Let all-divisors be the list obtained from divisors(n). Initialize an empty list called prime-list. For each element d in all-divisors: Run prime(d) to check whether d is a prime number. If prime(d) is true, add d to prime-list. Once all elements are examined, report prime-list as the final list ofRead more
Let all-divisors be the list obtained from divisors(n).
Initialize an empty list called prime-list.
For each element d in all-divisors:
Run prime(d) to check whether d is a prime number.
If prime(d) is true, add d to prime-list.
Once all elements are examined, report prime-list as the final list of prime divisors.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 11 The World of Algorithms Question Answer (2026-27)
Compute the list of factors: let L = divisors(n). Check the number of elements in L: If the total count of elements in L equals 2 (which means its only factors are 1 and n), report that n is prime. Otherwise (if it has fewer than 2 elements, such as n = 1 or more than 2), report that n is not prime.Read more
Compute the list of factors: let L = divisors(n).
Check the number of elements in L:
If the total count of elements in L equals 2 (which means its only factors are 1 and n), report that n is prime.
Otherwise (if it has fewer than 2 elements, such as n = 1 or more than 2), report that n is not prime.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 11 The World of Algorithms Question Answer (2026-27)
By the division lemma, write m = qn + r, where r = m mod n. If d divides both m and n, then m = ad and n = bd. Hence, r = m − qn = (a − qb)d, meaning d divides r. If d divides both n and r, then n = bd and r = cd. Thus, m = q(bd) + cd = (qb + c)d, meaning d divides m. For more NCERT SolutionsRead more
By the division lemma, write m = qn + r, where r = m mod n.
If d divides both m and n, then m = ad and n = bd. Hence, r = m − qn = (a − qb)d, meaning d divides r.
If d divides both n and r, then n = bd and r = cd. Thus, m = q(bd) + cd = (qb + c)d, meaning d divides m.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 11 The World of Algorithms Question Answer (2026-27)
(i) Bisecting PB creates 2 segments equal to AP, intercepting 2 segments of length √5 along QC, so QC = 2√5. (ii) Trisecting AP gives three 1/3-length intervals matching PB, so QC = √5/3. (iii) Using 1/3-sized units gives AP:PB = 3:2, making QC = 2/3 AQ = 2√5/3. (iv) Let AP/PB = m/n. Partitioning ABRead more
(i) Bisecting PB creates 2 segments equal to AP, intercepting 2 segments of length √5 along QC, so QC = 2√5. (ii) Trisecting AP gives three 1/3-length intervals matching PB, so QC = √5/3. (iii) Using 1/3-sized units gives AP:PB = 3:2, making QC = 2/3 AQ = 2√5/3. (iv) Let AP/PB = m/n. Partitioning AB into equal segments and drawing parallels intercepts equal subsegments on AC, proving AP/PB = AQ/QC.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
(i) The 4 points can be arranged cyclically in (4-1)!/2 = 3 distinct edge-connectivity cycles, giving 3 quadrilaterals. (ii) There are two cases: • Case 1 (points form a convex quadrilateral): Exactly 1 is convex and the other 2 are self-intersecting. • Case 2 (one point lies inside the triangle ofRead more
(i) The 4 points can be arranged cyclically in (4-1)!/2 = 3 distinct edge-connectivity cycles, giving 3 quadrilaterals. (ii) There are two cases: • Case 1 (points form a convex quadrilateral): Exactly 1 is convex and the other 2 are self-intersecting. • Case 2 (one point lies inside the triangle of the other three): None are convex; all 3 form non-convex dart shapes.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
Write an algorithm primedivisors(n) to compute the list of divisors of n that are prime numbers. (Hint: Compute divisors(n) and then filter out the primes in this list.)
Let all-divisors be the list obtained from divisors(n). Initialize an empty list called prime-list. For each element d in all-divisors: Run prime(d) to check whether d is a prime number. If prime(d) is true, add d to prime-list. Once all elements are examined, report prime-list as the final list ofRead more
Run prime(d) to check whether d is a prime number.
If prime(d) is true, add d to prime-list.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 11 The World of Algorithms Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-11/
See lessWrite an algorithm prime(n) to check if n is prime. (Hint: A prime number p has exactly two distinct factors, 1 and p. Can you make use of divisors(n) to write out prime(n)?)
Compute the list of factors: let L = divisors(n). Check the number of elements in L: If the total count of elements in L equals 2 (which means its only factors are 1 and n), report that n is prime. Otherwise (if it has fewer than 2 elements, such as n = 1 or more than 2), report that n is not prime.Read more
If the total count of elements in L equals 2 (which means its only factors are 1 and n), report that n is prime.
Otherwise (if it has fewer than 2 elements, such as n = 1 or more than 2), report that n is not prime.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 11 The World of Algorithms Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-11/
See lessAssume that m ≥ n. Verify that d divides m and n if and only if d divides both n and m mod n.
By the division lemma, write m = qn + r, where r = m mod n. If d divides both m and n, then m = ad and n = bd. Hence, r = m − qn = (a − qb)d, meaning d divides r. If d divides both n and r, then n = bd and r = cd. Thus, m = q(bd) + cd = (qb + c)d, meaning d divides m. For more NCERT SolutionsRead more
By the division lemma, write m = qn + r, where r = m mod n.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 11 The World of Algorithms Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-11/
See lessSuppose P is a point on side AB of △ABC and the line through P parallel to BC meets AC in point Q. For parts (i) to (iii), assume AP = 1, AQ = √5 and see Fig. 12.39. (i) If PB = 2 find QC. (Hint: See the dotted line) (ii) If PB = 1/3 find QC. (Hint: See the dotted lines) (iii) If PB = 2/3 find QC. (Hint: Divide both PB and AP suitably.) (iv) Show that if AP/PB is a rational number, then AP/PB = AQ/QC. In Grade 10, you will prove this equality without assuming AP/PB is rational. That will require a new idea.
(i) Bisecting PB creates 2 segments equal to AP, intercepting 2 segments of length √5 along QC, so QC = 2√5. (ii) Trisecting AP gives three 1/3-length intervals matching PB, so QC = √5/3. (iii) Using 1/3-sized units gives AP:PB = 3:2, making QC = 2/3 AQ = 2√5/3. (iv) Let AP/PB = m/n. Partitioning ABRead more
(i) Bisecting PB creates 2 segments equal to AP, intercepting 2 segments of length √5 along QC, so QC = 2√5. (ii) Trisecting AP gives three 1/3-length intervals matching PB, so QC = √5/3. (iii) Using 1/3-sized units gives AP:PB = 3:2, making QC = 2/3 AQ = 2√5/3. (iv) Let AP/PB = m/n. Partitioning AB into equal segments and drawing parallels intercepts equal subsegments on AC, proving AP/PB = AQ/QC.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessConsider 4 points A, B, C, D in the plane with no three collinear. Answer the following questions. Some answers may require you to consider different cases, depending on how the points are positioned in the plane. (i) How many different quadrilaterals do they form if the quadrilateral is allowed to be self-intersecting or non-convex? (ii) How many of these quadrilaterals are self-intersecting? How many are convex?
(i) The 4 points can be arranged cyclically in (4-1)!/2 = 3 distinct edge-connectivity cycles, giving 3 quadrilaterals. (ii) There are two cases: • Case 1 (points form a convex quadrilateral): Exactly 1 is convex and the other 2 are self-intersecting. • Case 2 (one point lies inside the triangle ofRead more
(i) The 4 points can be arranged cyclically in (4-1)!/2 = 3 distinct edge-connectivity cycles, giving 3 quadrilaterals. (ii) There are two cases: • Case 1 (points form a convex quadrilateral): Exactly 1 is convex and the other 2 are self-intersecting. • Case 2 (one point lies inside the triangle of the other three): None are convex; all 3 form non-convex dart shapes.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See less