Ncert Solution Class 9 Maths Ganita Manjari Exercise 10.1
How Quantities Combine :Understanding Data
Learning Outcome: By the end of this chapter, students will be able to calculate weighted averages, solve mixture problems, interpret different types of charts, and make informed conclusions from data.
Q1. 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?
Q 2. A farmer mixes three equal quantities of fertilisers. The first one contains 110 nitrogen, the second contains 950 nitrogen, and the third contains 360 nitrogen. What is the fraction of nitrogen in the mixture?
Ans 2. All three frequencies are taken in equal quantity, so they have equal weights.
So, the required mean is
\[ = \frac{\frac{1}{10}+\frac{9}{50}+\frac{3}{60}}{3} \] \[ = \frac{\frac{30+54+15}{300}}{3} \] \[ = \frac{99}{900} = \frac{11}{100} \]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 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
Ans 3.
Total quantity of alloy:
\[ = 9 + 5 + 17 = 31 \]Purity of Gold:
\[ = \frac{9 \times 12 + 5 \times 10 + 17 \times 11} {9 + 12 + 17} \] \[ = \frac{345}{31} = 11.129 \]So, purity of gold is 11.13 Varna.
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.
Ans 4.
Number of days in May = 31
Number of days in June = 30
Number of days in July = 31
Total rainfall in May = \(31 \times 3.5\text{ mm}\)
Total rainfall in June = \(30 \times 10\text{ mm}\)
Total rainfall in July = \(31 \times 8.7\text{ mm}\)
Combined average rainfall:
\[ = \frac{31 \times 3.5 + 30 \times 10 + 31 \times 8.7} {31+30+31} \] \[ = \frac{108.5+300+269.7}{92} \] \[ = \frac{678.2}{92} \] \[ = 7.37 \]Therefore, the combined average rainfall is 7.37 mm per day.
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.
Q. 5 (i)
Amount of spice in 1st case:
\[ = \frac{5}{100}\times 100 = 5 \]Amount of spice in 2nd case:
\[ = \frac{10}{100}\times 200 = 20 \]Amount of spice in 3rd case:
\[ = \frac{15}{100}\times 300 = 45 \]Total quantity of mixture:
\[ = 100+200+300=600\text{ ml} \]Combined percentage concentration is given by:
\[ =\frac{5+20+45}{600}\times100 \] \[ =\frac{70}{600}\times100 \] \[ =\frac{70}{6}\% \] \[ =11.666\% \]Therefore, the combined concentration is approximately 11.67%.
Q. 5 (ii)
Amount of spice in 1st case:
\[ =\frac{5}{100}\times300=15 \]Amount of spice in 2nd case:
\[ =\frac{10}{100}\times200=20 \]Amount of spice in 3rd case:
\[ =\frac{15}{100}\times100=15 \]Total quantity of mixture:
\[ =300+200+100=600 \]Combined percentage concentration is given by:
\[ =\frac{15+20+15}{300+200+100}\times100 \] \[ =\frac{50}{600}\times100 \] \[ =\frac{50}{6}\% \] \[ =8.333\% \]Therefore, the combined concentration is approximately 8.33%.