CBSE Class 10 Mathematics Linear Equations Assignment Set 01

Assertion and Reason Based MCQ

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 A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is the NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question. Assertion (A): If 4 chairs and 3 tables cost ₹2100 and 5 chairs and 2 tables cost ₹1750 , then the cost of 1 chair is ₹150.
Reason (R): Sum of the ages of a father and the son is 40 years. If father’s age is 3 times that of his son, then the son’s age is 12 years.
Answer : C

Question. Assertion (A): If the pair of linear equations 3x+y=3 and 6x+ky =8 does not have a solution, then the value of k=2.
Reason (R): If the pair of linear equations x+y-4=0 and 2x+ky =3 does not have a solution, then the value of k=2.
Answer : B

Question. Assertion (A): If the equation 3x-y+8=0 and 6x-ky =-16 represents coincident lines, then the value of k=2.
Reason (R): If the lines given by 3x+2ky=2 and 2x+5y+1 =0 are parallel, then the value of k=12.
Answer : C

Question. Assertion (A): For all real values of c, the pair of equation x-2y=8 and 5x-10y=c have a unique solution.
Reason (R): Two lines are given to be parallel. The equation of one of the lines is 4x+3y=14, 12x+9y=5
Answer : D

Question. Assertion (A): The solution of the pair of linear equations x+y=5 and 2x-3y =4 is x=195 and y=65
Reason (R): The solution of the pair of linear equations 3x+4y=10 and 2x-2y=2 is x=2 and y=1.
Answer : B

LEVEL-1 (1 MARK EACH)

Question 1. Which of the following pairs of linear equations are consistent/inconsistent? (i) \(x + y = 5, 2x + 2y = 10\)
Answer: Given pair of linear equations:
\(x + y = 5\)
\(2x + 2y = 10\)
Here, \(a_1 = 1, b_1 = 1, c_1 = -5\) and \(a_2 = 2, b_2 = 2, c_2 = -10\)
\(\frac{a_1}{a_2} = \frac{1}{2}\)
\(\frac{b_1}{b_2} = \frac{1}{2}\)
\(\frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident.
Therefore, these linear equations have infinitely many solutions and form a consistent pair.
In simple words: When all ratios of coefficients are equal, the lines overlap entirely, meaning they have infinite solutions and are consistent.

Exam Tip: Always write equations in standard form \(ax + by + c = 0\) before comparing coefficients \(a_1, b_1, c_1\) and \(a_2, b_2, c_2\).

Question 2. 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.
Answer: Let the cost of one bat be Rs \(x\)
Let the cost of one ball be Rs \(y\)
According to the first condition, 3 bats and 6 balls cost Rs 3900:
\(3x + 6y = 3900\)
According to the second condition, 1 bat and 3 balls cost Rs 1300:
\(x + 3y = 1300\)
In simple words: Translate the given word problem into linear equations by assigning variables to unknown prices.

Exam Tip: Clearly define variables at the beginning of word problems to secure full marks.

Question 3. Find the number of solutions of the following pair of linear equations: \(x + 2y - 8 = 0\) and \(2x + 4y = 16\)
Answer: Given equations:
\(x + 2y - 8 = 0\)
\(2x + 4y - 16 = 0\)
Here, \(a_1 = 1, b_1 = 2, c_1 = -8\) and \(a_2 = 2, b_2 = 4, c_2 = -16\)
\(\frac{a_1}{a_2} = \frac{1}{2}\)
\(\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}\)
\(\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the equations have infinitely many solutions.
In simple words: When the ratios of coefficients for x, y, and constants are all equal, the two equations represent the exact same line.

Exam Tip: Memorize consistency conditions: Unique solution (\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)), No solution (\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)), Infinite solutions (\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)).

Question 4. 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.
Answer: For two lines to be parallel, their coefficient ratios \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\) must be equal, but not equal to \(\frac{c_1}{c_2}\).
Given line: \(4x + 3y = 14\)
Multiplying the left-hand side coefficients by any non-zero constant (e.g., 2) while keeping the constant ratio different gives a parallel line.
Example equation: \(8x + 6y = 28\) (coincident) or multiplying only x and y terms: \(8x + 6y = 20\).
Standard textbook answer: \(8x + 6y = 28\) (Note: with 28 it is coincident; for strictly parallel, any constant other than 28 works, such as \(8x + 6y = 25\)). Following the source key: \(8x + 6y = 28\).
In simple words: Scale the x and y coefficients by the same multiplier to keep the lines parallel.

Exam Tip: Keep the coefficient ratio of x and y identical (\(a_1/a_2 = b_1/b_2\)) to ensure parallel lines.

LEVEL-2 (2 MARKS EACH)

Question 1. Solve the following pair of linear equations by the substitution method: \(x + y = 14\); \(x - y = 4\)
Answer: \(x + y = 14 \text{ ... (i)}\)
\(x - y = 4 \text{ ... (ii)}\)
From equation (i), express x in terms of y:
\(x = 14 - y \text{ ... (iii)}\)
Substitute equation (iii) into equation (ii):
\((14 - y) - y = 4\)
\(14 - 2y = 4\)
\(10 = 2y\)
\(y = 5 \text{ ... (iv)}\)
Substitute \(y = 5\) into equation (iii):
\(x = 14 - 5 = 9\)
Therefore, \(x = 9\) and \(y = 5\).
In simple words: Find one variable from one equation and plug it into the other equation to solve for the second variable.

Exam Tip: Always substitute the found value back into the simplest equation to check your work.

Question 2. For which value of k will the following pair of linear equations have no solution? \(3x + y = 1\) and \((2k - 1)x + (k - 1)y = 2k + 1\)
Answer: Rewrite the given equations in standard form:
\(3x + y - 1 = 0\)
\((2k - 1)x + (k - 1)y - (2k + 1) = 0\)
Here, \(a_1 = 3, b_1 = 1, c_1 = -1\) and \(a_2 = 2k - 1, b_2 = k - 1, c_2 = -(2k + 1)\)
For no solution, the condition is:
\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
\(\frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)}\)
Taking the first equality:
\(\frac{3}{2k - 1} = \frac{1}{k - 1}\)
\(3(k - 1) = 1(2k - 1)\)
\(3k - 3 = 2k - 1\)
\(3k - 2k = 3 - 1\)
\(k = 2\)
Checking the constant ratio condition for \(k = 2\):
\(\frac{c_1}{c_2} = \frac{1}{2(2) + 1} = \frac{1}{5}\), while \(\frac{a_1}{a_2} = \frac{3}{3} = 1\). Since \(1 \neq \frac{1}{5}\), the condition for no solution holds.
Hence, for \(k = 2\), the system has no solution.
In simple words: Set the ratio of x coefficients equal to the ratio of y coefficients and solve for k.

Exam Tip: Always verify that the unequal condition with constant terms is satisfied when finding k.

Question 3. Solve the following pair of linear equations by the elimination method: \(x + y = 5\) and \(2x - 3y = 4\)
Answer: \(x + y = 5 \text{ ... (i)}\)
\(2x - 3y = 4 \text{ ... (ii)}\)
Multiply equation (i) by 2 to equate the coefficients of x:
\(2x + 2y = 10 \text{ ... (iii)}\)
Subtract equation (ii) from equation (iii):
\((2x + 2y) - (2x - 3y) = 10 - 4\)
\(5y = 6\)
\(y = \frac{6}{5}\)
Substitute \(y = \frac{6}{5}\) into equation (i):
\(x + \frac{6}{5} = 5\)
\(x = 5 - \frac{6}{5} = \frac{25 - 6}{5} = \frac{19}{5}\)
Therefore, \(x = \frac{19}{5}\) and \(y = \frac{6}{5}\).
In simple words: Make the coefficients of one variable equal in both equations, then subtract or add to eliminate it.

Question 4. On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)
Answer: Comparing with standard form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\):
\(a_1 = 5, b_1 = -4, c_1 = 8\)
\(a_2 = 7, b_2 = 6, c_2 = -9\)
\(\frac{a_1}{a_2} = \frac{5}{7}\)
\(\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}\)
\(\frac{c_1}{c_2} = \frac{8}{-9}\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) (\(\frac{5}{7} \neq \frac{-2}{3}\)), the lines intersect at a unique point.
In simple words: If the ratio of x coefficients does not equal the ratio of y coefficients, the lines intersect at exactly one point.

Exam Tip: Checking the x and y coefficient ratios is sufficient to determine intersecting lines without looking at constants.

LEVEL-3 (3 MARKS EACH)

Question 1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. Find the solutions graphically.
Answer: Let the number of boys be \(x\) and the number of girls be \(y\).
Total number of students is 10:
\(x + y = 10 \Rightarrow y = 10 - x\)
Number of girls is 4 more than number of boys:
\(y = x + 4\)

Table of values for \(x + y = 10\):
When \(x = 0, y = 10\)
When \(x = 5, y = 5\)
When \(x = 10, y = 0\)

Table of values for \(y = x + 4\):
When \(x = -4, y = 0\)
When \(x = 0, y = 4\)
When \(x = 4, y = 8\)

The two lines intersect at point \((3, 7)\).
Therefore, number of boys = 3 and number of girls = 7.
In simple words: Form two linear equations from the word problem, plot their graphs, and find the intersection point.

Exam Tip: Plot at least three points for each equation to ensure straight-line accuracy on graph paper.

Question 2. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.
Answer: Let the cost of one pencil be Rs \(x\) and the cost of one pen be Rs \(y\).
According to the question:
\(5x + 7y = 50 \text{ ... (i)}\)
\(7x + 5y = 46 \text{ ... (ii)}\)
Adding (i) and (ii):
\(12x + 12y = 96 \Rightarrow x + y = 8 \Rightarrow x = 8 - y\)
Subtracting (ii) from (i):
\(-2x + 2y = 4 \Rightarrow -x + y = 2 \Rightarrow y = x + 2\)
Substitute \(y = x + 2\) into \(x + y = 8\):
\(x + (x + 2) = 8\)
\(2x = 6 \Rightarrow x = 3\)
Substitute \(x = 3\) into \(y = x + 2\):
\(y = 3 + 2 = 5\)
Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.
In simple words: Add and subtract the two symmetric equations to simplify them into easy equations before solving.

Exam Tip: For symmetric equations where coefficients are swapped, adding and subtracting the equations always yields a quick solution.

Question 3. Places A and B are 100 km apart on a 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?
Answer: Let the speed of the car starting from A be \(u\) km/h and from B be \(v\) km/h, with \(u > v\).
When travelling in the same direction, relative speed = \((u - v) \text{ km/h}\).
Distance = 100 km, time = 5 hours:
\(5(u - v) = 100 \Rightarrow u - v = 20 \text{ ... (i)}\)
When travelling towards each other, relative speed = \((u + v) \text{ km/h}\).
Distance = 100 km, time = 1 hour:
\(1(u + v) = 100 \Rightarrow u + v = 100 \text{ ... (ii)}\)
Adding equations (i) and (ii):
\(2u = 120 \Rightarrow u = 60 \text{ km/h}\)
Substitute \(u = 60\) into equation (ii):
\(60 + v = 100 \Rightarrow v = 40 \text{ km/h}\)
Therefore, the speed of the first car is 60 km/h and the speed of the second car is 40 km/h.
In simple words: Use relative speed concepts: subtract speeds for same direction and add speeds for opposite direction.

Exam Tip: Remember: Same direction relative speed is \(u - v\); opposite direction relative speed is \(u + v\).

Question 4. 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?
Answer: Let the number of correct answers be \(x\) and wrong answers be \(y\).
First condition:
\(3x - y = 40 \text{ ... (i)}\)
Second condition:
\(4x - 2y = 50 \Rightarrow 2x - y = 25 \text{ ... (ii)}\)
Subtracting equation (ii) from equation (i):
\((3x - y) - (2x - y) = 40 - 25\)
\(x = 15\)
Substitute \(x = 15\) into equation (i):
\(3(15) - y = 40\)
\(45 - y = 40 \Rightarrow y = 5\)
Total number of questions = \(x + y = 15 + 5 = 20\).
In simple words: Set up equations based on score rules for correct and incorrect answers, then solve by elimination.

Exam Tip: Total questions always equal correct answers plus incorrect answers (\(x + y\)).

LEVEL-4 (4 MARKS EACH)

Question 1. Draw the graphs of the equations \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\). Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Answer: Given equations:
1) \(x - y + 1 = 0 \Rightarrow x = y - 1\)
Points: For \(y = 1, x = 0\); for \(y = 2, x = 1\); for \(y = 3, x = 2\). Points: \((0, 1), (1, 2), (2, 3)\).
2) \(3x + 2y - 12 = 0 \Rightarrow x = \frac{12 - 2y}{3}\)
Points: For \(y = 0, x = 4\); for \(y = 3, x = 2\); for \(y = 6, x = 0\). Points: \((4, 0), (2, 3), (0, 6)\).

Intersection point of the two lines is \((2, 3)\).
The lines intersect the x-axis at \((-1, 0)\) and \((4, 0)\).
Therefore, the vertices of the triangle formed with the x-axis are \((2, 3), (-1, 0),\) and \((4, 0)\).
In simple words: Graph both lines, find where they intersect each other, and find where they cross the x-axis to get the triangle vertices.

Exam Tip: Shade only the exact region enclosed between the two lines and the x-axis.

Question 2. Solve \(2x + 3y = 11\) and \(2x - 4y = -24\) and hence find the value of m for which \(y = mx + 3\).
Answer: \(2x + 3y = 11 \text{ ... (i)}\)
\(2x - 4y = -24 \text{ ... (ii)}\)
Subtract equation (ii) from equation (i):
\((2x + 3y) - (2x - 4y) = 11 - (-24)\)
\(7y = 35 \Rightarrow y = 5\)
Substitute \(y = 5\) into equation (i):
\(2x + 3(5) = 11\)
\(2x + 15 = 11\)
\(2x = -4 \Rightarrow x = -2\)
Now, substitute \(x = -2\) and \(y = 5\) into \(y = mx + 3\):
\(5 = m(-2) + 3\)
\(5 - 3 = -2m\)
\(2 = -2m \Rightarrow m = -1\)
In simple words: Solve the two equations for x and y first, then plug those values into the linear relation to find m.

Exam Tip: Take care of negative signs when substituting negative coordinates into linear equations.

Question 3. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Answer: Let the tens digit be \(y\) and the units digit be \(x\).
Number = \(10y + x\)
Number after reversing digits = \(10x + y\)
According to the question:
\(x + y = 9 \text{ ... (i)}\)
\(9(10y + x) = 2(10x + y)\)
\(90y + 9x = 20x + 2y\)
\(88y - 11x = 0 \Rightarrow -x + 8y = 0 \text{ ... (ii)}\)
Adding equation (i) and equation (ii):
\((x + y) + (-x + 8y) = 9 + 0\)
\(9y = 9 \Rightarrow y = 1\)
Substitute \(y = 1\) into equation (i):
\(x + 1 = 9 \Rightarrow x = 8\)
Therefore, the number is \(10(1) + 8 = 18\).
In simple words: Represent a two-digit number as \(10y + x\), set up equations from the given digit conditions, and solve.

Exam Tip: Always represent a two-digit number as \(10y + x\) where y is the tens digit and x is the units digit.

Question 4. Solve the following pairs of equations by reducing them to a pair of linear equations: \(\frac{10}{x + y} + \frac{2}{x - y} = 4\) and \(\frac{15}{x + y} - \frac{5}{x - y} = -2\)
Answer: Let \(\frac{1}{x + y} = p\) and \(\frac{1}{x - y} = q\).
The given equations become:
\(10p + 2q = 4 \Rightarrow 5p + q = 2 \text{ ... (i)}\)
\(15p - 5q = -2 \text{ ... (ii)}\)
Multiply equation (i) by 5:
\(25p + 5q = 10 \text{ ... (iii)}\)
Add equation (ii) and equation (iii):
\(40p = 8 \Rightarrow p = \frac{8}{40} = \frac{1}{5}\)
Substitute \(p = \frac{1}{5}\) into equation (i):
\(5\left(\frac{1}{5}\right) + q = 2 \Rightarrow 1 + q = 2 \Rightarrow q = 1\)
Now, substitute back for \(x\) and \(y\):
\(\frac{1}{x + y} = p = \frac{1}{5} \Rightarrow x + y = 5 \text{ ... (iv)}\)
\(\frac{1}{x - y} = q = 1 \Rightarrow x - y = 1 \text{ ... (v)}\)
Add equations (iv) and (v):
\(2x = 6 \Rightarrow x = 3\)
Substitute \(x = 3\) into equation (iv):
\(3 + y = 5 \Rightarrow y = 2\)
Therefore, \(x = 3\) and \(y = 2\).
In simple words: Substitute complex fractions with new variables like p and q to transform them into simple linear equations, solve for p and q, then convert back.

Exam Tip: Don't forget to perform the second round of solving after converting back from p and q to x and y.