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

33 questions · 33 still being checked

End-of-Chapter Exercises 11–16 (part 7 of 7)

  1. Exercise 11

    NCERT_Question_Class9_Maths_Ch10_EoC_Q11 Look at the following graph. What do you notice? What do you wonder? Write your inferences.

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    Approximate shares (in \(\displaystyle \%\)) read from the chart:
    Age groupSeeingHearingSpeechMovementMental retardationMental illnessAny otherMultiple
    $\displaystyle 0$-$\displaystyle 19$$\displaystyle 18$$\displaystyle 20$$\displaystyle 9$$\displaystyle 13$$\displaystyle 8$$\displaystyle 2$$\displaystyle 21$$\displaystyle 9$
    $\displaystyle 20$-$\displaystyle 39$$\displaystyle 15$$\displaystyle 18$$\displaystyle 9$$\displaystyle 22$$\displaystyle 7$$\displaystyle 3$$\displaystyle 20$$\displaystyle 6$
    $\displaystyle 40$-$\displaystyle 59$$\displaystyle 19$$\displaystyle 18$$\displaystyle 8$$\displaystyle 23$$\displaystyle 5$$\displaystyle 4$$\displaystyle 17$$\displaystyle 6$
    $\displaystyle 60$ and above$\displaystyle 25$$\displaystyle 19$$\displaystyle 4$$\displaystyle 25$$\displaystyle 2$$\displaystyle 2$$\displaystyle 11$$\displaystyle 12$
    \[\text{Movement: } 13 \to 22 \to 23 \to 25, \qquad \text{Any other: } 21 \to 20 \to 17 \to 11 \]Notice: each bar totals \(\displaystyle 100\%\). With age, movement and seeing disabilities take a larger share; 'any other', speech and mental retardation take a smaller one. Multiple disability is highest at $\displaystyle 60$ and above.Wonder: how many persons each bar stands for, why movement disability grows with age, and what 'any other' contains.Inference: shares carry no counts, so a share rising does not mean that disability became more common. Other observations are valid.Answer: Movement (\(\displaystyle 13\%\to25\%\)) and seeing (\(\displaystyle 18\%\to25\%\)) take a growing share with age, 'any other' a shrinking one (\(\displaystyle 21\%\to11\%\)); the chart gives shares only, and other observations are equally valid.

  2. Exercise 12

    Individual project: Do at least one of the following.
    (i)
    Recall the previous day and fill in the approximate time spent lying down, sitting, and standing/moving around, etc. Ask at least 2\displaystyle 2 of your family members about their day and fill it in. Alternatively, you can choose to track just your own body-state for any one of the weekdays, Saturday, and Sunday. Visualise it using a stacked bar chart. Discuss with your class how you arrived at the estimated time spent in each state.
    (ii)
    Visualise your family's monthly expenditure using a 100\displaystyle 100% stacked bar chart.
    (a)
    First, identify the expense categories in your family budget.
    (b)
    Discuss with your family members to decide how you will collect and organise the data.
    (c)
    Collect expense data for at least 3\displaystyle 3 months.
    (d)
    Represent the data in a 100\displaystyle 100% stacked bar chart, with each bar showing how the total monthly expenditure is divided across the different categories.
    (e)
    Finally, write your observations and inferences based on the chart.

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    (i) A weekday, in hours:
    PersonLying downSittingStanding or moving
    Student$\displaystyle 9$$\displaystyle 7$$\displaystyle 8$
    Mother$\displaystyle 8$$\displaystyle 4$$\displaystyle 12$
    Father$\displaystyle 7$$\displaystyle 10$$\displaystyle 7$
    \[9+7+8 = 24, \qquad 8+4+12 = 24, \qquad 7+10+7 = 24 \]I listed each day in blocks (sleep, school, meals, chores) and adjusted them to total $\displaystyle 24$ h. All bars have length $\displaystyle 24$ h, so this stacked bar chart is also a \(\displaystyle 100\%\) one.(ii) Sample family, in ₹:(a), (b) Six categories; one person records bills in a notebook, totalling monthly.(c)
    MonthFoodHousingUtilitiesTransportEducationOtherTotal
    $\displaystyle 1$$\displaystyle 8000$$\displaystyle 5000$$\displaystyle 1600$$\displaystyle 2000$$\displaystyle 2000$$\displaystyle 1400$$\displaystyle 20000$
    $\displaystyle 2$$\displaystyle 9000$$\displaystyle 5000$$\displaystyle 2000$$\displaystyle 2500$$\displaystyle 2500$$\displaystyle 4000$$\displaystyle 25000$
    $\displaystyle 3$$\displaystyle 7000$$\displaystyle 5000$$\displaystyle 1400$$\displaystyle 2000$$\displaystyle 2000$$\displaystyle 2600$$\displaystyle 20000$
    (d) \[\text{share} = \frac{\text{category expense}}{\text{month total}} \times 100, \qquad \text{e.g. } \frac{8000}{20000} \times 100 = 40\% \]
    MonthFoodHousingUtilitiesTransportEducationOther
    $\displaystyle 1$\(\displaystyle 40\%\)\(\displaystyle 25\%\)\(\displaystyle 8\%\)\(\displaystyle 10\%\)\(\displaystyle 10\%\)\(\displaystyle 7\%\)
    $\displaystyle 2$\(\displaystyle 36\%\)\(\displaystyle 20\%\)\(\displaystyle 8\%\)\(\displaystyle 10\%\)\(\displaystyle 10\%\)\(\displaystyle 16\%\)
    $\displaystyle 3$\(\displaystyle 35\%\)\(\displaystyle 25\%\)\(\displaystyle 7\%\)\(\displaystyle 10\%\)\(\displaystyle 10\%\)\(\displaystyle 13\%\)
    (e) Housing is fixed at ₹$\displaystyle 5000$, so its share falls when the total rises; the extra ₹$\displaystyle 5000$ in month $\displaystyle 2$ shows only in the totals.Answer: Example: (i) three people, each bar $\displaystyle 24$ h long; (ii) three months of family expenses, housing at \(\displaystyle 25\%, 20\%, 25\%\) of the total. Other data and categories are equally valid.

  3. Exercise 13

    Small-group project: Make a group of 3\displaystyle 3-4\displaystyle 4 students. Choose one scenario to design a custom rating scheme by assigning appropriate weights:
    (a)
    shopping experience at a cloth store,
    (b)
    travel experience in a bus,
    (c)
    tourism experience of a nearby tourist spot,
    (d)
    clinic/hospital experience
    (i)
    Discuss and arrive at 4\displaystyle 4-6\displaystyle 6 aspects to rate. For each aspect chosen, justify why it matters for an overall experience.
    (ii)
    Discuss and decide the relative weights and justify.
    (iii)
    Collect or imagine ratings from at least 10\displaystyle 10 people on a scale of 1\displaystyle 1-5\displaystyle 5 (5\displaystyle 5 being very good).
    (iv)
    Compute each individual rating. Compute the overall average across individuals.
    (v)
    Make a 100\displaystyle 100% stacked bar chart using your data.
    (vi)
    Write a short note on the observations and inferences, what your rating system captures well and what it misses.

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    Scenario (b), a bus journey.(i) Safety (risk to life), punctuality (time lost), cleanliness (health), seat comfort (long trips), staff behaviour (help when needed).(ii) Weights by importance: \[\text{Safety} : \text{Punctuality} : \text{Cleanliness} : \text{Comfort} : \text{Staff} = 4 : 3 : 2 : 2 : 1 \] Safety matters most; staff behaviour rarely decides a journey.(iii) Ratings ($\displaystyle 1$ to $\displaystyle 5$) from ten riders (imagined), weights in brackets; the last column is from (iv):
    RiderSafety ($\displaystyle 4$)Punctuality ($\displaystyle 3$)Cleanliness ($\displaystyle 2$)Comfort ($\displaystyle 2$)Staff ($\displaystyle 1$)Rating
    $\displaystyle 1$$\displaystyle 5$$\displaystyle 4$$\displaystyle 3$$\displaystyle 3$$\displaystyle 4$$\displaystyle 4.00$
    $\displaystyle 2$$\displaystyle 4$$\displaystyle 3$$\displaystyle 2$$\displaystyle 3$$\displaystyle 5$$\displaystyle 3.33$
    $\displaystyle 3$$\displaystyle 5$$\displaystyle 5$$\displaystyle 4$$\displaystyle 4$$\displaystyle 4$$\displaystyle 4.58$
    $\displaystyle 4$$\displaystyle 3$$\displaystyle 2$$\displaystyle 2$$\displaystyle 1$$\displaystyle 3$$\displaystyle 2.25$
    $\displaystyle 5$$\displaystyle 4$$\displaystyle 4$$\displaystyle 3$$\displaystyle 3$$\displaystyle 4$$\displaystyle 3.67$
    $\displaystyle 6$$\displaystyle 5$$\displaystyle 3$$\displaystyle 3$$\displaystyle 2$$\displaystyle 4$$\displaystyle 3.58$
    $\displaystyle 7$$\displaystyle 4$$\displaystyle 2$$\displaystyle 1$$\displaystyle 2$$\displaystyle 3$$\displaystyle 2.58$
    $\displaystyle 8$$\displaystyle 5$$\displaystyle 4$$\displaystyle 4$$\displaystyle 4$$\displaystyle 5$$\displaystyle 4.42$
    $\displaystyle 9$$\displaystyle 3$$\displaystyle 3$$\displaystyle 2$$\displaystyle 2$$\displaystyle 3$$\displaystyle 2.67$
    $\displaystyle 10$$\displaystyle 4$$\displaystyle 4$$\displaystyle 3$$\displaystyle 4$$\displaystyle 4$$\displaystyle 3.83$
    (iv) \[\text{rating} = \frac{\sum w\,r}{\sum w}, \qquad \text{Rider 1: } \frac{4(5)+3(4)+2(3)+2(3)+1(4)}{4+3+2+2+1} = \frac{48}{12} = 4.00 \] \[\text{overall average} = \frac{48+40+55+27+44+43+31+53+32+46}{12 \times 10} = \frac{419}{120} \approx 3.49 \](v) Share of the ten riders giving each rating ($\displaystyle 5$ to $\displaystyle 1$):
    Aspect$\displaystyle 5$$\displaystyle 4$$\displaystyle 3$$\displaystyle 2$$\displaystyle 1$
    Safety\(\displaystyle 40\%\)\(\displaystyle 40\%\)\(\displaystyle 20\%\)\(\displaystyle 0\%\)\(\displaystyle 0\%\)
    Punctuality\(\displaystyle 10\%\)\(\displaystyle 40\%\)\(\displaystyle 30\%\)\(\displaystyle 20\%\)\(\displaystyle 0\%\)
    Cleanliness\(\displaystyle 0\%\)\(\displaystyle 20\%\)\(\displaystyle 40\%\)\(\displaystyle 30\%\)\(\displaystyle 10\%\)
    Comfort\(\displaystyle 0\%\)\(\displaystyle 30\%\)\(\displaystyle 30\%\)\(\displaystyle 30\%\)\(\displaystyle 10\%\)
    Staff\(\displaystyle 20\%\)\(\displaystyle 50\%\)\(\displaystyle 30\%\)\(\displaystyle 0\%\)\(\displaystyle 0\%\)
    (vi) Safety ($\displaystyle 4.2$) and staff behaviour ($\displaystyle 3.9$) score well; cleanliness ($\displaystyle 2.7$) and comfort ($\displaystyle 2.8$) lag. The weighted rating shows priorities but hides spread, since one rider gave cleanliness $\displaystyle 1$, and it misses fare and route.Answer: Bus journey rated on safety, punctuality, cleanliness, comfort and staff with weights \(\displaystyle 4:3:2:2:1\); overall average \(\displaystyle \approx 3.49\) out of 5. Other scenarios, aspects and weights are equally valid.

  4. Exercise 14

    Whole class project: Each student shares the average age of their family and the number of family members. Discuss among the class and come up with a way to find out the average age of all the families of the class.

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    Sample data for five families:
    StudentAverage ageMembersTotal age
    $\displaystyle 1$$\displaystyle 28$$\displaystyle 4$$\displaystyle 112$
    $\displaystyle 2$$\displaystyle 32$$\displaystyle 5$$\displaystyle 160$
    $\displaystyle 3$$\displaystyle 24$$\displaystyle 3$$\displaystyle 72$
    $\displaystyle 4$$\displaystyle 35$$\displaystyle 6$$\displaystyle 210$
    $\displaystyle 5$$\displaystyle 30$$\displaystyle 4$$\displaystyle 120$
    \[\text{class average} = \frac{n_1a_1 + n_2a_2 + \cdots + n_ka_k}{n_1 + n_2 + \cdots + n_k} \] \[\frac{112+160+72+210+120}{4+5+3+6+4} = \frac{674}{22} \approx 30.6 \text{ years} \] \[\frac{28+32+24+35+30}{5} = 29.8 \neq 30.6 \]Each family's average is weighted by its number of members. The plain mean of the averages differs, as it ignores family size. Siblings in the class share one family, counted once. Pooling every member's age gives the same value.Answer: Weighted mean of the family averages, with family sizes as weights: \(\displaystyle \dfrac{\sum n_ia_i}{\sum n_i}\) (sample data: \(\displaystyle \approx 30.6\) years).

  5. Exercise 15

    Given some data with corresponding weights, what would happen to the weighted average if all the weights are increased by a constant value, say 1\displaystyle 1? If required, experiment with some data. What do you observe? Justify your answer using algebra.

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    Let the weights be \(\displaystyle w_1,\dots,w_n\), with \(\displaystyle W = \sum w_i\), weighted average \(\displaystyle \bar{x}_w\) and ordinary mean \(\displaystyle \bar{x}\).\[\bar{x}_w = \frac{\sum w_ix_i}{W}, \qquad \bar{x} = \frac{\sum x_i}{n} \] \[x_{\text{new}} = \frac{\sum (w_i+1)x_i}{\sum (w_i+1)} = \frac{\sum w_ix_i + \sum x_i}{W+n} = \frac{W\bar{x}_w + n\bar{x}}{W+n} \] \[x_{\text{new}} - \bar{x}_w = \frac{n(\bar{x} - \bar{x}_w)}{W+n} \]Experiment: data \(\displaystyle 60, 64, 73\) with weights \(\displaystyle 3, 2, 5\).\[\bar{x}_w = \frac{673}{10} = 67.3, \qquad \bar{x} = \frac{197}{3} \approx 65.67 \] \[x_{\text{new}} = \frac{60(4)+64(3)+73(6)}{4+3+6} = \frac{870}{13} \approx 66.92 \] \[\text{check: } \frac{10(67.3)+3\left(\dfrac{197}{3}\right)}{13} = \frac{673+197}{13} = \frac{870}{13} \]The new average is a weighted mean of \(\displaystyle \bar{x}_w\) and \(\displaystyle \bar{x}\), so it lies between them: it moves toward the ordinary mean. It is unchanged only if \(\displaystyle \bar{x}_w = \bar{x}\), for example when all weights are equal. Unlike doubling the weights, adding a constant changes the weighted average in general.Answer: The weighted average moves toward the ordinary mean by \(\displaystyle \dfrac{n(\bar{x}-\bar{x}_w)}{W+n}\); it stays the same only if \(\displaystyle \bar{x}_w = \bar{x}\).

  6. Exercise 16

    A farm has some cows, sheep, and chickens. Last year the cows made up 60%\displaystyle 60 \%, the sheep 25%\displaystyle 25 \%, and the chickens 15%\displaystyle 15 \%. There was a decrease in the number of all three animals' population over the year. Choose the possibilities for the change in their respective shares of the population -
    (i)
    % of cows decreased, % of sheep decreased, % of chickens decreased
    (ii)
    % of cows increased, % of sheep increased, % of chickens increased
    (iii)
    % of cows remained the same, % of sheep remained the same, % of chickens remained the same
    (iv)
    % of cows decreased, % of sheep increased, % of chickens remained the same
    (v)
    % of cows increased, % of sheep increased, % of chickens decreased.

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    Shares always add up to \(\displaystyle 100\%\): \[60\% + 25\% + 15\% = 100\% \] Take last year's counts as \(\displaystyle 60, 25, 15\).(i) Not possible. If each share fell, the three new shares would add up to less than \(\displaystyle 60+25+15 = 100\), but they must total \(\displaystyle 100\%\).(ii) Not possible. If each share rose, the total would exceed \(\displaystyle 100\%\).(iii) Possible. Each population falls by the same fraction: \[(60, 25, 15) \to (48, 20, 12), \qquad \frac{48}{80} = 60\%, \quad \frac{20}{80} = 25\%, \quad \frac{12}{80} = 15\% \](iv) Possible. \[(60, 25, 15) \to (44, 24, 12), \qquad \frac{44}{80} = 55\%, \quad \frac{24}{80} = 30\%, \quad \frac{12}{80} = 15\% \] Every count fell; the cows' share fell, the sheep's rose, the chickens' stayed the same.(v) Possible. \[(60, 25, 15) \to (50, 22, 8), \qquad \frac{50}{80} = 62.5\%, \quad \frac{22}{80} = 27.5\%, \quad \frac{8}{80} = 10\% \] Every count fell; the cows' and sheep's shares rose, the chickens' fell.Answer: (iii), (iv) and (v) are possible; (i) and (ii) are not.