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Suvarna had purchased 1g gold at ₹15k (short for ₹15,000). This is shown by point O denoting the average price of gold that she possesses. Six different scenarios are given below for her next transaction. For each scenario estimate and mark the average price of gold she will have after the transaction. (i) Purchase 1g gold at ₹30k (ii) Purchase 2g gold at ₹30k (iii) Purchase 1g gold at ₹10k (iv) Purchase 10g gold at ₹10k (v) Purchase 0.5g gold at ₹30k (vi) Purchase 0.5g gold at ₹15k.
Calculating each weighted average: (i) [(1 × 15) + (1 × 30)]/2 = 22.5k (mark midway between 15 and 30). (ii) [15 + (2 × 30)]/3 = 25k. (iii) [(1 × 15) + (1 × 10)]/2 = 12.5k. (iv) [15 + (10 × 10)]/11 ≈ 10.45k (near 10k). (v) [15 + (0.5 × 30)]/1.5 = 20k. (vi) [15 + (0.5 × 15)]/1.5 = 15k (remains exactlRead more
Calculating each weighted average: (i) [(1 × 15) + (1 × 30)]/2 = 22.5k (mark midway between 15 and 30). (ii) [15 + (2 × 30)]/3 = 25k. (iii) [(1 × 15) + (1 × 10)]/2 = 12.5k. (iv) [15 + (10 × 10)]/11 ≈ 10.45k (near 10k). (v) [15 + (0.5 × 30)]/1.5 = 20k. (vi) [15 + (0.5 × 15)]/1.5 = 15k (remains exactly at O).
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-10/
See lessShreyas holds 25 shares of a company at an average price of ₹150 and Vaishnavi holds 5 shares of the same company at an average price of ₹150. (i) Shreyas buys 10 shares of this company at a price of ₹30 each. What is the average price per share for him after the purchase? (ii) Vaishnavi buys some shares of this company at a price of ₹30 each and the average price per share for her after the purchase is ₹70. How many shares did she buy?
(i) Shreyas originally spent 25 × 150 = ₹3750. Adding 10 shares at ₹30 adds ₹300, giving a total expenditure of ₹4050 for 35 shares. His new average is 4050/35 ≈ ₹115.71 per share. (ii) Vaishnavi initially held shares worth 5 × 150 = ₹750. Buying x shares at ₹30 gives (750 + 30x)/(5 + x) = 70. SolviRead more
(i) Shreyas originally spent 25 × 150 = ₹3750. Adding 10 shares at ₹30 adds ₹300, giving a total expenditure of ₹4050 for 35 shares. His new average is 4050/35 ≈ ₹115.71 per share. (ii) Vaishnavi initially held shares worth 5 × 150 = ₹750. Buying x shares at ₹30 gives (750 + 30x)/(5 + x) = 70. Solving gives 750 + 30x = 350 + 70x, so 40x = 400, meaning she bought 10 shares.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-10/
See lessHow would you compare the prime factorisations of two numbers?
To compare two factorisations, find each common prime factor present in both lists. For every shared prime, compare its exponents in the two numbers and choose the minimum exponent. Multiplying each common prime raised to this minimum exponent gives their GCD. (Taking the maximum exponent across allRead more
To compare two factorisations, find each common prime factor present in both lists. For every shared prime, compare its exponents in the two numbers and choose the minimum exponent. Multiplying each common prime raised to this minimum exponent gives their GCD. (Taking the maximum exponent across all primes appearing in either factorisation similarly gives their LCM.)
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 lessWe can also find the gcd of two numbers by computing prime factorisation of both the numbers. Try to write an algorithm to compute the prime factorisation of a number. The prime factorisation of 180 is 2² × 3² × 5¹. How would you represent this?
We can represent prime factorisation as a list of pairs, where each pair consists of (prime_factor, power). For the number 180 (2² × 3² × 5¹), the representation is [(2, 2), (3, 2), (5, 1)]. Alternatively, it can be represented as an expanded list of repeated prime factors in ascending order, such aRead more
We can represent prime factorisation as a list of pairs, where each pair consists of (prime_factor, power). For the number 180 (2² × 3² × 5¹), the representation is [(2, 2), (3, 2), (5, 1)]. Alternatively, it can be represented as an expanded list of repeated prime factors in ascending order, such as [2, 2, 3, 3, 5].
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 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 less