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

33 questions · 33 still being checked

End-of-Chapter Exercises 1–10 (part 6 of 7)

  1. Exercise 1

    In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6\displaystyle 6 runs in the first over making the run rate 6.
    (i)
    In the second over they scored 12\displaystyle 12 runs. What is the run rate now?
    (ii)
    After Over 19\displaystyle 19, their run rate was 6. What is the run rate after 20\displaystyle 20 overs given the team made 12\displaystyle 12 runs in the last over?

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    \[\text{run rate} = \frac{\text{total runs}}{\text{overs}} \](i) \[\text{run rate} = \frac{6+12}{1+1} = \frac{18}{2} = 9 \](ii) \[\text{runs after 19 overs} = 6\times 19 = 114 \] \[\text{run rate} = \frac{114+12}{19+1} = \frac{126}{20} = 6.3 \]Answer: (i) \(\displaystyle 9\) runs per over; (ii) \(\displaystyle 6.3\) runs per over.

  2. Exercise 2

    Five friends who play badminton surveyed the city and collected the price of shuttlecocks across brands and types (Nylon and Feather). The following is the list of the prices, in rupees (₹), that each one compiled. Yusuf: 40\displaystyle 40(N), 105\displaystyle 105(N), 383\displaystyle 383(F), 108\displaystyle 108(N), 165\displaystyle 165(F), 116\displaystyle 116(F) Srikanth: 194\displaystyle 194(N), 85\displaystyle 85(N), 93\displaystyle 93(N), 121\displaystyle 121(N) Kashvi: 49\displaystyle 49(N), 297\displaystyle 297(F), 105\displaystyle 105(N), 275\displaystyle 275(F), 40\displaystyle 40(N) Prasanna: 124\displaystyle 124(N), 333\displaystyle 333(F), 182\displaystyle 182(N), 258\displaystyle 258(F) Gracy: 220\displaystyle 220(F), 458\displaystyle 458(F), 129\displaystyle 129(F), 183\displaystyle 183(N) Can you describe a way they can work together to find the average price of a shuttlecock across types?
    (i)
    Each one calculated the average of the prices they had gathered. Suppose these are ay,as,ak,ap,ag\displaystyle a_{\mathrm{y}}, a_{\mathrm{s}}, a_{\mathrm{k}}, a_{\mathrm{p}}, a_{\mathrm{g}} respectively. Write an expression that gives the combined average.
    (ii)
    Suppose an,af\displaystyle a_{\mathrm{n}}, a_{\mathrm{f}} are the average prices of the nylon shuttlecocks and the feather shuttlecocks, respectively. Write an expression that gives the average price of a shuttlecock. Will this be equal to the answer we got using the method in the previous part?

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    (i) Each friend reports the average and how many prices they gathered; the combined average weights each average by that count. The counts are \(\displaystyle 6, 4, 5, 4, 4\), total \(\displaystyle 23\). \[6a_y = 917,\quad 4a_s = 493,\quad 5a_k = 766,\quad 4a_p = 897,\quad 4a_g = 990 \] \[\text{combined average} = \frac{6a_y + 4a_s + 5a_k + 4a_p + 4a_g}{6+4+5+4+4} \] \[= \frac{917+493+766+897+990}{23} = \frac{4063}{23} \approx 176.65 \](ii) There are \(\displaystyle 13\) nylon and \(\displaystyle 10\) feather prices, with total \(\displaystyle 1429\) and \(\displaystyle 2634\). \[\text{average} = \frac{13a_n + 10a_f}{13+10} = \frac{1429+2634}{23} = \frac{4063}{23} \approx 176.65 \] Yes, equal: both are the total of all prices divided by \(\displaystyle 23\). (The plain \(\displaystyle \frac{a_n+a_f}{2}\approx 186.66\) would not be.)Answer: (i) \(\displaystyle \dfrac{6a_y+4a_s+5a_k+4a_p+4a_g}{23}\), about ₹$\displaystyle 176.65$; (ii) \(\displaystyle \dfrac{13a_n+10a_f}{23}\), which is the same value.

  3. Exercise 3

    Shreyas holds 25\displaystyle 25 shares of a company at an average price of ₹150\displaystyle 150 and Vaishnavi holds 5\displaystyle 5 shares of the same company at an average price of ₹150.
    (i)
    Shreyas buys 10\displaystyle 10 shares of this company at a price of ₹30\displaystyle 30 each. What is the average price per share for him after the purchase?
    (ii)
    Vaishnavi buys some shares of this company at a price of ₹30\displaystyle 30 each and the average price per share for her after the purchase is ₹70. How many shares did she buy?

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    (i) \[\text{average} = \frac{25\times150 + 10\times30}{25+10} \] \[= \frac{3750+300}{35} = \frac{4050}{35} = \frac{810}{7} \approx 115.71 \](ii) Let Vaishnavi buy \(\displaystyle x\) shares. \[\frac{5\times150 + 30x}{5+x} = 70 \] \[750 + 30x = 350 + 70x \] \[40x = 400 \Rightarrow x = 10 \] \[\text{check: } \frac{750+300}{15} = 70 \]Answer: (i) about ₹$\displaystyle 115.71$ per share; (ii) \(\displaystyle 10\) shares.

  4. Exercise 4

    (Pṛthūdakasvāmī, commentary on Brahmagupta's Brāhmasphuṭasiddhānta, c. 864\displaystyle 864 CE) A "hasta" (meaning "forearm") refers to a unit of length measuring around 18\displaystyle 18 inches. A pool 30\displaystyle 30 hastas long is dug to different depths along its length. It is divided into 5\displaystyle 5 sections having lengths 4\displaystyle 4, 5\displaystyle 5, 6\displaystyle 6, 7\displaystyle 7, and 8\displaystyle 8 hastas, and is dug to depths of 9\displaystyle 9, 7\displaystyle 7, 7\displaystyle 7, 3\displaystyle 3, and 2\displaystyle 2 hastas, respectively. Find the mean depth of the pool.

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    The mean depth is the weighted mean of the depths, with the section lengths as weights. \[x = \frac{w_1x_1 + w_2x_2 + \cdots + w_5x_5}{w_1 + w_2 + \cdots + w_5} \] \[x = \frac{4\times9 + 5\times7 + 6\times7 + 7\times3 + 8\times2}{4+5+6+7+8} \] \[x = \frac{36+35+42+21+16}{30} = \frac{150}{30} = 5 \]Answer: mean depth \(\displaystyle =5\) hastas (about \(\displaystyle 5\times18=90\) inches).

  5. Exercise 5

    NCERT_Question_Class9_Maths_Ch10_EoC_Q5
    Suvarna had purchased 1\displaystyle 1 g gold at ₹15k (short for ₹15,000\displaystyle 15,000). This is shown by point O denoting the average price of gold that she possesses. Six different scenarios are given below for her next transaction. For each scenario estimate and mark the average price of gold she will have after the transaction.
    (i)
    Purchase 1g gold at ₹30k
    (ii)
    Purchase 2g gold at ₹30k
    (iii)
    Purchase 1g gold at ₹10k
    (iv)
    Purchase 10g gold at ₹10k
    (v)
    Purchase 0.5\displaystyle 0.5 g gold at ₹30k
    (vi)
    Purchase 0.5g gold at ₹15k

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    Weights are the grams bought; prices are in ₹k. \[\text{new average} = \frac{15\times1 + p\times g}{1+g} \](i) Equal grams, so exactly midway. \[\frac{15+30}{2} = 22.5 \](ii) More gold at 30k pulls it closer to 30k. \[\frac{15+60}{3} = 25 \](iii) Equal grams, so exactly midway. \[\frac{15+10}{2} = 12.5 \](iv) $\displaystyle 10$ g at 10k dominates, so just above 10k. \[\frac{15+100}{11} = \frac{115}{11} \approx 10.45 \](v) Half a gram at 30k moves it a third of the way from 15k. \[\frac{15+15}{1.5} = 20 \](vi) Same price as the average, so no change. \[\frac{15+7.5}{1.5} = 15 \]Answer: (i) 22.5k; (ii) 25k; (iii) 12.5k; (iv) about 10.45k; (v) 20k; (vi) 15k, at O.

  6. Exercise 6

    Given some data with corresponding weights, how would the weighted average change if all the weights are doubled? If required, experiment with some data. What do you observe? Justify your answer using algebra.

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    Doubling all the weights leaves the weighted average unchanged.Experiment: values \(\displaystyle 60, 70\) with weights \(\displaystyle 1, 3\), then \(\displaystyle 2, 6\). \[\frac{60\cdot1 + 70\cdot3}{1+3} = \frac{270}{4} = 67.5 \] \[\frac{60\cdot2 + 70\cdot6}{2+6} = \frac{540}{8} = 67.5 \]Algebra, for values \(\displaystyle x_i\) and weights \(\displaystyle w_i\): \[\frac{2w_1x_1 + \cdots + 2w_nx_n}{2w_1 + \cdots + 2w_n} = \frac{2\,(w_1x_1 + \cdots + w_nx_n)}{2\,(w_1 + \cdots + w_n)} = \frac{w_1x_1 + \cdots + w_nx_n}{w_1 + \cdots + w_n} \] The factor \(\displaystyle 2\) cancels, so only the ratio of the weights matters. Any common nonzero multiplier cancels the same way.Answer: the weighted average does not change (here \(\displaystyle 67.5\) both times).

  7. Exercise 7

    NCERT_Question_Class9_Maths_Ch10_EoC_Q7
    Answer the following questions based on the graph.
    (i)
    In 2003\displaystyle 2003, approximately how many total objects were found orbiting Earth in space? In what year did this number double?
    (ii)
    Find the approximate number and share of payload objects and other objects in the year 2025.
    (iii)
    What can you say about the number of payload objects over time, and the number of other objects over time? What can you say about the share of payload objects over time, and the share of other objects over time?

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    (i) The $\displaystyle 2003$ bar reaches the $\displaystyle 10,000$ line. \[2\times 10{,}000 = 20{,}000 \] The bars first reach $\displaystyle 20,000$ in about $\displaystyle 2021$ ($\displaystyle 2020$: about $\displaystyle 18,500$; $\displaystyle 2022$: about $\displaystyle 22,400$).(ii) The $\displaystyle 2025$ bar is about $\displaystyle 33,000$ tall, with about $\displaystyle 17,000$ payload objects. \[\text{other objects} \approx 33{,}000 - 17{,}000 = 16{,}000 \] \[\text{payload share} \approx \frac{17000}{33000} \approx 52\%, \qquad \text{other share} \approx \frac{16000}{33000} \approx 48\% \](iii) Values read from the graph: \[\begin{array}{|c|c|c|c|c|}\hline \text{Year} & \text{Payload} & \text{Other} & \text{Payload share} & \text{Other share} \\ \hline 1990 & 1{,}800 & 4{,}900 & 27\% & 73\% \\ 2000 & 2{,}300 & 6{,}900 & 25\% & 75\% \\ 2010 & 3{,}000 & 11{,}600 & 21\% & 79\% \\ 2020 & 4{,}700 & 13{,}800 & 25\% & 75\% \\ 2025 & 17{,}000 & 16{,}000 & 52\% & 48\% \\ \hline \end{array} \] Both counts keep rising: payload slowly, then steeply after $\displaystyle 2020$; other objects with a jump near 2007. Payload's share stays about a quarter (a fifth in $\displaystyle 2010$), then rises to about half; other objects' share falls correspondingly.Answer: (i) about $\displaystyle 10,000$; doubled in about 2021. (ii) payload about $\displaystyle 17,000$ ($\displaystyle 52$%), other about $\displaystyle 16,000$ ($\displaystyle 48$%). (iii) both counts rise; payload share climbs from about a quarter to about half.

  8. Exercise 8

    NCERT_Question_Class9_Maths_Ch10_EoC_Q8
    Observe the following infographic.
    (i)
    Identify the correct inference(s) from the statements given.
    (a)
    More schools have a playground in Haryana compared to Uttarakhand.
    (b)
    Approximately every 2\displaystyle 2 out of 3\displaystyle 3 schools in Arunachal Pradesh have a playground.
    (c)
    Punjab has the highest number of schools with a playground.
    (d)
    Suppose it is given that Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra.
    (ii)
    Using the information given, can we find the nation-wide percentage of schools with a playground? If not, what additional information is needed?

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    (i) \[\text{schools with a playground} = \text{percentage}\times\text{total schools} \]
    (a)
    Not valid: $\displaystyle 90$% and $\displaystyle 78$% are of different totals.
    (b)
    Valid:
    \[\frac{2}{3} \approx 0.67 \approx 68\% \]
    (c)
    Not valid: a percentage is not a count.
    (d)
    Valid: with \(\displaystyle M > T\) schools,
    \[0.93M > 0.93T > 0.74T \]
    (ii) No. The national percentage is a weighted average of the state percentages, with each state's number of schools as its weight.
    \[\text{national \%} = \frac{\sum (\text{schools}_i \times p_i)}{\sum \text{schools}_i} \]
    The total number of schools in each state/UT is needed.
    Answer: (i) (b) and (d); (ii) No, the number of schools in each state/UT is also needed.

  9. Exercise 9

    Decision Dilemma:
    (i)
    Which of the plays would you choose to watch based on the following chart?. NCERT_Question_Class9_Maths_Ch10_EoC_Q9_i
    (ii)
    The corresponding stacked bar chart is shown below. NCERT_Question_Class9_Maths_Ch10_EoC_Q9_ii Would you change your decision after looking at this chart? Why/ Why not? Discuss.

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    (i) Shares read from the chart (approximate):
    Play$\displaystyle 5$-star$\displaystyle 4$-star$\displaystyle 3$-star$\displaystyle 2$-star$\displaystyle 1$-star
    A\(\displaystyle 29\%\)\(\displaystyle 43\%\)\(\displaystyle 14\%\)\(\displaystyle 2\%\)\(\displaystyle 12\%\)
    B\(\displaystyle 24\%\)\(\displaystyle 47\%\)\(\displaystyle 15\%\)\(\displaystyle 3\%\)\(\displaystyle 11\%\)
    C\(\displaystyle 40\%\)\(\displaystyle 13\%\)\(\displaystyle 4\%\)\(\displaystyle 10\%\)\(\displaystyle 33\%\)
    \[\text{4- and 5-star: } A \approx 29+43 = 72\%, \quad B \approx 24+47 = 71\%, \quad C \approx 40+13 = 53\% \]A and B are liked by about $\displaystyle 7$ raters in $\displaystyle 10$; C is polarising, with one rater in three giving $\displaystyle 1$ star. A has the larger $\displaystyle 5$-star share, so I would pick Play A.(ii) Bar lengths give the counts (approximate):
    Play$\displaystyle 5$-star$\displaystyle 4$-star$\displaystyle 3$-star$\displaystyle 2$-star$\displaystyle 1$-starTotal
    A$\displaystyle 60$$\displaystyle 90$$\displaystyle 30$$\displaystyle 5$$\displaystyle 25$$\displaystyle 210$
    B$\displaystyle 220$$\displaystyle 440$$\displaystyle 135$$\displaystyle 30$$\displaystyle 100$$\displaystyle 925$
    C$\displaystyle 150$$\displaystyle 50$$\displaystyle 15$$\displaystyle 40$$\displaystyle 125$$\displaystyle 380$
    \[\text{4- and 5-star ratings: } A \approx 60+90 = 150, \quad B \approx 220+440 = 660, \quad C \approx 150+50 = 200 \] \[\bar{x}_A = \frac{5(60)+4(90)+3(30)+2(5)+1(25)}{210} \approx 3.7, \qquad \bar{x}_B = \frac{5(220)+4(440)+3(135)+2(30)+1(100)}{925} = \frac{3425}{925} \approx 3.7 \]Yes, I would switch to Play B: its \(\displaystyle 71\%\) rests on about $\displaystyle 925$ ratings, A's on only 210. Shares show quality; counts show how much evidence there is. Other reasoned choices are valid.Answer: (i) Play A (about \(\displaystyle 72\%\) give $\displaystyle 4$ or $\displaystyle 5$ stars, with the larger $\displaystyle 5$-star share); (ii) yes, switch to Play B, whose \(\displaystyle 71\%\) good ratings come from about $\displaystyle 925$ raters.

  10. Exercise 10

    A triathlon is an endurance race consisting of swimming 3.8\displaystyle 3.8 km, cycling 180\displaystyle 180 km, and running 42.2\displaystyle 42.2 km. Athletes compete for the fastest overall time, completing each segment in that order. The following table shows the finish time of 3\displaystyle 3 athletes in each segment. Table 10.4\displaystyle 10.4: Finish time of three athletes in each segment of a triathlon race (values in hh : mm)
    SwimmingCyclingRunning
    Athlete 1\displaystyle 101\displaystyle 01:08\displaystyle 0805\displaystyle 05:00\displaystyle 0003\displaystyle 03:15\displaystyle 15
    Athlete 2\displaystyle 201\displaystyle 01:05\displaystyle 0505\displaystyle 05:10\displaystyle 1003\displaystyle 03:35\displaystyle 35
    Athlete 3\displaystyle 301\displaystyle 01:22\displaystyle 2205\displaystyle 05:55\displaystyle 5503\displaystyle 03:50\displaystyle 50
    (i)
    What is a suitable representation of this data-a stacked bar chart or a 100\displaystyle 100% stacked bar chart? Why do you think so?
    (ii)
    Suppose a 100\displaystyle 100% stacked bar chart is drawn. Which of the following question(s) can be answered by looking at just that chart?
    (a)
    Who finished the race first?
    (b)
    Approximately what fraction of their race time did Athlete 1\displaystyle 1 spend cycling?
    (c)
    Which athlete took the longest for running?

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    (i) Total times in minutes:\[\text{Athlete 1: } 68+300+195 = 563 = 9\text{ h }23\text{ min} \] \[\text{Athlete 2: } 65+310+215 = 590 = 9\text{ h }50\text{ min} \] \[\text{Athlete 3: } 82+355+230 = 667 = 11\text{ h }7\text{ min} \]A stacked bar chart. The totals differ and decide the winner, so the bars must keep the actual times.(ii) Share of each athlete's own total time:
    AthleteSwimmingCyclingRunning
    $\displaystyle 1$\(\displaystyle 12.1\%\)\(\displaystyle 53.3\%\)\(\displaystyle 34.6\%\)
    $\displaystyle 2$\(\displaystyle 11.0\%\)\(\displaystyle 52.5\%\)\(\displaystyle 36.4\%\)
    $\displaystyle 3$\(\displaystyle 12.3\%\)\(\displaystyle 53.2\%\)\(\displaystyle 34.5\%\)
    (a) No: every bar has the same length, so the totals are lost.(b) Yes: \[\frac{300}{563} \approx 0.53 \] about half.(c) No: the shares point to Athlete $\displaystyle 2$ (\(\displaystyle 36.4\%\)), but \[195 < 215 < 230 \text{ min} \] shows Athlete $\displaystyle 3$ ran longest.Answer: (i) stacked bar chart; (ii) only (b), about \(\displaystyle 53\%\) of Athlete $\displaystyle 1$'s time was cycling.