation by 2:
3x - 2y = 8
3x - 2y = 8
Since the equations are the same, they represent the same line and are dependent. Therefore, there are infinitely many solutions.

Q11: Solve the pair of linear equations:  (3 Marks) 
2x + y = 7
x - 3y = -4
Ans: To solve the pair of linear equations, we can use the substitution or elimination method.
Using the elimination method:
Multiply the first equation by 3:
3(2x + y) = 3(7)
6x + 3y = 21
Now, add the two equations:
(6x + 3y) + (x - 3y) = 21 - 4
7x = 17
Divide by 7:
x = 17/7
Now, substitute the value of x into any equation to find y. Using the first equation:
2(17/7) + y = 7
34/7 + y = 7
Subtract 34/7 from both sides:
y = 7 - 34/7
y = (49 - 34)/7
y = 15/7
So, the solution to the pair of linear equations is x = 17/7 and y = 15/7.

Q12: The perimeter of a rectangle is 46 meters. The length is 4 meters more than twice the width. Find the dimensions of the rectangle using a pair of linear equations.  (5 Marks) 
Ans: Let's assume the width of the rectangle is x meters.
Then, the length of the rectangle is (2x + 4) meters.
The perimeter of the rectangle is given by:
Perimeter = 2(length + width)
Given that the perimeter is 46 meters, we can write the equation as:
2((2x + 4) + x) = 46
Simplify the equation:
2(3x + 4) = 46
Expand:
6x + 8 = 46
Now, subtract 8 from both sides:
6x = 38
Divide by 6:
x = 38/6
x = 19/3
So, the width of the rectangle is 19/3 meters. Now, find the length:
Length = 2x + 4
Length = 2(19/3) + 4
Length = 38/3 + 4
Length = (38 + 12)/3
Length = 50/3
So, the dimensions of the rectangle are width = 19/3 meters and length = 50/3 meters.

Q13: A two-digit number is 6 times the sum of its digits. If 9 is added to the number, the digits interchange their places. Find the number using a pair of linear equations.  (5 Marks)
Ans: Let's assume the two-digit number is represented as 10x + y, where x is the digit in the tens place, and y is the digit in the units place.
The given information can be translated into the following pair of linear equations:
Equation 1: 10x + y = 6(x + y)
Equation 2: 10x + y + 9 = 10y + x
Let's solve the equations to find the value of x and y.
Step 1: Solve Equation 1 for y in terms of x:
10x + y = 6x + 6y
Subtract 6x from both sides:
y = 6y - 4x
Now, bring all the terms involving y on one side:
y - 6y = -4x
Simplify:
-5y = -4x
Now, divide by -5:
y = (4/5)x
Step 2: Substitute the value of y from Equation 1 into Equation 2:
10x + (4/5)x + 9 = 10((4/5)x) + x
Now, solve for x:
50x + 4x + 45 = 8x + x
Combine like terms:
54x + 45 = 9x
Now, subtract 9x from both sides:
54x - 9x = -45
Simplify:
45x = -45
Now, divide by 45:
x = -1
Step 3: Substitute the value of x back into Equation 1 to find y:
10(-1) + y = 6(-1 + y)
-10 + y = -6 + 6y
Now, bring all the terms involving y on one side:
y - 6y = 6 - 10
Simplify:
-5y = -4
Now, divide by -5:
y = 4/5
Since x is the digit in the tens place, it cannot be negative. Therefore, we reject x = -1.
So, there is no valid solution for this problem. The two-digit number that satisfies the given conditions does not exist.