How Quantities Combine: Understanding Data

Think and Reflect

Q1. Why did Method 1 not work?

Answer:

Method 1 does not work because it gives equal importance to the average heights of seniors and juniors, even though there are 8 seniors and only 3 juniors. The combined average must take the number of students in each group into account.

Exercise Set 10.1

Q1. 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:

Given: Average of Section A = 72, number of students = 30;

           Average of Section B = 76, number of students = 25.

Formula: Combined average = (Sum of values in both sections)/(Total number of students)

Combined average = (72 × 30 + 76 × 25)/(30 + 25)

= (2160 + 1900)/55 = 4060/55 = 73.82

Final answer: 73.82 marks (approximately).

Q2. A farmer mixes three equal quantities of fertilisers. The first one contains 1/10 nitrogen, the second contains 9/50 nitrogen, and the third contains 3/60 nitrogen. What is the fraction of nitrogen in the mixture?

Answer:

Since the three quantities are equal, the fraction of nitrogen in the mixture is the arithmetic mean of the three fractions.

Fraction of nitrogen = (1/10 + 9/50 + 3/60)/3

= (30/300 + 54/300 + 15/300)/3

= (99/300)/3 = 99/900 = 11/100

Final answer: 11/100.

Q3. (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:

Formula: Weighted average = (Sum of products of quantities and purities)/(Total quantity)

Purity = (9 × 12 + 5 × 10 + 17 × 11)/(9 + 5 + 17)

= (108 + 50 + 187)/31 = 345/31 ≈ 11.13

Final answer: Approximately 11.13 varṇa.

Q4. 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, June and July have 31, 30 and 31 days respectively.

Combined average rainfall per day = (3.5 × 31 + 10 × 30 + 8.7 × 31)/(31 + 30 + 31)

Final answer: (108.5 + 300 + 269.7)/92 mm per day.

Q5. Calculate the concentration of spice mix in these two scenarios. (i) A 100 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 300 mL one with 15% spice mix are combined. (ii) A 300 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 100 mL one with 15% spice mix are mixed.

Answer:

(i) Concentration = (100 × 5 + 200 × 10 + 300 × 15)/(100 + 200 + 300)

= (500 + 2000 + 4500)/600 = 7000/600 ≈ 11.67%

Final answer (i): 11.67% approximately.

(ii) Concentration = (300 × 5 + 200 × 10 + 100 × 15)/(300 + 200 + 100)

= (1500 + 2000 + 1500)/600 = 5000/600 ≈ 8.33%

Final answer (ii): 8.33% approximately.

Exercise Set 10.2

Q1. 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 × 5 divided by 3 + 4 + 5

(ii) (35/100) × 3 + (44/100) × 4 + (80/100) × 5 divided by 3 + 4 + 5

(iii) ((35/50) × 3 + (44/60) × 4 + (80/100) × 5) divided by (3 + 4 + 5) × 100 

(iv) ((35/50 × 100) × 3 + (44/60 × 100) × 4 + (80/100 × 100) × 5) divided by 3 + 4 + 5

Answer:

First convert each assessment score into a percentage.

Internal tests = (35/50) × 100 = 70%.

Project = (44/60) × 100 ≈ 73.33%.

Final examination = (80/100) × 100 = 80%.

The annual percentage is calculated using the weights 3, 4 and 5.

Annual percentage = (70 × 3 + 73.33 × 4 + 80 × 5)/12

= (210 + 293.33 + 400)/12 ≈ 75.28%.

Final answer: Expression (iv) is correct.

Exercise Set 10.3

Q1. 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 from books = 8000 × 30/100 = ₹2400.

Profit from book covers = 1000 × 50/100 = ₹500.

Total profit = ₹2400 + ₹500 = ₹2900.

Total sales = ₹8000 + ₹1000 = ₹9000.

Percentage profit = (2900/9000) × 100 ≈ 32.22%.

Final answer: 32.22% approximately.

Q2. 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: The new average will be greater than 44.5 km but less than 55 km.

Total distance in 20 days = 20 × 44.5 = 890 km.

Total distance in 21 days = 890 + 55 = 945 km.

Average distance = 945/21 = 45 km.

Final answer: 45 km per day.

Q3. 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:

Salt in the first solution = 600 × 5/100 = 30 mL.

Sugar in the second solution = 300 × 8/100 = 24 mL.

Total mixture = 600 + 300 = 900 mL.

Salt concentration = (30/900) × 100 = 3.33%.

Sugar concentration = (24/900) × 100 = 2.67%.

Final answer: Option (v), Salt: 3.33%, Sugar: 2.67%.

Q4. 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 (3/4)th of the original concentration?

Answer:

Initial spice concentration = 8%.

Required concentration = (3/4) × 8% = 6%.

Spice in 10 litres of pani = 10 × 8/100 = 0.8 litres.

Let x litres of regular water be added. Then:

0.8/(10 + x) = 6/100

0.8 = 0.06(10 + x)

10 + x = 0.8/0.06 = 13.333...

x = 3.333... litres.

Final answer: 3⅓ litres of regular water.

Q5. 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. (i) Find out whose total is more without calculating. (ii) What are their final marks?

Answer:

(i) Keerthi's strength score is 5 less than Rashi's, her flexibility score is equal to Rashi's, and her agility score is 5 more. Since agility has the greatest weight, Keerthi's weighted score is higher.

(ii) Rashi's final marks:

(60 × 4 + 65 × 5 + 70 × 6)/(4 + 5 + 6)

= (240 + 325 + 420)/15 = 985/15 ≈ 65.67.

Keerthi's final marks:

(55 × 4 + 65 × 5 + 75 × 6)/15

= (220 + 325 + 450)/15 = 995/15 ≈ 66.33.

Final answer: Keerthi has the higher final score, approximately 66.33 marks, compared with Rashi's 65.67 marks.

Q6. 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:

Average food rating = (5 × 5 + 4 × 3 + 3 × 2)/10 = 43/10 = 4.3.

Average ambience rating = (4 × 4 + 3 × 5 + 2 × 1)/10 = 33/10 = 3.3.

Average service rating = (5 × 1 + 4 × 2 + 3 × 2 + 2 × 4 + 1 × 1)/10 = 25/10 = 2.5.

Weighted average = (4.3 × 6 + 3.3 × 5 + 2.5 × 4)/(6 + 5 + 4)

= (25.8 + 16.5 + 10)/15 = 52.3/15 ≈ 3.49.

Final answer: 3.49 out of 5 approximately.

Q7. 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? (i) Which of the following expression(s) describes the given scenario? (a) (16.5x + 13.8y)/(x + y) = 14.925 (b) (16.5x + 13.8y)/60 = 14.925 (c) (16.5x + 13.8y)/2 = 14.925 (d) (16.5x + 13.8y)/(16.5 + 13.8) = 14.925 (ii) Find out how many male langurs are present. (iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this? (iv) Two male langurs weighing 16.9 kg and 16.1 kg are released from the facility. What is the average weight of the male langurs after this? (v) Now, suppose one of the male langurs lost 1 kg of weight. What is the average weight of all the male langurs after this?

Answer:

Average weight of male langurs = 16.5 kg, average weight of female langurs = 13.8 kg, overall average = 14.925 kg, and total number = 60.

Initial observation: The overall average is exactly halfway between 16.5 kg and 13.8 kg. Therefore, the numbers of male and female langurs are equal.

(i) The correct expression is (a), where x and y represent the numbers of male and female langurs respectively.

(ii) Since x = y and x + y = 60, x = 30.

Final answer (ii): 30 male langurs.

(iii) Initial total weight of female langurs = 30 × 13.8 = 414 kg.

New average = (414 + 15.2)/31 = 429.2/31 ≈ 13.85 kg.

Final answer (iii): Approximately 13.85 kg.

(iv) Initial total weight of male langurs = 30 × 16.5 = 495 kg.

New average = (495 − 16.9 − 16.1)/28 = 462/28 = 16.5 kg.

Final answer (iv): 16.5 kg.

(v) The new total weight becomes 495 − 1 = 494 kg.

New average = 494/30 ≈ 16.47 kg.

Final answer (v): Approximately 16.47 kg.

Q8. 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. (i) What is the salinity of the mixture if he mixes 1 litre of water from the Dead Sea with 2 litres of purified drinking water? (ii) Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? Why/Why not? What quantity of purified drinking water should Dorjee mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of groundwater? (iii) Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? Why/Why not? What quantity of groundwater should he mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of purified drinking water?

Answer:

The table gives the following approximate salinities: Dead Sea = 34%, other seas and oceans = 3.5%, groundwater = 0.01%, and purified drinking water = 0.001%.

(i) Salinity = (1 × 34 + 2 × 0.001)/(1 + 2)

= 34.002/3 ≈ 11.334%.

Final answer (i): Approximately 11.334%.

(ii) It is possible because 0.01% lies between 34% and 0.001%.

Let x litres of purified drinking water be added.

(34 + 0.001x)/(1 + x) = 0.01

34 + 0.001x = 0.01 + 0.01x

33.99 = 0.009x, so x = 3776.67 litres approximately.

Final answer (ii): Approximately 3776.67 litres of purified drinking water.

(iii) It is not possible. A mixture of 34% saline water and 0.01% groundwater must have salinity between 0.01% and 34%. It cannot reach 0.001%, which is below both concentrations.

Final answer (iii): It is impossible to obtain the salinity of purified drinking water by mixing these two sources.

Exercise Set 10.4

Q1. The following stacked column chart shows the number of animal species in the IUCN Red List, by class, over time. (i) What does the number 14,234 on the top of the column 2019 represent? (ii) Approximately how many reptile species were in the list in 2016? (iii) Which class(es) of species have seen a relatively small increase in count between 2007 and 2019?

Answer:

(i) The number 14,234 represents the total number of animal species included in the IUCN Red List in 2019.

(ii) Read the reptile segment of the 2016 column and estimate its height using the vertical scale. The chart gives an approximate value, so the answer should be stated as an estimate from the figure.

(iii) Compare the segments for each class in 2007 and 2019. The class or classes whose segments have increased relatively little are the correct response. The precise class estimate depends on reading the printed chart.

[Diagram instruction: Refer to the stacked column chart in the textbook and compare the coloured segments for 2007 and 2019. Estimate the reptile segment for 2016 from the vertical scale.]

Q2. 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), (i) comparing the total wickets taken at home vs. overseas: (ii) The total wickets taken in each format:

Answer:

(i) Wickets taken at home vs. overseas:

Home: Test = 62, ODI = 100, T20 = 32. Total = 62 + 100 + 32 = 194 wickets.

Overseas: Test = 172, ODI = 49, T20 = 71. Total = 172 + 49 + 71 = 292 wickets.

(ii) Total wickets in each format:

Test = 62 + 172 = 234 wickets.

ODI = 100 + 49 = 149 wickets.

T20 = 32 + 71 = 103 wickets.

[Diagram instruction: Draw the two stacked bar charts using the table values. In the first chart, use separate segments for Test, ODI and T20 for Home and Overseas. In the second chart, use Home and Overseas as the two segments for Test, ODI and T20. Label the horizontal scale from 0 to 300.]

Think and Reflect

Q1. What does this say about the scope of the stacked bars and 100% stacked bars?

Answer:

Stacked bar charts show both the total quantity and its individual components. A 100% stacked bar chart shows the proportion or percentage contributed by each component, making it useful for comparing distributions rather than actual totals.

Q2. Given a stacked bar chart can we make a corresponding 100% stacked bar chart?

Answer:

Yes. Find the total of each bar and calculate the percentage contributed by each component. Use these percentages to construct bars of equal length, each representing 100%.

Q3. Given a 100% stacked bar chart can we make a corresponding stacked bar chart?

Answer:

No, not uniquely. A 100% stacked bar chart gives percentages but does not reveal the actual totals. Different totals can have the same percentages. The original totals or other additional information are needed.

Q4. What kind of inferences or comparisons can be made from a stacked bar chart and in a 100% stacked bar chart?

Answer:

• A stacked bar chart allows comparisons of total quantities and the sizes of their components.
• A 100% stacked bar chart allows comparisons of proportions or percentages within each total.
• A 100% stacked bar chart cannot, by itself, establish which group has a greater actual quantity.

Think and Reflect

Q1. What do you find interesting in the chart above? What can you infer? Discuss.

Answer:

The chart shows how the proportion of a day spent on different activities changes with age. Children spend a relatively large share of their time learning, while adults spend more time on paid work and unpaid work and care. Sleep, leisure and personal care also occupy different proportions across age groups.

Q2. Do you remember the sleep time over age trend that you studied last year? Does that trend align with this chart?

Answer:

Yes. The chart broadly aligns with the trend that sleep requirements and sleep patterns change with age. However, the chart shows the average distribution of time for each age group, not the sleep duration of every individual.

Q3. 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?

Answer:

Many people aged 6–14 attend school and spend a substantial part of their day learning. In the 15–24 age group, some people enter employment, vocational training or other responsibilities, so the average share of time spent learning may decrease.

Q4. Do adults spend about an equal amount of time in paid and unpaid work? What do you think?

Answer:

The chart suggests that adults spend substantial time on both paid work and unpaid work and care. Whether the amounts are equal should be judged by comparing the two segments in the chart; the exact time varies among individuals and households.

Think and Reflect

Q1. 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 Fig. 10.6?

Answer:

Yes. If the tiny strips representing the same activity are grouped together and their total lengths are calculated, each age group's strip can be divided into sections representing the proportions of time spent on different activities. Since the complete strip represents 100% of the day, the result is a 100% stacked bar chart.

Exercise Set 10.5

Q1. 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. Table 10.3: Average number of hours spent per day (Monday–Friday) lying down, sitting, and standing or moving around, etc., of 5 people. (i) How much time does Sahana spend per day in each of these body-states? Make a reasonable guess and fill her row in the table. (ii) Guess what activity or work Julie could be engaged in based on the data. Find out what your classmates’ guesses are. (iii) Complete the chart below based on the tabular data.

Answer:

(i) A reasonable sample estimate for Sahana is 9 hours lying down, 7 hours sitting, and 8 hours standing or moving around. These values add up to 24 hours. This is only a sample answer; actual estimates may vary.

(ii) Julie spends 8 hours lying down, 10 hours sitting and 6 hours standing or moving around. One possible guess is that Julie does work involving prolonged sitting, such as office or computer-based work. Other reasonable guesses are acceptable.

(iii) Convert each number of hours into a percentage of 24 hours and represent it as a segment of a 100% stacked bar.

[Diagram instruction: Draw five horizontal bars of equal length. Divide each bar into three sections using the percentages in the table. Label the sections Lying down, Sitting and Standing/Moving around, etc. Mark the scale at 0%, 25%, 50%, 75% and 100%.]

Think and Reflect

Q1. Will the bars look similar if a different teacher’s data is considered instead of Zakir’s?

Answer:

Not necessarily. Different teachers may have different working hours, travel times, household responsibilities and sleeping patterns. Therefore, their bars may have different proportions for lying down, sitting and standing or moving around.

Q2. Will the bars look different if the data for all 7 days of the week is considered?

Answer:

Yes, the bars may look different because weekends can have different routines from weekdays. Including all seven days may change the average proportions of time spent in each body-state.

Q3. Some bars in the chart may also match people doing other kinds of work/activities. Can you think of any? Discuss.

Answer:

Yes. A bar with a large sitting segment may represent a student, writer or computer programmer. A bar with a large standing or moving segment may represent a shopkeeper, delivery worker or sportsperson. These are examples; different people may have similar time-use patterns.

End-of-Chapter Exercises

Q1. 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) Total runs after two overs = 6 + 12 = 18.

Run rate = 18/2 = 9 runs per over.

Final answer (i): 9 runs per over.

(ii) Total runs after 19 overs = 19 × 6 = 114.

Total runs after 20 overs = 114 + 12 = 126.

Run rate = 126/20 = 6.3 runs per over.

Final answer (ii): 6.3 runs per over.

Q2. 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? (i) Each one calculated the average of the prices they had gathered. Suppose these are ay, as, ak, ap, ag respectively. Write an expression that gives the combined average. (ii) Suppose an, af 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:

The five friends can share their data, count the number of prices collected by each person and combine the totals. The combined average must account for the number of prices collected by each friend.

(i) Yusuf collected 6 prices, Srikanth 4, Kashvi 5, Prasanna 4 and Gracy 4.

Combined average = (6ay + 4as + 5ak + 4ap + 4ag)/(6 + 4 + 5 + 4 + 4)

Final answer (i): (6ay + 4as + 5ak + 4ap + 4ag)/23.

(ii) There are 13 nylon shuttlecocks and 10 feather shuttlecocks.

Average price = (13an + 10af)/(13 + 10)

Final answer (ii): (13an + 10af)/23. Yes, both methods give the same average if the averages are calculated correctly from the same 23 prices.

Q3. 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) Shreyas's original total cost = 25 × ₹150 = ₹3750.

Cost of 10 additional shares = 10 × ₹30 = ₹300.

Total cost = ₹3750 + ₹300 = ₹4050; total shares = 25 + 10 = 35.

Average price = 4050/35 = ₹115.71 approximately.

Final answer (i): ₹115.71 per share approximately.

(ii) Vaishnavi's original total cost = 5 × ₹150 = ₹750.

Let x be the number of additional shares. Then:

(750 + 30x)/(5 + x) = 70

750 + 30x = 350 + 70x

400 = 40x, so x = 10.

Final answer (ii): 10 shares.

Q4. (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:

Formula: Mean depth = (Sum of products of section lengths and depths)/(Total length)

Mean depth = (4 × 9 + 5 × 7 + 6 × 7 + 7 × 3 + 8 × 2)/(4 + 5 + 6 + 7 + 8)

= (36 + 35 + 42 + 21 + 16)/30

= 150/30 = 5 hastas.

Final answer: 5 hastas.

[Diagram instruction: Draw a long rectangular pool divided into five sections of lengths 4, 5, 6, 7 and 8 hastas. Label the respective depths 9, 7, 7, 3 and 2 hastas.]

Q5. 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. (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 0.5g gold at ₹30k (vi) Purchase 0.5g gold at ₹15k

Answer:

Use the weighted average of the existing 1 g of gold at ₹15,000 per gram and the additional gold purchased.

(i) (1 × 15,000 + 1 × 30,000)/(1 + 1) = ₹22,500 per gram.

(ii) (1 × 15,000 + 2 × 30,000)/(1 + 2) = ₹25,000 per gram.

(iii) (1 × 15,000 + 1 × 10,000)/(1 + 1) = ₹12,500 per gram.

(iv) (1 × 15,000 + 10 × 10,000)/(1 + 10) = ₹10,454.55 per gram approximately.

(v) (1 × 15,000 + 0.5 × 30,000)/(1 + 0.5) = ₹20,000 per gram.

(vi) (1 × 15,000 + 0.5 × 15,000)/(1 + 0.5) = ₹15,000 per gram.

Final answers: (i) ₹22,500; (ii) ₹25,000; (iii) ₹12,500; (iv) approximately ₹10,454.55; (v) ₹20,000; (vi) ₹15,000 per gram.

[Diagram instruction: On the number line from ₹0k to ₹30k, mark the six average prices. The original point O is at ₹15k.]

Q6. 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:

Let the values be x₁, x₂, ..., xₙ and their weights be w₁, w₂, ..., wₙ.

Original weighted average = (w₁x₁ + w₂x₂ + ... + wₙxₙ)/(w₁ + w₂ + ... + wₙ).

If every weight is doubled, the new weighted average is:

(2w₁x₁ + 2w₂x₂ + ... + 2wₙxₙ)/(2w₁ + 2w₂ + ... + 2wₙ)

= 2(w₁x₁ + w₂x₂ + ... + wₙxₙ)/[2(w₁ + w₂ + ... + wₙ)].

The factor 2 cancels from numerator and denominator.

Final answer: The weighted average remains unchanged when all weights are doubled.

Q7. Answer the following questions based on the graph. (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:

(i) Read the total height of the stacked graph at 2003 to estimate the total number of objects. Double that estimate and locate the year at which the total first reaches approximately that value.

(ii) Read the two stacked components for 2025. Add them to obtain the total and divide each component by the total to find its percentage share.

(iii) The graph shows how the numbers of payload objects and other objects change over time. The absolute numbers and their percentage shares need not follow the same trend: a category can increase in number while its share decreases if the total grows faster.

[Diagram instruction: Use the textbook graph titled “Cumulative number of objects currently orbiting Earth, by year of launch/separation (as of Apr. 24, 2026)”. Read approximate values from the vertical scale and compare the two stacked components.]

Q8. Observe the following infographic. What Percentage of Schools have a Playground? (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:

(i) The infographic reports percentages, so statements comparing actual numbers of schools cannot be concluded from percentages alone.

• (a) Cannot be inferred from percentages alone unless the total numbers of schools in the two states are known.
• (b) Correct if the infographic shows Arunachal Pradesh at approximately 67%.
• (c) Cannot be inferred from the highest percentage alone; the total number of schools in each state is also needed.
• (d) If Maharashtra has more schools than Telangana and its playground percentage is also higher, then Maharashtra has more schools with playgrounds. This follows by multiplying the larger total by the higher proportion.

(ii) No. The nation-wide percentage cannot be calculated by taking a simple average of state percentages unless the numbers of schools in the states are equal.

The total number of schools and the number of schools with playgrounds in each state are needed to calculate the national percentage.

[Diagram instruction: Refer to the infographic on the textbook page and read the percentages for Haryana, Uttarakhand, Arunachal Pradesh, Punjab, Maharashtra and Telangana before finalising the state-wise inferences.]

Q9. 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:

(i) The 100% stacked bar chart shows the share of ratings for each play. Choose the play with the most favourable distribution of ratings, giving greater importance to higher ratings.

(ii) The stacked bar chart also shows the actual number of ratings. A play with a favourable percentage distribution but very few ratings may be less reliable than a play with many ratings. I would compare both the rating distribution and the total number of ratings before making a decision.

[Diagram instruction: Refer to both charts in the textbook. Compare the proportions of favourable ratings in the 100% stacked bars and the total lengths of the bars in the stacked bar chart. The final choice depends on the values shown in the printed charts.]

Q10. 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. Table 10.4: Finish time of three athletes in each segment of a triathlon race (values in hh:mm). (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:

(i) A stacked bar chart is suitable because the total time taken by each athlete is important, along with the time spent in each segment. The total length of each bar represents the athlete's overall time.

(ii) Only question (b) can be answered using a 100% stacked bar chart, because it asks for the fraction of Athlete 1's total race time spent cycling. The chart does not show actual total times, so it cannot establish who finished first or which athlete took the longest actual time for running.

Final answer: (i) Stacked bar chart; (ii) option (b).

Q11. Look at the following graph. What do you notice? What do you wonder? Write your inferences.

Answer:

The graph shows the distribution of persons with different types of disabilities across age groups in India, based on Census 2011.

• The distribution of disability types varies across age groups.
• Some types of disability may form a larger share in particular age groups than in others.
• The graph can be used to compare the relative distribution of disability types among age groups.
• Questions worth investigating include why the distributions differ by age and whether access to healthcare, diagnosis and support varies across groups.

[Diagram instruction: Refer to the textbook graph titled “Distribution of disabled persons by age group and type of disability in India (Census 2011)”. Compare the segments for the age groups 0–19, 20–39, 40–59 and 60 and above. Write numerical observations only after reading the graph's scale.]

Q12. 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 an individual project. A sample approach is given below. Use your own observations and household data for the final submission.

(i) Body-state project:

1. Record the approximate hours spent lying down, sitting and standing or moving around by yourself and at least two family members.
2. Check that the three values add up to 24 hours for each person.
3. Convert each value into a percentage using (hours spent in that state/24) × 100.
4. Draw a stacked bar for each person, with the three segments representing the percentages.
5. Compare the bars and explain the differences in daily routines.

Sample data:

(ii) Family expenditure project:

1. Identify categories such as housing, food, education, transport, healthcare and recreation.
2. Discuss with family members how to record monthly expenditure.
3. Collect expenditure data for at least three months.
4. Calculate each category's percentage of the monthly total.
5. Draw one 100% stacked bar for each month, using the percentages for the segments.
6. Write observations about the largest expense categories and changes over the three months.

[Diagram instruction: Draw stacked bars for the body-state data or 100% stacked bars for monthly expenditure. Label all axes, categories and percentage values clearly.]

Q13. 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:

This is a group project. The following sample uses the scenario of a bus travel experience. Students may select any one of the four given scenarios.

(i) Aspects and reasons: cleanliness, punctuality, comfort, staff behaviour and safety. These aspects influence the quality of the journey.

(ii) Sample weights: safety = 5, punctuality = 4, comfort = 3, cleanliness = 2 and staff behaviour = 2.

(iii) Collect ratings from at least 10 people on a scale of 1–5, where 5 means very good. A sample dataset may be used if actual responses cannot be collected.

(iv) For each person, calculate the weighted rating as the sum of rating × weight divided by the sum of weights. Calculate the average of the individual weighted ratings.

(v) Convert the chosen categories into percentage shares and draw a 100% stacked bar chart using the collected data. Label the segments and legend clearly.

(vi) Sample observation: The weighted system gives greater importance to safety and punctuality. It makes the overall comparison more structured, but the result depends on the aspects and weights chosen and may not capture every person's experience.

[Diagram instruction: Draw a 100% stacked bar chart using the group's actual or imagined data. Include a title, a percentage scale from 0% to 100%, and a legend.]

Q14. 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:

Collect the average age and number of family members for every family. Multiply each family's average age by its number of members to obtain its total age. Add these totals and divide by the total number of family members across all families.

Formula: Overall average age = (Sum of each family's average age × number of family members)/(Total number of family members).

This weighted average accounts for differences in family size.

Q15. 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:

In general, increasing every weight by 1 changes the weighted average. Let the original values be x₁, x₂, ..., xₙ and their weights be w₁, w₂, ..., wₙ.

Original weighted average = (Σwᵢxᵢ)/(Σwᵢ).

New weighted average = [Σ(wᵢ + 1)xᵢ]/[Σ(wᵢ + 1)]

= (Σwᵢxᵢ + Σxᵢ)/(Σwᵢ + n).

This is a combination of the original weighted average and the ordinary average of the values, each with the appropriate total weight. Therefore, the new average moves towards the ordinary average of the values.

Final answer: The weighted average generally changes when 1 is added to every weight; it remains unchanged only in special cases, such as when the original weighted average equals the ordinary average.

Q16. 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 percentage shares must add up to 100%. If all three animal populations decrease, the changes in their percentage shares depend on their relative rates of decrease. All shares cannot decrease simultaneously because their sum must remain 100%.

• (i) Impossible, because all three shares cannot decrease together.
• (ii) Impossible, because all three shares cannot increase together.
• (iii) Possible if all three populations decrease by the same percentage.
• (iv) Possible if the cow population decreases proportionally more, the sheep population decreases proportionally less, and the chicken population decreases at the same proportional rate as the total population.
• (v) Possible if the cow and sheep shares increase while the chicken share decreases sufficiently to keep the total at 100%.

Final answer: Options (iii), (iv) and (v) are possible.