NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 10 How Quantities Combine Understanding Data End of Chapter Exercises

End-of-Chapter Exercises

Question 1. In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6.
(i) In the second over they scored 12 runs. What is the run rate now?
(ii) After Over 19, their run rate was 6. What is the run rate after 20 overs given the team made 12 runs in the last over?
Answer:
(i) Run rate = \( \frac{6 + 12}{2} = \frac{18}{2} = 9 \). 9 runs per over.
(ii) Runs after 19 overs = 6 × 19 = 114.
Run rate after 20 overs = \( \frac{114 + 12}{20} = \frac{126}{20} = 6.3 \). 6.3 runs per over.

Teacher's Note:
1. One big over lifts the rate a lot after 1 over, but only a little after 19 overs.
2. Recover total runs first, then divide.

Question 2. Five friends who play badminton surveyed the city and collected the price of shuttlecocks across brands and types (Nylon and Feather). The prices (₹) each one compiled are:
Yusuf: 40(N), 105(N), 383(F), 108(N), 165(F), 116(F)
Srikanth: 194(N), 85(N), 93(N), 121(N)
Kashvi: 49(N), 297(F), 105(N), 275(F), 40(N)
Prasanna: 124(N), 333(F), 182(N), 258(F)
Gracy: 220(F), 458(F), 129(F), 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 a_y, a_s, a_k, a_p, a_g respectively. Write an expression that gives the combined average.
(ii) Suppose a_n, a_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?
Answer:
Working together: each friend reports how many prices they collected and their average (or total). The combined average is the weighted mean of their averages, weighted by these counts.
Counts: Yusuf 6, Srikanth 4, Kashvi 5, Prasanna 4, Gracy 4 (total 23); nylon 13, feather 10.
(i) Combined average = \( \frac{6a_y + 4a_s + 5a_k + 4a_p + 4a_g}{23} \).
(ii) Average price = \( \frac{13a_n + 10a_f}{23} \).
Yes, the two are equal: both numerators are the sum of the same 23 prices and both denominators are 23; the prices are only grouped differently (by friend or by type).
Check: friend totals 917 + 493 + 766 + 897 + 990 = 4063; nylon total 1429 and feather total 2634, also 4063. Average = 4063 ÷ 23 ≈ ₹176.65.
Note: the simple mean \( \frac{a_n + a_f}{2} = \frac{109.92 + 263.4}{2} \approx 186.66 \) would be wrong, since there are more nylon than feather shuttlecocks.

Teacher's Note:
1. However the data is grouped, weight each group average by its count.
2. Counting carefully (13 nylon, 10 feather) is the key step.

Question 3. Shreyas holds 25 shares of a company at an average price of ₹150 and Vaishnavi holds 5 shares of the same company at an average price of ₹150.
(i) Shreyas buys 10 shares of this company at a price of ₹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 each and the average price per share for her after the purchase is ₹70. How many shares did she buy?
Answer:
(i) Average = \( \frac{25 \times 150 + 10 \times 30}{25 + 10} = \frac{3750 + 300}{35} = \frac{4050}{35} \approx 115.71 \). About ₹115.71 per share.
(ii) Let her buy x shares: \( \frac{5 \times 150 + 30x}{5 + x} = 70 \).
\( 750 + 30x = 350 + 70x \Rightarrow 40x = 400 \Rightarrow x = 10 \).
Check: \( \frac{750 + 300}{15} = \frac{1050}{15} = 70 \) ✓. She bought 10 shares.

Question 4. (Pṛthūdakasvāmī, commentary on Brahmagupta's Brāhmasphuṭasiddhānta, c. 864 CE) A 'hasta' (meaning 'forearm') is a unit of length of about 18 inches. A pool 30 hastas long is dug to different depths along its length. It is divided into 5 sections having lengths 4, 5, 6, 7, and 8 hastas, and is dug to depths of 9, 7, 7, 3, and 2 hastas, respectively. Find the mean depth of the pool.
Answer:
The mean depth is the weighted mean of the depths, with the section lengths as weights (Brahmagupta's rule).
Length × depth: 4 × 9 = 36; 5 × 7 = 35; 6 × 7 = 42; 7 × 3 = 21; 8 × 2 = 16. Sum = 150; total length = 30.
Mean depth = \( \frac{150}{30} = 5 \).
The mean depth is 5 hastas (about 5 × 18 = 90 inches).

Question 5. Suvarna had purchased 1 g gold at ₹15k (short for ₹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 1 g gold at ₹30k
(ii) Purchase 2 g gold at ₹30k
(iii) Purchase 1 g gold at ₹10k
(iv) Purchase 10 g gold at ₹10k
(v) Purchase 0.5 g gold at ₹30k
(vi) Purchase 0.5 g gold at ₹15k
Answer:
In every case (prices in ₹ thousand per gram): new average = \( \frac{1 \times 15 + (\text{grams bought}) \times (\text{price})}{1 + \text{grams bought}} \).
(i) \( \frac{15 + 30}{2} = 22.5 \): equal amounts, so exactly midway.
(ii) \( \frac{15 + 60}{3} = 25 \): twice as much at 30, so two-thirds of the way to 30.
(iii) \( \frac{15 + 10}{2} = 12.5 \): midway between 15 and 10.
(iv) \( \frac{15 + 100}{11} = \frac{115}{11} \approx 10.45 \): far more gold at 10, so very close to 10.
(v) \( \frac{15 + 15}{1.5} = 20 \): less gold at 30, so only one-third of the way to 30.
(vi) \( \frac{15 + 7.5}{1.5} = 15 \): same price, so no change.

(i) ₹22.5k; (ii) ₹25k; (iii) ₹12.5k; (iv) ≈ ₹10.45k; (v) ₹20k; (vi) ₹15k (stays at O).

Teacher's Note:
1. The new average moves towards the new price in proportion to the quantity bought.
2. Buying at the same price never changes the average.

Question 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.
Answer:
Experiment: marks 60, 64, 73 with weights 3, 2, 5 give \( \frac{180 + 128 + 365}{10} = 67.3 \).
Doubled weights 6, 4, 10: \( \frac{360 + 256 + 730}{20} = \frac{1346}{20} = 67.3 \).
Observation: the weighted average does not change.
Algebra: \( \frac{2w_1x_1 + 2w_2x_2 + \cdots + 2w_nx_n}{2w_1 + 2w_2 + \cdots + 2w_n} = \frac{2(w_1x_1 + \cdots + w_nx_n)}{2(w_1 + \cdots + w_n)} = \frac{w_1x_1 + \cdots + w_nx_n}{w_1 + \cdots + w_n} \).
The factor 2 cancels. The same holds for any non-zero multiplier k: only the ratio of the weights matters, which is why weights are often given as a ratio such as 3 : 2 : 5.

Teacher's Note:
1. Scaling all weights keeps their ratio, so the mean stays put.
2. One numerical check plus the general algebra makes a complete answer.

Question 7. Answer the following questions based on the graph (cumulative number of objects currently orbiting Earth, by year of launch; *payload: satellites, spacecraft etc.; **other: mission parts, debris, rocket stages).
(i) In 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?
Answer:
(Values are read approximately from the graph.)
(i) The 2003 column reaches about 10,000 objects. Double that is about 20,000, which the total first reaches around 2020–2021, roughly 18 years later.
(ii) In 2025 the whole column is about 33,000, of which payload is about 17,000.
Payload ≈ 17,000, share ≈ \( \frac{17000}{33000} \times 100\% \approx 52\% \); other objects ≈ 16,000, share ≈ 48%.
So in 2025 the objects are roughly half payload and half other objects.
(iii) Numbers: payload objects grew slowly until about 2018 (to around 4,000–5,000), then shot up after 2019 (to about 17,000), most likely due to large satellite constellations. Other objects grew steadily, with sudden jumps around 2007 and 2009 (break-ups in orbit), reaching about 16,000.
Shares: from about 1970 to 2018 payload was only about one-fifth to one-quarter of all objects, and other objects about three-quarters or more. After 2019 the payload share climbed sharply to about one-half, and the share of other objects fell to about one-half.

Teacher's Note:
1. Read totals from the top of the column and parts from the piece lengths.
2. A share can change quickly when one part grows much faster than the other.

Question 8. Observe the infographic 'What Percentage of Schools have a Playground?' (a map of India giving, for each state, the percentage of its schools that have a playground). The values used in this question are shown below.
(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 out of 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?
Answer:
The map gives only percentages for each state, not the number of schools.
(a) Not a valid inference. Haryana's percentage (90%) is higher than Uttarakhand's (78%), but without the number of schools we cannot compare counts; a state with many more schools could have more playgrounds.
(b) Valid. Arunachal Pradesh shows 68%, close to \( \frac{2}{3} \) ≈ 66.7%, so about 2 in every 3 schools have a playground.
(c) Not a valid inference. Punjab (98%) has one of the highest percentages, but the number depends on how many schools it has. A large state with a lower percentage (e.g. Uttar Pradesh, 76%) could have many more.
(d) Valid. If Telangana has T schools and Maharashtra M > T, then 0.93M > 0.93T > 0.74T, so Maharashtra has more schools with a playground.
(ii) No. The national percentage is a weighted mean of the state percentages, weighted by the number of schools: \( \frac{p_1n_1 + p_2n_2 + \cdots + p_kn_k}{n_1 + n_2 + \cdots + n_k} \).
We need the number of schools in each state/UT. A plain average of percentages would be wrong because states differ greatly in size.
Answer: (i) (b) and (d); (ii) no, the number of schools in each state/UT is needed.

Teacher's Note:
1. Percentages alone never compare counts across groups.
2. In (d), a larger base and a larger percentage together settle it.

Question 9. Decision Dilemma: (i) Which of the plays would you choose to watch based on the following chart?
(ii) The corresponding stacked bar chart is shown below. Would you change your decision after looking at this chart? Why/Why not? Discuss.

Answer:
(Values are read approximately from the charts.)
(i) Shares: Play A: 5★ ≈ 29%, 4★ ≈ 43% (4★ or 5★ ≈ 72%), 1★ ≈ 11%. Play B: 5★ ≈ 24%, 4★ ≈ 48% (≈ 72%), 1★ ≈ 10%. Play C: 5★ ≈ 40%, 4★ ≈ 13% (≈ 53%), 1★ ≈ 32%.
Play C has the biggest 5★ share, but almost one-third of its viewers gave just 1★: it splits opinion. A and B both have about 72% good ratings and few bad ones; A has a slightly bigger 5★ share.
Based on the 100% chart, choose Play A (Play B is almost as good).
(ii) The stacked chart shows actual counts: Play A has only about 210 ratings, Play C about 380, and Play B about 930, more than four times Play A.
Many would now switch to Play B: its equally good share comes from far more viewers, so it is more reliable (and more people chose to see it). Play A's shares rest on few ratings and could change with a handful of new reviews. Some may still prefer A for its slightly higher 5★ share; the lesson is that a 100% chart hides how many people are behind each bar.

Teacher's Note:
1. A share is only as trustworthy as the number of ratings behind it.
2. Use the 100% chart for shares and the stacked chart for sample size.

Question 10. A triathlon is an endurance race consisting of swimming 3.8 km, cycling 180 km, and running 42.2 km. Athletes compete for the fastest overall time, completing each segment in that order. The finish time of 3 athletes in each segment (Table 10.4, hh : mm) is:
Athlete 1: swimming 01:08, cycling 05:00, running 03:15
Athlete 2: 01:05, 05:10, 03:35
Athlete 3: 01:22, 05:55, 03:50
(i) What is a suitable representation of this data — a stacked bar chart or a 100% stacked bar chart? Why do you think so?
(ii) Suppose a 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 spend cycling?
(c) Which athlete took the longest for running?
Answer:
In minutes: Athlete 1: 68, 300, 195 (total 563 = 9 h 23 min); Athlete 2: 65, 310, 215 (590 = 9 h 50 min); Athlete 3: 82, 355, 230 (667 = 11 h 07 min).
(i) A stacked bar chart. In a race the key fact is the total time (who finished first), along with each segment. A stacked bar shows the total as the bar length and the segments as pieces; a 100% chart makes all bars equal and hides the totals.

(ii) The 100% stacked bar chart:

(a) Cannot be answered: all bars have the same length, so total times are lost.
(b) Can be answered: Athlete 1's cycling piece is \( \frac{300}{563} \times 100\% \approx 53\% \), just over half the bar.
(c) Cannot be answered: the chart shows only each athlete's own share. Athlete 2's running share (36.4%) is bigger than Athlete 3's (34.5%), yet Athlete 3 ran for longest (3 h 50 min).
Answer: (i) stacked bar chart; (ii) only (b), about half (≈ 53%).

Teacher's Note:
1. Totals ⇒ stacked; shares ⇒ 100% stacked.
2. A bigger share does not mean a bigger amount when totals differ.

Question 11. Look at the following graph. What do you notice? What do you wonder? Write your inferences.
Distribution of disabled persons by age group and type of disability in India (Census 2011). *Recorded in the Census as 'mental retardation'.
Answer:
This is a 100% stacked bar chart: each bar shows how the disabled persons of one age group are split among the types of disability (values approximate).
What we notice:
1. Seeing disability is a bigger share for 60 & above (about 25%) than for the younger groups (about 15–19%); eyesight often weakens with age.
2. Movement disability rises steadily with age, from about 13% (0–19) to about 25% (60 & above).
3. Hearing disability is a large share (about 18–20%) in every age group.
4. Speech and intellectual disability are larger shares among the young (about 9% and 8% for 0–19) and small among the elderly (about 4% and 2%).
5. Multiple disability is highest for 60 & above (about 12%), and 'any other' is large for the younger groups (about 20%).
What we wonder:
1. These are shares, not numbers: which age group actually has the most disabled persons? The 100% chart cannot say.
2. What does 'any other' include, and why is it so large?
3. How would the chart differ for rural and urban areas, or for men and women, and how has it changed since 2011?
Inference: the type of disability depends strongly on age. Age-related conditions (seeing, movement, multiple) grow with age, while conditions present from birth or childhood (speech, intellectual) form a larger share among the young. Support services should be planned differently for different age groups.

Question 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 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% 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 months.
(d) Represent the data in a 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.
Answer:
This is a project, so answers will differ. A sample is given.
(i) Sample data for one weekday (hours): Myself: lying 8, sitting 9, standing/moving 7; Mother: 7, 5, 12; Grandfather: 9, 8, 7 (each totals 24).
Estimating: lying down = sleep + rest; sitting = school/office desk time, meals, study, TV, seated travel; standing/moving = walking, housework, cooking, games. The three must add to 24 hours.

Observation: Mother stands/moves the most (housework); Grandfather lies down the most.
(ii) Sample:
(a) Categories: Food & groceries, Rent/Housing, Education, Transport, Utilities (electricity, water, phone), Others.
(b) Keep a notebook or use bills and UPI history; record every expense under a category; total each category at month-end.
(c) Data (₹): Month 1: 9000, 12000, 5000, 3000, 2500, 3500 (total 35,000). Month 2: 9500, 12000, 5000, 3200, 2800, 2500 (total 35,000). Month 3: 10000, 12000, 8000, 3000, 3000, 4000 (total 40,000).
(d) Each amount as a percentage of its month's total, e.g. Month 1 rent = \( \frac{12000}{35000} \times 100\% \approx 34\% \).

(e) Rent is the largest share every month, but its share dropped in Month 3 (34% to 30%) although the amount did not change, because the total rose. Education jumped to 20% in Month 3 (perhaps admission or book fees). Food stays about a quarter of the budget.

Question 13. Small-group project: Make a group of 3–4 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 4–6 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 people on a scale of 1–5 (5 being very good).
(iv) Compute each individual rating. Compute the overall average across individuals.
(v) Make a 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.
Answer:
Sample for scenario (b), bus travel:
(i) and (ii) Safety (careful driving, safe vehicle; most important), weight 4. Punctuality (reaching on time), weight 3. Comfort (seats, crowding, ventilation), weight 2. Cleanliness (hygiene), weight 2. Staff behaviour (polite, helpful crew), weight 1. Total weight 12.
(iii) Imagined ratings from 10 passengers (number giving 5, 4, 3, 2, 1 stars) and averages:
Safety: 2, 4, 3, 1, 0 → 3.7; Punctuality: 1, 3, 4, 1, 1 → 3.2; Comfort: 2, 5, 2, 1, 0 → 3.8; Cleanliness: 4, 4, 1, 1, 0 → 4.1; Staff: 3, 4, 2, 1, 0 → 3.9.
(iv) One passenger (Safety 4, Punctuality 3, Comfort 4, Cleanliness 5, Staff 4): \( \frac{4 \times 4 + 3 \times 3 + 4 \times 2 + 5 \times 2 + 4 \times 1}{12} = \frac{47}{12} \approx 3.92 \).
The average of all 10 individual ratings equals the weighted mean of the aspect averages: \( \frac{3.7 \times 4 + 3.2 \times 3 + 3.8 \times 2 + 4.1 \times 2 + 3.9 \times 1}{12} = \frac{44.1}{12} \approx 3.68 \).
(v)

(vi) The bus does well on cleanliness and staff but poorly on punctuality. Because safety and punctuality carry high weights, the overall 3.68 is a little below the plain average of the five aspects (18.7 ÷ 5 = 3.74). The scheme reflects what passengers value most, but misses fare, route coverage and frequency, and the weights are our own judgement; another group might choose differently.

Question 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.
Answer:
Families have different sizes, so the family averages must not simply be averaged. Instead:
1. For each family, total age = average age × number of members.
2. Add the total ages of all families.
3. Add the numbers of members of all families.
4. Divide: average age = \( \frac{a_1n_1 + a_2n_2 + \cdots + a_kn_k}{n_1 + n_2 + \cdots + n_k} \), where a_i is the average age and n_i the size of family i (a weighted mean with family sizes as weights).
Example with 4 students: (38 years, 4 members), (30, 5), (45, 3), (28, 6). Totals: 152 + 150 + 135 + 168 = 605 years over 18 people.
Average age = 605 ÷ 18 ≈ 33.6 years. (The plain average of the four averages, 141 ÷ 4 = 35.25, would be wrong.)

Question 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? If required, experiment with some data. What do you observe? Justify your answer using algebra.
Answer:
Experiment: 60, 64, 73 with weights 3, 2, 5 give 67.3. Adding 1 to each weight (4, 3, 6): \( \frac{240 + 192 + 438}{13} = \frac{870}{13} \approx 66.92 \).
The simple mean is \( \frac{60 + 64 + 73}{3} \approx 65.67 \).
Observation: unlike doubling, adding a constant usually changes the weighted average, pulling it towards the simple mean (here from 67.3 towards 65.67), because the weights become more nearly equal (3 : 2 : 5 becomes 4 : 3 : 6).
Algebra: let W = w₁ + … + w_n, S_w = w₁x₁ + … + w_nx_n, S = x₁ + … + x_n.
New mean = \( \frac{(w_1 + 1)x_1 + \cdots + (w_n + 1)x_n}{(w_1 + 1) + \cdots + (w_n + 1)} = \frac{S_w + S}{W + n} = \frac{W \cdot \frac{S_w}{W} + n \cdot \frac{S}{n}}{W + n} \).
So the new value is itself a weighted mean of the old weighted mean (weight W) and the simple mean (weight n); it lies between them.
It is unchanged only if the weighted mean already equals the simple mean (e.g. all weights equal, or all data equal). Larger added constants push it ever closer to the simple mean.

Teacher's Note:
1. Multiplying weights keeps their ratio; adding to them changes it.
2. The key algebraic step rewrites the new mean as a mix of two means.

Question 16. A farm has some cows, sheep, and chickens. Last year the cows made up 60%, the sheep 25%, and the chickens 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.
Answer:
The three shares always add up to 100%, so they cannot all fall or all rise.
(i) Impossible: three smaller shares would total less than 100%.
(ii) Impossible: three larger shares would total more than 100%.
Take last year as 60 cows, 25 sheep, 15 chickens (total 100).
(iii) Possible. 48 cows, 20 sheep, 12 chickens (total 80): all numbers fall, shares stay \( \frac{48}{80} = 60\%, \ \frac{20}{80} = 25\%, \ \frac{12}{80} = 15\% \).
(iv) Possible. 44 cows, 24 sheep, 12 chickens (total 80): shares 55% (down), 30% (up), 15% (same).
(v) Possible. 50 cows, 20 sheep, 5 chickens (total 75): shares ≈ 66.7% (up), ≈ 26.7% (up), ≈ 6.7% (down).
Possible: (iii), (iv) and (v). Impossible: (i) and (ii). A fall in number does not decide the change in share; that depends on how much the total falls.

Teacher's Note:
1. Shares are tied together by the 100% total.
2. One concrete example is enough to show a case is possible.