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Naman Academy For Class 9
Chapter 10: How Quantities Combine: Understanding DataAll 5 Questions on This Page

Class 9 Maths Chapter 10 Exercise Set 10.1 Solutions

Step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 10 Exercise Set 10.1 (Questions 1–5). Covers combined averages, average of averages, weighted mean, mixtures, concentrations, and historic gold purity problems.

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Key Concepts for Exercise Set 10.1

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1

Average of Averages (Weighted Mean)

When combining two or more groups with different averages and group sizes, each group's average must be weighted by its size. The combined average is the total sum divided by the total count.

Combined average = (an + bm) / (n + m)
2

Weighted Mean Formula for Mixtures

When quantities with different concentrations or values are combined, each quantity acts as a weight. The combined concentration is the sum of (weight × concentration) divided by the total weight.

Combined concentration = (w₁x₁ + w₂x₂ + … + wₖxₖ) / (w₁ + w₂ + … + wₖ)
3

Equal Quantities (Arithmetic Mean of Concentrations)

When all component quantities are equal (w₁ = w₂ = w₃), the weights cancel out, and the mixture concentration equals the simple arithmetic average of the concentrations.

Fraction in mixture = (x₁ + x₂ + x₃) / 3
4

Weighting Daily Rates by Month Length

To find the combined daily average over several months, each month's daily average must be multiplied by the exact number of days in that month (e.g. May = 31, June = 30, July = 31).

Combined average = Total rainfall / Total number of days

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

For two collections with averages a and b, and sizes n and m, respectively, the combined average is:

Combined average = an + bmn + m
Given:

Section A: a = 72, n = 30

Section B: b = 76, m = 25

Substituting these values into the formula:
Combined average = 72 × 30 + 76 × 2530 + 25
= 2160 + 190055
= 406055
= 81211
= 73 911
≈ 73.82
Answer:73 911 ≈ 73.82 marks

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

Solution

The three quantities of fertilisers are equal. Therefore, we can use the weighted-mean formula with equal weights.

For quantities with concentrations x₁, x₂, x₃ and respective quantities w₁, w₂, w₃:

Fraction in mixture = w₁x₁ + w₂x₂ + w₃x₃w₁ + w₂ + w₃

Since the quantities are equal, let:

w₁ = w₂ = w₃ = 1

The concentrations of nitrogen are:

x₁ = 110, x₂ = 950, x₃ = 360

Substituting into the formula:
Fraction of nitrogen = 110 + 950 + 3603

The LCM of 10, 50, 60 is 300.

= 30300 + 54300 + 153003
= 993003
= 99900
= 11100
Answer:11100

(Śrīdharācārya, Pāṭigaṇita, c. 750 CE) In ancient India, Varṇa was the measure of gold purity. A purity of 16 varṇa meant pure gold; in general, a purity of k varṇa meant that the gold-alloy was k/16 gold and the rest impurities. (Now the term used is karat; 16 Varna = 24 karat.) 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

The quantities and purities of the three pieces are different. Therefore, we use the weighted-mean formula.

For quantities w₁, w₂, w₃ having purities x₁, x₂, x₃, respectively:

Combined purity = w₁x₁ + w₂x₂ + w₃x₃w₁ + w₂ + w₃
Given:

First piece: Quantity w₁ = 9 units, purity x₁ = 12 varṇa.

Second piece: Quantity w₂ = 5 units, purity x₂ = 10 varṇa.

Third piece: Quantity w₃ = 17 units, purity x₃ = 11 varṇa.

Substituting the values:
Combined purity = 9 × 12 + 5 × 10 + 17 × 119 + 5 + 17
= 108 + 50 + 18731
= 34531
= 11 431
≈ 11.13
Answer:11 431 varṇa ≈ 11.13 varṇa

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
Given:
Average rainfall per day in May = 3.5 mm
Number of days in May = 31
Average rainfall per day in June = 10 mm
Number of days in June = 30
Average rainfall per day in July = 8.7 mm
Number of days in July = 31

The combined average is the total rainfall divided by the total number of days.

Using the weighted-mean formula:

Combined average = Total rainfall in all three monthsTotal number of days

First, calculate the total rainfall in each month:

May:

3.5 × 31 = 108.5 mm

June:

10 × 30 = 300 mm

July:

8.7 × 31 = 269.7 mm
Therefore, the required expression is:
31 × 3.5 + 30 × 10 + 31 × 8.731 + 30 + 31
Evaluating the expression:
Combined average = 108.5 + 300 + 269.792
= 678.292
≈ 7.37 mm per day
Answer:31(3.5) + 30(10) + 31(8.7)92 ≈ 7.37 mm per day

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) First scenario

The general formula for the concentration of a mixture is:

Combined concentration = w₁x₁ + w₂x₂ + w₃x₃w₁ + w₂ + w₃

Here, w₁, w₂, w₃ are the volumes in mL, and x₁, x₂, x₃ are the numerical percentage values.

Given:

w₁ = 100, w₂ = 200, w₃ = 300

x₁ = 5, x₂ = 10, x₃ = 15

Substituting into the formula:
Concentration = 100 × 5 + 200 × 10 + 300 × 15100 + 200 + 300%
= 500 + 2000 + 4500600%
= 7000600%
= 353%
= 11 23%

Answer (i):

11 23% ≈ 11.67%

(ii) Second scenario
Given:

w₁ = 300, w₂ = 200, w₃ = 100

x₁ = 5, x₂ = 10, x₃ = 15

Substituting into the same formula:
Concentration = 300 × 5 + 200 × 10 + 100 × 15300 + 200 + 100%
= 1500 + 2000 + 1500600%
= 5000600%
= 253%
= 8 13%

Answer (ii):

8 13% ≈ 8.33%
Answer:(i) 11 23% ≈ 11.67% | (ii) 8 13% ≈ 8.33%

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