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Class 8 Maths: Number Play, 8 practice questions with answers

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These are 8 practice questions on Chapter 5, Number Play, of the new NCERT Class 8 Maths textbook, Ganita Prakash, Part 1. Each has a worked answer and the textbook page it comes from. They cover checking a claim with parity, checking a multiplication with digital roots, why some numbers are always multiples of 11, the digital root of a sum, which sums of consecutive numbers are multiples of their count, and using a remainder to find a seat. 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 none of the numbers is taken from the chapter, but 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 8, Part 1, Chapter 5, printed pages 112 to 134):

Parity and digital roots

Question 1. The houses on a lane are numbered 49, 52, 63 and 78. Raju puts ‘+’ and ‘–’ signs between these numbers, in this order, and says one of his answers is 137. Without trying the signs, show that he must have made a mistake.

Answer: every such expression is even, but 137 is odd. All the expressions 49 ± 52 ± 63 ± 78 have the same parity, because changing one sign changes the value by twice that number, an even amount. One of them is 49 + 52 + 63 + 78 = 242, which is even, so all of them are even. 137 is odd, so it cannot be one of Raju's answers. (Parity of a ± b ± c ± d, pages 113 to 115.)

Question 2. Ravi works out 52 × 41 and writes 2142. Use digital roots to check his answer, then find the correct product.

Answer: 2142 is wrong; 52 × 41 = 2132. The digital root of 52 is 5 + 2 = 7, and of 41 is 4 + 1 = 5. The digital root of the product must be the digital root of 7 × 5 = 35, which is 3 + 5 = 8. The digital root of 2142 is 2 + 1 + 4 + 2 = 9, not 8, so Ravi's answer is wrong. Working it out: 52 × 41 = 2080 + 52 = 2132, and 2 + 1 + 3 + 2 = 8, which matches. (Digital roots and remainders on division by 9, pages 124, 125 and 130; checking calculations with digital roots, page 131.)

Question 3. Without adding the numbers, find the digital root of 2011 + 3101. Then check by adding.

Answer: 9. The digital root of 2011 is 2 + 0 + 1 + 1 = 4, and of 3101 is 3 + 1 + 0 + 1 = 5. A digital root is the remainder on division by 9 (with 9 for a multiple of 9), and remainders add, so the sum has digital root 4 + 5 = 9. Check: 2011 + 3101 = 5112, and 5 + 1 + 1 + 2 = 9. So 5112 is also a multiple of 9. (Digital roots, page 130; remainders on division by 9, pages 124 and 125; if a divides M and N, it divides M + N, page 119.)

Question 4. Without dividing, find what remainder 3051 × 4127 leaves when it is divided by 9.

Answer: 0. The digits of 3051 add to 3 + 0 + 5 + 1 = 9, so 3051 is a multiple of 9. Every multiple of a multiple of 9 is again a multiple of 9, so 3051 × 4127 is divisible by 9 and the remainder is 0. Check: 3051 × 4127 = 12591477 = 9 × 1399053. (Divisibility by 9, page 125; multiples of a number divisible by k, page 120.)

Elevens, consecutive numbers and a seat

Question 5. Why is every 4-digit number of the form abba, which reads the same forwards and backwards, a multiple of 11? Explain with the alternating sum of the digits, then check your reason with 5885.

Answer: for abba the alternating sum is always 0. Starting from the units digit, the alternating sum of abba is a − b + b − a = 0, and a number whose alternating sum is 0 (or a multiple of 11) is a multiple of 11. So every number of the form abba is a multiple of 11. Check: 5 − 8 + 8 − 5 = 0, and 5885 = 11 × 535. (The shortcut for divisibility by 11, pages 127 to 129.)

Question 6. Show that a two-digit number plus the number made by reversing its digits is always a multiple of 11. Give one example.

Answer: (10a + b) + (10b + a) = 11(a + b). If the tens digit is a and the units digit is b, the number is 10a + b and its reverse is 10b + a. Their sum is 11a + 11b = 11(a + b), which always has 11 as a factor. Example: 52 + 25 = 77 = 11 × 7, and 5 + 2 = 7. (Writing a number by place value, page 123; multiples, page 119.)

Question 7. Is the sum of any 5 consecutive numbers always a multiple of 5? Is the sum of any 4 consecutive numbers ever a multiple of 4? Check with 81, 82, 83, 84 and 85.

Answer: yes for 5 (it is 5 times the middle number); never for 4 (it always leaves remainder 2). Five consecutive numbers are n, n + 1, n + 2, n + 3, n + 4, with sum 5n + 10 = 5(n + 2), so the sum is 5 times the middle number. Four consecutive numbers have sum 4n + 6 = 4(n + 1) + 2, which always leaves remainder 2 on division by 4. Check: 81 + 82 + 83 + 84 + 85 = 415 = 5 × 83, and 81 + 82 + 83 + 84 = 330 = 4 × 82 + 2. (Sums of consecutive numbers, page 112; writing numbers with remainders in algebra, page 121.)

Question 8. In a large hall, seats are numbered 1, 2, 3, … across rows of 9 seats each, so seats 1 to 9 make row 1, the next 9 seats make row 2, and so on. Using digit sums, find which row seat 4768 is in, and its position in that row.

Answer: row 530, seat 7 in the row. The digits of 4768 add to 25, and 2 + 5 = 7, so 4768 leaves remainder 7 on division by 9. That means 4768 = 9 × 529 + 7: there are 529 full rows before it, and it is the 7th seat of the next row, row 530. (Remainders on division by 9 from digit sums, pages 124 and 125; numbers of the form 9k + r, page 121.)

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Sources

All opened and read on 22 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 111.

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.