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Class 9 Maths: The World of Algorithms, 8 practice questions with answers

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These are 8 practice questions on Chapter 11, The World of Algorithms, of the new NCERT Class 9 Maths textbook, Ganita Manjari, Part II (First Edition, August 2026). Each has a worked answer and the textbook page it comes from. They cover running the column-addition algorithm step by step, why it is faster than counting, the divisors algorithm, the first gcd algorithm and how to make it check fewer numbers, Euclid's subtraction algorithm, Āryabhaṭa's division algorithm, why gcd(m, n) = gcd(n, m − n), and how much faster division is than subtraction. 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 larger 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 Manjari, Grade 9, Part II, Chapter 11, printed pages 37 to 50):

Adding and counting

Question 1. Run the addition algorithm on 2969 + 3974. For each column, write the column sum, the digit written down and the carry.

Answer: 6943.

There are no more digits and the carry is 0, so nothing is written to the left. The answer is 6943. (Addition algorithm, pages 37 and 38.)

Question 2. Using the chapter's reasoning, how do counting dots and adding digit by digit compare if we move from adding two 3-digit numbers to adding two 7-digit numbers?

Answer: about 10,000 times as many dots to count, but only 4 more columns to add. Each extra digit makes the numbers about ten times larger, so four extra digits mean about 10 × 10 × 10 × 10 = 10,000 times as many dots. The digit-by-digit algorithm only needs one more column for each extra digit, so 4 more columns. (Speed of the addition algorithm, page 38.)

Divisors and the gcd

Question 3. Run the divisors algorithm for n = 3127, given that 3127 = 53 × 59 and that 53 and 59 are prime. What list do you get, and how many numbers does the algorithm check?

Answer: [1, 53, 59, 3127]; it checks all 3127 numbers from 1 to 3127. The list starts empty. 1 divides 3127, so the list becomes [1]. The next number that divides 3127 is 53, then 59, and finally 3127 itself; no other number divides it, because its only prime factors are 53 and 59. The list is in increasing order because the numbers were checked from 1 upwards. (Divisors algorithm, page 40.)

Question 4. 4189 = 59 × 71, where 59 and 71 are prime. Use the first gcd algorithm to find gcd(4189, 3127). How many numbers does it check in all? How many would the improved scans check?

Answer: gcd = 59; 7316 checks, or 4189 with a combined scan, or 3127 if we stop at the smaller number. divisors(4189) = [1, 59, 71, 4189] and divisors(3127) = [1, 53, 59, 3127]. The common divisors are [1, 59], and the rightmost element is the gcd: 59. Listing the divisors of each number separately checks 4189 + 3127 = 7316 numbers. Scanning both together needs max(4189, 3127) = 4189 checks, and since a common divisor cannot be larger than the smaller number, scanning up to min(4189, 3127) = 3127 is enough. (First gcd algorithm, pages 42 and 43; improving the algorithm, page 44.)

Euclid and Āryabhaṭa

Question 5. Use Euclid's subtraction algorithm to find gcd(671, 427). How many times does the problem get reduced?

Answer: gcd = 61, after 6 reductions.

(Euclid's subtraction algorithm, pages 46 and 47.)

Question 6. Use Āryabhaṭa's division algorithm to find gcd(4189, 3127). Compare the answer with Question 4.

Answer: 59, the same as in Question 4, after 4 reductions.

The first algorithm checked thousands of numbers; this one needed only 4 reductions. (Āryabhaṭa's division algorithm, pages 48 and 49.)

Question 7. In Question 6, the first step replaced 4189 and 3127 by 3127 and 1062. Using 4189 = 59 × 71 and 3127 = 59 × 53, show that 59 also divides 1062, as the chapter's argument for gcd(m, n) = gcd(n, m − n) says it must.

Answer: 1062 = 4189 − 3127 = 59 × 71 − 59 × 53 = 59 × (71 − 53), a whole number of 59s, so 59 divides 1062. In the chapter's terms, if blocks of size 59 tile both 4189 and 3127, they also tile the difference. (Why gcd(m, n) = gcd(n, m − n), page 46.)

Question 8. How many reductions does Euclid's subtraction algorithm need to find gcd(2003, 5), and how many does the division algorithm need?

Answer: subtraction needs 404 reductions; division needs 4. The gcd is 1. By subtraction, 5 is taken away from 2003 again and again, one reduction at a time, until 3 is left and the last few small steps finish the work: 404 reductions in all. By division: 2003 mod 5 = 3, so gcd(2003, 5) → gcd(5, 3) → gcd(3, 2) → gcd(2, 1) → gcd(1, 0), which is 4 reductions, and the answer is 1. The subtraction method's work grows with the size of the number, the division method's with its number of digits. (Subtraction can be slow, page 48; division algorithm, pages 48 and 49.)

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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 + 36.

The algorithms 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.