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NCERT Solutions · Class 9 Mathematics How Quantities Combine: Understanding Data

33 questions · 33 still being checked

Exercise Set 10.1 1–5 (part 1 of 7)

  1. Exercise 1

    The average score of students on a test in Section A is 72\displaystyle 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\displaystyle 30 students and Section B has 25\displaystyle 25 students?

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    Each section's average is weighted by its number of students.\[\text{Combined average} = \frac{72\times 30 + 76\times 25}{30+25} \]\[= \frac{2160 + 1900}{55} = \frac{4060}{55} = \frac{812}{11} \approx 73.82 \]Answer: \(\displaystyle \dfrac{812}{11} \approx 73.82 \)

  2. Exercise 2

    A farmer mixes three equal quantities of fertilisers. The first one contains 110\displaystyle \frac{1}{10} nitrogen, the second contains 950\displaystyle \frac{9}{50} nitrogen, and the third contains 360\displaystyle \frac{3}{60} nitrogen. What is the fraction of nitrogen in the mixture?

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    The quantities are equal, so take \(\displaystyle y\) of each; \(\displaystyle \tfrac{3}{60} = \tfrac{1}{20} \).\[\text{fraction} = \frac{\dfrac{1}{10}y + \dfrac{9}{50}y + \dfrac{1}{20}y}{y+y+y} \]\[= \frac{\dfrac{10+18+5}{100}\,y}{3y} = \frac{33}{300} = \frac{11}{100} \]Answer: \(\displaystyle \dfrac{11}{100} = 0.11 \) nitrogen

  3. Exercise 3

    (Śrīdharācārya, Pātīgaṇita, c. 750\displaystyle 750 CE) In ancient India, Varṇa was the measure of gold purity. A purity of 16\displaystyle 16 varna meant pure gold; in general, a purity of k varna meant that the gold-alloy was k/16\displaystyle \mathrm{k} / 16 gold and the rest impurities. (Now the term used is karat; 16\displaystyle 16 Varna = 24\displaystyle 24 karat.) Suppose a goldsmith melts together three pieces of gold: 9\displaystyle 9 units at 12\displaystyle 12 varna, 5\displaystyle 5 units at 10\displaystyle 10 varna, and 17\displaystyle 17 units at 11\displaystyle 11 varna. Find the purity in varna of the combined gold.

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    This solution has not been cross-checked against the answer printed in NCERT.

    Pure gold in each piece is \(\displaystyle \tfrac{k}{16} \) of its units.\[\text{pure gold} = \tfrac{12}{16}(9) + \tfrac{10}{16}(5) + \tfrac{11}{16}(17) = \tfrac{108+50+187}{16} = \tfrac{345}{16} \]\[\text{total units} = 9+5+17 = 31 \]\[\frac{k}{16} = \frac{345/16}{31} \;\Rightarrow\; k = \frac{345}{31} \approx 11.13 \]Answer: \(\displaystyle \dfrac{345}{31} \approx 11.13 \) varna

  4. Exercise 4

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

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    This solution has not been cross-checked against the answer printed in NCERT.

    May, June and July have $\displaystyle 31$, $\displaystyle 30$ and $\displaystyle 31$ days, so each monthly average is weighted by its number of days.\[\text{Combined average} = \frac{3.5\times 31 + 10\times 30 + 8.7\times 31}{31+30+31} \]\[= \frac{108.5 + 300 + 269.7}{92} = \frac{678.2}{92} \approx 7.37 \text{ mm} \]Answer: \(\displaystyle \dfrac{3.5\times 31 + 10\times 30 + 8.7\times 31}{31+30+31} \approx 7.37 \) mm per day

  5. Exercise 5

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

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    (i) Spice in the mixture divided by the volume of the mixture.\[\frac{100(0.05) + 200(0.10) + 300(0.15)}{100+200+300} \]\[= \frac{5 + 20 + 45}{600} = \frac{70}{600} = \frac{7}{60} \approx 0.1167 \](ii) The same volumes, weighted the other way round.\[\frac{300(0.05) + 200(0.10) + 100(0.15)}{300+200+100} \]\[= \frac{15 + 20 + 15}{600} = \frac{50}{600} = \frac{1}{12} \approx 0.0833 \]Answer: (i) \(\displaystyle \approx 11.67\% \); (ii) \(\displaystyle \approx 8.33\% \)