USP IX The World of Algorithms Test Paper with Solutions
Class IXMaths📋 Sample PaperCh 11: The World of AlgorithmsNCERT book: Ganita Manjari
Updated 22 Oct 2026
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Class 9 Maths Ganita Manjari Chapter 11: The World of Algorithms – Test Paper with Solutions
Part II • CBSE 2026-27 • 40 Marks • 60 Minutes
18 QuestionsStep-by-step solutionsFree
This free practice test paper with solutions for Class 9 Maths Chapter 11, The World of Algorithms (NCERT Ganita Manjari, Part II, session 2026-27) has 18 questions for 40 marks and is made to be finished in about 60 minutes. The paper has MCQs, short answer questions, a case study and long answer questions. Try every question on your own first. Then tap Show Answer to see the full working, with the marks for each step, and compare it with your own solution.
What this test paper covers
Adding numbers by algorithm: columns, carry and the final carry
Divisors and the first gcd algorithm: lists of divisors and common divisors
Better gcd algorithms: scanning up to the smaller number and the most-recent-common-divisor method
Euclid and Aryabhata: the subtraction algorithm and the division algorithm using mod
lcm and gcd: how they are linked
How the paper is arranged
Section
Type of questions
Questions
Marks each
Total
A
Multiple choice questions
8
1
8
B
Short answer questions
3
2
6
C
Short answer questions
4
3
12
D
Case study
1
4
4
E
Long answer questions
2
5
10
Total
18
40
Try each question first, then tap Show Answer to check your working.
SECTION AMultiple Choice Questions (1 mark each)8 marks
1
When two numbers are added column by column, the largest possible sum in one column (including the carry) is:
1 mark
A9
B18
C19
D20
Answer and working
The largest digit is 9, so the largest column sum is 9 + 9 + 1 = 19. The carry coming in is at most 1.
Answer: (C) 19
2
The value of 47 mod 6 is:
1 mark
A7
B1
C6
D5
Answer and working
47 = 6 × 7 + 5, so the remainder is 5.
Answer: (D) 5
3
To find divisors(40) with the basic algorithm, how many numbers are checked?
1 mark
A40
B20
C39
D80
Answer and working
The basic algorithm checks every number j from 1 to n. For n = 40 it checks 40 numbers.
Answer: (A) 40
4
In the long-division method for gcd, the last non-zero divisor is:
1 mark
Athe quotient
Bthe gcd
Cthe remainder
Dthe lcm
Answer and working
Divide the larger number by the smaller, then repeat with the divisor and the remainder. The last non-zero divisor is the gcd.
Answer: (B) the gcd
5
A common divisor of two numbers m and n can never be larger than:
1 mark
Am + n
Bm × n
Cthe smaller of m and n
Dthe larger of m and n
Answer and working
A divisor of the smaller number cannot be larger than that number. So a common divisor is at most min(m, n).
Answer: (C) the smaller of m and n
6
The value of lcm(6, 9) × gcd(6, 9) is:
1 mark
A54
B15
C18
D3
Answer and working
lcm(6, 9) = 18 and gcd(6, 9) = 3, so the product is 18 × 3 = 54. This is also 6 × 9.
Answer: (A) 54
7
The idea of replacing gcd(m, n) by gcd(n, m mod n) belongs to:
1 mark
AEuclid’s subtraction algorithm
Bthe lcm algorithm
Ccolumn addition
DAryabhata’s division algorithm
Answer and working
It uses the remainder (m mod n), which is Aryabhata’s division algorithm. Euclid’s algorithm uses subtraction: gcd(m, n) = gcd(n, m − n).
Answer: (D) Aryabhata’s division algorithm
8
A 3-digit number in a sum is replaced by a 5-digit number. Counting dots needs about how many times more work, and how many more columns does the column method need?
1 mark
A10 times, 1 more column
B100 times, 2 more columns
C100 times, 1 more column
D10 times, 2 more columns
Answer and working
Each extra digit gives 10 times as many dots, so 2 extra digits give 100 times as many dots. But the column method needs only 2 more columns.
Answer: (B) 100 times, 2 more columns
SECTION BShort Answer Questions (2 marks each)6 marks
9
Add 7385 + 946 with the column algorithm. Treat the missing thousands digit of 946 as 0. Write the sum and the carry for each column.
2 marks
Answer and working
Column
Working
Write
Carry
Units
5 + 6 = 11
1
1
Tens
8 + 4 + 1 = 13
3
1
Hundreds
3 + 9 + 1 = 13
3
1
Thousands
7 + 0 + 1 = 8
8
0
1 mark
Sum = 8331. Check: 7385 + 946 = 8331.
1 mark
Answer: 8331
10
Find (a) 83 mod 9 and (b) 250 mod 16.
2 marks
Answer and working
(a) 83 = 9 × 9 + 2, so 83 mod 9 = 2.
1 mark
(b) 250 = 16 × 15 + 10, so 250 mod 16 = 10.
1 mark
Answer: (a) 2 (b) 10
11
Use Euclid’s subtraction algorithm to find gcd(35, 21). Write every reduction.
The last non-zero divisor is 13, so gcd(299, 221) = 13.
1 mark
Answer: 13
14
Divisors of a number come in pairs. Take n = 50.
3 marks
(a)Write all the pairs of divisors.[1]
(b)Say up to which number we need to check, and why.[2]
Answer and working
(a) The pairs are (1, 50), (2, 25) and (5, 10).
1 mark
(b) We need to check only the numbers 1 to 7. In each pair the smaller number is at most √50 ≈ 7.07.
1 mark
Reason: if both numbers of a pair were bigger than √50, their product would be bigger than 50.
1 mark
Answer: (a) three pairs as above (b) check up to 7
15
Counting one dot per second, answer the following for the sum 3600 + 2400.
3 marks
(a)How long does it take to count the dots? Give the answer in minutes.[2]
(b)The column algorithm takes 15 seconds per column. How long does it take to add these two numbers?[1]
Answer and working
(a) Number of dots = 3600 + 2400 = 6000, so the time is 6000 seconds.
1 mark
6000 ÷ 60 = 100 minutes (1 hour 40 minutes).
1 mark
(b) Both numbers have 4 digits, so there are 4 columns: 4 × 15 = 60 seconds, that is 1 minute.
1 mark
Answer: (a) 100 minutes (b) 1 minute
SECTION DCase Study (4 marks)4 marks
16
Meena wants to cover a rectangular floor, 390 cm long and 234 cm wide, with square tiles of the same size, without cutting any tile. She wants the tiles to be as big as possible.
4 marks
(i)Which mathematical idea gives the side of the largest square tile?[1]
(ii)Use Aryabhata’s division algorithm to find the side of the tile.[2]
(iii)How many tiles does Meena need?[1]
Answer and working
(i) The greatest common divisor (gcd or hcf). The side of the largest square tile must divide both 390 and 234.
1 mark
(ii) 390 = 234 × 1 + 156 234 = 156 × 1 + 78 156 = 78 × 2 + 0 The gcd is 78, so the side of the tile is 78 cm.
2 marks
(iii) Along the length: 390 ÷ 78 = 5 tiles. Along the width: 234 ÷ 78 = 3 tiles. Number of tiles = 5 × 3 = 15.
1 mark
Answer: (i) gcd (ii) 78 cm (iii) 15 tiles
SECTION ELong Answer Questions (5 marks each)10 marks
17
Compare the two gcd algorithms for gcd(11, 2).
5 marks
(a)Use Euclid’s subtraction algorithm. Write every reduction and count them.[2]
(b)Use Aryabhata’s division algorithm. Write every division and count them.[2]
(c)Which method is better for very large numbers? Give a reason.[1]
(c) The division method is better. Its number of steps grows only with the number of digits, but the subtraction method needs about k steps for gcd(2k + 1, 2), so its steps grow with the value of the number.
(b) common-divisors = [1, 2, 3, 4, 6, 12]. The gcd is the rightmost element, so gcd(24, 36) = 12.
2 marks
(c) The numbers of divisors(24) that are not in divisors(36) are 8 and 24.
1 mark
Answer: (a) lists as above (b) 12 (c) 8 and 24
Key ideas from Chapter 11
Algorithm. An algorithm is a step-by-step procedure that gives the right answer if followed exactly. It is built from basic steps we can do directly, and we must also justify why it works.
Adding by columns. Add from right to left and carry 1 whenever a column total is 10 or more. The carry is always 0 or 1, because the largest column sum is 9 + 9 + 1 = 19. If the last column makes a carry, write it on the left.
Divisors and gcd. divisors(n) checks every number from 1 to n, so the list is in increasing order. The gcd is the rightmost element of the list of common divisors. A common divisor is never larger than min(m, n).
Euclid and Aryabhata. For m ≥ n, Euclid’s subtraction gives gcd(m, n) = gcd(n, m − n), and Aryabhata’s division gives gcd(m, n) = gcd(n, m mod n). Stop when you reach gcd(m, 0) = m. In long division, the last non-zero divisor is the gcd. Division needs far fewer steps than subtraction.
lcm and gcd. lcm(m, n) × gcd(m, n) = m × n.
Common mistakes to avoid
Forgetting the final carry, so 473 + 695 becomes 168 instead of 1168.
Choosing the leftmost common divisor (always 1) instead of the rightmost as the gcd.
Writing gcd(m, 0) = 0. The correct answer is m.
Frequently asked questions
What is an algorithm?
An algorithm is a step-by-step procedure to arrive at an answer. Anyone who follows the steps exactly gets the right answer. Column addition and Euclid’s method for the gcd are examples.
What does m mod n mean?
m mod n is the remainder when m is divided by n. For example, 17 mod 5 = 2.
How does Euclid’s subtraction algorithm find the gcd?
It replaces gcd(m, n) by gcd(n, m − n) again and again, because both pairs have the same common divisors. It stops at gcd(m, 0), and the answer is m.
Why is Aryabhata’s division method faster than subtraction?
Repeated subtraction can take very many steps for large numbers. Division with remainder does all those subtractions in one step, so the number of steps grows only with the number of digits.
How are lcm and gcd related?
The product of the lcm and the gcd of two numbers is equal to the product of the two numbers. For 6 and 9, the lcm is 18, the gcd is 3, and 18 × 3 = 54 = 6 × 9.
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