Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 11: The World of Algorithms (0904CH11)

Ganita Manjari Part II · 17 questions · 5 exercises

Chapter code: 0904ch11 · Book: Ganita Manjari Part II · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part II Chapter 11. Every exercise has a clear, step-by-step answer for chapter code 0904CH11.

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Exercise Set 11.1(5 questions)

Add two 4-digit numbers using the stated right-to-left column algorithm. Follow the steps exactly and check the result.

School Level Solution

Take 4738 and 6957.Units: 8+7=15; write 5, carry 1. Tens: 3+5+1=9; write 9, carry 0. Hundreds: 7+9=16; write 6, carry 1. Thousands: 4+6+1=11; write 1, carry 1. No digits remain, so the final carry 1 is written at the left.Therefore 4738+6957=11695. The result agrees with direct addition.

What happens if a 5-digit number is added to a 3-digit number? Do the stated steps handle this correctly?

School Level Solution

The column algorithm must line up digits by place value from the right. For example, 54321+789 is written as 54321+00789. The columns then give 10 in the units (write 0, carry 1), 9 in tens, 1 in hundreds, 4 in thousands and 5 in ten-thousands, so the answer is 55110. If the algorithm requires a digit in every column, the shorter number must be padded by leading zeros; otherwise the written steps need this clarification.

Why is it important to align the columns from right to left?

School Level Solution

In the decimal place-value system, a digit’s value depends on its position. Aligning from the right places units above units, tens above tens, and so on. Then each column adds quantities of the same place value, and each carry is transferred to the next place. Misalignment would add different place values together and give an incorrect sum.

In Step 3 of the addition algorithm, why can the carry not be more than 1?

School Level Solution

Each digit is at most 9 and the previous carry is either 0 or 1. Thus the largest possible column sum is 9+9+1=19. The carry is the number of tens in that sum, which can only be 0 or 1. It can never be 2 or more.

What happens if the fifth step (writing the final carry) is omitted? Give examples where the algorithm still works and fails.

School Level Solution

For 1234+2345=3579, no carry remains after the thousands column, so omitting Step 5 does not change the result.For 4738+6957, the thousands column gives 4+6+1=11, leaving a carry of 1 after all displayed columns are used. If Step 5 is omitted, the written answer would be 1695 instead of 11695. Hence the final-carry step is essential whenever a carry remains.

Exercise Set 11.2(3 questions)

Describe algorithms to find elements in List 1 that are absent from List 2, and elements in List 2 that are absent from List 1. The lists are in increasing order.

School Level Solution

Algorithm for List 1 \ List 2 (indices i=j=0): 1. While i is within List 1 and j is within List 2: if L1[i] < L2[j], output L1[i] and increment i; if L1[i] > L2[j], increment j; if equal, increment both i and j. 2. Output all remaining elements of L1.Algorithm for List 2 \ List 1: exchange the roles of the lists. If duplicate values are allowed, first specify whether “elements” means distinct values or occurrences; for set difference, output each distinct missing value once. The increasing order makes these scans efficient.

Describe an algorithm to compute the least common multiple of two positive integers.

School Level Solution

A correct algorithm is: ```text LCM(m,n): g := GCD(m,n) return (m / g) * n ``` Here `GCD` can be Euclid’s division algorithm. Since m and n share exactly their gcd as common prime-factor content, dividing m by g before multiplying by n avoids counting shared factors twice. The answer is the least positive integer divisible by both m and n.

Divisors occur in pairs. (i) How many numbers between 1 and n must be checked to find all divisors in pairs? (ii) Would a gcd algorithm based on paired divisors work as described?

School Level Solution

(i) If j divides n, the paired factor is n/j. At least one of the pair is at most √n, since if both were larger their product would exceed n. Therefore check j=1,2,…,⌊√n⌋. Whenever j divides n, record both j and n/j (record only one copy if j=√n).(ii) Yes, the common-divisor method still works if the divisors are paired. However, the list may no longer be in increasing order. So one cannot automatically report the rightmost/last common divisor as the gcd; instead compare all common divisors and report the largest, or use a descending search that stops at the first common divisor.

Exercise Set 11.3(2 questions)

How would the divisors(n) algorithm change if it examined numbers from n down to 1?

School Level Solution

Initialize an empty list D. For j = n, n−1, …, 1, if n mod j = 0, append j to D. Because the scan is descending, D contains the divisors in decreasing order. The first divisor found is n and the last is 1.

In the improved gcd algorithm, what happens if common divisors are examined from min(m,n) down to 1?

School Level Solution

Set s=min(m,n). For j=s,s−1,…,1, test whether m mod j=0 and n mod j=0. The first j satisfying both conditions is the greatest common divisor, because every larger candidate has already been checked and rejected. Thus return j immediately; no list of common divisors is needed.

End-of-Chapter Exercises(5 questions)

Use the improved Euclid division algorithm to compute (i) gcd(375,825), (ii) gcd(51000,81000), (iii) gcd(1789287,237656), (iv) gcd(2587392,157656).

School Level Solution

(i) 825=2(375)+75; 375=5(75)+0, so gcd=75.(ii) 81000=1(51000)+30000; 51000=1(30000)+21000; 30000=1(21000)+9000; 21000=2(9000)+3000; 9000=3(3000)+0. Hence gcd=3000.(iii) Euclid’s algorithm gives successive remainders until the final non-zero remainder is 1, so gcd=1 (the numbers are coprime).(iv) The successive remainders terminate at 24, with the next remainder 0, so gcd=24.

Assume m≥n. Verify that d divides m and n if and only if d divides both n and m mod n.

School Level Solution

Write an algorithm prime(n) to check whether n is prime, using divisors(n).

School Level Solution

```text prime(n): if n < 2: return false D := divisors(n) if length(D) = 2: return true else: return false ``` A prime integer has exactly two positive divisors, 1 and itself. A number below 2 is not prime. Hence the test correctly distinguishes primes from composites.

Write an algorithm primedivisors(n) to compute the prime divisors of n.

School Level Solution

```text primedivisors(n): D := divisors(n) P := empty list for each d in D: if prime(d): append d to P return P ``` For example, divisors(180) = [1,2,3,4,5,6,9,10,12,15,18,20,30,36,45,60,90,180], so primedivisors(180) = [2,3,5]. The algorithm checks every divisor and keeps precisely the prime ones.

Write an algorithm for prime factorisation. (i) How can the prime factorisation 180=2²×3²×5 be represented? (ii) How can two prime factorizations be compared?

School Level Solution

```text primeFactorisation(n): F := empty list of (prime, exponent) pairs p := 2 while p*p <= n: e := 0 while n mod p = 0: n := n / p e := e + 1 if e > 0: append (p,e) to F p := p + 1 if n > 1: append (n,1) to F return F ```(i) Store 180=2²×3²×5 as [(2,2),(3,2),(5,1)], meaning prime 2 has exponent 2, prime 3 exponent 2 and prime 5 exponent 1.(ii) For two numbers, compare the exponents of each prime, treating a missing prime as exponent 0. The gcd uses the minimum exponent for each prime; the lcm uses the maximum exponent. If every exponent in one factorisation is no larger than the corresponding exponent in the other, the first number divides the second.

In-text and Think and Reflect(2 questions)

Explain why the right-to-left column addition algorithm works, and show how to adapt it to add decimal fractions.

School Level Solution

In base ten, 10 units of any place value equal 1 unit of the next place value. Adding a column gives a digit sum; write its remainder after division by 10 and carry the quotient to the next column. This preserves the total value at every step. For decimals, align decimal points and pad with zeros: 12.50 + 3.78 = 16.28. Equivalently, multiply both numbers by 100, add 1250 + 378 = 1628, then divide by 100 to obtain 16.28.

Use the divisor algorithm to list all positive divisors of 15, 135 and 775. What happens to the amount of work as the numbers grow?

School Level Solution

Check positive integers for exact divisibility and list each divisor once. Since 15 = 3 × 5, its divisors are 1, 3, 5, 15. Since 135 = 3³ × 5, its divisors are 1, 3, 5, 9, 15, 27, 45, 135. Since 775 = 5² × 31, its divisors are 1, 5, 25, 31, 155, 775. A naive search that tests every integer up to n requires work proportional to n; testing factor pairs only up to √n can reduce the number of candidates substantially.

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