CBSE Class 10 Pair of Linear Equations in Two Variables Sure Shot Questions Set 01

CBSE Class 10 Pair of Linear Equations Sure Shot Questions Set A. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Multiple Choice Questions

Question. A pair of linear equations \( a_1x + b_1y + c_1 = 0 \); \( a_2x + b_2y + c_2 = 0 \) is said to be inconsistent, if
(a) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
(b) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
(c) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
(d) \( \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \)
Answer: (b)

Question. Graphically, the pair of equations \( 7x – y = 5 \); \( 21x – 3y = 10 \) represents two lines which are
(a) intersecting at one point
(b) parallel
(c) intersecting at two points
(d) coincident
Answer: (b)

Question. The pair of equations \( 3x – 5y = 7 \) and \( – 6x + 10y = 7 \) have
(a) a unique solution
(b) infinite solutions
(c) no solution
(d) two solutions
Answer: (c)

Question. If a pair of linear equations is consistent, then the lines will be
(a) always coincident
(b) parallel
(c) always intersecting
(d) intersecting or coincident
Answer: (d)

Question. The pair of equations \( x = 0 \) and \( x = 5 \) has
(a) no solution
(b) one solution
(c) two solutions
(d) infinite solutions
Answer: (a)

Question. The pair of equation \( x = –4 \) and \( y = –5 \) graphically represents lines which are intersecting
(a) at \( (–5, –4) \)
(b) at \( (–4, –5) \)
(c) at \( (5, 4) \)
(d) at \( (4, 5) \)
Answer: (b)

Question. For what value of \( k \), do the equations \( 2x – 3y + 10 = 0 \) and \( 3x + ky + 15 = 0 \) represent coincident lines
(a) \( \left( \frac{-9}{2} \right) \)
(b) \( -11 \)
(c) \( \frac{9}{2} \)
(d) \( -7 \)
Answer: (a)

Question. If the lines given by \( 2x + ky = 1 \) and \( 3x – 5y = 7 \) are parallel, then the value of \( k \) is
(a) \( \frac{-10}{3} \)
(b) \( \frac{10}{3} \)
(c) \( -13 \)
(d) \( -7 \)
Answer: (a)

Question. One equation of a pair of dependent linear equations is \( 2x + 5y = 3 \). The second equation will be
(a) \( 2x + 5y = 6 \)
(b) \( 3x + 5y = 3 \)
(c) \( –10x – 25y + 15 = 0 \)
(d) \( 10x + 25y = 15 \)
Answer: (c)

Question. If \( x = a, y = b \) is the solution of the equations \( x + y = 5 \) and \( 2x – 3y = 4 \), then the values of \( a \) and \( b \) are respectively
(a) \( 6, –1 \)
(b) \( 2, 3 \)
(c) \( 1, 4 \)
(d) \( \frac{19}{5}, \frac{6}{5} \)
Answer: (d)

Question. The graph of \( x = –2 \) is a line parallel to the
(a) x-axis
(b) y-axis
(c) both x- and y-axis
(d) none of these
Answer: (b)

Question. The graph of \( y = 4x \) is a line
(a) parallel to x-axis
(b) parallel to y-axis
(c) perpendicular to y-axis
(d) passing through the origin
Answer: (d)

Question. The graph of \( y = 5 \) is a line parallel to the
(a) x-axis
(b) y-axis
(c) both axis
(d) none of these
Answer: (a)

Question. Two equations in two variables taken together are called
(a) linear equations
(b) quadratic equations
(c) simultaneous equations
(d) none of these
Answer: (c)

Question. If \( am \neq bl \) then the system of equations \( ax + by = c, lx + my = n \), has
(a) a unique solution
(b) no solution
(c) infinitely many solutions
(d) none of these
Answer: (a)

Question. If in the equation \( x + 2y = 10 \), the value of \( y \) is 6, then the value of \( x \) will be
(a) –2
(b) 2
(c) 4
(d) 5
Answer: (a)

Question. The graph of the equation \( 2x + 3y = 5 \) is a
(a) vertical line
(b) straight line
(c) horizontal line
(d) none of these
Answer: (b)

Question. The value of \( k \), for which equations \( 3x + 5y = 0 \) and \( kx + 10y = 0 \) has a non-zero solution is
(a) 6
(b) 0
(c) 2
(d) 5
Answer: (a)

Question. The value of \( k \), for which the system of equations \( x + (k + 1)y = 5 \) and \( (k + 1)x + 9y = 8k – 1 \) has infinitely many solutions is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (a)

Question. The value of \( k \) for which the equations \( (3k + 1)x + 3y = 2 \); \( (k^2 + 1)x + (k – 2)y = 5 \) has no solution, then \( k \) is equal to
(a) 2
(b) 3
(c) 1
(d) –1
Answer: (d)

Question. The pair of equations \( x = a \) and \( y = b \) graphically represents lines which are
(a) parallel
(b) intersecting at \( (b, a) \)
(c) coincident
(d) intersecting at \( (a, b) \)
Answer: (d)

Question. Asha has only Rs 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Rs 1 and Rs 2 coins are, respectively
(a) 35 & 15
(b) 15 & 35
(c) 35 & 20
(d) 25 & 25
Answer: (d)

Question. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages of the son and the father are, respectively
(a) 4 & 24
(b) 5 & 30
(c) 6 & 36
(d) 3 & 24
Answer: (c)

Question. The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is
(a) 27
(b) 72
(c) 45
(d) 36
Answer: (d)

Assertion—Reason Questions

Direction: In the following questions, a statement of Assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.

Question. Assertion: The value of \( q = \pm 2 \), if \( x = 3, y = 1 \) is the solution of the line \( 2x + y – q^2 – 3 = 0 \).
Reason: The solution of the line will satisfy the equation of the line.
Answer: (a)
Explanation: As \( x = 3 \) and \( y = 1 \) is the solution of \( 2x + y – q^2 – 3 = 0 \).
When \( x = 3 \) and \( y = 1 \), \( 2 \times 3 + 1 – q^2 – 3 = 0 \)
\( \Rightarrow 4 – q^2 = 0 \)
\( \Rightarrow 4 = q^2 \therefore q = \pm 2 \)
So, both A and R are correct and R explains A.

Question. Assertion: If the pair of lines are coincident, then we say that pair of lines is consistent and it has a unique solution.
Reason: If the pair of lines are parallel, then the pair has no solution and is called inconsistent pair of equations.
Answer: (d)
Explanation: If the lines are coincident, then it has infinite number of solutions.

Question. Assertion: \( x + y – 4 = 0 \) and \( 2x + ky – 3 = 0 \) has no solution if \( k = 2 \).
Reason: \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) are consistent if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Answer: (b)
Explanation: For assertion, given equation has no solution if \( \frac{1}{2} = \frac{1}{k} \neq \frac{-4}{-3} \) i.e., \( \frac{4}{3} \). When \( k = 2 \), \( \frac{1}{2} = \frac{1}{2} \neq \frac{4}{3} \)

Question. Assertion: If \( kx – y – 2 = 0 \) and \( 6x – 2y – 3 = 0 \) are inconsistent, then \( k = 3 \).
Reason: \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) are inconsistent if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Answer: (a)
Explanation: We have, \( kx – y – 2 = 0 \) and \( 6x – 2y – 3 = 0 \)
When \( k = 3 \), \( \frac{3}{6} = \frac{-1}{-2} \neq \frac{-2}{-3} \) i.e. \( \frac{1}{2} = \frac{1}{2} \neq \frac{2}{3} \)...[\( \because \frac{k}{6} = \frac{3}{6} \)]

Question. Assertion: The linear equations \( x – 2y – 3 = 0 \) and \( 3x + 4y – 20 = 0 \) have exactly one solution.
Reason: The linear equations \( 2x + 3y – 9 = 0 \) and \( 4x + 6y – 18 = 0 \) have a unique solution.
Answer: (c)
Explanation: We have, \( x – 2y – 3 = 0 \) and \( 3x + 4y – 20 = 0 \)
Here, \( \frac{1}{3} \neq \frac{-2}{4} \neq \frac{-3}{-20} \) [\(\because\) Intersecting lines have 1 solution]

Case Based Questions

I. A test consists of ‘True’ or ‘False’ questions. One mark is awarded for every correct answer while 1/4 mark is deducted for every wrong answer. A student knew answers to some of the questions. Rest of the questions he attempted by guessing. He answered 120 questions and got 90 marks.
Type of Questions: True/false
Marks given for correct answers: 1
Marks deducted for wrong answers: 0.25

Question. If answers to all questions that he attempted by guessing were wrong, how many questions did he answer correctly?
(a) 96
(b) 86
(c) 76
(d) 106
Answer: (a)
Explanation: Let the no. of questions whose answers are known to the student x and questions attempted by guessing be y.
\( x + y = 120 \) ...(i)
\( x - \frac{1}{4}y = 90 \) ...(ii)
Solving (i) and (ii), we get \( x = 96 \) and \( y = 24 \)
∴ No. of questions whose answer are known = 96

Question. How many questions did he guess?
(a) 12
(b) 18
(c) 24
(d) 26
Answer: (c)

Question. If answers to all the questions he attempted by guessing were wrong and answered 80 correctly, then how many marks did he get?
(a) 50
(b) 70
(c) 80
(d) 20
Answer: (b)
Explanation: Total no. of questions = 96 + 24 = 120
Marks = \( 80 - \frac{1}{4} \) of 40 = 70

Question. If answers to all the questions he attempted by guessing were wrong, then how many questions did answer correctly to score 95 marks?
(a) 100
(b) 200
(c) 250
(d) 150
Answer: (a)
Explanation: \( x - \frac{1}{4} \) of \( (120 - x) = 95 \)
\( \Rightarrow 5x = 380 + 120 \Rightarrow 5x = 500 \therefore x = 100 \)

It is common that Governments revise travel fares from time to time based on various factors such as inflation (a general increase in prices and fall in the purchasing value of money) on different types of vehicles like Autos, Rickshaws, Taxis, Radio cabs etc. The Auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations:
City A: 10km costs Rs 75, 15km costs Rs 110
City B: 8km costs Rs 91, 14km costs Rs 145

Question. If the fixed charges of auto rickshaw be Rs x and the running charges be Rs y per km, the pair of linear equations representing the situation in city A is
(a) \( x + 10y = 110, x + 15y = 75 \)
(b) \( x + 10y = 75, x + 15y = 110 \)
(c) \( 10x + y = 110, 15x + y = 75 \)
(d) \( 10x + y = 75, 15x + y = 110 \)
Answer: (b)

Question. A person travels a distance of 50 km in City A. The amount he has to pay is
(a) 155
(b) 255
(c) 355
(d) 455
Answer: (c)
Explanation: Solving equations from (i), we get \( x = 5 \) and \( y = 7 \)
Now, fare for 50 km = \( x + 50y = 5 + 50(7) = Rs 355 \)

Question. What will a person have to pay for travelling a distance of 30km in City B?
(a) Rs 185
(b) Rs 289
(c) Rs 275
(d) Rs 305
Answer: (b)
Explanation: \( x + 8y = 91 \) ...(i)
\( x + 14y = 145 \) ...(ii)
Subtracting (i) from (ii): \( 6y = 54 \Rightarrow y = 9 \)
Putting \( y = 9 \) in (i), we get \( x = 19 \)
Now, fare for 30 km = \( x + 30y = 19 + 30 \times 9 = Rs 289 \)

A part of monthly hostel charges in a college is fixed and the remaining depends on the number of days one has taken food in the mess. When a student Anu takes food for 25 days, she has to pay Rs 4500 as hostel charges, whereas another student Bindu who takes food for 30 days has to pay Rs 5200 as hostel charges. Considering the fixed charges per month by Rs x and the cost of food per day by Rs y, then answer the following questions:

Question. Represent algebraically the situation faced by both Anu and Bindu.
(a) \( x + 25y = 4500, x + 30y = 5200 \)
(b) \( 25x + y = 4500, 30x + y = 5200 \)
(c) \( x – 25y = 4500, x – 30y = 5200 \)
(d) \( 25x – y = 4500, 30x – y = 5200 \)
Answer: (a)

Question. The system of linear equations, represented by above situations has
(a) No solution
(b) Unique solution
(c) Infinitely many solutions
(d) None of these
Answer: (b)
Explanation: Using equations from point (i), \( a_1 = 1, b_1 = 25, c_1 = -4500 \) and \( a_2 = 1, b_2 = 30, c_2 = -5200 \).
\( \frac{a_1}{a_2} = 1, \frac{b_1}{b_2} = \frac{25}{30} = \frac{5}{6} \).
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), system has unique solution.

Question. The cost of food per day is
(a) Rs 120
(b) Rs 130
(c) Rs 140
(d) Rs 1300
Answer: (c)
Explanation: We have, \( x + 25y = 4500 \) ...(i) and \( x + 30y = 5200 \) ...(ii)
Subtracting (i) and (ii), we get \( 5y = 700 \Rightarrow y = 140 \).
∴ Cost of food per day is Rs 140.

Question. The fixed charges per month for the hostel is
(a) Rs 1500
(b) Rs 1200
(c) Rs 1000
(d) Rs 1300
Answer: (c)
Explanation: \( x + 25 \times 140 = 4500 \Rightarrow x = 4500 - 3500 = 1000 \).
∴ Fixed charges per month for the hostel is Rs 1000.

Question. If Bindu takes food for 20 days, then what amount she has to pay?
(a) Rs 4000
(b) Rs 3500
(c) Rs 3600
(d) Rs 3800
Answer: (d)
Explanation: Food charges for 20 days = \( 1000 + 20 \times 140 = Rs 3800 \)

V. Mr. Manoj Jindal arranged a lunch partly for some of his friends. The expense of the lunch is partly constant and partly proportional to the number of guests. The expenses amount to Rs 650 for 7 guests and Rs 970 for 11 guests. Denote the constant expense by x and proportional expense per person by y and answer the following questions:

Question. Represent both the situations algebraically.
(a) \( x + 7y = 650, x + 11y = 970 \)
(b) \( x – 7y = 650, x – 11y = 970 \)
(c) \( x + 11y = 650, x + 7y = 970 \)
(d) \( 11x + 7y = 650, 11x – 7y = 970 \)
Answer: (a)

Question. Proportional expense for each person is
(a) Rs 50
(b) Rs 80
(c) Rs 90
(d) Rs 100
Answer: (b)
Explanation: Subtracting equations (i) from (ii), we get \( 4y = 320 \therefore y = 80 \)

Question. The fixed (or constant) expense for the party is
(a) Rs 50
(b) Rs 80
(c) Rs 90
(d) Rs 100
Answer: (c)
Explanation: Putting \( y = 80 \) in equation (i), we get \( x + 7 \times 80 = 650 \Rightarrow x = 650 - 560 = Rs 90 \)

Question. If there are 15 guests at the lunch party, then what amount Mr. Jindal has to pay?
(a) Rs 1,250
(b) Rs 1,430
(c) Rs 1,350
(d) Rs 1,290
Answer: (d)
Explanation: Amount = \( Rs (90 + 15 \times 80) = Rs 1290 \)

Question. The system of linear equations representing both the situations will have
(a) a unique solution
(b) no solution
(c) infinitely solutions
(d) none of these
Answer: (a)
Explanation: \( x + 7y - 650 = 0; x + 11y - 970 = 0 \)
\( a_1 = 1, b_1 = 7, c_1 = 650; a_2 = 1, b_2 = 11, c_2 = 970 \)
\( \frac{a_1}{a_2} = 1, \frac{b_1}{b_2} = \frac{7}{11} \). Here, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Thus, system has a unique solution.

Please click the link below to download CBSE Class 10 Pair of Linear Equations Sure Shot Questions Set A.