Class 9 Mathematics Ganita Manjari Part 2 NCERT Solutions
Chapter-10: How Quantities Combine: Understanding Data
Exercise Set 10.1
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:
Average score of section A = 72
Numberof students in section A = 30
Total score of Section A = 72×30 = 2160
Average score of section B = 76
Numberof students in section B = 25
Total score of Section B = 76×25 = 1900
Combined Average = \frac{\text{Combined total score}}{\text{Total students}}
= \dfrac{2160+1900}{30+25}
= \dfrac{4060}{55}=\dfrac{812}{11} \approx 73.82
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:
Three equal quantities of fertilisers contain the following fractions of nitrogen:
First fertiliser =\dfrac{1}{10}
Second fertiliser =\dfrac{9}{50}
Third fertiliser =\dfrac{3}{60}
Since the three quantities are equal, the fraction of nitrogen in the mixture is the average of the three fractions.
So, fraction of nitrogen =\frac{\text{Total nitrogen}}{\text{Number of fertilisers}}
= \dfrac{\frac{1}{10}+\frac{9}{50}+\frac{3}{60}}{3}
= \dfrac{30+54+15}{300}×\dfrac{1}{3}
= \dfrac{99}{300}×\dfrac{1}{3}=\dfrac{99}{900}
= \dfrac{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.
Solution:
Total purity product = (9×12)+(5×10)+(17×11)
= 108+50+187=345
Total units =9+5+17=31
Combined purity = \frac{\text{Total purity product}}{\text{total units}}
=\dfrac{345}{31} \approx 11.13
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:
Average rainfall in may = 3.5 mm/day
Average rainfall in june = 10 mm/day
Average rainfall in may = 8.7 mm/day
Combined average = \frac{\text{Total rainfall}}{\text{Total days}}
Combined average = \dfrac{(3.5×31)+(10×30)+(8.7×31)}{31+30+31}
which is the required expression for their combined average.
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.
Solution:
Total volume of kadha = 100+200+300=600 mL
Total quantity of spice mix = (100×\dfrac{5}{100})+(200×\dfrac{10}{100})+(300×\dfrac{15}{100})
= 5+20+45=70 mL
Concentration = \frac{\text{Total quantity of spice mix}}{\text{Total volume of kadha}}×100\%
= \dfrac{70}{600}×100\% =\dfrac{70}{6}\% =\dfrac{35}{3}\% \approx 11.67\%
(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:
Total volume of kadha = 300+200+100=600 mL
Total quantity of spice mix = (300×\dfrac{5}{100})+(200×\dfrac{10}{100})+(100×\dfrac{15}{100})
= 15+20+15=50 mL
Concentration = \frac{\text{Total quantity of spice mix}}{\text{Total volume of kadha}}×100\%
= \dfrac{50}{600}×100\% =\dfrac{50}{6}\% =\dfrac{25}{3}\% \approx 8.33\%