NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 10 How Quantities Combine Understanding Data Exercise 10.1

Exercise Set 10.1

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?
Answer:
Total score of Section A = 72 × 30 = 2160.
Total score of Section B = 76 × 25 = 1900.
Combined average = \( \frac{2160 + 1900}{30 + 25} = \frac{4060}{55} \approx 73.82 \).
The combined average is about 73.82. It sits nearer 72 because Section A is the larger section.

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?
Answer:
Let each quantity be p. Since the quantities are equal, the weighted mean becomes a simple mean:
\( \frac{\frac{1}{10}p + \frac{9}{50}p + \frac{3}{60}p}{3p} = \frac{\frac{1}{10} + \frac{9}{50} + \frac{3}{60}}{3} \)
Using the LCM 300: \( \frac{1}{10} = \frac{30}{300}, \ \frac{9}{50} = \frac{54}{300}, \ \frac{3}{60} = \frac{15}{300} \), so the sum is \( \frac{99}{300} = \frac{33}{100} \).
Fraction of nitrogen = \( \frac{33}{100} \div 3 = \frac{11}{100} \).
The mixture is \( \frac{11}{100} \) nitrogen, i.e. 11%.

Question 3. (Śrīdharācārya, Pāṭīgaṇita, 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 varṇa = 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.
Answer:
The purity of the melted gold is the weighted mean of the purities, the weights being the amounts of gold.
Purity = \( \frac{9 \times 12 + 5 \times 10 + 17 \times 11}{9 + 5 + 17} = \frac{108 + 50 + 187}{31} = \frac{345}{31} \).
\( \frac{345}{31} = 11\frac{4}{31} \approx 11.13 \).
The combined gold has a purity of \( 11\frac{4}{31} \) ≈ 11.13 varṇa.

Teacher's Note:
1. The largest piece (17 units at 11 varṇa) pulls the answer close to 11.
2. Amounts of gold, not purities, go in the denominator.

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.
Answer:
The months have different lengths: May 31 days, June 30 days, July 31 days, so each daily average must be weighted by its number of days.
Total rainfall: May = 3.5 × 31 mm, June = 10 × 30 mm, July = 8.7 × 31 mm.
Combined average = \( \frac{3.5 \times 31 + 10 \times 30 + 8.7 \times 31}{31 + 30 + 31} \)
Value: \( \frac{108.5 + 300 + 269.7}{92} = \frac{678.2}{92} \approx 7.37 \) mm per day.

Teacher's Note:
1. The trap is to divide 3.5 + 10 + 8.7 by 3.
2. Days in each month are the weights.

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.
Answer:
(i) Concentration = \( \frac{100 \times 5 + 200 \times 10 + 300 \times 15}{100 + 200 + 300}\% = \frac{500 + 2000 + 4500}{600}\% = \frac{7000}{600}\% \) ≈ 11.67%.
(ii) Concentration = \( \frac{300 \times 5 + 200 \times 10 + 100 \times 15}{600}\% = \frac{1500 + 2000 + 1500}{600}\% = \frac{5000}{600}\% \) ≈ 8.33%.
In (i) the biggest volume is the strongest one, so the mix leans towards 15%; in (ii) the biggest volume is the weakest one, so it leans towards 5%.

Teacher's Note:
1. Same three strengths, different amounts, different answers.
2. The result always lies between the lowest and highest concentrations.