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Class 7 Maths: Expressions using Letter-Numbers, 8 practice questions with answers

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These are 8 practice questions on Chapter 4, Expressions using Letter-Numbers, of the new NCERT Class 7 Maths textbook, Ganita Prakash, Part 1. Each has a worked answer and the textbook page it comes from. They cover writing a formula for a draining tank, values of expressions with negative numbers, simplifying a perimeter, a bracket with a minus sign, showing a pattern always holds, a border of tiles, a number trick explained with algebra, and a fare formula. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script, and no question retells a worked example, exercise or in-text question from the chapter. No practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Prakash, Grade 7, Part 1, Chapter 4, printed pages 81 to 105):

Writing and using expressions

Question 1. A water tank on a roof holds 500 litres. When the tap below it is left open, the level falls by 12 litres every hour. (a) Write a formula for the litres left in the tank after h hours. (b) How much is left after 5 hours? (c) After how many hours does the gauge show 260 litres?

Answer: (a) 500 − 12h; (b) 440 litres; (c) 20 hours. (a) In h hours the tank loses h × 12 litres, which is written 12h, so the litres left are 500 − 12h. (b) For h = 5: 500 − 12 × 5 = 500 − 60 = 440 litres. (c) The tank must have lost 500 − 260 = 240 litres. At 12 litres an hour, that takes 240 ÷ 12 = 20 hours. Check: 500 − 12 × 20 = 500 − 240 = 260. (Formulas from situations, pages 82 to 84; writing 12 × h as 12h, page 87.)

Question 2. In a cold-storage room, the temperature in °C, m minutes after the cooling is switched on, is given by 12 − 2m. (a) Find the temperature after 4, 6 and 10 minutes. (b) After how many minutes is it −6 °C? (c) Find the value of 3y − 2 when y = −4 and when y = −1.

Answer: (a) 4 °C, 0 °C and −8 °C; (b) 9 minutes; (c) −14 and −5. (a) Replace m by each number: 12 − 2 × 4 = 4; 12 − 2 × 6 = 0; 12 − 2 × 10 = 12 − 20 = −8. (b) To fall from 12 to −6 the temperature must drop 12 − (−6) = 18 degrees, at 2 degrees a minute: 18 ÷ 2 = 9 minutes. Check: 12 − 2 × 9 = −6. (c) 3y means 3 × y. For y = −4: 3 × (−4) − 2 = −12 − 2 = −14. For y = −1: 3 × (−1) − 2 = −3 − 2 = −5. (Values of algebraic expressions, pages 86 and 87; the formula is a made-up one for practice, not a real cooling rule.)

Question 3. A triangle has sides of length 2a + 5, a − 3 and 4a + 1 cm. (a) Write its perimeter in simplified form. (b) Check your answer by replacing a by 4 in both the unsimplified sum of the sides and your simplified expression.

Answer: (a) 7a + 3 cm; (b) both give 31 cm. (a) The perimeter is (2a + 5) + (a − 3) + (4a + 1). Terms may be added in whatever order is easiest, so group the like terms: 2a + a + 4a = 7a, and 5 − 3 + 1 = 3. The perimeter is 7a + 3 cm. (b) For a = 4 the sides are 13, 1 and 17 cm, and 13 + 1 + 17 = 31. The simplified form gives 7 × 4 + 3 = 31. The two match, as they must for equal expressions. (Simplifying by adding like terms, and checking with values, pages 88 to 90.)

Question 4. Two bamboo poles are cut for a stall. The first is 5k + 8 metres long and the second is 2k − 3 metres long, where k is a whole number bigger than 1. (a) Write an expression for how much longer the first pole is, and simplify it. (b) Check your answer with k = 4.

Answer: (a) 3k + 11 metres; (b) for k = 4, both ways give 23 m. (a) The difference is (5k + 8) − (2k − 3). A minus sign outside a bracket changes the sign of every term inside, so this is 5k + 8 − 2k + 3. Grouping like terms: 5k − 2k = 3k and 8 + 3 = 11, so the first pole is 3k + 11 metres longer. Taking away −3 means adding 3. (b) For k = 4 the poles are 5 × 4 + 8 = 28 m and 2 × 4 − 3 = 5 m, which differ by 23 m; and 3 × 4 + 11 = 23. (Brackets with a minus sign outside, pages 85 and 86; like terms, pages 88 to 90.)

Question 5. Take any two odd numbers that come one after the other, such as 7 and 9, or 21 and 23. (a) Add each pair. What do you notice? (b) Using 2n + 1 and 2n + 3 for two such odd numbers, show that this always happens.

Answer: (a) 16 and 44, both multiples of 4; (b) the sum is 4n + 4 = 4(n + 1), always a multiple of 4. (a) 7 + 9 = 16 = 4 × 4, and 21 + 23 = 44 = 4 × 11. (b) For any whole number n, 2n + 1 is odd, and the next odd number is 2 more, 2n + 3. Their sum is (2n + 1) + (2n + 3) = 2n + 2n + 1 + 3 = 4n + 4. By the distributive property, 4n + 4 = 4(n + 1), which is 4 times a whole number, so the sum is always a multiple of 4, whatever the pair. (Simplifying, pages 88 and 89; using algebra to show that a pattern always holds, page 99.)

Patterns and formulas

Question 6. A square swimming pool is n metres long on each side. A border of square tiles, each 1 metre on a side, is laid all the way round it, touching the pool. (a) How many tiles go round a pool of side 5 m? (b) Write an expression for the number of tiles for a pool of side n m, and simplify it. (c) A border uses 68 tiles. What is the side of the pool?

Answer: (a) 24; (b) 4n + 4; (c) 16 m. (a) Along each of the 4 sides there are 5 tiles, which makes 20, and 1 more tile fills each of the 4 corners: 20 + 4 = 24. (b) The same counting gives n + n + n + n for the sides and 4 for the corners: n + n + n + n + 4 = 4n + 4. (c) 4n + 4 = 68 means 4n = 64, so n = 16: a pool 16 m on a side. Check: 4 × 16 + 4 = 68. (Writing and simplifying an expression for a pattern, pages 84 and 88 to 90.)

Question 7. Here is a number trick. Think of a number. Add 5. Multiply by 2. Subtract 4. Halve the result. Finally subtract the number you first thought of. (a) Try it with 7 and with 30. (b) Use a letter-number to show that the answer is always the same.

Answer: (a) 3 both times; (b) the steps give n + 3, and n + 3 − n = 3. (a) With 7: 7 + 5 = 12, × 2 = 24, − 4 = 20, halved = 10, − 7 = 3. With 30: 35, 70, 66, 33, and 33 − 30 = 3. (b) Call the number n. Adding 5 gives n + 5. Multiplying by 2 gives 2(n + 5) = 2n + 10, by the distributive property. Subtracting 4 gives 2n + 6. Halving gives n + 3, since half of 2n is n and half of 6 is 3. Subtracting n leaves 3, for every starting number. (Distributive property, page 89; simplifying expressions, pages 88 to 90.)

Question 8. An auto-rickshaw charges ₹30 for the first kilometre and ₹15 for each kilometre after that. (a) Write an expression for the fare for a ride of d km, where d is at least 1, and simplify it. (b) What is the fare for 6 km? (c) How far can you ride for ₹180?

Answer: (a) 30 + 15(d − 1) = 15d + 15; (b) ₹105; (c) 11 km. (a) After the first kilometre there are d − 1 more, each costing ₹15, so the fare is 30 + 15(d − 1). By the distributive property, 15(d − 1) = 15d − 15, so the fare is 30 + 15d − 15 = 15d + 15. (b) For d = 6: 15 × 6 + 15 = 90 + 15 = ₹105. (c) 15d + 15 = 180 gives 15d = 165, so d = 11 km. Check: 30 + 15 × 10 = 180. (Formulas from situations, pages 82 to 84; distributive property and simplifying, pages 89 and 90.)

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Sources

All opened and read on 1 November 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 80 for this chapter.

The rules are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.

Practise with the worksheet: Letter-Number Expressions, Class 7 worksheet. Printable, in English, Hindi and Marathi, with an answer key.