Data and statistics: Questions and Answers

57 data and statistics questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• In a badminton academy, there is a group of 11 trainees — 8 seniors and 3 juniors. Their heights and average heights (in cm) are given in the table below. Find the average height of the whole group.
  Two students calculate the average height of the whole group in two ways:
  Method 1: 165.5 + 149.332 = 314.832 = 157.415
  Method 2: 165 + 169 + 164 + 167 + 170 + 159 + 164 + 166 + 146 + 149 + 15311 = 177211 = 161.09
  Whose calculation gives the correct average height of the whole group?Answer: The average height of the whole group is 161.09 cm. Method 2 is correct: it adds all 11 heights and divides by 11.
• Why did Method 1 not work?Answer: Method 1 adds the two averages and divides by 2, so it treats the seniors’ average and the juniors’ average equally. But there are 8 seniors and only 3 juniors, so the seniors’ average should count much more.
• Shreyas calculated it differently as (165.5 × 8) + (149.33 × 3)8 + 3 = 1324 + 44811 = 177211 = 161.09. Do you understand why this also works? Can you see why 165.5 × 8 gives the sum of the heights of all the seniors and 149.33 × 3 gives the sum of the heights of all the juniors?Answer: Since average = sumnumber of values, we get sum = average × number of values. So 165.5 × 8 = 1324 cm is the total height of the 8 seniors and 149.33 × 3 ≈ 448 cm is the total of the 3 juniors. Adding them gives the total of all 11 heights, and dividing by 11 gives the correct average, 161.09 cm.
• Jaspreet recently learnt cycling. She has explored different routes in her town. She tracked how much time she cycled on weekdays over the last 3 weeks. Find the mean time spent cycling per weekday over the last 3 weeks.Answer: Mean time = 64 + 79 + 9015 = 23315 ≈ 15.53 minutes per weekday. Averaging the three weekly averages, 12.8 + 15.8 + 183, gives the same 15.53 because every week has 5 days.
• We saw earlier how Method 1 (badminton example) did not produce the correct value. Why does Method 1 give the correct answer in this case? When does Method 1 work and when does it not work?Answer: Here every week has the same number of days (5), so ap + bp + cpp + p + p = a + b + c3. Method 1 (averaging the averages) works when all the collections are the same size; it does not work when the sizes differ, as with 8 seniors and 3 juniors.
• Two glasses of equal quantities of lemonade are prepared. One glass has 10% jaggery and the other has 20% jaggery. If we mix the lemonade from both glasses, what is the concentration of jaggery in the mixture?Answer: The mixture has 15% jaggery, midway between 10% and 20%, because the two glasses hold equal quantities.
• A bowl of 500 mL lemonade has 10% jaggery. A glass of 200 mL lemonade has 20% jaggery. If we mix the lemonade from the bowl and the glass, what is the concentration of jaggery in the mixture?
  Can you estimate what part of the mixture is jaggery? Is it 15%, or is it more or less? Why do you think so?Answer: Estimate: less than 15%, closer to 10%, because more of the mixture comes from the 10% bowl. Exactly: 500 × 0.1 + 200 × 0.2500 + 200 = 90700 ≈ 0.13, i.e. about 13% (12.86%).
• Brass is an alloy of copper and zinc. Batch A of brass weighs 200 kg of which 70% is copper. Batch B weighs 120 kg of which 50% is copper. Batch C is 45% copper. When all three batches are combined, we get an alloy that is 55% copper. What is the weight of Batch C?Answer: Batch C weighs 240 kg. From 0.55 = 200 × 0.7 + 120 × 0.5 + 0.45y200 + 120 + y we get 0.55(320 + y) = 200 + 0.45y, so y = 240.
• 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?Answer: Combined average = 72 × 30 + 76 × 2530 + 25 = 406055 ≈ 73.82.
• A farmer mixes three equal quantities of fertilisers. The first one contains 110 nitrogen, the second contains 950 nitrogen, and the third contains 360 nitrogen. What is the fraction of nitrogen in the mixture?Answer: Equal quantities, so the fraction is the simple mean: 13(110 + 950 + 360) = 13 × 33100 = 11100 (that is 0.11 or 11%).
• (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.Answer: Purity = 9 × 12 + 5 × 10 + 17 × 119 + 5 + 17 = 34531 ≈ 11.13 varṇa (exactly 11431 varṇa).
• 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.Answer: May and July have 31 days and June has 30, so the combined average = 3.5 × 31 + 10 × 30 + 8.7 × 3131 + 30 + 31 mm per day ≈ 7.37 mm per day.
• Calculate the concentration of spice mix in these two scenarios.Answer: (i) 7000600 ≈ 11.67% (ii) 5000600 ≈ 8.33%. Same three concentrations, but the answer moves towards whichever one comes in the largest quantity.
• Rehmat’s marks in Maths are as follows: 60% in internal tests, 64% in the project, 73% in the final exam. The annual percentage score is calculated by combining the internals, project, and final exam in the ratio 3 : 2 : 5. What is Rehmat’s annual score in Maths?Answer: Annual score = 60 × 3 + 64 × 2 + 73 × 53 + 2 + 5 = 67310 = 67.3%.
• Savitri’s marks in Kashmiri are as follows: 35 out of 50 in internal tests, 44 out of 60 in the project, and 80 out of 100 in the final exam. The annual percentage score is calculated by combining the internals, project, and final exam in the ratio 3 : 4 : 5. Which of the following expression(s) gives her annual score (as a percentage) in Kashmiri?
  (i) 35 × 3 + 44 × 4 + 80 × 53 + 4 + 5
  (ii) (35/100) × 3 + (44/100) × 4 + (80/100) × 53 + 4 + 5
  (iii) (35/50) × 3 + (44/60) × 4 + (80/100) × 53 + 4 + 5 × 100
  (iv) ((35/50) × 100) × 3 + ((44/60) × 100) × 4 + ((80/100) × 100) × 53 + 4 + 5Answer: (iii) and (iv). Both first turn each mark into a fraction of its own maximum (35 out of 50, 44 out of 60, 80 out of 100) and then take the weighted mean with weights 3, 4, 5. Her annual score ≈ 75.28%.
• A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales?Answer: Profit = 0.3 × 8000 + 0.5 × 1000 = ₹2900 on sales of ₹9000, so profit % = 29009000 × 100 ≈ 32.22%.
• A white stork’s migration is tracked. The average daily distance travelled, calculated over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily distance travelled over these 21 days? Make a guess before you calculate.Answer: Guess: a little more than 44.5 km. Exactly: 20 × 44.5 + 5521 = 94521 = 45 km per day.
• A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture?
  (i) Salt: 5%, Sugar: 8%   (ii) Salt: 13%, Sugar: 3%   (iii) Salt: 6%, Sugar: 6%
  (iv) Salt: 5.55%, Sugar: 8.88%   (v) Salt: 3.33%, Sugar: 2.67%   (vi) Salt: 4.1%, Sugar: 7.08%Answer: (v) Salt: 3.33%, Sugar: 2.67%. Salt = 30 mL and sugar = 24 mL, each now spread through 900 mL.
• At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into the 10 litres of pani so that the spice level is reduced to (34)^th of the original concentration?Answer: Target = 34 × 8% = 6%. Spice = 0.8 L, so 0.810 + x = 0.06 ⇒ x = 103 ≈ 3.33 litres of water.
• A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio 4 : 5 : 6. Rashi has scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively.Answer: (i) Keerthi: she is 5 marks lower in strength (weight 4) but 5 marks higher in agility (weight 6), which counts more. (ii) Rashi = 98515 ≈ 65.67, Keerthi = 99515 ≈ 66.33.
• A restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting data is shown in the table below. What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?Answer: Averages: food 4.3, ambience 3.3, service 2.8. Combined = 6 × 4.3 + 5 × 3.3 + 4 × 2.815 = 53.515 ≈ 3.57.
• The following table shows the weight data of langurs in an animal facility. Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number?Answer: More female langurs (14.925 is closer to 13.8 than to 16.5). (i) (a) (and (b) as well if its printed minus sign is a slip). (ii) 25 males (and 35 females). (iii) ≈ 13.84 kg. (iv) 16.5 kg (unchanged). (v) ≈ 16.46 kg.
• Dorjee has collected 1 litre of water from the Dead Sea! Using the information in the table, answer the following questions. A calculator can be used if necessary.Answer: (i) 0.34 + 2 × 0.000013 ≈ 11.33%. (ii) Yes, since 0.01% lies between 0.001% and 34%; he needs about 3777 litres (3776.67 L) of purified water. (iii) No: both waters are saltier than 0.001%, so any mixture is at least 0.01% salty.
• The following table shows the average weekly expenditure, in rupees (₹), of three families across categories. What would a cluster-column chart for this data look like?Answer: It is a set of clusters of columns standing side by side on one scale (0 to 2000 ₹). There are two choices: cluster by category (6 clusters, each with 3 columns for Families A, B, C) or cluster by family (3 clusters, each with 6 columns, one per category).
• Answer the following questions based on the charts above. Try to answer each question by looking at Choice 1 first, and then Choice 2. Which was easier?Answer: 1. Family C. 2. Family A. 3. About ₹1300 (₹1280). 4. Housing. 5. Healthcare (₹1770, slightly more than food, ₹1700). Choice 1 is easier for 1 and 2 (comparing families); Choice 2 is easier for 4 and 5 (comparing categories within one family).
• Find the answers to the following questions.Answer: (i) Family C (₹7040). (ii) About ₹7000 (₹7040). (iii) About 13–14% (9507040 × 100 ≈ 13.5%).
• Was it as straightforward to answer these questions as the earlier questions? Is there a visualisation that makes it simpler to answer such questions? Look at the following visualisation of the same data. Try to understand how the chart is organised and answer the three questions above.Answer: No: with cluster charts we had to add six columns for each total. A stacked bar chart makes it simple, because the whole length of each bar is the family’s total. (i) Family C (shortest bar). (ii) About ₹7000. (iii) The housing piece is about one-seventh of Family C’s bar, about 13–14%.
• Now, use the stacked bar chart to answer the five questions shown earlier. Is it possible? Is it easy? If not, what made it harder than before?Answer: It is possible but harder. The answers are the same (1. Family C, 2. Family A, 3. about ₹1300, 4. Housing, 5. Healthcare), but only the first piece of each bar starts at 0; the other pieces start at different places, so their lengths must be worked out as end − start and compared without a common base.
• The following stacked column chart shows the number of animal species in the IUCN Red List, by class, over time.Answer: (i) The total number of animal species (all classes together) on the list in 2019: 14,234. (ii) About 1,100 reptile species in 2016. (iii) Mammals and birds grew the least (only by about 150–250 each).
• The wickets taken by a bowler in International Cricket matches till 2025 are shown in the table. Complete the given stacked bar charts (the bar lengths can be approximate),Answer: (i) Home: 62 + 100 + 32 = 194; Overseas: 172 + 49 + 71 = 292. (ii) Test: 62 + 172 = 234; ODI: 100 + 49 = 149; T20: 32 + 71 = 103.
• Try answering the following question based on the expenditure stacked bar chart you saw earlier: Which family has the smallest share of their total expenditure towards healthcare? Was it easy? If not, what visualisation can help answer such questions easily?Answer: Family B (12107286 ≈ 16.6%, against 21.0% for A and 18.5% for C). It is not easy from the stacked bar, because the bars have different lengths. Pie charts or a 100% stacked bar chart show shares directly.
• The average category-wise daily electricity consumption of a house, in kWh, in 2005 and 2025 is shown in the table below. The corresponding stacked bar chart is also shown. Make the corresponding 100% stacked bar chart.Answer: 2005 (total 1000): lighting 36%, cooling 30%, kitchen 12%, other 22%. 2025 (total 2000): lighting 11%, cooling 40%, kitchen 23%, other 26%. Draw two bars of equal length split in these percentages.
• What do you observe? What do you find interesting in the data/graphs? Discuss with others.Answer: Total use doubled (1000 → 2000). Cooling became the biggest user (30% → 40%, 300 → 800). Lighting fell both in amount (360 → 220) and share (36% → 11%). Kitchen and other appliances grew in amount and share.
• The 100% stacked bar below compares the proportion of flowers blooming in Fatima’s garden and Naveen’s garden across seasons. Which of the following statements can be inferred from this chart?
  (Hint: Find out the blooms per season in each garden if both the gardens had the same total number of blooms over the year. Similarly find out what happens if the totals are very different.)Answer: Only (i) can be inferred. (ii) and (iii) compare numbers across gardens and (iv) is about totals, but a 100% stacked bar shows only fractions within each garden, not actual numbers.
• 1. What does this say about the scope of the stacked bars and 100% stacked bars?
  2. Given a stacked bar chart can we make a corresponding 100% stacked bar chart?
  3. Given a 100% stacked bar chart can we make a corresponding stacked bar chart?
  4. What kind of inferences or comparisons can be made from a stacked bar chart and in a 100% stacked bar chart?Answer: 1. Stacked bars show amounts (and totals); 100% stacked bars show only shares. 2. Yes: divide each piece by its bar’s total. 3. No, not without the totals: many stacked charts give the same 100% chart. 4. Stacked: compare totals and actual amounts; 100%: compare proportions within and across bars.
• The following chart shows the time spent on different activities in a day by the average Indian. What do you notice? What do you wonder about?Answer: Sleep takes the biggest part (34%, about 8 hours), then leisure, social & travel (27%, about 6½ hours). Paid work (10%) and learning (5%) look small. You may notice that this does not match your own day (you hardly do paid work, and your parents spend no time learning): the strip is an average over people of all ages, genders, regions and backgrounds, so it hides those differences.
• 1. What do you find interesting in the chart above? What can you infer? Discuss.
  2. Do you remember the sleep time over age trend that you studied last year? Does that trend align with this chart?
  3. Can you explain why the learning time of people in the age group 15 – 24 has reduced significantly compared to that of the age group 6 – 14?
  4. Do adults spend about an equal amount of time in paid and unpaid work? What do you think?Answer: 1. Learning is large only for children (22%) and almost zero for adults; work grows with age till 59; leisure is steady (26%) and largest for the elderly (33%). 2. Yes: sleep falls from children (38%, about 9 h) to adults (32%, about 7½ h), then rises for the elderly. 3. Many 15–24-year-olds have left school or college and started work or household duties. 4. On average, yes (14% ≈ 3.4 h paid, 15% ≈ 3.6 h unpaid), but the average hides that many people do mostly one kind.
• What type of chart is this — a stacked bar chart or a 100% stacked bar chart?Answer: Both. Each piece is an amount of time (a stacked bar), and every column has the same total, 24 hours, so the pieces are also the shares of the day (a 100% stacked bar).
• Do you remember the ‘What Can A Strip Say?’ activity from last year? In each strip, if we club the tiny strips belonging to each activity together, will we get a 100% stacked bar chart like the one in the figure?Answer: Yes. Each strip is one whole day (24 hours). Joining the small pieces of the same activity keeps the total length and gives one piece per activity whose length is that activity’s share of the day: a 100% stacked bar. Several people’s strips side by side make a chart like the age-group chart.
• The smart watch data of 5 people was tracked from Monday to Friday. The recorded data tells us how many hours each person spent lying down, sitting, and standing or moving around, etc., as shown in the following table.Answer: (i) A reasonable guess for Sahana: lying down 9 h, sitting 10 h, standing/moving 5 h (total 24 h). (ii) Julie sits 10 h and moves only 6 h: an office worker (for example, a bank clerk or a software engineer) or a driver. (iii) Convert each row to percentages of 24 h and draw the bars (Sahana: 38%, 42%, 21%).
• 1. Will the bars look similar if a different teacher’s data is considered instead of Zakir’s?
  2. Will the bars look different if the data for all 7 days of the week is considered?
  3. Some bars in the chart may also match people doing other kinds of work/activities. Can you think of any? Discuss.Answer: 1. Roughly similar (teachers stand and move a lot), but not identical: it depends on the subject, number of classes, travel and habits. 2. Yes, probably: weekends usually have more lying down and sitting and less standing for working people and students; a patient’s bar may change little. 3. Raghu’s bar fits a waiter, shop assistant or traffic police; Julie’s fits a driver or office worker; Pavani’s fits anyone who is bedridden.
• 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.Answer: (i) 6 + 122 = 9 runs per over. (ii) 19 × 6 + 1220 = 12620 = 6.3 runs per over.
• 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(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?Answer: Each friend reports the sum (or average) of their prices and how many they found; they add the sums and divide by the total count, 23. (i) 6a_y + 4a_s + 5a_k + 4a_p + 4a_g23. (ii) 13a_n + 10a_f23; yes, both equal 406323 ≈ ₹176.65.
• 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.Answer: (i) 25 × 150 + 10 × 3035 = 405035 ≈ ₹115.71. (ii) 5 × 150 + 30n5 + n = 70 ⇒ n = 10 shares.
• (Pṛthūdakasvāmī, commentary on Brahmagupta’s Brāhmasphuṭasiddhānta, c. 864 CE) A “hasta” (meaning “forearm”) refers to a unit of length measuring around 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: Mean depth = 4 × 9 + 5 × 7 + 6 × 7 + 7 × 3 + 8 × 24 + 5 + 6 + 7 + 8 = 15030 = 5 hastas.
• Suvarna had purchased 1g 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.Answer: (i) ₹22.5k (ii) ₹25k (iii) ₹12.5k (iv) ≈ ₹10.45k (v) ₹20k (vi) ₹15k (stays at O).
• 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: It does not change. Doubling every weight doubles both the numerator and the denominator: 2w₁x₁ + … + 2w_nx_n2w₁ + … + 2w_n = w₁x₁ + … + w_nx_nw₁ + … + w_n.
• Answer the following questions based on the graph.Answer: (i) About 10,000 objects in 2003; it doubled to about 20,000 in about 2021–2022. (ii) 2025: about 33,000 objects, roughly 16,500 payload (≈ 50%) and 16,500 other (≈ 50%). (iii) Payload objects grew slowly for decades, then shot up after about 2019; other objects grew steadily, with jumps in 2007 and 2009. The payload share stayed near 20–25% until about 2019 and rose to about 50% by 2025, so the share of other objects fell from about 75–80% to about 50%.
• Observe the following infographic.Answer: (i) Correct inferences: (b) and (d). (a) and (c) compare numbers of schools, which percentages alone cannot do. (ii) No; we also need the number of schools in each state (or the number with playgrounds), to take a weighted mean.
• Decision Dilemma:Answer: (i) Play A or Play B: about 71% of their viewers gave 4 or 5 stars and few gave 1 star; Play C splits opinion (about 40% gave 5★ but a third gave 1★). (ii) Many would switch to Play B: it has about 925 ratings against about 210 for A, so its good rating is much more reliable. Play C still splits opinion.
• 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 following table shows the finish time of 3 athletes in each segment.Answer: (i) A stacked bar chart: the race is decided by total time, which a stacked bar shows as the bar’s length, along with each segment. (ii) From a 100% chart only (b) can be answered (cycling ≈ 53%, just over half of Athlete 1’s time); (a) and (c) need actual times.
• Look at the following graph. What do you notice? What do you wonder? Write your inferences.Answer: Notice: with age, the shares of seeing (≈ 18% → 25%) and movement (≈ 13% → 25%) disabilities rise, while speech (≈ 9% → 4%), intellectual disability (≈ 8% → 2%) and ‘any other’ (≈ 21% → 11%) fall; hearing stays near 18–20%. Wonder: how many people are in each group, and why? Inference: age-related problems (eyesight, joints) dominate in old age; the chart shows shares, not numbers.
• Individual project: Do at least one of the following.Answer: A model answer for both: (i) record hours lying down, sitting and standing/moving for yourself and 2 family members (each adding to 24 h) and draw one stacked bar per person; (ii) list expense categories, collect 3 months of expenses, convert each month to percentages of its total and draw one 100% stacked bar per month, then write what you notice.
• 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 experienceAnswer: Model answer for (d) a clinic: aspects cleanliness, waiting time, doctor’s care, staff behaviour, cost with weights 2 : 3 : 4 : 2 : 1. Each person’s rating = 2C + 3W + 4D + 2S + 1K12; for 10 sample people these range from 2.42 to 4.42 and the overall average is 435120 ≈ 3.63.
• 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: Multiply each family’s average age by its number of members to get the family’s total age; add all the totals and divide by the total number of members: average age = a₁n₁ + a₂n₂ + …n₁ + n₂ + … (a weighted mean). Simply averaging the family averages would be wrong unless all families are the same size.
• 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: It usually changes: it moves towards the ordinary (unweighted) mean of the data. It stays the same only if the weighted average already equals the ordinary mean (for example, when all weights are equal).
• 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: (iii), (iv) and (v) are possible. (i) and (ii) are impossible, because the three shares must always add up to 100%: they cannot all fall or all rise.

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