CBSE Class 10 · Maths
CBSE Class 10 Maths Pair of Linear Equations in Two Variables Competency Based Questions
Help your child excel in CBSE Class 10 Maths with these expert-verified Pair of Linear Equations in Two Variables competency based questions — covering MCQs, case studies, and assertion-reason types. Practising these questions builds the problem-solving skills and conceptual clarity needed to score high in board exams.
CBSE Class 10 Maths Pair of Linear Equations in Two Variables — Questions with Solutions
The pair of linear equations $2x=5y+6$ and $15y=6x-18$ represents two lines which are:
Solution
Option (C) is correct.
From the given equations:
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3},\quad \frac{b_{1}}{b_{2}}=\frac{-5}{-15}=\frac{1}{3},\quad \frac{c_{1}}{c_{2}}=\frac{-6}{-18}=\frac{1}{3}$
Since $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$, the lines are coincident.
From the given equations:
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3},\quad \frac{b_{1}}{b_{2}}=\frac{-5}{-15}=\frac{1}{3},\quad \frac{c_{1}}{c_{2}}=\frac{-6}{-18}=\frac{1}{3}$
Since $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$, the lines are coincident.
The point of intersection of the line represented by $3x-y=3$ and $y$-axis is given by
Solution
Option (A) is correct.
At the $y$-axis, $x=0$. Substituting:
$$3(0)-y=3 \Rightarrow y=-3$$
Hence the line cuts the $y$-axis at $(0,-3)$.
At the $y$-axis, $x=0$. Substituting:
$$3(0)-y=3 \Rightarrow y=-3$$
Hence the line cuts the $y$-axis at $(0,-3)$.
3 chairs and 1 table cost ₹900; whereas 5 chairs and 3 tables cost ₹2,100. If the cost of 1 chair is ₹$x$ and the cost of 1 table is ₹$y$, the situation is represented algebraically as
Solution
Option (C) is correct.
Cost of one chair $=x$, cost of one table $=y$
$$3x+y=₹900$$
$$5x+3y=₹2100$$
Cost of one chair $=x$, cost of one table $=y$
$$3x+y=₹900$$
$$5x+3y=₹2100$$
The lines representing the given pair of linear equations are non-intersecting. Which of the following statements is true?
Solution
Option (B) is correct.
Non-intersecting lines are parallel (no solution):
$$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
Non-intersecting lines are parallel (no solution):
$$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
The lines $k_1$, $k_2$ and $k_3$ represent three different equations. The solution of $k_1$ and $k_3$ is $x=3,\,y=0$; the solution of $k_2$ and $k_3$ is $x=4,\,y=1$.
Which of these is the equation of line $k_3$?
Which of these is the equation of line $k_3$?
Solution
Option (A) is correct.
Both points $(3,0)$ and $(4,1)$ lie on $k_3$. Verify: $3-0=3$ ✓ and $4-1=3$ ✓.
So the equation of $k_3$ is $x-y=3$.
Both points $(3,0)$ and $(4,1)$ lie on $k_3$. Verify: $3-0=3$ ✓ and $4-1=3$ ✓.
So the equation of $k_3$ is $x-y=3$.
Graphically, the pair of equations $6x-3y+10=0$ and $2x-y+9=0$ represents two lines which are
Solution
Option (D) is correct.
$\frac{a_1}{a_2}=\frac{6}{2}=3,\quad \frac{b_1}{b_2}=\frac{-3}{-1}=3,\quad \frac{c_1}{c_2}=\frac{10}{9}$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the lines are parallel.
$\frac{a_1}{a_2}=\frac{6}{2}=3,\quad \frac{b_1}{b_2}=\frac{-3}{-1}=3,\quad \frac{c_1}{c_2}=\frac{10}{9}$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the lines are parallel.
Two lines are given to be parallel. The equation of one of the lines is $3x-2y=5$. The equation of the second line can be
Solution
Option (C) is correct.
For parallel lines: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
Checking option (C): $\frac{3}{-12}=\frac{-2}{8}\neq\frac{5}{7}$, i.e. $-\frac{1}{4}=-\frac{1}{4}\neq\frac{5}{7}$ ✓
For parallel lines: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
Checking option (C): $\frac{3}{-12}=\frac{-2}{8}\neq\frac{5}{7}$, i.e. $-\frac{1}{4}=-\frac{1}{4}\neq\frac{5}{7}$ ✓
Shown below are graphs of the lines $y-2x=0$, $x+y=6$ and $px+qy=r$.
Which of these is the solution for the pair of equations $x+y=6$ and $px+qy=r$?
Which of these is the solution for the pair of equations $x+y=6$ and $px+qy=r$?
Solution
Option (B) is correct.
From the graph, the lines $x+y=6$ and $px+qy=r$ intersect at point $P$ where $x=4$ and $y=2$.
From the graph, the lines $x+y=6$ and $px+qy=r$ intersect at point $P$ where $x=4$ and $y=2$.
What is the value of $k$ such that the following pair of equations have infinitely many solutions?
$x-2y=3$ and $-3x+ky=-9$.
$x-2y=3$ and $-3x+ky=-9$.
Solution
Option (D) is correct.
For infinitely many solutions: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{1}{-3}=\frac{-2}{k}=\frac{-3}{9}$
$\Rightarrow \frac{1}{-3}=\frac{-2}{k}\Rightarrow k=6$
Verification: $\frac{-2}{6}=-\frac{1}{3}=\frac{-3}{9}$ ✓
For infinitely many solutions: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{1}{-3}=\frac{-2}{k}=\frac{-3}{9}$
$\Rightarrow \frac{1}{-3}=\frac{-2}{k}\Rightarrow k=6$
Verification: $\frac{-2}{6}=-\frac{1}{3}=\frac{-3}{9}$ ✓
The pair of linear equations $\frac{3x}{2}+\frac{5y}{3}=7$ and $9x+10y=14$ is
Solution
Option (B) is correct.
Rewriting the first equation: $\frac{3}{2}x+\frac{5}{3}y=7$, multiply by 6: $9x+10y=42$.
Now $\frac{a_1}{a_2}=\frac{9}{9}=1,\quad\frac{b_1}{b_2}=\frac{10}{10}=1,\quad\frac{c_1}{c_2}=\frac{42}{14}=3$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the system is inconsistent (parallel lines, no solution).
Rewriting the first equation: $\frac{3}{2}x+\frac{5}{3}y=7$, multiply by 6: $9x+10y=42$.
Now $\frac{a_1}{a_2}=\frac{9}{9}=1,\quad\frac{b_1}{b_2}=\frac{10}{10}=1,\quad\frac{c_1}{c_2}=\frac{42}{14}=3$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the system is inconsistent (parallel lines, no solution).
If the system of equations $3x+y=1$ and $(2k-1)x+(k-1)y=2k+1$ is inconsistent, then $k=$
Solution
Option (D) is correct.
For inconsistency: $\frac{3}{2k-1}=\frac{1}{k-1}\neq\frac{-1}{-(2k+1)}$
$\Rightarrow 3(k-1)=2k-1\Rightarrow 3k-3=2k-1\Rightarrow k=2$
For inconsistency: $\frac{3}{2k-1}=\frac{1}{k-1}\neq\frac{-1}{-(2k+1)}$
$\Rightarrow 3(k-1)=2k-1\Rightarrow 3k-3=2k-1\Rightarrow k=2$
If the lines given by $3x+2ky=2$ and $2x+5y+1=0$ are parallel, then the value of $k$ is
Solution
Option (C) is correct.
For parallel lines: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
$\frac{3}{2}=\frac{2k}{5}\Rightarrow 4k=15\Rightarrow k=\frac{15}{4}$
For parallel lines: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
$\frac{3}{2}=\frac{2k}{5}\Rightarrow 4k=15\Rightarrow k=\frac{15}{4}$
The values of $x$ and $y$ satisfying the two equations $32x+33y=34$ and $33x+32y=31$ respectively are
Solution
Option (A) is correct.
Subtracting equation (ii) from (i): $-x+y=3$, so $y=3+x$.
Substituting into (i): $32x+33(3+x)=34\Rightarrow 65x=-65\Rightarrow x=-1$
Then $y=3+(-1)=2$. Hence $x=-1,\,y=2$.
Subtracting equation (ii) from (i): $-x+y=3$, so $y=3+x$.
Substituting into (i): $32x+33(3+x)=34\Rightarrow 65x=-65\Rightarrow x=-1$
Then $y=3+(-1)=2$. Hence $x=-1,\,y=2$.
In $\triangle ABC$, $\angle A=x°$, $\angle B=(3x-2)°$, $\angle C=y°$. Also $\angle C-\angle B=9°$. The sum of the greatest and the smallest angles of this triangle is
Solution
Option (A) is correct.
Sum of angles: $x+(3x-2)+y=180°\Rightarrow 4x+y=182$ …(i)
$\angle C-\angle B=9°\Rightarrow y-(3x-2)=9°\Rightarrow y-3x=7°$ …(ii)
Subtracting (ii) from (i): $7x=175°\Rightarrow x=25°$; then $y=82°$
$\angle A=25°,\;\angle B=73°,\;\angle C=82°$
Sum of greatest and smallest $=82°+25°=107°$.
Sum of angles: $x+(3x-2)+y=180°\Rightarrow 4x+y=182$ …(i)
$\angle C-\angle B=9°\Rightarrow y-(3x-2)=9°\Rightarrow y-3x=7°$ …(ii)
Subtracting (ii) from (i): $7x=175°\Rightarrow x=25°$; then $y=82°$
$\angle A=25°,\;\angle B=73°,\;\angle C=82°$
Sum of greatest and smallest $=82°+25°=107°$.
The sum of the digits of a two-digit number is 9. If 27 is subtracted from the number, its digits are interchanged. Which of these is the product of the digits of the number?
Solution
Option (C) is correct.
Let tens digit $=x$, units digit $=y$. Then $x+y=9$ …(i)
$10x+y-27=10y+x\Rightarrow 9x-9y=27\Rightarrow x-y=3$ …(ii)
Solving: $x=6,\,y=3$. Product $=6\times3=18$.
Let tens digit $=x$, units digit $=y$. Then $x+y=9$ …(i)
$10x+y-27=10y+x\Rightarrow 9x-9y=27\Rightarrow x-y=3$ …(ii)
Solving: $x=6,\,y=3$. Product $=6\times3=18$.
What is the value of $q$ if $\frac{p}{2}+3q=6$ and $2p-2q=10$?
Solution
Option (A) is correct.
From equation (i): $p+6q=12$ …(i)
From equation (ii): $p-q=5$ …(ii)
Subtracting (ii) from (i): $7q=7\Rightarrow q=1$.
From equation (i): $p+6q=12$ …(i)
From equation (ii): $p-q=5$ …(ii)
Subtracting (ii) from (i): $7q=7\Rightarrow q=1$.
If a pair of linear equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ has a unique solution, then which of the following is true?
Solution
Option (B) is correct.
For a unique solution: $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\Rightarrow a_1b_2\neq a_2b_1$.
For a unique solution: $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\Rightarrow a_1b_2\neq a_2b_1$.
The pair of equations
$$3x-5y=7 \quad \text{and} \quad -6x+10y=7$$
have:
$$3x-5y=7 \quad \text{and} \quad -6x+10y=7$$
have:
Solution
Option (C) is correct.
$\frac{a_1}{a_2}=\frac{3}{-6}=-\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{-5}{10}=-\frac{1}{2},\quad\frac{c_1}{c_2}=\frac{7}{7}=1$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the lines are parallel and there is no solution.
$\frac{a_1}{a_2}=\frac{3}{-6}=-\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{-5}{10}=-\frac{1}{2},\quad\frac{c_1}{c_2}=\frac{7}{7}=1$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the lines are parallel and there is no solution.
If a pair of linear equations is consistent, then the lines will be
Solution
Option (D) is correct.
A consistent system has at least one solution. This occurs when:
– Lines intersect → unique solution
– Lines are coincident → infinitely many solutions
Parallel lines (no solution) are inconsistent.
A consistent system has at least one solution. This occurs when:
– Lines intersect → unique solution
– Lines are coincident → infinitely many solutions
Parallel lines (no solution) are inconsistent.
The pair of equations
$$x=0 \quad \text{and} \quad x=5$$
has:
$$x=0 \quad \text{and} \quad x=5$$
has:
Solution
Option (A) is correct.
$x=0$ is a vertical line at the $y$-axis; $x=5$ is a vertical line at $x=5$. These are parallel vertical lines that never intersect — so there is no solution.
$x=0$ is a vertical line at the $y$-axis; $x=5$ is a vertical line at $x=5$. These are parallel vertical lines that never intersect — so there is no solution.
A test consists of True / False questions. One mark is awarded for every correct answer, while $\frac{1}{4}$ mark is deducted for every wrong answer.
A student knew answers to some of the questions. The rest he attempted by guessing. He answered $120$ questions and got $90$ marks.
(i) If all questions attempted by guessing were wrong, how many questions did he answer correctly?
(ii) How many questions did he guess?
(iii) If all guessed answers were wrong and he answered $80$ correctly, how many marks did he get?
(iv) If all guessed answers were wrong, how many correct answers are needed to score $95$ marks?
A student knew answers to some of the questions. The rest he attempted by guessing. He answered $120$ questions and got $90$ marks.
(i) If all questions attempted by guessing were wrong, how many questions did he answer correctly?
(ii) How many questions did he guess?
(iii) If all guessed answers were wrong and he answered $80$ correctly, how many marks did he get?
(iv) If all guessed answers were wrong, how many correct answers are needed to score $95$ marks?
Answer
Let correct answers $= x$ and wrong answers $= y$.
Since total questions $= 120$:
$$x + y = 120 \quad \text{(1)}$$
Total marks $= 90$:
$$x – \frac{1}{4}y = 90 \quad \text{(2)}$$
(i) From (1): $y = 120 – x$. Substituting into (2):
$$x – \frac{1}{4}(120 – x) = 90 \Rightarrow \frac{5x}{4} = 120 \Rightarrow x = 96$$
The student answered 96 questions correctly.
(ii) Wrong answers (all guessed): $y = 120 – 96 = \mathbf{24}$
(iii) With 80 correct: wrong answers $= 120 – 80 = 40$
$$\text{Marks} = 80 – \frac{1}{4}\times40 = 80 – 10 = \mathbf{70}$$
(iv) Let correct answers $= x$; wrong $= 120 – x$:
$$x – \frac{1}{4}(120 – x) = 95 \Rightarrow \frac{5x}{4} = 125 \Rightarrow x = \mathbf{100}$$
Since total questions $= 120$:
$$x + y = 120 \quad \text{(1)}$$
Total marks $= 90$:
$$x – \frac{1}{4}y = 90 \quad \text{(2)}$$
(i) From (1): $y = 120 – x$. Substituting into (2):
$$x – \frac{1}{4}(120 – x) = 90 \Rightarrow \frac{5x}{4} = 120 \Rightarrow x = 96$$
The student answered 96 questions correctly.
(ii) Wrong answers (all guessed): $y = 120 – 96 = \mathbf{24}$
(iii) With 80 correct: wrong answers $= 120 – 80 = 40$
$$\text{Marks} = 80 – \frac{1}{4}\times40 = 80 – 10 = \mathbf{70}$$
(iv) Let correct answers $= x$; wrong $= 120 – x$:
$$x – \frac{1}{4}(120 – x) = 95 \Rightarrow \frac{5x}{4} = 125 \Rightarrow x = \mathbf{100}$$
Amit is planning to buy a house and the layout is given below. The areas of the two bedrooms and the kitchen together is $95\ \text{sq.m.}$
(i) Form the pair of linear equations from this situation.
(ii) Find the length of the outer boundary of the layout.
(iii) Find the area of each bedroom and the kitchen.
(iv) Find the area of the living room.
(v) Find the cost of laying tiles in the kitchen at ₹50 per sq.m.
(i) Form the pair of linear equations from this situation.
(ii) Find the length of the outer boundary of the layout.
(iii) Find the area of each bedroom and the kitchen.
(iv) Find the area of the living room.
(v) Find the cost of laying tiles in the kitchen at ₹50 per sq.m.
Answer
(i) Area of two bedrooms $= 10x$ sq.m., area of kitchen $= 5y$ sq.m.:
$$10x + 5y = 95 \Rightarrow 2x + y = 19 \quad \text{(1)}$$
Also: $x + 2 + y = 15 \Rightarrow x + y = 13 \quad \text{(2)}$
(ii) Outer boundary: $12 + 15 + 12 + 15 = \mathbf{54\ m}$
(iii) Subtracting (2) from (1): $x = 6\ m$; then $y = 7\ m$
Area of each bedroom $= 5 \times 6 = \mathbf{30\ sq.m.}$
Area of kitchen $= 5 \times 7 = \mathbf{35\ sq.m.}$
(iv) Total layout area $= 15 \times 7 = 105$ sq.m. (Note: two bedrooms = 60 sq.m.)
Area of living room $= 105 – 30 – 30 – 35 = 10$ sq.m. (or using layout dimensions)
Living room area $= 105 – 95 = \mathbf{10\ sq.m.}$
(v) Cost $= 50 \times 35 = \mathbf{₹1750}$
$$10x + 5y = 95 \Rightarrow 2x + y = 19 \quad \text{(1)}$$
Also: $x + 2 + y = 15 \Rightarrow x + y = 13 \quad \text{(2)}$
(ii) Outer boundary: $12 + 15 + 12 + 15 = \mathbf{54\ m}$
(iii) Subtracting (2) from (1): $x = 6\ m$; then $y = 7\ m$
Area of each bedroom $= 5 \times 6 = \mathbf{30\ sq.m.}$
Area of kitchen $= 5 \times 7 = \mathbf{35\ sq.m.}$
(iv) Total layout area $= 15 \times 7 = 105$ sq.m. (Note: two bedrooms = 60 sq.m.)
Area of living room $= 105 – 30 – 30 – 35 = 10$ sq.m. (or using layout dimensions)
Living room area $= 105 – 95 = \mathbf{10\ sq.m.}$
(v) Cost $= 50 \times 35 = \mathbf{₹1750}$
Dipesh bought 3 notebooks and 2 pens for ₹80. Lokesh bought 4 notebooks and 3 pens for ₹110. Let the cost of one notebook be ₹$x$ and one pen be ₹$y$.
(i) Which set of equations describes the given problem?
(A) $2x+3y=80$ and $3x+4y=110$
(B) $3x+2y=80$ and $4x+3y=110$
(C) $2x+3y=80$ and $4x+3y=110$
(D) $3x+2y=80$ and $3x+4y=110$
(ii) Whether the estimation of Ramesh (notebook ₹25, pen ₹2.50) and Amar (notebook ₹16, pen ₹16) is applicable for Lokesh?
(iii) What is the exact cost of the notebook?
(iv) What is the exact cost of the pen?
(v) What is the total cost if they purchase 15 notebooks and 12 pens?
(i) Which set of equations describes the given problem?
(A) $2x+3y=80$ and $3x+4y=110$
(B) $3x+2y=80$ and $4x+3y=110$
(C) $2x+3y=80$ and $4x+3y=110$
(D) $3x+2y=80$ and $3x+4y=110$
(ii) Whether the estimation of Ramesh (notebook ₹25, pen ₹2.50) and Amar (notebook ₹16, pen ₹16) is applicable for Lokesh?
(iii) What is the exact cost of the notebook?
(iv) What is the exact cost of the pen?
(v) What is the total cost if they purchase 15 notebooks and 12 pens?
Answer
(i) Option (B): $3x+2y=80$ and $4x+3y=110$
(ii) Option (D): Both estimations are wrong.
Ramesh’s estimate (₹25, ₹2.50): Lokesh would pay $4(25)+3(2.5)=₹107.50\neq₹110$
Amar’s estimate (₹16, ₹16): Lokesh would pay $4(16)+3(16)=₹112\neq₹110$
(iii) Option (B): ₹20
Solving: $3x+2y=80$ and $4x+3y=110\Rightarrow x=20$
(iv) Option (A): ₹10
Substituting $x=20$: $3(20)+2y=80\Rightarrow y=10$
(v) Option (C): ₹420
$15\times20 + 12\times10 = 300+120 = ₹420$
(ii) Option (D): Both estimations are wrong.
Ramesh’s estimate (₹25, ₹2.50): Lokesh would pay $4(25)+3(2.5)=₹107.50\neq₹110$
Amar’s estimate (₹16, ₹16): Lokesh would pay $4(16)+3(16)=₹112\neq₹110$
(iii) Option (B): ₹20
Solving: $3x+2y=80$ and $4x+3y=110\Rightarrow x=20$
(iv) Option (A): ₹10
Substituting $x=20$: $3(20)+2y=80\Rightarrow y=10$
(v) Option (C): ₹420
$15\times20 + 12\times10 = 300+120 = ₹420$
Mr. R. K. Agrawal owns an amusement park in Delhi. The ticket charge is ₹150 for children and ₹400 for adults. On one day, 480 tickets were sold and ₹1,34,500 was collected. Let children’s tickets $= x$ and adults’ tickets $= y$.
(i) Which system of equations models the problem?
(A) $x+y=480$ and $3x+8y=2690$
(B) $x+2y=480$ and $3x+4y=2690$
(C) $x+y=480$ and $3x+4y=2690$
(D) $x+2y=480$ and $3x+8y=2690$
(ii) How many children attended?
(iii) How many adults attended?
(iv) How much is collected if 300 children and 350 adults attended?
(v) On another day, total attendees were 750 and total collection was ₹2,12,500. Find the number of children and adults.
(i) Which system of equations models the problem?
(A) $x+y=480$ and $3x+8y=2690$
(B) $x+2y=480$ and $3x+4y=2690$
(C) $x+y=480$ and $3x+4y=2690$
(D) $x+2y=480$ and $3x+8y=2690$
(ii) How many children attended?
(iii) How many adults attended?
(iv) How much is collected if 300 children and 350 adults attended?
(v) On another day, total attendees were 750 and total collection was ₹2,12,500. Find the number of children and adults.
Answer
(i) Option (A): $x+y=480$ and $3x+8y=2690$
(Dividing $150x+400y=134500$ by 50 gives $3x+8y=2690$)
(ii) Option (C): 230 children
Solving: $x+y=480$ and $3x+8y=2690\Rightarrow x=230$
(iii) Option (A): 250 adults ($y=480-230=250$)
(iv) Option (D): ₹1,85,000
$150\times300+400\times350=45000+140000=₹185000$
(v) Option (B): (350, 400)
$x+y=750$ and $3x+8y=4250\Rightarrow x=350,\,y=400$
(Dividing $150x+400y=134500$ by 50 gives $3x+8y=2690$)
(ii) Option (C): 230 children
Solving: $x+y=480$ and $3x+8y=2690\Rightarrow x=230$
(iii) Option (A): 250 adults ($y=480-230=250$)
(iv) Option (D): ₹1,85,000
$150\times300+400\times350=45000+140000=₹185000$
(v) Option (B): (350, 400)
$x+y=750$ and $3x+8y=4250\Rightarrow x=350,\,y=400$
A part of monthly hostel charges is fixed, and the rest depends on the number of days food is taken. Anu takes food for 25 days and pays ₹4500. Bindu takes food for 30 days and pays ₹5200. Let fixed charges $= ₹x$ and cost of food per day $= ₹y$.
(i) Represent algebraically the situation:
(A) $x+25y=4500,\;x+30y=5200$
(B) $25x+y=4500,\;30x+y=5200$
(C) $x-25y=4500,\;x-30y=5200$
(D) $25x-y=4500,\;30x-y=5200$
(ii) The system has:
(A) No solution (B) Unique solution (C) Infinitely many solutions (D) None of these
(iii) The cost of food per day is:
(A) ₹120 (B) ₹130 (C) ₹140 (D) ₹1300
(iv) The fixed charges per month is:
(A) ₹1500 (B) ₹1200 (C) ₹1000 (D) ₹1300
(v) If Bindu takes food for 20 days, how much does she pay?
(A) ₹4000 (B) ₹3500 (C) ₹3600 (D) ₹3800
(i) Represent algebraically the situation:
(A) $x+25y=4500,\;x+30y=5200$
(B) $25x+y=4500,\;30x+y=5200$
(C) $x-25y=4500,\;x-30y=5200$
(D) $25x-y=4500,\;30x-y=5200$
(ii) The system has:
(A) No solution (B) Unique solution (C) Infinitely many solutions (D) None of these
(iii) The cost of food per day is:
(A) ₹120 (B) ₹130 (C) ₹140 (D) ₹1300
(iv) The fixed charges per month is:
(A) ₹1500 (B) ₹1200 (C) ₹1000 (D) ₹1300
(v) If Bindu takes food for 20 days, how much does she pay?
(A) ₹4000 (B) ₹3500 (C) ₹3600 (D) ₹3800
Answer
(i) Option (A): $x+25y=4500$ and $x+30y=5200$
(ii) Option (B): Unique solution (the coefficients of $x$ and $y$ are different, so lines intersect)
(iii) Option (C): ₹140
$(x+30y)-(x+25y)=5200-4500\Rightarrow 5y=700\Rightarrow y=140$
(iv) Option (C): ₹1000
$x+25(140)=4500\Rightarrow x=1000$
(v) Option (D): ₹3800
$x+20y=1000+20(140)=1000+2800=₹3800$
(ii) Option (B): Unique solution (the coefficients of $x$ and $y$ are different, so lines intersect)
(iii) Option (C): ₹140
$(x+30y)-(x+25y)=5200-4500\Rightarrow 5y=700\Rightarrow y=140$
(iv) Option (C): ₹1000
$x+25(140)=4500\Rightarrow x=1000$
(v) Option (D): ₹3800
$x+20y=1000+20(140)=1000+2800=₹3800$
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The pair of equations $x+2y-5=0$ and $-4x-8y+20=0$ have infinitely many solutions.
Reason (R): If $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, then the pair of equations has infinitely many solutions (coincident lines).
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The pair of equations $x+2y-5=0$ and $-4x-8y+20=0$ have infinitely many solutions.
Reason (R): If $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, then the pair of equations has infinitely many solutions (coincident lines).
Answer
Option (A) is correct.
$a_1=1,\,b_1=2,\,c_1=-5$ and $a_2=-4,\,b_2=-8,\,c_2=20$
$$\frac{1}{-4}=\frac{2}{-8}=\frac{-5}{20}=-\frac{1}{4}$$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, the lines are coincident → infinitely many solutions.
Both A and R are true and R correctly explains A.
$a_1=1,\,b_1=2,\,c_1=-5$ and $a_2=-4,\,b_2=-8,\,c_2=20$
$$\frac{1}{-4}=\frac{2}{-8}=\frac{-5}{20}=-\frac{1}{4}$$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, the lines are coincident → infinitely many solutions.
Both A and R are true and R correctly explains A.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The graph of $3x+2y=12$ and $5x-2y=4$ gives a pair of intersecting lines.
Reason (R): If $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$, the equations represent intersecting lines.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The graph of $3x+2y=12$ and $5x-2y=4$ gives a pair of intersecting lines.
Reason (R): If $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$, the equations represent intersecting lines.
Answer
Option (A) is correct.
$\frac{a_1}{a_2}=\frac{3}{5},\quad\frac{b_1}{b_2}=\frac{2}{-2}=-1$
Since $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$, the lines intersect at exactly one point.
Both A and R are true and R correctly explains A.
$\frac{a_1}{a_2}=\frac{3}{5},\quad\frac{b_1}{b_2}=\frac{2}{-2}=-1$
Since $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$, the lines intersect at exactly one point.
Both A and R are true and R correctly explains A.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the pair of lines are coincident, then the pair is consistent and has a unique solution.
Reason (R): If the pair of lines are parallel, then the pair has no solution and is called an inconsistent pair.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the pair of lines are coincident, then the pair is consistent and has a unique solution.
Reason (R): If the pair of lines are parallel, then the pair has no solution and is called an inconsistent pair.
Answer
Option (D) is correct.
Coincident lines have infinitely many solutions (not a unique solution), so Assertion (A) is false.
Parallel lines have no solution and are indeed inconsistent, so Reason (R) is true.
Coincident lines have infinitely many solutions (not a unique solution), so Assertion (A) is false.
Parallel lines have no solution and are indeed inconsistent, so Reason (R) is true.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The linear equations $x-2y-3=0$ and $3x+4y-20=0$ have exactly one solution.
Reason (R): The linear equations $2x+3y-9=0$ and $4x+6y-18=0$ have a unique solution.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The linear equations $x-2y-3=0$ and $3x+4y-20=0$ have exactly one solution.
Reason (R): The linear equations $2x+3y-9=0$ and $4x+6y-18=0$ have a unique solution.
Answer
Option (C) is correct.
For Assertion: $\frac{1}{3}\neq\frac{-2}{4}$, so lines intersect → one solution → Assertion is true.
For Reason: The second equation is $2\times$ the first, so they are coincident (infinitely many solutions, not unique) → Reason is false.
For Assertion: $\frac{1}{3}\neq\frac{-2}{4}$, so lines intersect → one solution → Assertion is true.
For Reason: The second equation is $2\times$ the first, so they are coincident (infinitely many solutions, not unique) → Reason is false.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The graphical representation of $x+2y=3$ and $2x+4y+7=0$ gives a pair of coincident lines.
Reason (R): If $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$, the equations represent intersecting lines.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The graphical representation of $x+2y=3$ and $2x+4y+7=0$ gives a pair of coincident lines.
Reason (R): If $\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$, the equations represent intersecting lines.
Answer
Option (D) is correct.
$\frac{a_1}{a_2}=\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{2}{4}=\frac{1}{2},\quad\frac{c_1}{c_2}=\frac{-3}{7}$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the lines are parallel (not coincident) → Assertion is false.
The Reason correctly states the condition for intersecting lines → Reason is true.
$\frac{a_1}{a_2}=\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{2}{4}=\frac{1}{2},\quad\frac{c_1}{c_2}=\frac{-3}{7}$
Since $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$, the lines are parallel (not coincident) → Assertion is false.
The Reason correctly states the condition for intersecting lines → Reason is true.
A bookstore gives books on rent. There is a fixed charge for the first two days and an additional charge per subsequent day. Amruta paid ₹22 for 6 days; Radhika paid ₹16 for 4 days. Let fixed charge $= ₹x$ and additional charge per day $= ₹y$.
(i) The algebraic representation for Radhika’s payment is:
(A) $x-4y=16$ (B) $x+4y=16$ (C) $x-2y=16$ (D) $x+2y=16$
(ii) The algebraic representation for Amruta’s payment is:
(A) $x-2y=11$ (B) $x-2y=22$ (C) $x+4y=22$ (D) $x-4y=11$
(iii) What are the fixed charges for a book?
(A) ₹9 (B) ₹13 (C) ₹10 (D) ₹15
(iv) What are the additional charges per subsequent day?
(A) ₹6 (B) ₹5 (C) ₹4 (D) ₹3
(v) What is the total amount paid by both if they each kept the book 2 more days?
(A) ₹35 (B) ₹52 (C) ₹50 (D) ₹58
(i) The algebraic representation for Radhika’s payment is:
(A) $x-4y=16$ (B) $x+4y=16$ (C) $x-2y=16$ (D) $x+2y=16$
(ii) The algebraic representation for Amruta’s payment is:
(A) $x-2y=11$ (B) $x-2y=22$ (C) $x+4y=22$ (D) $x-4y=11$
(iii) What are the fixed charges for a book?
(A) ₹9 (B) ₹13 (C) ₹10 (D) ₹15
(iv) What are the additional charges per subsequent day?
(A) ₹6 (B) ₹5 (C) ₹4 (D) ₹3
(v) What is the total amount paid by both if they each kept the book 2 more days?
(A) ₹35 (B) ₹52 (C) ₹50 (D) ₹58
Answer
(i) Option (D): $x+2y=16$
Radhika keeps for 4 days → pays fixed charge + 2 extra days: $x+2y=16$
(ii) Option (C): $x+4y=22$
Amruta keeps for 6 days → pays fixed charge + 4 extra days: $x+4y=22$
(iii) Option (C): ₹10
From $x+2y=16$ and $x+4y=22$: subtracting gives $2y=6\Rightarrow y=3$; then $x=10$
(iv) Option (D): ₹3 ($y=3$)
(v) Option (C): ₹50
Additional cost for 2 more days each: $2y=6$ per person.
Total $= 22+16+6+6=₹50$
Radhika keeps for 4 days → pays fixed charge + 2 extra days: $x+2y=16$
(ii) Option (C): $x+4y=22$
Amruta keeps for 6 days → pays fixed charge + 4 extra days: $x+4y=22$
(iii) Option (C): ₹10
From $x+2y=16$ and $x+4y=22$: subtracting gives $2y=6\Rightarrow y=3$; then $x=10$
(iv) Option (D): ₹3 ($y=3$)
(v) Option (C): ₹50
Additional cost for 2 more days each: $2y=6$ per person.
Total $= 22+16+6+6=₹50$
Auto charges in a city comprise a fixed charge plus a charge per km. Study the following:
| City | Distance (km) | Amount (₹) |
|:—:|:—:|:—:|
| City A | 10 | 75 |
| City A | 15 | 110 |
| City B | 8 | 91 |
| City B | 14 | 145 |
Situation 1 (City A): Fixed charge $= ₹x$, running charge $= ₹y$ per km.
(i) The pair of linear equations is:
(A) $x+10y=110,\;x+15y=75$ (B) $x+10y=75,\;x+15y=110$
(C) $10x+y=110,\;15x+y=75$ (D) $10x+y=75,\;15x+y=110$
(ii) A person travels 50 km. The amount to pay is:
(A) ₹155 (B) ₹255 (C) ₹355 (D) ₹455
Situation 2 (City B):
(iii) What will a person pay for 30 km?
(A) ₹185 (B) ₹289 (C) ₹275 (D) ₹305
(iv) The graph of lines representing City B conditions is:
(A)
(B)
(C)
(D)
| City | Distance (km) | Amount (₹) |
|:—:|:—:|:—:|
| City A | 10 | 75 |
| City A | 15 | 110 |
| City B | 8 | 91 |
| City B | 14 | 145 |
Situation 1 (City A): Fixed charge $= ₹x$, running charge $= ₹y$ per km.
(i) The pair of linear equations is:
(A) $x+10y=110,\;x+15y=75$ (B) $x+10y=75,\;x+15y=110$
(C) $10x+y=110,\;15x+y=75$ (D) $10x+y=75,\;15x+y=110$
(ii) A person travels 50 km. The amount to pay is:
(A) ₹155 (B) ₹255 (C) ₹355 (D) ₹455
Situation 2 (City B):
(iii) What will a person pay for 30 km?
(A) ₹185 (B) ₹289 (C) ₹275 (D) ₹305
(iv) The graph of lines representing City B conditions is:
(A)
(B)
(C)
(D)
Answer
(i) Option (B): $x+10y=75$ and $x+15y=110$
(ii) Option (C): ₹355
Solving: $x+10y=75$ and $x+15y=110$
$\Rightarrow 5y=35\Rightarrow y=7$; then $x=75-70=5$
For 50 km: $5+50\times7=5+350=₹355$
(iii) Option (B): ₹289
For City B: $x+8y=91$ and $x+14y=145$
$\Rightarrow 6y=54\Rightarrow y=9$; then $x=19$
For 30 km: $19+30\times9=19+270=₹289$
(iv) Option (C) — the graph showing two intersecting lines.
(ii) Option (C): ₹355
Solving: $x+10y=75$ and $x+15y=110$
$\Rightarrow 5y=35\Rightarrow y=7$; then $x=75-70=5$
For 50 km: $5+50\times7=5+350=₹355$
(iii) Option (B): ₹289
For City B: $x+8y=91$ and $x+14y=145$
$\Rightarrow 6y=54\Rightarrow y=9$; then $x=19$
For 30 km: $19+30\times9=19+270=₹289$
(iv) Option (C) — the graph showing two intersecting lines.
The total cost of Snowden Ice Cream Parlour is divided into fixed cost ($x$) and variable cost ($y$). The total cost was ₹27,500 after selling 150 units and ₹32,500 after selling 250 units.
(i) Frame the equations that represent the total cost in terms of fixed and variable costs.
(ii) Find the fixed cost. Show your work.
(iii) Another new flavour has fixed cost ₹10,000, variable cost ₹40 per unit, and selling price ₹60 per unit. Find the number of units at the break-even point.
(i) Frame the equations that represent the total cost in terms of fixed and variable costs.
(ii) Find the fixed cost. Show your work.
(iii) Another new flavour has fixed cost ₹10,000, variable cost ₹40 per unit, and selling price ₹60 per unit. Find the number of units at the break-even point.
Answer
(i) Forming the equations:
$$x + 150y = 27500 \quad \text{(1)}$$
$$x + 250y = 32500 \quad \text{(2)}$$
(ii) Subtracting (1) from (2):
$$100y = 5000 \Rightarrow y = 50$$
Substituting in (1): $x + 150(50) = 27500 \Rightarrow x = \mathbf{₹20{,}000}$
(iii) Let units sold $= n$ and revenue $= m$:
$$m = 60n \quad \text{(revenue)}$$
$$m = 10000 + 40n \quad \text{(cost)}$$
At break-even: $60n = 10000 + 40n \Rightarrow 20n = 10000 \Rightarrow n = \mathbf{500\ units}$
$$x + 150y = 27500 \quad \text{(1)}$$
$$x + 250y = 32500 \quad \text{(2)}$$
(ii) Subtracting (1) from (2):
$$100y = 5000 \Rightarrow y = 50$$
Substituting in (1): $x + 150(50) = 27500 \Rightarrow x = \mathbf{₹20{,}000}$
(iii) Let units sold $= n$ and revenue $= m$:
$$m = 60n \quad \text{(revenue)}$$
$$m = 10000 + 40n \quad \text{(cost)}$$
At break-even: $60n = 10000 + 40n \Rightarrow 20n = 10000 \Rightarrow n = \mathbf{500\ units}$
Two schools P and Q decided to award prizes for Hockey (₹$x$ per student) and Cricket (₹$y$ per student). School P awarded ₹9,500 to 5 (Hockey) and 4 (Cricket) students. School Q awarded ₹7,370 to 4 (Hockey) and 3 (Cricket) students.
(i) Represent the information algebraically.
(ii) What is the prize amount for Hockey?
(iii) What is the total prize amount if there are 2 students each from both games?
(i) Represent the information algebraically.
(ii) What is the prize amount for Hockey?
(iii) What is the total prize amount if there are 2 students each from both games?
Answer
(i) Algebraic representation:
$$5x + 4y = 9500 \quad \text{(i)}$$
$$4x + 3y = 7370 \quad \text{(ii)}$$
(ii) Using elimination (multiply (i) by 3, (ii) by 4):
$$15x + 12y = 28500$$
$$16x + 12y = 29480$$
Subtracting: $x = 980$; then $y = 1150$
Prize for Hockey: ₹980 per student
(iii) Total prize for 2 students each:
$$2(980 + 1150) = 2 \times 2130 = \mathbf{₹4260}$$
$$5x + 4y = 9500 \quad \text{(i)}$$
$$4x + 3y = 7370 \quad \text{(ii)}$$
(ii) Using elimination (multiply (i) by 3, (ii) by 4):
$$15x + 12y = 28500$$
$$16x + 12y = 29480$$
Subtracting: $x = 980$; then $y = 1150$
Prize for Hockey: ₹980 per student
(iii) Total prize for 2 students each:
$$2(980 + 1150) = 2 \times 2130 = \mathbf{₹4260}$$
A coaching institute runs two batches. In Batch I: 20 poor and 5 rich children; total fees ₹9,000/month. In Batch II: 5 poor and 25 rich children; total fees ₹26,000/month. Each poor child pays ₹$x$/month and each rich child pays ₹$y$/month.
(i) Represent the information in terms of $x$ and $y$.
(ii) Find the monthly fee paid by a poor child.
(iii) If Batch II has 10 poor and 20 rich children, what is the total monthly fees from Batch II?
(i) Represent the information in terms of $x$ and $y$.
(ii) Find the monthly fee paid by a poor child.
(iii) If Batch II has 10 poor and 20 rich children, what is the total monthly fees from Batch II?
Answer
(i) Algebraic representation:
$$20x + 5y = 9000 \quad \text{(Batch I)}$$
$$5x + 25y = 26000 \quad \text{(Batch II)}$$
(ii) Multiply (i) by 5: $100x + 25y = 45000$
Subtract Batch II equation: $95x = 19000 \Rightarrow x = \mathbf{₹200}$
(iii) Finding $y$: $20(200)+5y=9000\Rightarrow 5y=5000\Rightarrow y=₹1000$
Total fees for 10 poor + 20 rich:
$$10\times200 + 20\times1000 = 2000 + 20000 = \mathbf{₹22{,}000}$$
$$20x + 5y = 9000 \quad \text{(Batch I)}$$
$$5x + 25y = 26000 \quad \text{(Batch II)}$$
(ii) Multiply (i) by 5: $100x + 25y = 45000$
Subtract Batch II equation: $95x = 19000 \Rightarrow x = \mathbf{₹200}$
(iii) Finding $y$: $20(200)+5y=9000\Rightarrow 5y=5000\Rightarrow y=₹1000$
Total fees for 10 poor + 20 rich:
$$10\times200 + 20\times1000 = 2000 + 20000 = \mathbf{₹22{,}000}$$
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the pair of linear equations $3x+y=3$ and $6x+ky=8$ does not have a solution, then $k=2$.
Reason (R): If the pair of linear equations $x+y-4=0$ and $2x+ky=3$ does not have a solution, then $k=2$.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the pair of linear equations $3x+y=3$ and $6x+ky=8$ does not have a solution, then $k=2$.
Reason (R): If the pair of linear equations $x+y-4=0$ and $2x+ky=3$ does not have a solution, then $k=2$.
Answer
Option (B) is correct.
For Assertion: $\frac{3}{6}=\frac{1}{k}\neq\frac{-3}{-8}\Rightarrow k=2$ → Assertion is true.
For Reason: $\frac{1}{2}=\frac{1}{k}\neq\frac{4}{3}\Rightarrow k=2$ → Reason is also true.
Both are true, but R deals with a different pair of equations — it is not the correct explanation of A.
For Assertion: $\frac{3}{6}=\frac{1}{k}\neq\frac{-3}{-8}\Rightarrow k=2$ → Assertion is true.
For Reason: $\frac{1}{2}=\frac{1}{k}\neq\frac{4}{3}\Rightarrow k=2$ → Reason is also true.
Both are true, but R deals with a different pair of equations — it is not the correct explanation of A.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): For all real values of $c$, the pair $x-2y=8$ and $5x-10y=c$ has a unique solution.
Reason (R): Two lines are parallel. One line is $4x+3y=14$; the other is $12x+9y=5$.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): For all real values of $c$, the pair $x-2y=8$ and $5x-10y=c$ has a unique solution.
Reason (R): Two lines are parallel. One line is $4x+3y=14$; the other is $12x+9y=5$.
Answer
Option (D) is correct.
For Assertion: $\frac{a_1}{a_2}=\frac{1}{5}$ and $\frac{b_1}{b_2}=\frac{-2}{-10}=\frac{1}{5}$, so $\frac{a_1}{a_2}=\frac{b_1}{b_2}$ — the system can never have a unique solution → Assertion is false.
For Reason: $\frac{4}{12}=\frac{3}{9}=\frac{1}{3}\neq\frac{14}{5}$, so the lines are indeed parallel → Reason is true.
For Assertion: $\frac{a_1}{a_2}=\frac{1}{5}$ and $\frac{b_1}{b_2}=\frac{-2}{-10}=\frac{1}{5}$, so $\frac{a_1}{a_2}=\frac{b_1}{b_2}$ — the system can never have a unique solution → Assertion is false.
For Reason: $\frac{4}{12}=\frac{3}{9}=\frac{1}{3}\neq\frac{14}{5}$, so the lines are indeed parallel → Reason is true.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If $3x-y+8=0$ and $6x-ky=-16$ represent coincident lines, then $k=2$.
Reason (R): If the lines $3x+2ky=2$ and $2x+5y+1=0$ are parallel, then $k=15$.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If $3x-y+8=0$ and $6x-ky=-16$ represent coincident lines, then $k=2$.
Reason (R): If the lines $3x+2ky=2$ and $2x+5y+1=0$ are parallel, then $k=15$.
Answer
Option (C) is correct.
For Assertion (coincident lines): $\frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}\Rightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2$ → Assertion is true.
For Reason (parallel lines): $\frac{3}{2}=\frac{2k}{5}\Rightarrow4k=15\Rightarrow k=\frac{15}{4}\neq15$ → Reason is false.
For Assertion (coincident lines): $\frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}\Rightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2$ → Assertion is true.
For Reason (parallel lines): $\frac{3}{2}=\frac{2k}{5}\Rightarrow4k=15\Rightarrow k=\frac{15}{4}\neq15$ → Reason is false.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
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): The cost of 1 table is ₹500.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
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): The cost of 1 table is ₹500.
Answer
Option (A) is correct.
Let chair $= x$, table $= y$:
$$4x+3y=2100 \quad \text{(i)},\qquad 5x+2y=1750 \quad \text{(ii)}$$
$(i)\times2-(ii)\times3$: $8x+6y-15x-6y=4200-5250\Rightarrow-7x=-1050\Rightarrow x=150$
Substituting: $y=500$
Assertion is true (chair = ₹150), Reason is true (table = ₹500), and R correctly explains A.
Let chair $= x$, table $= y$:
$$4x+3y=2100 \quad \text{(i)},\qquad 5x+2y=1750 \quad \text{(ii)}$$
$(i)\times2-(ii)\times3$: $8x+6y-15x-6y=4200-5250\Rightarrow-7x=-1050\Rightarrow x=150$
Substituting: $y=500$
Assertion is true (chair = ₹150), Reason is true (table = ₹500), and R correctly explains A.
Assertion–Reason Question
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The solution of $x+y=5$ and $2x-3y=4$ is $x=\frac{19}{5}$ and $y=\frac{6}{5}$.
Reason (R): The value of $10(x+y)$ is 40.
Mark the correct choice:
(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 NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The solution of $x+y=5$ and $2x-3y=4$ is $x=\frac{19}{5}$ and $y=\frac{6}{5}$.
Reason (R): The value of $10(x+y)$ is 40.
Answer
Option (C) is correct.
From $x+y=5$: $2x+2y=10$. Subtracting $2x-3y=4$: $5y=6\Rightarrow y=\frac{6}{5}$; then $x=\frac{19}{5}$.
Assertion is true ($x=\frac{19}{5},\,y=\frac{6}{5}$).
But $10(x+y)=10\times5=50\neq40$ → Reason is false.
From $x+y=5$: $2x+2y=10$. Subtracting $2x-3y=4$: $5y=6\Rightarrow y=\frac{6}{5}$; then $x=\frac{19}{5}$.
Assertion is true ($x=\frac{19}{5},\,y=\frac{6}{5}$).
But $10(x+y)=10\times5=50\neq40$ → Reason is false.
Frequently Asked Questions
What is the chapter 'Pair of Linear Equations in Two Variables' about in CBSE Class 10 Maths? ⌄
This chapter teaches your child how to represent real-life situations using two linear equations with two unknowns and solve them using methods like substitution, elimination, and cross-multiplication. It also covers graphical interpretation — understanding when lines are intersecting, parallel, or coincident.
How many marks does this chapter carry in the CBSE Class 10 board exam? ⌄
This chapter is part of the 'Algebra' unit, which collectively carries 20 marks in the CBSE Class 10 Maths board exam. Questions from this chapter typically appear as MCQs, case-based questions, and short/long answer questions.
What are the most important topics in Pair of Linear Equations in Two Variables? ⌄
The key topics include graphical representation of linear equations, conditions for consistency/inconsistency (using a₁/a₂, b₁/b₂, c₁/c₂ ratios), algebraic methods of solution (substitution and elimination), and real-life word problems involving two variables.
What are common mistakes students make in this chapter? ⌄
Common mistakes include incorrectly identifying which ratio condition applies (unique solution, no solution, or infinitely many), sign errors while applying elimination, and misinterpreting word problems while forming equations. Practising a variety of question types helps avoid these pitfalls.
How does Angle Belearn help students master Pair of Linear Equations in Two Variables? ⌄
Angle Belearn provides CBSE-aligned competency-based questions with detailed step-by-step solutions, helping your child build both conceptual understanding and exam readiness. Our questions cover all formats — MCQs, case studies, and assertion-reason — exactly as expected in the board exam.
