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Pair of Linear Equations in Two Variables: 4 marks Questions (CBSE Class 10)

9 different 4 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)

Q364 marksCase StudyPair of Linear Equations in Two Variables CBSE 2026 · Basic 430/5/1

Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories.
To find the number of calories burned per minute on each machine, answer the following :
(i) Represent the above situation in terms of a pair of linear equations in two variables.
(ii) Show that the equations have unique solution.
(iii) (a) Solve both equations to find the values of the variables using elimination method.
OR (b) Solve both equations to find the values of the variables using substitution method.

Show answer & solution

Answer: (i)

15 x + 30 y = 435

and

30 x + 40 y = 690

, i.e.

x + 2 y = 29

and

3 x + 4 y = 69

, where x and y are calories burned per minute on the exercise bicycle and the double cross walker (ii)

\frac{a _{1}}{a _{2}} = \frac{1}{3} \neq = \frac{b _{1}}{b _{2}} = \frac{1}{2}

, so a unique solution (iii)

x = 11

y = 9

(by either method)

(i) Let x and y be the calories burned per minute on the exercise bicycle and the double cross walker.
$15 x + 30 y = 435 \Rightarrow x + 2 y = 29$ ; $30 x + 40 y = 690 \Rightarrow 3 x + 4 y = 69$
(ii) For $x + 2 y = 29$ and $3 x + 4 y = 69$ : $\frac{a _{1}}{a _{2}} = \frac{1}{3}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{4} = \frac{1}{2}$ ; these are unequal, so the pair has a unique solution.
(iii)(a) Elimination: multiply $x + 2 y = 29$ by 2: $2 x + 4 y = 58$ . Subtract from $3 x + 4 y = 69$ : $x = 11$ . Then $2 y = 29 - 11 = 18$ , $y = 9$ .
(iii)(b) Substitution: $x = 29 - 2 y$ ; $3 (29 - 2 y) + 4 y = 69 \Rightarrow 87 - 2 y = 69 \Rightarrow y = 9$ , $x = 29 - 18 = 11$ .
So 11 calories per minute on the exercise bicycle and 9 calories per minute on the double cross walker.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q364 marksCase StudyPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/4/1

A telecommunication company came up with two plans– plan A and plan B for its customers.
The plans are represented by linear equations where ‘ $t$ ’ represents the time (in minutes) bought and ‘C’ represents the cost. The equations are :
Plan A : $3 C = 20 t$
Plan B : $3 C = 10 t + 300$
Based on above information, answer the following questions :
(i) If you purchase plan B, how much initial amount you have to pay ? (1)
(ii) Charu purchased plan A. How many minutes she bought for ₹ 250 ? (1)
(iii) (a) At how many minutes, do both the plans charge the same amount? What is that amount? (2)
OR
(b) Which plan is better if you want to buy 60 minutes? Give reason for your answer. (2)

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Answer: (i) ₹ 100 (ii) 37.5 minutes (iii) (a) 30 minutes, ₹ 200 OR (b) Plan B, as it costs ₹ 300 against ₹ 400 for plan A

(i) Initial amount is the cost at $t = 0$ : $3 C = 300$ , so $C = ₹ 100$ .
(ii) $3 \times 250 = 20 t$ , so $t = \frac{750}{20} = 37.5$ minutes.
(iii) (a) $20 t = 10 t + 300$ , so $t = 30$ minutes; $C = \frac{20 \times 30}{3} = ₹ 200$ .
(iii) (b) For $t = 60$ : plan A, $C = \frac{1200}{3} = ₹ 400$ ; plan B, $C = \frac{600 + 300}{3} = ₹ 300$ . Plan B is cheaper, so it is better.

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q364 marksCase StudyPair of Linear Equations in Two Variables CBSE 2025 · Standard 30/2/1

A school is organizing a grand cultural event to show the talent of its students. To accommodate the guests, the school plans to rent chairs and tables from a local supplier. It finds that rent for each chair is ₹ 50 and for each table is ₹ 200. The school spends ₹ 30,000 for renting the chairs and tables. Also, the total number of items (chairs and tables) rented are 300.
If the school rents ‘ $x$ ’ chairs and ‘ $y$ ’ tables, answer the following questions :
(i) Write down the pair of linear equations representing the given information. (1)
(ii) (a) Find the number of chairs and number of tables rented by the school. (2)
OR (ii) (b) If the school wants to spend a maximum of ₹ 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent. (2)
(iii) What is maximum number of tables that can be rented in ₹ 30,000 if no chairs are rented ? (1)

Show answer & solution

Answer: (i)

x + y = 300

and

50 x + 200 y = 30000

(ii) (a) 200 chairs and 100 tables; OR (b) 220 chairs and 80 tables (iii) 150 tables

(i) Number of items: $x + y = 300$ ; cost: $50 x + 200 y = 30000$ , i.e. $x + 4 y = 600$
(ii) (a) Subtract: $(x + 4 y) - (x + y) = 600 - 300$ , so $3 y = 300$ , $y = 100$ ; $x = 200$
200 chairs and 100 tables
(ii) (b) $x + y = 300$ and $50 x + 200 y = 27000$ , i.e. $x + 4 y = 540$
$3 y = 240$ , so $y = 80$ , $x = 220$ : 220 chairs and 80 tables
(iii) Tables only: $200 y = 30000$ , so $y = 150$ tables

Also asked in: 2025 Standard 30/2/2, 2025 Standard 30/2/3

Q384 marksCase StudyPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/4/1

Deepankar bought 3 notebooks and 2 pens for ₹ 80 and his friend Suryansh bought 4 notebooks and 3 pens for ₹ 110 from the school bookshop.
Based on the above information, answer the following questions.
(i) If the price of one notebook be ₹ $x$ and the price of one pen be ₹ $y$ , write the given situation algebraically. (1)
(ii) (a) What is the price of one notebook ? (2)
OR (b) What is the price of one pen ? (2)
(iii) What is the total amount to be paid by Suryansh, if he purchases 6 notebooks and 3 pens ? (1)

Show answer & solution

Answer: (i)

3 x + 2 y = 80

4 x + 3 y = 110

(ii) (a) ₹20 OR (b) ₹10 (iii) ₹150

(i) $3 x + 2 y = 80$ and $4 x + 3 y = 110$ .
(ii) Multiply the first equation by 3 and the second by 2: $9 x + 6 y = 240$ , $8 x + 6 y = 220$ .
Subtracting, $x = 20$ . Then $3 (20) + 2 y = 80 \Rightarrow y = 10$ .
(a) One notebook costs ₹20. (b) One pen costs ₹10.
(iii) $6 \times 20 + 3 \times 10 = 120 + 30 = 150$ , so ₹150.

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q364 marksCase StudyPair of Linear Equations in Two Variables CBSE 2024 · Standard 30/5/1

Essel World is one of India’s largest amusement parks that offers a diverse range of thrilling rides, water attractions and entertainment options for visitors of all ages. The park is known for its iconic “Water Kingdom” section, making it a popular destination for family outings and fun-filled adventure. The ticket charges for the park are ₹ 150 per child and ₹ 250 per adult.
On a day, the cashier of the park found that 300 tickets were sold and an amount of ₹ 55,000 was collected.
Based on the above, answer the following questions :
(i) If the number of children visited be $x$ and the number of adults visited be $y$ , then write the given situation algebraically. (1)
(ii) (a) How many children visited the amusement park that day ? (2)
OR
(b) How many adults visited the amusement park that day ? (2)
(iii) How much amount will be collected if 250 children and 100 adults visit the amusement park ? (1)

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Answer: (i)

x + y = 300

150 x + 250 y = 55000

(ii) (a) 200 children OR (b) 100 adults (iii) ₹ 62,500

(i) $x + y = 300$ and $150 x + 250 y = 55000$ , i.e. $3 x + 5 y = 1100$ .
Substituting $x = 300 - y$ : $900 - 3 y + 5 y = 1100 \Rightarrow y = 100$ , $x = 200$ .
(ii) (a) 200 children visited.
(ii) (b) 100 adults visited.
(iii) $250 \times 150 + 100 \times 250 = 37500 + 25000 = ₹ 62, 500$ .

Also asked in: 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q374 marksCase StudyPair of Linear Equations in Two Variables CBSE 2023 · Basic 430/6/1

Lokesh, a production manager in Mumbai, hires a taxi everyday to go to his office. The taxi charges in Mumbai consists of a fixed charges together with the charges for the distance covered. His office is at a distance of 10 km from his home. For a distance of 10 km to his office, Lokesh paid ₹ 105. While coming back home, he took another route. He covered a distance of 15 km and the charges paid by him were ₹ 155.
Based on the above information, answer the following questions :
(i) What are the fixed charges ? (1)
(ii) What are the charges per km ? (1)
(iii) If fixed charges are ₹ 20 and charges per km are ₹ 10, then how much Lokesh have to pay for travelling a distance of 10 km ? (2)
OR (iii) Find the total amount paid by Lokesh for travelling 10 km from home to office and 25 km from office to home. [Fixed charges and charges per km are as in (i) & (ii). (2)

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Answer: (i) ₹ 5 (ii) ₹ 10 (iii) ₹ 120 OR (iii) ₹ 360

Let fixed charge = ₹ $x$ and charge per km = ₹ $y$ .
$x + 10 y = 105$ and $x + 15 y = 155$ . Subtracting, $5 y = 50$ , so $y = 10$ and $x = 5$ .
(i) Fixed charges = ₹ 5.
(ii) Charges per km = ₹ 10.
(iii) $20 + 10 \times 10 =$ ₹ 120.
OR (iii) Home to office: $5 + 10 \times 10 = 105$ ; office to home: $5 + 25 \times 10 = 255$ . Total = ₹ 360.

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q364 marksCase StudyPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/4/1

Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey ₹ $x$ per student and Cricket ₹ $y$ per student. School ‘P’ decided to award a total of ₹ 9,500 for the two games to 5 and 4 students respectively; while school ‘Q’ decided to award ₹ 7,370 for the two games to 4 and 3 students respectively.
Based on the above information, answer the following questions :
(i) Represent the following information algebraically (in terms of $x$ and $y$ ). (1)
(ii) (a) What is the prize amount for hockey ? (2)
OR (b) Prize amount on which game is more and by how much ? (2)
(iii) What will be the total prize amount if there are 2 students each from two games ? (1)

Show answer & solution

Answer: (i)

5 x + 4 y = 9500

and

4 x + 3 y = 7370

(ii) (a) ₹980 OR (b) Cricket, by ₹170 (iii) ₹4,260

(i) School P: $5 x + 4 y = 9500$ ; School Q: $4 x + 3 y = 7370$
(ii) (a) Multiply the first equation by 3 and the second by 4: $15 x + 12 y = 28500$ and $16 x + 12 y = 29480$
Subtract: $x = 980$ , so the prize for hockey is ₹980.
(ii) (b) $4 y = 9500 - 5 (980) = 4600 \Rightarrow y = 1150$
Cricket prize is more, by $1150 - 980 =$ ₹170.
(iii) Total $= 2 x + 2 y = 2 (980 + 1150) =$ ₹4,260

Also asked in: 2023 Standard 30/4/2

Q374 marksCase StudyPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/4/3

Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey ₹ $x$ per student and Cricket ₹ $y$ per student. School ‘P’ decided to award a total of ₹ 9,500 for the two games to 5 and 4 students respectively; while school ‘Q’ decided to award ₹ 7,370 for the two games to 4 and 3 students respectively.
Based on the given information, answer the following questions :
(i) Represent the following information algebraically (in terms of $x$ and $y$ ). (1)
(ii) (a) What is the prize amount for hockey ? (2)
OR (b) Prize amount on which game is more and by how much ? (2)
(iii) What will be the total prize amount if there are 2 students each from two games ? (1)

Show answer & solution

Answer: (i)

5 x + 4 y = 9500

and

4 x + 3 y = 7370

(ii) (a) ₹980 OR (b) Cricket, by ₹170 (iii) ₹4,260

(i) School P: $5 x + 4 y = 9500$ ; School Q: $4 x + 3 y = 7370$
(ii) (a) Multiply the first equation by 3 and the second by 4: $15 x + 12 y = 28500$ and $16 x + 12 y = 29480$
Subtract: $x = 980$ , so the prize for hockey is ₹980.
(ii) (b) $4 y = 9500 - 5 (980) = 4600 \Rightarrow y = 1150$
Cricket prize is more, by $1150 - 980 =$ ₹170.
(iii) Total $= 2 x + 2 y = 2 (980 + 1150) =$ ₹4,260

Q364 marksCase StudyPair of Linear Equations in Two Variables CBSE 2023 · Standard 30/6/1

A coaching institute of Mathematics conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, there are 20 poor and 5 rich children, whereas in batch II, there are 5 poor and 25 rich children. The total monthly collection of fees from batch I is ₹ 9000 and from batch II is ₹ 26,000. Assume that each poor child pays ₹ $x$ per month and each rich child pays ₹ y per month.
Based on the above information, answer the following questions :
(i) Represent the information given above in terms of $x$ and y. (1)
(ii) Find the monthly fee paid by a poor child. (2)
OR Find the difference in the monthly fee paid by a poor child and a rich child.
(iii) If there are 10 poor and 20 rich children in batch II, what is the total monthly collection of fees from batch II ? (1)

Show answer & solution

Answer: (i)

20 x + 5 y = 9000

5 x + 25 y = 26000

(ii) ₹ 200; OR: ₹ 800 (iii) ₹ 22,000

(i) Batch I: $20 x + 5 y = 9000$ , i.e. $4 x + y = 1800$ . Batch II: $5 x + 25 y = 26000$ , i.e. $x + 5 y = 5200$ .
(ii) From $4 x + y = 1800$ , $y = 1800 - 4 x$ .
$x + 5 (1800 - 4 x) = 5200$ , so $- 19 x = - 3800$ , x = 200.
Monthly fee of a poor child = ₹ 200 (and y = 1000).
OR: Rich child pays ₹ 1000, so the difference = 1000 – 200 = ₹ 800.
(iii) $10 \times 200 + 20 \times 1000 = 2000 + 20000 = ₹ 22, 000$

Also asked in: 2023 Standard 30/6/2, 2023 Standard 30/6/3

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