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Write 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 lessThe edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm³)?
For the largest right circular cylinder cut out of a cube of edge r cm, the circular base must fit tightly within the square face of side r cm, giving diameter r cm and radius r/2 cm. The height of the cylinder equals the cube edge, h = r cm. Thus, volume V = π(radius)²(height) = π(r/2)²(r) = (πr³)/Read more
For the largest right circular cylinder cut out of a cube of edge r cm, the circular base must fit tightly within the square face of side r cm, giving diameter r cm and radius r/2 cm. The height of the cylinder equals the cube edge, h = r cm. Thus, volume V = π(radius)²(height) = π(r/2)²(r) = (πr³)/4 cm³.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
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