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Pair of Linear Equations in Two Variables: 3 marks Questions (CBSE Class 10)

40 different 3 marks questions on Pair of Linear Equations in Two Variables from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (68)2 marks (30)3 marks (40)4 marks (9)5 marks (29)

Q263 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/4/1

In a class test, Veer scored 6 more than twice as many marks as Kevin scored. If one of them had scored 4 more marks, their total score would have been 40. Find the marks obtained by Veer and Kevin.

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Answer: Veer: 26 marks, Kevin: 10 marks

Let Veer score x marks and Kevin score y marks.
$x = 2 y + 6$ ... (1)
If one of them scored 4 more, total = 40: $x + y + 4 = 40 \Rightarrow x + y = 36$ ... (2)
Substituting (1) in (2): $2 y + 6 + y = 36 \Rightarrow 3 y = 30 \Rightarrow y = 10$
$x = 2 \times 10 + 6 = 26$
Veer scored 26 and Kevin scored 10.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q26 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/4/1

Solve the linear equations $3 x + y = 14$ and $y = 2$ graphically.

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Answer:

x = 4

y = 2

For $3 x + y = 14$ : points (4, 2), (3, 5), (5, $- 1$ ).
For $y = 2$ : points (0, 2), (2, 2), (4, 2); a line parallel to the x-axis.
Plot both lines on the same axes.
They intersect at (4, 2).
Solution: $x = 4$ , $y = 2$ .

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q303 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/5/1

Solve the system of linear equations : $x = 4$ and $3 x - 2 y = 6$ graphically.

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Answer:

x = 4

y = 3

$x = 4$ is a vertical line through $(4, 0)$ .
For $3 x - 2 y = 6$ : $y = \frac{3 x - 6}{2}$ . Points: $(0, - 3)$ , $(2, 0)$ , $(4, 3)$ .
Plot both lines on the same axes.
The lines intersect at $(4, 3)$ .
Solution: $x = 4$ , $y = 3$

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/5/2

Use graphical method to solve the system of linear equations : $y = - 3$ and $x + 2 y = 4$ .

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Answer:

x = 10

y = - 3

$y = - 3$ is a horizontal line through $(0, - 3)$ .
For $x + 2 y = 4$ : $x = 4 - 2 y$ . Points: $(4, 0)$ , $(0, 2)$ , $(10, - 3)$ .
Plot both lines on the same axes.
The lines intersect at $(10, - 3)$ .
Solution: $x = 10$ , $y = - 3$

Q273 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/5/3

Use graphical method to solve the system of linear equations : $x = - 3$ and $5 x - 2 y = - 5$ .

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Answer:

x = - 3

y = - 5

$x = - 3$ is a vertical line through $(- 3, 0)$ .
For $5 x - 2 y = - 5$ : $y = \frac{5 x + 5}{2}$ . Points: $(- 1, 0)$ , $(1, 5)$ , $(- 3, - 5)$ .
Plot both lines on the same axes.
The lines intersect at $(- 3, - 5)$ .
Solution: $x = - 3$ , $y = - 5$

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/1/1

Solve the following system of equations graphically :
$x + 3 y = 6$ ; $2 x - 3 y = 12$

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Answer:

x = 6

y = 0

(the lines intersect at (6, 0))

For $x + 3 y = 6$ : points (0, 2), (3, 1), (6, 0).
For $2 x - 3 y = 12$ : points (0, $- 4$ ), (3, $- 2$ ), (6, 0).
Plot both lines on the same axes.
They intersect at (6, 0), so $x = 6$ , $y = 0$ .

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q28 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/1/1

$x$ and $y$ are complementary angles such that $x : y = 1 : 2$ . Express the given information as a system of linear equations in two variables and hence solve it.

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Answer:

x + y = 9 0^{\circ}

y = 2 x

;

x = 3 0^{\circ}

y = 6 0^{\circ}

Complementary: $x + y = 90$ ... (1)
$x : y = 1 : 2$ gives $y = 2 x$ ... (2)
Substitute (2) in (1): $3 x = 90$ , so $x = 3 0^{\circ}$ .
$y = 6 0^{\circ}$ .

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/2/1

Solve graphically the following pair of linear equations :
$2 x - y = 2$ and $4 x - y = 4$
Also, write the coordinates of the points where the lines represented by these equations cut the y-axis.

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Answer: Solution

x = 1

y = 0

; the lines cut the y-axis at (0,

- 2

) and (0,

- 4

$2 x - y = 2 \Rightarrow y = 2 x - 2$ : points (0, $- 2$ ), (1, 0), (2, 2).
$4 x - y = 4 \Rightarrow y = 4 x - 4$ : points (0, $- 4$ ), (1, 0), (2, 4).
Plot both lines; they intersect at (1, 0), so $x = 1$ , $y = 0$ .
Put $x = 0$ : the first line cuts the y-axis at (0, $- 2$ ) and the second at (0, $- 4$ ).

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q28 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/2/1

An academy offering cricket coaching bought 10 bats and 5 balls for ₹ 32,500. Later, the academy bought 2 bats and 8 balls for ₹ 10,000. If there is no change in the cost of the bat and of the ball, find the cost of 1 bat and 1 ball.

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Answer: Cost of 1 bat = ₹ 3,000; cost of 1 ball = ₹ 500

Let a bat cost ₹ x and a ball ₹ y.
$10 x + 5 y = 32500 \Rightarrow 2 x + y = 6500$ ...(1)
$2 x + 8 y = 10000 \Rightarrow x + 4 y = 5000$ ...(2)
From (1), $y = 6500 - 2 x$ . Substitute in (2): $x + 26000 - 8 x = 5000 \Rightarrow 7 x = 21000 \Rightarrow x = 3000$ .
$y = 6500 - 6000 = 500$ .
So a bat costs ₹ 3,000 and a ball costs ₹ 500.

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/3/1

A fraction becomes $\frac{1}{3}$ , when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ , when 8 is added to its denominator. Find the fraction.

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Answer:

\frac{5}{12}

Let the fraction be $\frac{x}{y}$ .
$\frac{x - 1}{y} = \frac{1}{3} \Rightarrow y = 3 x - 3$ ... (1)
$\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4 x = y + 8$ ... (2)
Substituting (1) in (2): $4 x = 3 x - 3 + 8$ , so $x = 5$ and $y = 12$ .
The fraction is $\frac{5}{12}$ .

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q28 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/3/1

Find the value of $k$ for which the following pair of linear equations will have infinitely many solutions :
$k x + 3 y - (k - 3) = 0$ and $12 x + k y - k = 0$
Hence, find any two solutions of the given pair of equations.

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Answer:

k = 6

; two solutions, e.g.

(0, 1)

and

(1, - 1)

For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$ .
$\frac{k}{12} = \frac{3}{k} \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6$ .
$\frac{3}{k} = \frac{k - 3}{k} \Rightarrow k - 3 = 3 \Rightarrow k = 6$ . So $k = 6$ .
The equations become $6 x + 3 y - 3 = 0$ and $12 x + 6 y - 6 = 0$ , both equivalent to $2 x + y = 1$ .
Two solutions: $x = 0, y = 1$ and $x = 1, y = - 1$ .

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q273 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/4/1

The monthly incomes of two persons are in the ratio 9 : 7 and their monthly expenditures are in the ratio 4 : 3. If each saved ₹ 5,000, express the given situation algebraically as a system of linear equations in two variables. Hence, find their respective monthly incomes.

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Answer:

9 x - 4 y = 5000

7 x - 3 y = 5000

; incomes ₹ 45,000 and ₹ 35,000

Let incomes be ₹ $9 x$ and ₹ $7 x$ , expenditures ₹ $4 y$ and ₹ $3 y$ .
$9 x - 4 y = 5000$ ... (1), $7 x - 3 y = 5000$ ... (2)
(1) $\times 3$ : $27 x - 12 y = 15000$ ; (2) $\times 4$ : $28 x - 12 y = 20000$
Subtract: $x = 5000$ ; then $4 y = 45000 - 5000 = 40000$ , $y = 10000$
Incomes: $9 \times 5000 =$ ₹ 45,000 and $7 \times 5000 =$ ₹ 35,000

Q293 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/4/2

The two angles of a right angled triangle other than $9 0^{\circ}$ are in the ratio $2 : 3$ . Express the given situation algebraically as a system of linear equations in two variables and hence solve it.

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Answer:

x + y = 90

3 x - 2 y = 0

; angles

3 6^{\circ}

and

5 4^{\circ}

Let the two acute angles be $x$ and $y$ (in degrees).
$x + y = 90$ ... (1) (angle sum $18 0^{\circ}$ with one angle $9 0^{\circ}$ )
$x : y = 2 : 3 \Rightarrow 3 x - 2 y = 0$ ... (2)
From (1), $y = 90 - x$ ; then $3 x - 180 + 2 x = 0 \Rightarrow x = 36$
$y = 54$ . The angles are $3 6^{\circ}$ and $5 4^{\circ}$ .

Q313 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/4/3

The perimeter of a rectangle is 70 cm. The length of the rectangle is 5 cm more than twice is breadth. Express the given situation as a system of linear equations in two variables and hence solve it.

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Answer:

l + b = 35

l - 2 b = 5

; length

= 25

cm, breadth

= 10

Let length $= l$ cm and breadth $= b$ cm.
$2 (l + b) = 70 \Rightarrow l + b = 35$ ... (1)
$l = 2 b + 5 \Rightarrow l - 2 b = 5$ ... (2)
Subtract (2) from (1): $3 b = 30 \Rightarrow b = 10$
$l = 25$ . Length 25 cm, breadth 10 cm.

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/5/1

Solve the following system of equations graphically :
$2 x - y - 2 = 0$
$- 4 x + y + 4 = 0$
Also, find the absolute difference between the ordinates of the points where given lines cut y $-$ axis.

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Answer:

x = 1

y = 0

; absolute difference of y-intercept ordinates

= 2

Line 1: $y = 2 x - 2$ ; points (0, $- 2$ ), (1, 0), (2, 2)
Line 2: $y = 4 x - 4$ ; points (0, $- 4$ ), (1, 0), (2, 4)
Plotting, the lines intersect at (1, 0), so $x = 1$ , $y = 0$ .
Lines cut the y-axis at (0, $- 2$ ) and (0, $- 4$ ).
Absolute difference of ordinates $= ∣ - 2 - (- 4) ∣ = 2$

Q303 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/5/2

Solve the following system of equations graphically :
$2 x + y = 5$ and $4 x - y = 7$ . Hence, write the coordinates of the points where given lines meet y-axis.

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Answer:

x = 2

y = 1

; the lines meet the y-axis at (0, 5) and (0,

- 7

)

Line 1: $y = 5 - 2 x$ ; points (0, 5), (1, 3), (2, 1)
Line 2: $y = 4 x - 7$ ; points (0, $- 7$ ), (1, $- 3$ ), (2, 1)
Plotting, the lines intersect at (2, 1), so $x = 2$ , $y = 1$ .
Line 1 meets the y-axis at (0, 5); line 2 meets it at (0, $- 7$ ).

Q313 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/5/3

Solve the following system of equations graphically :
$2 x + 3 y = 6$
$x + y - 1 = 0$
Also, find the sum of ordinates of the points where given lines meet y axis.

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Answer:

x = - 3

y = 4

; sum of ordinates

= 3

Line 1: $2 x + 3 y = 6$ ; points (0, 2), (3, 0), ( $- 3$ , 4)
Line 2: $x + y = 1$ ; points (0, 1), (1, 0), ( $- 3$ , 4)
Plotting, the lines intersect at ( $- 3$ , 4), so $x = - 3$ , $y = 4$ .
The lines meet the y-axis at (0, 2) and (0, 1).
Sum of ordinates $= 2 + 1 = 3$

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/6/1

Check whether the following pair of equations is consistent or not. If consistent, solve graphically
$x + 3 y = 6$
$3 y - 2 x = - 12$

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Answer: Consistent (unique solution);

x = 6

y = 0

Write as $x + 3 y - 6 = 0$ and $- 2 x + 3 y + 12 = 0$ .
$\frac{a _{1}}{a _{2}} = - \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = 1$ ; since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the pair is consistent with a unique solution.
Line 1 points: $(0, 2)$ , $(3, 1)$ , $(6, 0)$
Line 2 points: $(0, - 4)$ , $(3, - 2)$ , $(6, 0)$
Plotting, the lines intersect at $(6, 0)$ , so $x = 6$ , $y = 0$ .

Q303 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/6/2

Check whether the following system of equations is consistent or not.
If consistent, solve graphically
$x - 2 y + 4 = 0, 2 x - y - 4 = 0$

Show answer & solution

Answer: Consistent (unique solution);

x = 4

y = 4

$\frac{a _{1}}{a _{2}} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 2}{- 1} = 2$ ; since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the system is consistent with a unique solution.
Line 1 ( $x - 2 y + 4 = 0$ ) points: $(- 4, 0)$ , $(0, 2)$ , $(4, 4)$
Line 2 ( $2 x - y - 4 = 0$ ) points: $(2, 0)$ , $(0, - 4)$ , $(4, 4)$
Plotting, the lines intersect at $(4, 4)$ , so $x = 4$ , $y = 4$ .

Q263 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/6/3

Check whether the given system of equations is consistent or not. If consistent, solve graphically.
$x - 2 y = 0$
$2 x + y = 0$

Show answer & solution

Answer: Consistent (unique solution);

x = 0

y = 0

$\frac{a _{1}}{a _{2}} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 2}{1} = - 2$ ; since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the system is consistent with a unique solution.
Line 1 ( $x = 2 y$ ) points: $(0, 0)$ , $(2, 1)$ , $(4, 2)$
Line 2 ( $y = - 2 x$ ) points: $(0, 0)$ , $(1, - 2)$ , $(- 1, 2)$
Both lines pass through the origin, so they intersect at $(0, 0)$ : $x = 0$ , $y = 0$ .

Q313 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/1/2

The greater of two supplementary angles exceeds the smaller by $1 8^{\circ}$ . Find measures of these two angles.

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Answer:

9 9^{\circ}

and

8 1^{\circ}

Let the angles be $x$ and $y$ with $x > y$ .
$x + y = 18 0^{\circ}$ and $x - y = 1 8^{\circ}$ .
Adding: $2 x = 19 8^{\circ}$ , so $x = 9 9^{\circ}$ and $y = 8 1^{\circ}$ .

Also asked in: 2024 Basic 430/1/3

Q293 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/1/1

Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now ?

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Answer: Rashmi is 42 years and Nazma is 16 years old.

Let the present ages of Rashmi and Nazma be $x$ and $y$ years.
Three years ago: $x - 3 = 3 (y - 3)$ , so $x - 3 y = - 6$ ... (1)
Ten years later: $x + 10 = 2 (y + 10)$ , so $x - 2 y = 10$ ... (2)
(2) $-$ (1): $y = 16$ .
From (2): $x = 10 + 32 = 42$ .
Rashmi is 42 years old and Nazma is 16 years old.

Q273 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/1/2

In a chemistry lab, there is some quantity of 50% acid solution and some quantity of 25% acid solution. How much of each should be mixed to make 10 litres of 40% acid solution ?

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Answer: 6 litres of 50% solution and 4 litres of 25% solution

Let $x$ litres of 50% solution and $y$ litres of 25% solution be mixed.
$x + y = 10$ ... (1)
Acid content: $0.5 x + 0.25 y = 0.4 \times 10 = 4$ , i.e. $2 x + y = 16$ ... (2)
(2) $-$ (1): $x = 6$ ; then $y = 4$ .
So 6 litres of 50% solution and 4 litres of 25% solution are needed.

Q263 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/1/3

The sum of the digits of a 2-digit number is 14. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.

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Answer: 68

Let the tens digit be $x$ and the units digit be $y$ ; the number is $10 x + y$ .
$x + y = 14$ ... (1)
$(10 y + x) - (10 x + y) = 18$ , so $9 y - 9 x = 18$ , i.e. $y - x = 2$ ... (2)
Adding (1) and (2): $2 y = 16$ , $y = 8$ ; then $x = 6$ .
The number is 68.

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/3/1

Solve the following system of linear equations graphically :
$x - y + 1 = 0$
$x + y = 5$

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Answer:

x = 2

y = 3

Line 1: $y = x + 1$ ; points (0, 1), (1, 2), (2, 3).
Line 2: $y = 5 - x$ ; points (0, 5), (5, 0), (2, 3).
Plot both lines on the same axes.
They intersect at (2, 3).
So the solution is $x = 2$ , $y = 3$ .

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/4/1

Rehana went to a bank to withdraw ₹ 2,000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Rehana got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 did she receive.

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Answer: 10 notes of ₹ 50 and 15 notes of ₹ 100

Let the number of ₹ 50 notes be $x$ and of ₹ 100 notes be $y$ .
$x + y = 25$ and $50 x + 100 y = 2000$ , i.e. $x + 2 y = 40$ .
Subtracting: $y = 15$ , so $x = 10$ .
She received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Q303 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/4/3

A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay ₹ 1,500 as hostel charges while another student B, who takes food for 26 days, pays ₹ 1,800. Find the fixed charges and the cost of food.

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Answer: Fixed charges ₹ 500; cost of food ₹ 50 per day

Let the fixed charge be ₹ $x$ and the cost of food per day be ₹ $y$ .
$x + 20 y = 1500$ and $x + 26 y = 1800$ .
Subtracting: $6 y = 300$ , so $y = 50$ .
$x = 1500 - 1000 = 500$ .
Fixed charges ₹ 500 and cost of food ₹ 50 per day.

Q283 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/4/1

A lending library has a fixed charge for first three days and an additional charge for each day thereafter. Rittik paid ₹ 27 for a book kept for 7 days and Manmohan paid ₹ 21 for a book kept for 5 days. Find the fixed charges and the charge for each extra day.

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Answer: Fixed charge ₹ 15; charge for each extra day ₹ 3.

Let the fixed charge be ₹ $x$ and the charge per extra day be ₹ $y$ .
7 days = 3 days + 4 extra days: $x + 4 y = 27$ .
5 days = 3 days + 2 extra days: $x + 2 y = 21$ .
Subtracting: $2 y = 6$ , so $y = 3$ .
Then $x = 21 - 6 = 15$ .

Also asked in: 2023 Basic 430/4/3

Q28 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/4/1

Find the values of ‘a’ and ‘b’ for which the system of linear equations
$3 x + 4 y = 12$ , $(a + b) x + 2 (a - b) y = 24$
has infinite number of solutions.

Show answer & solution

Answer:

a = 5

b = 1

For infinitely many solutions: $\frac{3}{a + b} = \frac{4}{2 ( a - b )} = \frac{12}{24} = \frac{1}{2}$ .
$\frac{3}{a + b} = \frac{1}{2} \Rightarrow a + b = 6$ .
$\frac{4}{2 ( a - b )} = \frac{1}{2} \Rightarrow a - b = 4$ .
Adding: $2 a = 10$ , so $a = 5$ and $b = 1$ .

Also asked in: 2023 Basic 430/4/3

Q313 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/4/2

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction ?

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Answer:

\frac{3}{5}

Let the fraction be $\frac{x}{y}$ .
$\frac{x + 1}{y - 1} = 1 \Rightarrow x - y = - 2$ ... (1)
$\frac{x}{y + 1} = \frac{1}{2} \Rightarrow 2 x - y = 1$ ... (2)
Subtracting (1) from (2): $x = 3$ ; then $y = 5$ .
The fraction is $\frac{3}{5}$ .

Q31 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/4/2

For which value of ‘k’ will the following pair of linear equations have no solution ?
$3 x + y = 1$
$(2 k - 1) x + (k - 1) y = 2 k + 1$

Show answer & solution

Answer:

k = 2

For no solution: $\frac{3}{2 k - 1} = \frac{1}{k - 1} \neq = \frac{1}{2 k + 1}$ .
$3 (k - 1) = 2 k - 1 \Rightarrow k = 2$ .
Check: $\frac{1}{k - 1} = 1$ and $\frac{1}{2 k + 1} = \frac{1}{5}$ , which are unequal.
So $k = 2$ .

Q263 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/5/1

Sabina went to a bank ATM to withdraw ₹ 2,000. She received ₹ 50 and ₹ 100 notes only. If Sabina got 25 notes in all, how many notes of ₹ 50 and ₹ 100 did she receive ?

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Answer: 10 notes of ₹ 50 and 15 notes of ₹ 100

Let the number of ₹ 50 notes be $x$ and ₹ 100 notes be $y$ .
$x + y = 25$ ... (1)
$50 x + 100 y = 2000$ , i.e. $x + 2 y = 40$ ... (2)
(2) $-$ (1): $y = 15$ ; then $x = 10$ .
She received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q26 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/5/1

Five years ago, Amit was thrice as old as Baljeet. Ten years hence, Amit shall be twice as old as Baljeet. What are their present ages ?

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Answer: Amit: 50 years, Baljeet: 20 years

Let present ages of Amit and Baljeet be $x$ and $y$ years.
$x - 5 = 3 (y - 5)$ , so $x = 3 y - 10$ ... (1)
$x + 10 = 2 (y + 10)$ , so $x = 2 y + 10$ ... (2)
From (1) and (2): $3 y - 10 = 2 y + 10$ , so $y = 20$ and $x = 50$ .
Amit is 50 years and Baljeet is 20 years old.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q263 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/1/1

Half of the difference between two numbers is 2. The sum of the greater number and twice the smaller number is 13. Find the numbers.

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Answer: 7 and 3

Let the greater number be x and the smaller y.
$\frac{x - y}{2} = 2$ , so $x - y = 4$ ... (1)
$x + 2 y = 13$ ... (2)
Subtract (1) from (2): $3 y = 9$ , $y = 3$ .
$x = 4 + 3 = 7$ .
The numbers are 7 and 3.

Also asked in: 2023 Standard 30/1/2

Q313 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/1/3

Jaya scored 40 marks in a test getting 3 marks for each correct answer and losing 1 mark for each incorrect answer. Had 4 marks being awarded for each correct answer and 2 marks were deducted for each incorrect answer then Jaya again would have scored 40 marks. How many questions were there in the Test ?

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Answer: 40 questions

Let x be the number of correct answers and y the number of incorrect answers.
$3 x - y = 40$ ... (1)
$4 x - 2 y = 40$ , i.e. $2 x - y = 20$ ... (2)
Subtract (2) from (1): x = 20; then y = 3(20) – 40 = 20.
Total questions = x + y = 40.

Q303 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/2/1

If the system of linear equations
$2 x + 3 y = 7$ and $2 a x + (a + b) y = 28$
have infinite number of solutions, then find the values of ‘a’ and ‘b’.

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Answer:

a = 4

b = 8

For infinitely many solutions $\frac{2}{2 a} = \frac{3}{a + b} = \frac{7}{28} = \frac{1}{4}$
$\frac{1}{a} = \frac{1}{4}$ gives $a = 4$
$\frac{3}{a + b} = \frac{1}{4}$ gives $a + b = 12$ , so $b = 8$

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q30 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/2/1

If $217 x + 131 y = 913$ and
$131 x + 217 y = 827$ ,
then solve the equations for the values of x and y.

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Answer:

x = 3

y = 2

Adding: $348 x + 348 y = 1740$ , so $x + y = 5$
Subtracting: $86 x - 86 y = 86$ , so $x - y = 1$
Solving: $x = 3$ , $y = 2$

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q293 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/5/1

Two people are 16 km apart on a straight road. They start walking at the same time. If they walk towards each other with different speeds, they will meet in 2 hours. Had they walked in the same direction with same speeds as before, they would have met in 8 hours. Find their walking speeds.

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Answer: 5 km/h and 3 km/h

Let the speeds be x km/h and y km/h, x > y.
Towards each other: $2 (x + y) = 16$ , so $x + y = 8$ .
Same direction: $8 (x - y) = 16$ , so $x - y = 2$ .
Adding: 2x = 10, x = 5; then y = 3.
Speeds are 5 km/h and 3 km/h.

Q273 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/5/2

A 2-digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

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Answer: 42

Let the tens digit be x and the units digit be y; the number is 10x + y.
$10 x + y = 7 (x + y)$ gives $3 x = 6 y$ , i.e. $x = 2 y$ .
$(10 x + y) - (10 y + x) = 18$ gives $x - y = 2$ .
So $2 y - y = 2$ , y = 2, x = 4.
The number is 42.

Q273 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/5/3

A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator. It becomes $\frac{1}{4}$ when 8 is added to the denominator. Find the fraction.

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Answer:

\frac{5}{12}

Let the fraction be $\frac{x}{y}$ .
$\frac{x - 1}{y} = \frac{1}{3}$ gives $3 x - y = 3$ ... (1)
$\frac{x}{y + 8} = \frac{1}{4}$ gives $4 x - y = 8$ ... (2)
(2) – (1): x = 5; then y = 3(5) – 3 = 12.
The fraction is $\frac{5}{12}$ .

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