Ex 10.1 Class 9 Maths Ganita Manjari Solutions | Sid Classes

Solve Ganita Manjari Class 9 Ex 10.1: weighted averages and mixtures. Clear steps for CBSE exams. Sid Classes.

Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.1 Weighted Averages and Mixtures. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough makes weighted means, fertilizer mixtures, ancient gold purity calculations, rainfall averages, and solution concentrations entirely clear.

🔑 Key Concepts & Formulas:

  • Weighted Average (Mean): \(\text{Weighted Average} = \frac{\sum (w_i \cdot x_i)}{\sum w_i}\), where \(w_i\) represents weights (such as quantities, student counts, or days) and \(x_i\) represents individual values.
  • Mixture Concentration: The ratio of the total amount of solute/active ingredient to the total volume of the mixture.

NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.1 Solutions

Question 1

The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has 30 students and Section B has 25 students?

Solution:

* Total score of Section A \(= 30 \times 72 = 2160\).
* Total score of Section B \(= 25 \times 76 = 1900\).
* Total number of students \(= 30 + 25 = 55\).

Combined Average \(= \frac{\text{Total score of A} + \text{Total score of B}}{\text{Total students}} = \frac{2160 + 1900}{55} = \frac{4060}{55} = \frac{812}{11} \approx \mathbf{73.82}\).

Answer: The combined average score of both sections is \(\mathbf{\frac{812}{11}}\) (or approximately \(\mathbf{73.82}\)).

Question 2

A farmer mixes three equal quantities of fertilisers. The first one contains \(\frac{1}{10}\) nitrogen, the second contains \(\frac{9}{50}\) nitrogen, and the third contains \(\frac{3}{60}\) nitrogen. What is the fraction of nitrogen in the mixture?

Solution:

Since equal quantities are mixed, we take the simple arithmetic mean of the nitrogen fractions:
Fraction \(= \frac{1}{3} \left(\frac{1}{10} + \frac{9}{50} + \frac{3}{60}\right)\).

Simplify the fractions: \(\frac{3}{60} = \frac{1}{20}\).
Find the LCM of denominators \((10, 50, 20)\), which is \(100\):
* \(\frac{1}{10} = \frac{10}{100}\)
* \(\frac{9}{50} = \frac{18}{100}\)
* \(\frac{1}{20} = \frac{5}{100}\)

Sum \(= \frac{10 + 18 + 5}{100} = \frac{33}{100}\).
Mixture fraction \(= \frac{1}{3} \times \frac{33}{100} = \frac{11}{100} = \mathbf{0.11}\).

Answer: The fraction of nitrogen in the mixture is \(\mathbf{\frac{11}{100}}\) (or \(0.11\)).

Question 3

(Śrīdharācārya, Pātīganita, c. 750 CE) In ancient India, varṇa was the measure of gold purity... Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varṇa, 5 units at 10 varṇa, and 17 units at 11 varṇa. Find the purity in varṇa of the combined gold.

Solution:

Using the weighted average formula for gold purity:
* Total varṇa-units \(= (9 \times 12) + (5 \times 10) + (17 \times 11) = 108 + 50 + 187 = 345\).
* Total units of gold \(= 9 + 5 + 17 = 31\).

Combined Purity \(= \frac{345}{31} \approx \mathbf{11.13}\text{ varṇa}\).

Answer: The purity of the combined gold is \(\mathbf{\frac{345}{31}}\) varṇa (or approximately \(11.13\) varṇa).

Question 4

The average rainfall per day in the months of May, June, and July in a certain location are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression that gives their combined average.

Solution:

Accounting for the number of days in each month (May = 31 days, June = 30 days, July = 31 days), the combined weighted average rainfall per day across the three months is expressed as:
$$\text{Combined Average} = \frac{3.5(31) + 10(30) + 8.7(31)}{31 + 30 + 31} = \frac{108.5 + 300 + 269.7}{92} = \frac{678.2}{92} \approx \mathbf{7.37}\text{ mm}$$
*(Note: If calculated as a simple unweighted monthly mean: \(\frac{3.5 + 10 + 8.7}{3} = \frac{22.2}{3} = 7.4\text{ mm}\).)*

Question 5

Calculate the concentration of spice mix in these two scenarios:
(i) A 100 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 300 mL one with 15% spice mix are combined.
(ii) A 300 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 100 mL one with 15% spice mix are mixed.

Solution:

(i) Scenario 1:
* Total volume \(= 100 + 200 + 300 = 600\text{ mL}\).
* Total spice mix \(= (100 \times 0.05) + (200 \times 0.10) + (300 \times 0.15) = 5 + 20 + 45 = 70\text{ mL}\).
* Concentration \(= \frac{70}{600} \times 100 = \mathbf{11.67\%}\).

(ii) Scenario 2:
* Total volume \(= 300 + 200 + 100 = 600\text{ mL}\).
* Total spice mix \(= (300 \times 0.05) + (200 \times 0.10) + (100 \times 0.15) = 15 + 20 + 15 = 50\text{ mL}\).
* Concentration \(= \frac{50}{600} \times 100 = \mathbf{8.33\%}\).

Answer: (i) \(\mathbf{11.67\%}\); (ii) \(\mathbf{8.33\%}\).

Frequently Asked Questions (FAQs)

Q1: How do you calculate the weighted average of multiple groups?
A: Multiply each group's average by its respective weight (size), sum them up, and divide by the total sum of weights.

Q2: Why do month lengths matter when calculating average rainfall over multiple months?
A: Because months have differing numbers of days (e.g., 30 vs 31 days), a weighted average based on day count gives the true daily mean.

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