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Pair of Linear Equations in Two Variables: 5 marks Questions (CBSE Class 10)

29 different 5 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)

Q355 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Basic 430/4/1

The difference between two numbers is 12. The greater number is 6 less than twice the smaller one.
(i) Representing the above situation, frame two linear equations in two variables.
(ii) Show that the equations have unique solution.
(iii) Solve the equations and hence find the numbers.

Show answer & solution

Answer: (i)

x - y = 12

x - 2 y = - 6

(ii) unique solution (iii) 30 and 18

(i) Let the greater number be x and the smaller be y: $x - y = 12$ and $x = 2 y - 6$ , i.e. $x - 2 y = - 6$ .
(ii) $\frac{a _{1}}{a _{2}} = \frac{1}{1} = 1$ , $\frac{b _{1}}{b _{2}} = \frac{- 1}{- 2} = \frac{1}{2}$ ; since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect and there is a unique solution.
(iii) Subtracting the second equation from the first: $y = 18$ .
Then $x = 12 + 18 = 30$ .
The numbers are 30 and 18.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q35 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Basic 430/4/1

Solve the following equations graphically :
$x + y = 7$ and $2 x - 5 y = 7$

Show answer & solution

Answer:

x = 6

, y = 1

For $x + y = 7$ : points (0, 7), (7, 0), (6, 1).
For $2 x - 5 y = 7$ : points (1, – 1), (6, 1), (– 4, – 3).
Plot both lines on the same axes.
The lines intersect at (6, 1).
So $x = 6$ , $y = 1$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q325 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/1/1

Determine graphically, the coordinates of vertices of a triangle whose equations are $2 x - 3 y + 6 = 0$ ; $2 x + 3 y - 18 = 0$ and $x = 0$ . Also, find the area of this triangle.

Show answer & solution

Answer: Vertices

(0, 2)

(0, 6)

(3, 4)

; area = 6 sq. units

$2 x - 3 y + 6 = 0$ gives $y = \frac{2 x + 6}{3}$ : points $(0, 2)$ , $(3, 4)$ , $(- 3, 0)$ .
$2 x + 3 y - 18 = 0$ gives $y = \frac{18 - 2 x}{3}$ : points $(0, 6)$ , $(3, 4)$ , $(9, 0)$ .
$x = 0$ is the $y$ -axis.
Plotting, the two lines meet at $(3, 4)$ and cut the $y$ -axis at $(0, 2)$ and $(0, 6)$ .
Vertices: $(0, 2)$ , $(0, 6)$ , $(3, 4)$ .
Base on the $y$ -axis $= 6 - 2 = 4$ ; height = distance of $(3, 4)$ from the $y$ -axis = 3.
Area $= \frac{1}{2} \times 4 \times 3 = 6$ sq. units.

Q355 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/1/2

Two sections, A and B, of class X contributed a total of ₹ 1500 for the Uttarakhand flood victims. The contribution from X-A was ₹ 100 less than that of X-B. Graphically, find the amounts contributed by both sections.

Show answer & solution

Answer: X-A: ₹ 700, X-B: ₹ 800

Let X-A contribute ₹ $x$ and X-B contribute ₹ $y$ .
$x + y = 1500$ and $x = y - 100$ , i.e. $x - y = - 100$ .
Points for $x + y = 1500$ : $(0, 1500)$ , $(1500, 0)$ , $(700, 800)$ .
Points for $x - y = - 100$ : $(0, 100)$ , $(400, 500)$ , $(700, 800)$ .
Plotting both lines, they intersect at $(700, 800)$ .
So X-A contributed ₹ 700 and X-B contributed ₹ 800.

Q345 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/1/3

Draw the graph of the pair of linear equations $x - y + 2 = 0$ and $4 x - y - 4 = 0$ . Calculate the area of the triangle formed by the lines so drawn and the $x$ -axis.

Show answer & solution

Answer: Lines meet at

(2, 4)

; vertices

(2, 4)

(- 2, 0)

(1, 0)

; area = 6 sq. units

$x - y + 2 = 0$ , i.e. $y = x + 2$ : points $(- 2, 0)$ , $(0, 2)$ , $(2, 4)$ .
$4 x - y - 4 = 0$ , i.e. $y = 4 x - 4$ : points $(1, 0)$ , $(0, - 4)$ , $(2, 4)$ .
Plotting, the lines intersect at $(2, 4)$ and meet the $x$ -axis at $(- 2, 0)$ and $(1, 0)$ .
Base on the $x$ -axis $= 1 - (- 2) = 3$ ; height $= 4$ .
Area $= \frac{1}{2} \times 3 \times 4 = 6$ sq. units.

Q325 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/2/1

Aarush bought 2 pencils and 3 chocolates for ₹ 11 and Tanish bought 1 pencil and 2 chocolates for ₹ 7 from the same shop. Represent this situation in the form of a pair of linear equations. Find the price of 1 pencil and 1 chocolate, graphically.

Show answer & solution

Answer: 2x + 3y = 11, x + 2y = 7; pencil ₹ 1, chocolate ₹ 3

Let the price of 1 pencil be ₹ x and of 1 chocolate be ₹ y.
Equations: $2 x + 3 y = 11$ and $x + 2 y = 7$ .
Points on $2 x + 3 y = 11$ : (1, 3), (4, 1), (-2, 5).
Points on $x + 2 y = 7$ : (1, 3), (3, 2), (7, 0).
Plot both lines on the same axes; they intersect at (1, 3).
So 1 pencil costs ₹ 1 and 1 chocolate costs ₹ 3.

Also asked in: 2026 Standard 30/2/2

Q345 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/2/3

Represent the following pair of linear equations graphically and hence comment on the condition of consistency of this pair :
x - 5y = 6; 2x - 10y = 12

Show answer & solution

Answer: The lines coincide; the pair is consistent (dependent) with infinitely many solutions.

Points on $x - 5 y = 6$ : (6, 0), (1, -1), (-4, -2).
Points on $2 x - 10 y = 12$ : (6, 0), (1, -1), (11, 1).
Plotting, both equations give the same line (coincident lines).
Also $\frac{a _{1}}{a _{2}} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 5}{- 10} = \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{6}{12} = \frac{1}{2}$ , all equal.
So the pair is consistent and dependent, with infinitely many solutions.

Q345 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/3/1

Solve the following system of equations graphically :
$x - 2 y = 3$ , $3 x - 8 y = 7$

Show answer & solution

Answer:

x = 5

y = 1

$x - 2 y = 3$ : points (3, 0), (5, 1), (1, – 1)
$3 x - 8 y = 7$ : points (5, 1), (– 3, – 2), (1, – 0.5)
Plot both lines on the same axes
The lines intersect at (5, 1)
Check: $5 - 2 = 3$ and $15 - 8 = 7$
Solution: $x = 5$ , $y = 1$

Also asked in: 2026 Standard 30/3/3

Q34 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/3/1

Five years ago, Adil was thrice as old as Bharat. Ten years later Adil shall be twice as old as Bharat. To know the present ages of Adil and Bharat :
(i) form the linear equations representing the above information.
(ii) show that the system of equations is consistent with unique solution.
(iii) find the present ages of Adil and Bharat.

Show answer & solution

Answer: (i)

x - 3 y = - 10

x - 2 y = 10

(ii)

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

, so unique solution (iii) Adil 50 years, Bharat 20 years

(i) Let present ages: Adil $= x$ , Bharat $= y$ years
$x - 5 = 3 (y - 5)$ gives $x - 3 y = - 10$
$x + 10 = 2 (y + 10)$ gives $x - 2 y = 10$
(ii) $\frac{a _{1}}{a _{2}} = \frac{1}{1} = 1$ , $\frac{b _{1}}{b _{2}} = \frac{- 3}{- 2} = \frac{3}{2}$
Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the system is consistent with a unique solution
(iii) Subtract: $(x - 2 y) - (x - 3 y) = 10 - (- 10)$ , so $y = 20$
$x = 10 + 2 \times 20 = 50$
Adil is 50 years, Bharat is 20 years

Also asked in: 2026 Standard 30/3/3

Q355 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/3/2

Solve the following system of equations graphically :
$2 x + 3 y = 5$ , $- 3 x + y = - 2$

Show answer & solution

Answer:

x = 1

y = 1

$2 x + 3 y = 5$ : points (1, 1), (– 2, 3), (4, – 1)
$- 3 x + y = - 2$ , i.e. $y = 3 x - 2$ : points (0, – 2), (1, 1), (2, 4)
Plot both lines on the same axes
The lines intersect at (1, 1)
Check: $2 + 3 = 5$ and $- 3 + 1 = - 2$
Solution: $x = 1$ , $y = 1$

Q35 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/3/2

The sum of the digits of a 2-digit number is 11. The number obtained by interchanging its digits exceeds the given number by 9. To know the number :
(i) form the linear equations representing the above situation.
(ii) verify that the equations have a unique solution.
(iii) solve the equations to get the given 2-digit number.

Show answer & solution

Answer: (i)

x + y = 11

y - x = 1

(ii)

\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}

, so unique solution (iii) 56

(i) Let tens digit $= x$ , units digit $= y$ ; number $= 10 x + y$
$x + y = 11$
$(10 y + x) - (10 x + y) = 9$ gives $9 y - 9 x = 9$ , i.e. $- x + y = 1$
(ii) $\frac{a _{1}}{a _{2}} = \frac{1}{- 1} = - 1$ , $\frac{b _{1}}{b _{2}} = \frac{1}{1} = 1$
Since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the equations have a unique solution
(iii) Adding: $2 y = 12$ , so $y = 6$ , $x = 5$
The number is 56 (check: $65 - 56 = 9$ )

Q325 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/5/1

Determine graphically whether the following pair of linear equations
$2 x + 3 y = 12$ and $x - y = 1$
has unique solution or infinitely many solutions.

Show answer & solution

Answer: Unique solution: the lines intersect at (3, 2).

For $2 x + 3 y = 12$ : points (0, 4), (3, 2), (6, 0).
For $x - y = 1$ : points (0, –1), (1, 0), (3, 2).
Plot both lines on the same axes.
The lines intersect at exactly one point, (3, 2).
Also $\frac{2}{1} \neq = \frac{3}{- 1}$ , so the lines are not parallel.
Hence the pair has a unique solution, $x = 3$ , $y = 2$ .

Q345 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/5/2

The sum of a 2-digit number and the number obtained by reversing the order of its digits, is 121. The two digits differ by 3.
(i) Represent the above information in the form of pair of linear equations.
(ii) Show that the equations have unique solution.
(iii) Solve the equations and find the number.

Show answer & solution

Answer: (i)

x + y = 11

x - y = 3

(tens digit

x

, units digit

y

) (ii)

\frac{1}{1} \neq = \frac{1}{- 1}

, so unique solution (iii) 74 (or 47 if the units digit is the larger one)

(i) Let the tens digit be $x$ and the units digit be $y$ . The number is $10 x + y$ and the reversed number is $10 y + x$ .
$(10 x + y) + (10 y + x) = 121$ gives $11 x + 11 y = 121$ , i.e. $x + y = 11$ .
The digits differ by 3: $x - y = 3$ (taking the tens digit as the larger).
(ii) $\frac{a _{1}}{a _{2}} = \frac{1}{1} = 1$ and $\frac{b _{1}}{b _{2}} = \frac{1}{- 1} = - 1$ ; since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the pair has a unique solution.
(iii) Adding: $2 x = 14$ , so $x = 7$ and $y = 4$ . The number is 74.
(If instead $y - x = 3$ , then $x = 4$ , $y = 7$ and the number is 47.)

Q345 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/5/3

Solve the following pair of equations using graphical method :
$3 x - 4 y + 3 = 0$ and $- 2 x + 5 y = 9$

Show answer & solution

Answer:

x = 3

y = 3

For $3 x - 4 y + 3 = 0$ , i.e. $y = \frac{3 x + 3}{4}$ : points (–1, 0), (3, 3), (7, 6).
For $- 2 x + 5 y = 9$ , i.e. $y = \frac{2 x + 9}{5}$ : points (–2, 1), (3, 3), (8, 5).
Plot both lines on the same axes.
The lines intersect at (3, 3).
Hence the solution is $x = 3$ , $y = 3$ .

Q325 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/6/1

Solve the following pair of linear equations by graphical method :
$2 x + y = 9$ and $x - 2 y = 2$

Show answer & solution

Answer:

x = 4

y = 1

(the lines intersect at (4, 1))

For $2 x + y = 9$ : points (0, 9), (3, 3), (4, 1).
For $x - 2 y = 2$ : points (2, 0), (0, $- 1$ ), (4, 1).
Plot both lines on the same axes.
The lines intersect at (4, 1).
Check: $2 (4) + 1 = 9$ and $4 - 2 (1) = 2$ . So $x = 4$ , $y = 1$ .

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q32 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/6/1

Nidhi received simple interest of ₹ 1,200 when invested ₹ $x$ at 6% p.a. and ₹ y at 5% p.a. for 1 year. Had she invested ₹ $x$ at 3% p.a. and ₹ y at 8% p.a. for that year, she would have received simple interest of ₹ 1,260. Find the values of $x$ and y.

Show answer & solution

Answer:

x

= ₹ 10,000, y = ₹ 12,000

$\frac{6 x}{100} + \frac{5 y}{100} = 1200$ , so $6 x + 5 y = 120000$ ... (1)
$\frac{3 x}{100} + \frac{8 y}{100} = 1260$ , so $3 x + 8 y = 126000$ ... (2)
(2) $\times$ 2: $6 x + 16 y = 252000$ ... (3)
(3) $-$ (1): $11 y = 132000$ , so $y = 12000$ .
From (1): $6 x = 120000 - 60000 = 60000$ , so $x = 10000$ .

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q325 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/1/1

Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of $8%$ per annum and $9%$ per annum, respectively. He received ₹ 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received ₹ 20 more as annual interest. How much money did he invest in each scheme ?

Show answer & solution

Answer: Scheme A: ₹ 12,000; Scheme B: ₹ 10,000

Let the amounts in schemes A and B be ₹ $x$ and ₹ $y$ .
$\frac{8 x}{100} + \frac{9 y}{100} = 1860$ , i.e. $8 x + 9 y = 186000$ ... (1)
After interchanging: $\frac{9 x}{100} + \frac{8 y}{100} = 1880$ , i.e. $9 x + 8 y = 188000$ ... (2)
Adding: $17 (x + y) = 374000$ , so $x + y = 22000$ ... (3)
Subtracting (1) from (2): $x - y = 2000$ ... (4)
From (3) and (4): $x = 12000$ , $y = 10000$ .
Check: $960 + 900 = 1860$ and $1080 + 800 = 1880$ .
He invested ₹ 12,000 in scheme A and ₹ 10,000 in scheme B.

Also asked in: 2025 Standard 30/1/2

Q335 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/1/3

A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of $24$ . If three times the number of red balls exceeds the number of blue balls by $20$ , find the number of red and blue balls.

Show answer & solution

Answer: Red balls

= 40

, blue balls

= 100

Let the number of red balls be $x$ and blue balls be $y$ .
$\frac{10}{100} x + \frac{20}{100} y = 24$ , i.e. $x + 2 y = 240$ ... (1)
$3 x - y = 20$ , i.e. $y = 3 x - 20$ ... (2)
Substitute (2) in (1): $x + 6 x - 40 = 240$ , so $7 x = 280$ and $x = 40$ .
$y = 3 (40) - 20 = 100$ .
Check: $4 + 20 = 24$ and $120 - 100 = 20$ .
Red balls $= 40$ , blue balls $= 100$ .

Q345 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/3/2

A man lent a part of his money at 10% p.a. and the rest at 15% p.a. His income at the end of the year is ₹ 1,900. If he had interchanged the rate of interest on the two sums, he would have earned ₹ 200 more. Find the amount lent in both cases.

Show answer & solution

Answer: ₹ 10,000 at 10% p.a. and ₹ 6,000 at 15% p.a.

Let ₹ $x$ be lent at 10% and ₹ $y$ at 15%.
$\frac{10 x}{100} + \frac{15 y}{100} = 1900$ , i.e. $2 x + 3 y = 38000$ ... (1)
After interchanging: $\frac{15 x}{100} + \frac{10 y}{100} = 2100$ , i.e. $3 x + 2 y = 42000$ ... (2)
Adding: $5 x + 5 y = 80000$ , so $x + y = 16000$ . Subtracting (1) from (2): $x - y = 4000$ .
$x = 10000$ , $y = 6000$ .
Check: $1000 + 900 = 1900$ and $1500 + 600 = 2100 = 1900 + 200$ .

Q335 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/3/3

The students of a class are made to stand equally in rows. If 3 students are extra in each row, there would be 1 row less. If 3 students are less in a row, there would be 2 more rows. Find the number of students in the class.

Show answer & solution

Answer: 36

Let there be $x$ rows with $y$ students in each row; total $= x y$ .
$(y + 3) (x - 1) = x y$ gives $3 x - y = 3$ ... (1)
$(y - 3) (x + 2) = x y$ gives $2 y - 3 x = 6$ ... (2)
Adding (1) and (2): $y = 9$ ; then $3 x = 12$ , $x = 4$ .
Number of students $= 4 \times 9 = 36$ .

Q355 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/3/1

Using graphical method, solve the following pair of equations :
$x + 2 y = 8$ and $3 x - 2 y = 12$

Show answer & solution

Answer:

x = 5

y = \frac{3}{2}

For $x + 2 y = 8$ : points (0, 4), (8, 0), (2, 3).
For $3 x - 2 y = 12$ : points (0, $- 6$ ), (4, 0), (2, $- 3$ ).
Plot both lines on the same axes; they intersect at (5, 1.5).
Check: $5 + 3 = 8$ and $15 - 3 = 12$ .
Solution: $x = 5$ , $y = \frac{3}{2}$ .

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q35 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/3/1

The sum of the digits of a 2-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Show answer & solution

Answer: 18

Let the tens digit be $x$ and the units digit be $y$ ; the number is $10 x + y$ .
$x + y = 9$ ... (1)
$9 (10 x + y) = 2 (10 y + x)$ gives $88 x = 11 y$ , i.e. $y = 8 x$ ... (2)
From (1) and (2), $9 x = 9$ , so $x = 1$ and $y = 8$ .
The number is 18. (Check: $9 \times 18 = 162 = 2 \times 81$ .)

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3, 2024 Basic 430/5/3

Q335 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/5/1

The ratio of monthly incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹ 2,000 per month, then find their monthly incomes.

Show answer & solution

Answer: ₹ 18,000 and ₹ 14,000

Let the incomes be ₹ 9x and ₹ 7x and the expenditures ₹ 4y and ₹ 3y.
$9 x - 4 y = 2000$ ... (1) and $7 x - 3 y = 2000$ ... (2).
(1) $\times$ 3: $27 x - 12 y = 6000$ ; (2) $\times$ 4: $28 x - 12 y = 8000$ .
Subtracting: $x = 2000$ .
Incomes: $9 \times 2000 =$ ₹ 18,000 and $7 \times 2000 =$ ₹ 14,000.
Check: $y = 4000$ , expenditures ₹ 16,000 and ₹ 12,000, savings ₹ 2,000 each.

Q325 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/5/2

A fraction becomes $\frac{3}{4}$ , if 2 is added to the numerator and 1 is added to the denominator. If 2 is subtracted from the numerator and 1 is subtracted from the denominator, it becomes $\frac{1}{2}$ . Find the fraction.

Show answer & solution

Answer:

\frac{7}{11}

Let the fraction be $\frac{x}{y}$ .
$\frac{x + 2}{y + 1} = \frac{3}{4}$ gives $4 x + 8 = 3 y + 3$ , i.e. $4 x - 3 y = - 5$ ... (1).
$\frac{x - 2}{y - 1} = \frac{1}{2}$ gives $2 x - 4 = y - 1$ , i.e. $y = 2 x - 3$ ... (2).
Substituting (2) in (1): $4 x - 6 x + 9 = - 5$ , so $x = 7$ and $y = 11$ .
The fraction is $\frac{7}{11}$ . Check: $\frac{9}{12} = \frac{3}{4}$ and $\frac{5}{10} = \frac{1}{2}$ .

Q325 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/2/1

Using graphical method, solve the following system of equations :
$3 x + y + 4 = 0$ and $3 x - y + 2 = 0$

Show answer & solution

Answer:

x = - 1

y = - 1

$3 x + y + 4 = 0$ gives $y = - 3 x - 4$ : points $(0, - 4)$ , $(- 1, - 1)$ , $(- 2, 2)$ .
$3 x - y + 2 = 0$ gives $y = 3 x + 2$ : points $(0, 2)$ , $(- 1, - 1)$ , $(1, 5)$ .
Plot both sets of points and draw the two lines.
The lines intersect at $(- 1, - 1)$ .
So the solution is $x = - 1$ , $y = - 1$ .

Also asked in: 2024 Standard 30/2/2

Q32 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/2/1

Tara 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 wrong answer, then Tara would have scored 50 marks. Assuming that Tara attempted all questions, find the total number of questions in the test.

Show answer & solution

Answer: 20 questions

Let the right answers be $x$ and wrong answers be $y$ .
$3 x - y = 40$ ... (1)
$4 x - 2 y = 50$ , i.e. $2 x - y = 25$ ... (2)
Subtracting (2) from (1): $x = 15$ . Then $y = 3 \times 15 - 40 = 5$ .
Total questions $= 15 + 5 = 20$ .

Also asked in: 2024 Standard 30/2/2

Q335 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/2/2

If the length of a rectangle is reduced by 5 cm and its breadth is increased by 2 cm, then the area of the rectangle is reduced by 80 cm $^{2}$ . However, if we increase the length by 10 cm and decrease the breadth by 5 cm, its area is increased by 50 cm $^{2}$ . Find the length and breadth of the rectangle.

Show answer & solution

Answer: Length = 40 cm, breadth = 30 cm

Let the length be $x$ cm and breadth $y$ cm.
$(x - 5) (y + 2) = x y - 80$ gives $2 x - 5 y - 10 = - 80$ , i.e. $2 x - 5 y = - 70$ ... (1)
$(x + 10) (y - 5) = x y + 50$ gives $- 5 x + 10 y - 50 = 50$ , i.e. $x - 2 y = - 20$ ... (2)
From (2), $x = 2 y - 20$ . In (1): $4 y - 40 - 5 y = - 70$ , so $y = 30$ .
Then $x = 60 - 20 = 40$ .
Check: $35 \times 32 = 1120 = 1200 - 80$ and $50 \times 25 = 1250 = 1200 + 50$ .

Q355 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/2/3

Using graphical method, solve the following system of equations :
$3 x - 2 y = 10$ and $5 x + 3 y = 4$

Show answer & solution

Answer:

x = 2

y = - 2

$3 x - 2 y = 10$ gives $y = \frac{3 x - 10}{2}$ : points $(0, - 5)$ , $(2, - 2)$ , $(4, 1)$ .
$5 x + 3 y = 4$ gives $y = \frac{4 - 5 x}{3}$ : points $(- 1, 3)$ , $(2, - 2)$ , $(5, - 7)$ .
Plot both sets of points and draw the two lines.
The lines intersect at $(2, - 2)$ .
So the solution is $x = 2$ , $y = - 2$ .

Q35 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/2/3

If three times the greater of two numbers is divided by the smaller one, we get 4 as the quotient and 3 as the remainder. Also, if seven times the smaller number is divided by greater one, we get 5 as the quotient and 1 as the remainder. Find the numbers.

Show answer & solution

Answer: 25 and 18

Let the greater number be $x$ and the smaller $y$ .
Dividend = divisor $\times$ quotient + remainder:
$3 x = 4 y + 3$ , i.e. $3 x - 4 y = 3$ ... (1)
$7 y = 5 x + 1$ , i.e. $5 x - 7 y = - 1$ ... (2)
(1) $\times 5$ $-$ (2) $\times 3$ : $- 20 y + 21 y = 15 + 3$ , so $y = 18$ .
From (1): $3 x = 72 + 3 = 75$ , so $x = 25$ .
The numbers are 25 and 18. (Check: $75 = 4 \times 18 + 3$ ; $126 = 5 \times 25 + 1$ .)

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