NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 How Quantities Combine: Understanding Data
Data is an important part of our everyday lives, from comparing marks and tracking expenses to understanding survey results and identifying trends. Chapter 10 – How Quantities Combine: Understanding Data in NCERT Class 9 Maths Ganita Manjari Part 2 introduces students to the systematic collection, organisation, representation, and interpretation of data.
The NCERT Solutions for Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine: Understanding Data provide clear, step-by-step explanations for all the questions and exercises given in the chapter. These solutions help students understand key concepts, interpret data accurately, apply mathematical reasoning, and develop confidence in solving data-based problems. They are useful for regular practice, homework, revision, and exam preparation.
NCERT Class 9 Maths Ganita Manjari Chapter 10 Solutions
NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.1
1. 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:
Average score of Section A = 72
Number of students in Section A = 30
Total score of Section A = 72 × 30 = 2160
Average score of Section B = 76
Number of students in Section B = 25
Total score of Section B = 76 × 25 = 1900
Total score of both sections = 2160 + 1900 = 4060
Total number of students = 30 + 25 = 55
Combined Average = 4060 ÷ 55 = 73.82
Therefore, the combined average score of both sections is approximately 73.82.
2. 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 farmer mixes equal quantities of all three fertilisers, the fraction of nitrogen in the mixture is the average of their nitrogen fractions.
First fertiliser = 1/10
Second fertiliser = 9/50
Third fertiliser = 3/60 = 1/20
Taking LCM = 100:
1/10 = 10/100
9/50 = 18/100
1/20 = 5/100
Total nitrogen fraction = 10/100 + 18/100 + 5/100 = 33/100
Since there are 3 equal quantities:
Fraction of nitrogen in the mixture = (33/100) ÷ 3 = 33/300 = 11/100
Therefore, the fraction of nitrogen in the mixture is 11/100.
3. (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 Varṇa = 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:
Since the three pieces have different weights, we use a weighted average.
Total gold content in varṇa-units:
9 × 12 = 108
5 × 10 = 50
17 × 11 = 187
Total = 108 + 50 + 187 = 345
Total weight of gold alloy:
9 + 5 + 17 = 31 units
Therefore,
Combined purity = 345/31 varṇa
= 11 4/31 varṇa
≈ 11.13 varṇa
Therefore, the purity of the combined gold is approximately 11.13 varṇa.
4. 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:
Solution:
The average rainfall per day is:
- May: 3.5 mm for 31 days
- June: 10 mm for 30 days
- July: 8.7 mm for 31 days
Since the months have different numbers of days, we use a weighted average.
Total rainfall in May:
31 × 3.5 = 108.5 mm
Total rainfall in June:
30 × 10 = 300 mm
Total rainfall in July:
31 × 8.7 = 269.7 mm
Therefore, total rainfall over the three months:
108.5 + 300 + 269.7 = 678.2 mm
Total number of days:
31 + 30 + 31 = 92 days
Combined average rainfall:
678.2 ÷ 92 ≈ 7.37 mm per day
Therefore, the combined average rainfall for May, June, and July is approximately 7.37 mm per day.
5. 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:
The concentration of spice mix in the final mixture is:
Concentration = (Total amount of spice mix / Total volume of kashayam) × 100
(i)
For 100 mL kashayam with 5% spice mix:
Spice mix = 5% of 100 = 5 mL
For 200 mL kashayam with 10% spice mix:
Spice mix = 10% of 200 = 20 mL
For 300 mL kashayam with 15% spice mix:
Spice mix = 15% of 300 = 45 mL
Total spice mix = 5 + 20 + 45 = 70 mL
Total volume = 100 + 200 + 300 = 600 mL
Concentration = (70 / 600) × 100
= 11.67%
Therefore, the concentration of spice mix in scenario (i) is approximately 11.67%.
(ii)
For 300 mL kashayam with 5% spice mix:
Spice mix = 5% of 300 = 15 mL
For 200 mL kashayam with 10% spice mix:
Spice mix = 10% of 200 = 20 mL
For 100 mL kashayam with 15% spice mix:
Spice mix = 15% of 100 = 15 mL
Total spice mix = 15 + 20 + 15 = 50 mL
Total volume = 300 + 200 + 100 = 600 mL
Concentration = (50 / 600) × 100
= 8.33%
Therefore, the concentration of spice mix in scenario (ii) is approximately 8.33%.
NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.2
1. 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)/(3 + 4 + 5)
(ii) [(35/100) × 3 + (44/100) × 4 + (80/100) × 5]/(3 + 4 + 5)
(iii) [(35/50) × 3 + (44/60) × 4 + (80/100) × 5]/(3 + 4 + 5) × 100
(iv) [(35/50 × 100) × 3 + (44/60 × 100) × 4 + (80/100 × 100) × 5]/(3 + 4 + 5)
Answer:
Since the three components have different maximum marks, we must first convert each score into a percentage before applying the ratio 3 : 4 : 5.
Internal tests:
35/50 × 100 = 70%
Project:
44/60 × 100 = 73.33%
Final exam:
80/100 × 100 = 80%
Now, taking the weighted average:
Annual score
= [(70 × 3) + (73.33 × 4) + (80 × 5)] / (3 + 4 + 5)
= (210 + 293.33 + 400) / 12
= 903.33 / 12
≈ 75.28%
Now checking the given expressions:
- (i) Incorrect, because it uses the raw marks directly even though the maximum marks are different.
- (ii) Incorrect, because 35, 44, and 80 are not all marks out of 100.
- (iii) Correct, because each score is first written as a fraction of its respective maximum marks.
- (iv) Correct, because each score is first converted into a percentage before taking the weighted average.
Therefore, the correct expressions are (iii) and (iv), and Savitri’s annual score is approximately 75.28%.
NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.3
1. 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 selling books:
30% of ₹8000
= 30/100 × 8000
= ₹2400
Profit from selling book covers:
50% of ₹1000
= 50/100 × 1000
= ₹500
Total profit:
₹2400 + ₹500 = ₹2900
Total sales:
₹8000 + ₹1000 = ₹9000
Percentage profit on total sales:
(2900 / 9000) × 100
= 32.22%
Therefore, the percentage of profit on the total sales is approximately 32.22%.
2. 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: Since the stork travelled 55 km on the 21st day, which is more than the previous average of 44.5 km, the new average should be slightly more than 44.5 km.
Total distance travelled in the first 20 days:
44.5 × 20 = 890 km
Distance travelled on the 21st day:
55 km
Total distance travelled in 21 days:
890 + 55 = 945 km
Average daily distance:
945 ÷ 21 = 45 km
Therefore, the average daily distance travelled over 21 days is 45 km.
3. 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?
Answer:
Salt in the first solution:
5% of 600 mL
= 5/100 × 600
= 30 mL
Sugar in the second solution:
8% of 300 mL
= 8/100 × 300
= 24 mL
Total volume of the mixture:
600 + 300 = 900 mL
Salt concentration in the mixture:
(30 / 900) × 100 = 3.33%
Sugar concentration in the mixture:
(24 / 900) × 100 = 2.67%
Therefore, the mixture contains 3.33% salt and 2.67% sugar.
4. 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:
Original concentration of spice = 8%
Required concentration = 3/4 of 8%
= 3/4 × 8%
= 6%
Amount of spice in 10 litres of pani:
8% of 10
= 8/100 × 10
= 0.8 litre
Let the quantity of water added be x litres.
After adding water, total volume:
10 + x litres
Since the amount of spice remains unchanged:
0.8 / (10 + x) × 100 = 6
0.8 / (10 + x) = 0.06
0.8 = 0.06(10 + x)
0.8 = 0.6 + 0.06x
0.2 = 0.06x
x = 0.2 / 0.06
= 10/3
= 3.33 litres
Therefore, approximately 3.33 litres of regular water should be added.
5. 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:
Solution:
The marks for strength, flexibility, and agility are combined in the ratio 4 : 5 : 6.
(i) Find out whose total is more without calculating.
Both Rashi and Keerthi have the same score in flexibility: 65.
Rashi scores 5 marks more in strength, while Keerthi scores 5 marks more in agility.
Since agility has a higher weight (6) than strength (4), Keerthi’s advantage contributes more to the final score.
Therefore, Keerthi’s total is higher.
(ii) What are their final marks?
For Rashi:
Final marks
= (60 × 4 + 65 × 5 + 70 × 6) / (4 + 5 + 6)
= (240 + 325 + 420) / 15
= 985 / 15
= 65.67
For Keerthi:
Final marks
= (55 × 4 + 65 × 5 + 75 × 6) / 15
= (220 + 325 + 450) / 15
= 995 / 15
= 66.33
Therefore, Rashi’s final marks are 65.67 and Keerthi’s final marks are 66.33.
6. 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:
First, calculate the average rating for each metric.
Food
Average rating
= (5 × 5 + 4 × 3 + 3 × 2 + 2 × 0 + 1 × 0) / 10
= (25 + 12 + 6) / 10
= 43 / 10
= 4.3
Ambience
Average rating
= (5 × 0 + 4 × 4 + 3 × 5 + 2 × 1 + 1 × 0) / 10
= (16 + 15 + 2) / 10
= 33 / 10
= 3.3
Service
Average rating
= (5 × 1 + 4 × 2 + 3 × 2 + 2 × 4 + 1 × 1) / 10
= (5 + 8 + 6 + 8 + 1) / 10
= 28 / 10
= 2.8
Now, the metrics are combined in the ratio:
Food : Ambience : Service = 6 : 5 : 4
Overall average rating:
= (4.3 × 6 + 3.3 × 5 + 2.8 × 4) / (6 + 5 + 4)
= (25.8 + 16.5 + 11.2) / 15
= 53.5 / 15
≈ 3.57
Therefore, the restaurant’s average rating is approximately 3.57 out of 5.
7. 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:
Solution:
The average weight of male langurs is 16.5 kg, while that of female langurs is 13.8 kg. The overall average, 14.925 kg, is closer to 13.8 kg than to 16.5 kg.
Therefore, there are more female langurs than male langurs.
(i) Which expression describes the given scenario?
Let:
x = number of male langurs
y = number of female langurs
Since the combined average weight is 14.925 kg,
(16.5x + 13.8y) / (x + y) = 14.925
Therefore, the correct expression is:
(a) (16.5x + 13.8y)/(x + y) = 14.925
(ii) Find the number of male langurs.
Total number of langurs = 60
Therefore,
x + y = 60
So,
y = 60 − x
Using the weighted-average equation:
(16.5x + 13.8y) / 60 = 14.925
Substituting y = 60 − x:
16.5x + 13.8(60 − x) = 14.925 × 60
16.5x + 828 − 13.8x = 895.5
2.7x = 67.5
x = 25
Therefore, there are 25 male langurs and 35 female langurs.
(iii) A female langur weighing 15.2 kg is admitted. Find the new average weight of the female langurs.
Number of female langurs = 35
Their total weight:
35 × 13.8 = 483 kg
After adding the new langur:
New total weight = 483 + 15.2 = 498.2 kg
New number of female langurs = 36
New average:
498.2 / 36 ≈ 13.84 kg
Therefore, the new average weight of the female langurs is approximately 13.84 kg.
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released. Find the new average weight.
Original total weight of male langurs:
25 × 16.5 = 412.5 kg
Weight of the two released langurs:
16.9 + 16.1 = 33 kg
Remaining total weight:
412.5 − 33 = 379.5 kg
Remaining number of male langurs:
25 − 2 = 23
New average:
379.5 / 23 = 16.5 kg
Therefore, the average weight of the remaining male langurs is 16.5 kg.
(v) One of the male langurs loses 1 kg. Find the new average.
After part (iv), there are 23 male langurs with a total weight of 379.5 kg.
If one langur loses 1 kg:
New total weight = 379.5 − 1 = 378.5 kg
New average:
378.5 / 23 ≈ 16.46 kg
Therefore, the new average weight of the male langurs is approximately 16.46 kg.
8. 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:
Solution:
From the table:
- Salinity of Dead Sea water ≈ 34%
- Salinity of groundwater ≈ 0.01%
- Salinity of purified drinking water ≈ 0.001%
(i) Mixing 1 litre of Dead Sea water with 2 litres of purified drinking water
Total volume = 1 + 2 = 3 litres
Amount of salt contributed:
Dead Sea water = 34% of 1 L
Purified water = 0.001% of 2 L
Therefore,
Salinity of mixture
= [34 × 1 + 0.001 × 2] / 3
= 34.002 / 3
= 11.334%
Therefore, the salinity of the mixture is approximately 11.334%.
(ii) Can Dead Sea water and purified drinking water be mixed to get the salinity of groundwater?
Yes.
The salinity of groundwater, 0.01%, lies between the salinities of Dead Sea water (34%) and purified drinking water (0.001%). Therefore, an appropriate mixture can have a salinity of 0.01%.
Let x litres of purified drinking water be mixed with 1 litre of Dead Sea water.
Required salinity = 0.01%
So,
(34 × 1 + 0.001x) / (1 + x) = 0.01
34 + 0.001x = 0.01 + 0.01x
33.99 = 0.009x
x = 33.99 / 0.009
x ≈ 3776.67 litres
Therefore, Dorjee would need to mix approximately 3776.67 litres of purified drinking water with 1 litre of Dead Sea water.
(iii) Can Dead Sea water and groundwater be mixed to get the salinity of purified drinking water?
No, it is not possible.
Dead Sea water has a salinity of 34%, and groundwater has a salinity of 0.01%. Any mixture of the two must have a salinity between 0.01% and 34%.
However, purified drinking water has a salinity of only 0.001%, which is lower than the salinity of both liquids being mixed.
Therefore, no quantity of groundwater can be mixed with 1 litre of Dead Sea water to obtain a salinity of 0.001%.
NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.4
1. 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) What does 14,234 represent?
The number 14,234 represents the total number of animal species across all the given classes that were on the IUCN Red List in 2019.
(ii) Approximately how many reptile species were in the list in 2016?
From the stacked column for 2016, the reptiles section represents approximately 1,200 species.
Therefore, about 1,200 reptile species were on the list in 2016.
(iii) Which classes have seen a relatively small increase between 2007 and 2019?
Comparing the sizes of the segments for 2007 and 2019, reptiles and birds show relatively small increases compared with several of the other classes.
Therefore, reptiles and birds have seen a relatively small increase in count between 2007 and 2019.
2.
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:
From the table:
| Format | Home | Overseas |
|---|---|---|
| Test | 62 | 172 |
| ODI | 100 | 49 |
| T20 | 32 | 71 |
(i) Comparing total wickets taken at home vs. overseas
For Home:
62 + 100 + 32 = 194 wickets
To draw the stacked bar:
- Test: 0 to 62
- ODI: 62 to 162
- T20: 162 to 194
For Overseas:
172 + 49 + 71 = 292 wickets
To draw the stacked bar:
- Test: 0 to 172
- ODI: 172 to 221
- T20: 221 to 292
Therefore, total wickets taken: Home = 194 and Overseas = 292.
(ii) Total wickets taken in each format
Test:
62 + 172 = 234 wickets
Stacked bar:
- Home: 0 to 62
- Overseas: 62 to 234
ODI:
100 + 49 = 149 wickets
Stacked bar:
- Home: 0 to 100
- Overseas: 100 to 149
T20:
32 + 71 = 103 wickets
Stacked bar:
- Home: 0 to 32
- Overseas: 32 to 103
Therefore, the totals are: Test = 234 wickets, ODI = 149 wickets, and T20 = 103 wickets.
NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.5
1. 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.
Solution:
Each person’s three activities together account for 24 hours in a day.
(i) How much time does Sahana spend in each body-state?
This question is based on a reasonable estimate, so different answers are possible.
For example, since Sahana is a student, we may assume that she spends:
- Lying down: 8 hours
- Sitting: 8 hours
- Standing/moving around: 8 hours
Total = 8 + 8 + 8 = 24 hours
So, one possible entry for Sahana is:
8 hours, 8 hours, 8 hours
(ii) What activity or work could Julie be engaged in?
Julie spends:
- 8 hours lying down
- 10 hours sitting
- 6 hours standing or moving around
Since she spends a relatively large amount of time sitting, Julie could be engaged in a desk-based job, such as an office worker, accountant, or computer professional.
This is only a reasonable guess; other answers are also possible.
(iii) Complete the 100% stacked bar chart.
Convert the hours into percentages using:
Percentage = (Number of hours / 24) × 100
| Person | Lying down | Sitting | Standing/Moving |
|---|---|---|---|
| Pavani | 83.33% | 12.50% | 4.17% |
| Raghu | 29.17% | 16.67% | 54.17% |
| Zakir | 33.33% | 25% | 41.67% |
| Sahana* | 33.33% | 33.33% | 33.33% |
| Julie | 33.33% | 41.67% | 25% |
*For Sahana, these percentages correspond to the suggested 8 h, 8 h, 8 h estimate.
These percentages can be used to complete each 100% stacked bar in the chart.
NCERT Class 9 Maths Ganita Manjari Chapter 10 End-of-Chapter Exercises
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?
Solution:
Run rate is calculated as:
Run rate = Total runs scored ÷ Total overs played
(i) Run rate after the second over
Runs scored in the first over = 6
Runs scored in the second over = 12
Total runs after 2 overs:
6 + 12 = 18
Run rate:
18 ÷ 2 = 9
Therefore, the run rate after 2 overs is 9 runs per over.
(ii) Run rate after 20 overs
Run rate after 19 overs = 6
So, total runs after 19 overs:
19 × 6 = 114
Runs scored in the 20th over = 12
Total runs after 20 overs:
114 + 12 = 126
Run rate after 20 overs:
126 ÷ 20 = 6.3
Therefore, the run rate after 20 overs is 6.3 runs per over.
2. Five friends who play badminton surveyed the city and collected the price of shuttlecocks across brands and types (Nylon and Feather). The following is the list of the prices, in rupees (₹), that each one compiled.
- Yusuf: 40(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:
Solution:
To find the average price of a shuttlecock across both types, the five friends can each find the total price and number of shuttlecocks in their list. They can then add all the totals and divide by the total number of shuttlecocks collected.
The numbers of prices collected are:
- Yusuf: 6
- Srikanth: 4
- Kashvi: 5
- Prasanna: 4
- Gracy: 4
Total number of shuttlecocks = 6 + 4 + 5 + 4 + 4 = 23
(i) Expression for the combined average
Let their individual averages be a_y, a_s, a_k, a_p and a_g.
Since they collected different numbers of prices, we must use a weighted average:
Combined average = (6a_y + 4a_s + 5a_k + 4a_p + 4a_g) / 23
Therefore,
Answer: (6a_y + 4a_s + 5a_k + 4a_p + 4a_g) / 23
Using the actual data, the total of all prices is ₹4063.
So,
Average price = 4063 / 23
≈ ₹176.65
(ii) Expression using nylon and feather shuttlecocks
First count the shuttlecocks of each type.
Number of nylon shuttlecocks = 13
Number of feather shuttlecocks = 10
Let:
a_n = average price of nylon shuttlecocks
a_f = average price of feather shuttlecocks
Then,
Average price = (13a_n + 10a_f) / 23
For the given data:
Average price of nylon shuttlecocks
= 1429 / 13
≈ ₹109.92
Average price of feather shuttlecocks
= 2634 / 10
= ₹263.40
Therefore,
Average price
= (13 × 109.92 + 10 × 263.40) / 23
≈ ₹176.65
Yes, this is equal to the answer obtained in part (i). Both methods use the same 23 prices; they simply group the data differently.
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:
Solution:
(i) Shreyas
Shreyas initially has 25 shares at an average price of ₹150.
Total initial cost:
25 × 150 = ₹3750
He buys 10 more shares at ₹30 each:
10 × 30 = ₹300
Total cost after purchase:
₹3750 + ₹300 = ₹4050
Total number of shares:
25 + 10 = 35
New average price per share:
4050 ÷ 35 = ₹115.71 approximately
Therefore, Shreyas’s new average price per share is approximately ₹115.71.
(ii) Vaishnavi
Vaishnavi initially has 5 shares at an average price of ₹150.
Initial total cost:
5 × 150 = ₹750
Let the number of shares she buys at ₹30 each be x.
Her new average price is ₹70, so:
(750 + 30x) / (5 + x) = 70
750 + 30x = 350 + 70x
400 = 40x
x = 10
Therefore, Vaishnavi bought 10 shares.
4. (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:
Since the pool has different depths over sections of different lengths, we use a weighted average.
Lengths of the sections = 4, 5, 6, 7, 8 hastas
Corresponding depths = 9, 7, 7, 3, 2 hastas
Mean depth:
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
Therefore, the mean depth of the pool is 5 hastas.
5. 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:
Suvarna initially has 1 g of gold at an average price of ₹15k per gram.
For each transaction:
New average price = Total amount spent ÷ Total quantity of gold
(i) Purchase 1 g gold at ₹30k
Total cost = 15 + 30 = ₹45k
Total gold = 1 + 1 = 2 g
Average price = 45 ÷ 2 = ₹22.5k per g
So, mark the point at 22.5.
(ii) Purchase 2 g gold at ₹30k per g
Cost of new gold = 2 × 30 = ₹60k
Total cost = 15 + 60 = ₹75k
Total gold = 1 + 2 = 3 g
Average price = 75 ÷ 3 = ₹25k per g
So, mark the point at 25.
(iii) Purchase 1 g gold at ₹10k
Total cost = 15 + 10 = ₹25k
Total gold = 2 g
Average price = 25 ÷ 2 = ₹12.5k per g
So, mark the point at 12.5.
(iv) Purchase 10 g gold at ₹10k per g
Cost of new gold = 10 × 10 = ₹100k
Total cost = 15 + 100 = ₹115k
Total gold = 1 + 10 = 11 g
Average price = 115 ÷ 11 ≈ ₹10.45k per g
So, mark the point slightly above 10.
(v) Purchase 0.5 g gold at ₹30k per g
Cost of new gold = 0.5 × 30 = ₹15k
Total cost = 15 + 15 = ₹30k
Total gold = 1 + 0.5 = 1.5 g
Average price = 30 ÷ 1.5 = ₹20k per g
So, mark the point at 20.
(vi) Purchase 0.5 g gold at ₹15k per g
Cost of new gold = 0.5 × 15 = ₹7.5k
Total cost = 15 + 7.5 = ₹22.5k
Total gold = 1.5 g
Average price = 22.5 ÷ 1.5 = ₹15k per g
So, the average remains at 15.
Final answers:
(i) ₹22.5k
(ii) ₹25k
(iii) ₹12.5k
(iv) ₹10.45k approximately
(v) ₹20k
(vi) ₹15k
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:
The weighted average does not change if all the weights are doubled.
Suppose the data values are x₁, x₂, ..., xₙ with corresponding weights w₁, w₂, ..., wₙ.
The weighted average is:
Weighted average = (x₁w₁ + x₂w₂ + ... + xₙwₙ) / (w₁ + w₂ + ... + wₙ)
If all the weights are doubled, the new weighted average becomes:
[x₁(2w₁) + x₂(2w₂) + ... + xₙ(2wₙ)] / (2w₁ + 2w₂ + ... + 2wₙ)
Taking 2 common from the numerator and denominator:
= 2(x₁w₁ + x₂w₂ + ... + xₙwₙ) / 2(w₁ + w₂ + ... + wₙ)
The 2 cancels out, giving:
(x₁w₁ + x₂w₂ + ... + xₙwₙ) / (w₁ + w₂ + ... + wₙ)
This is the original weighted average.
Therefore, doubling all the weights does not change the weighted average. In fact, multiplying all weights by the same non-zero number leaves the weighted average unchanged.
7. 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) Total objects in 2003 and when this number doubled
From the graph, in 2003 there were approximately 9,500 objects orbiting Earth.
Double of 9,500 is approximately:
9,500 × 2 = 19,000
The total number of objects reaches about 19,000 in 2019.
Therefore, there were about 9,500 objects in 2003, and this number approximately doubled by 2019.
(ii) Number and share of payload and other objects in 2025
From the 2025 bar:
- Total objects ≈ 35,000
- Payload objects ≈ 16,500–17,000
- Other objects ≈ 18,000
Approximate share of payload objects:
(16,500 / 35,000) × 100 ≈ 47%–48%
Approximate share of other objects:
(18,000 / 35,000) × 100 ≈ 52%–53%
Therefore, in 2025, roughly 48% were payload objects and 52% were other objects.
(iii) What trends can be observed?
The number of both payload objects and other objects has increased over time.
For many years, the number of payload objects increased gradually, but it has risen very sharply in recent years. The number of other objects has also increased steadily over time.
In terms of their shares, other objects formed a much larger share of the total for many years. However, the share of payload objects has increased considerably in recent years. By 2025, the two categories account for nearly equal shares, with other objects still slightly higher.
8. Observe the following infographic.
(i) Identify the correct inference(s) from the statements given.
(a) More schools have a playground in Haryana compared to Uttarakhand.
(b) Approximately every 2 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) Identify the correct inference(s)
(a) More schools have a playground in Haryana compared to Uttarakhand.
Haryana: 90%
Uttarakhand: 78%
Although Haryana has a higher percentage, the infographic does not tell us the total number of schools in each state. Therefore, we cannot conclude that the number of schools with playgrounds is higher in Haryana.
So, (a) cannot be inferred.
(b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground.
Arunachal Pradesh: 68%
2 out of 3 = 2/3 × 100 ≈ 66.7%
This is close to 68%.
So, (b) is correct.
(c) Punjab has the highest number of schools with a playground.
Punjab has 98% of schools with playgrounds, but this tells us only the percentage, not the actual number of schools.
So, (c) cannot be inferred.
(d) Suppose Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra.
Maharashtra: 93%
Telangana: 74%
If Maharashtra also has more total schools than Telangana, and a higher percentage of them have playgrounds, then Maharashtra must have a greater number of schools with playgrounds.
So, (d) is correct.
Therefore, the correct inferences are (b) and (d).
(ii) Can we find the nationwide percentage?
No. The nationwide percentage cannot be found from the percentages alone because different states and Union Territories have different numbers of schools.
We would additionally need the total number of schools in each state/UT. Then we could calculate the number of schools with playgrounds in each region and find:
Nationwide percentage = (Total schools with playgrounds across India / Total schools across India) × 100
Thus, the number of schools in each state and Union Territory is the additional information required.
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.
i) Which play would you choose based on the first chart?
The first chart shows the percentage distribution of ratings for each play.
A reasonable choice would be Play A, because a large proportion of its ratings are 4-star and 5-star, while comparatively few are 1-star or 2-star ratings.
Play B also has a very similar proportion of favourable ratings, whereas Play C has a noticeably larger share of low ratings.
Therefore, based only on the percentage distribution, Play A could be chosen.
(ii) Would you change your decision after looking at the second chart?
Possibly, yes. The second chart shows the actual number of ratings, not just their percentages.
It shows that Play B has received far more ratings than Play A, while still having a high proportion of 4-star and 5-star ratings. Hence, one may prefer Play B, because its rating pattern is based on a much larger number of viewers.
The two charts highlight different information:
- The 100% stacked bar chart compares the proportion of each type of rating.
- The stacked bar chart also shows the total number of ratings received by each play.
Thus, after considering both the quality and the number of ratings, one may change the choice from Play A to Play B. However, the final choice can vary depending on whether one gives more importance to rating proportions or to the number of viewers who rated the play.
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 following table shows the finish time of 3 athletes in each segment.
(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) Suitable representation
A stacked bar chart is more suitable.
This is because the total length of each bar represents the athlete’s total race time, while the individual sections show the time spent swimming, cycling, and running. This allows us to compare both the overall finishing times and the time spent on each segment.
A 100% stacked bar chart would make every athlete’s bar the same length, so the actual total race times could not be compared.
(ii) Questions that can be answered using only a 100% stacked bar chart
(a) Who finished the race first?
No. Since every bar is scaled to 100%, the total race times are not shown.
(b) Approximately what fraction of their race time did Athlete 1 spend cycling?
Yes. A 100% stacked bar chart shows what proportion of the total time was spent on each activity.
For Athlete 1:
Total time = 1 h 8 min + 5 h + 3 h 15 min
= 9 h 23 min = 563 min
Fraction spent cycling = 300/563 ≈ 53%
So, Athlete 1 spent a little over half of the total race time cycling.
(c) Which athlete took the longest for running?
No. The chart would show the proportion of each athlete’s total time spent running, not the actual running time.
Therefore, only question (b) can be answered from the 100% stacked bar chart.
11. Look at the following graph. What do you notice? What do you wonder? Write your inferences.
Solution:
The graph is a 100% stacked bar chart showing how different types of disability are distributed within four age groups in India, according to Census 2011. Since every bar represents 100%, the graph compares proportions, not the actual number of persons.
Some observations and inferences are:
- Seeing-related disability forms a noticeably larger share in the 60 years and above age group than in the younger age groups.
- Movement-related disability also becomes more prominent in the older age groups.
- Hearing-related disability forms a fairly large share across all age groups, though its proportion is lower among those aged 60 and above than in some younger groups.
- Speech-related disability accounts for a relatively small share in every age group.
- The category labelled “mental retardation” in the Census 2011 graph (now generally referred to as intellectual disability) forms a larger share among younger age groups and a smaller share among older age groups.
- Multiple disability has a comparatively high share in the 60 years and above group.
- The category “any other” accounts for a substantial share among the younger and middle-age groups but is much smaller for those aged 60 and above.
What we may wonder
We may wonder why seeing, movement, and multiple disabilities form larger shares among older people, whether this is related to ageing, and whether the pattern has changed since the 2011 Census.
We may also want to know the actual number of persons in each age group. This graph alone cannot tell us that because it shows only percentages.
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:
Solution:
Since this is an individual project, the actual data will differ from person to person. The following can be used as a sample project.
(i) Time spent in different body-states
I estimated the time spent by myself and two family members during a typical day.
| Person | Lying down | Sitting | Standing/Moving | Total |
|---|---|---|---|---|
| Myself | 8 hours | 9 hours | 7 hours | 24 hours |
| Mother | 7.5 hours | 8.5 hours | 8 hours | 24 hours |
| Father | 7 hours | 10 hours | 7 hours | 24 hours |
A stacked bar chart can be drawn with one bar for each person. Each bar should extend from 0 to 24 hours and be divided into the three body-states.
I estimated the times by recalling sleeping hours, time spent studying or working while sitting, meal and travel time, and time spent walking, exercising or doing household activities.
Observations: My father spent the most time sitting, while my mother spent the most time standing or moving around. All three spent around 7–8 hours lying down. The stacked bar chart makes it easy to compare how each person's 24-hour day is divided.
Since this is an individual project, the actual expenditure will vary from family to family. The following is a sample project showing how it can be completed.
(a) Identify the expense categories
The monthly family expenditure can be divided into categories such as house rent, food and groceries, utilities, education, transport, and other expenses.
(b) Decide how to collect and organise the data
The expenditure data can be collected by discussing monthly spending with family members and referring to bills, receipts, bank statements, and digital payment records. The expenses can then be grouped under the selected categories and recorded month-wise in a table.
(c) Collect expense data for at least 3 months
Suppose the following expenditure was recorded:
| Expense category | January | February | March |
|---|---|---|---|
| House rent | ₹15,000 | ₹15,000 | ₹15,000 |
| Food & groceries | ₹12,000 | ₹14,000 | ₹13,000 |
| Utilities | ₹5,000 | ₹6,000 | ₹5,500 |
| Education | ₹8,000 | ₹10,000 | ₹9,000 |
| Transport | ₹5,000 | ₹6,000 | ₹5,500 |
| Other expenses | ₹5,000 | ₹9,000 | ₹7,000 |
| Total expenditure | ₹50,000 | ₹60,000 | ₹55,000 |
(d) Represent the data using a 100% stacked bar chart
To draw the 100% stacked bar chart, first convert every expense into a percentage of that month's total expenditure:
Percentage expenditure = (Amount spent on category / Total monthly expenditure) × 100
| Expense category | January | February | March |
|---|---|---|---|
| House rent | 30% | 25% | 27.27% |
| Food & groceries | 24% | 23.33% | 23.64% |
| Utilities | 10% | 10% | 10% |
| Education | 16% | 16.67% | 16.36% |
| Transport | 10% | 10% | 10% |
| Other expenses | 10% | 15% | 12.73% |
| Total | 100% | 100% | 100% |
100% stacked bar chart showing the share of each expense category over three months.
(e) Each bar represents 100% of the expenditure for one month, and the different sections show the proportion spent on each category.
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:
Small-Group Project: Clinic/Hospital Experience
For this sample project, we chose (d) Clinic/Hospital Experience.
(i) Aspects to rate and why they matter
We selected five aspects:
| Aspect | Why it matters |
|---|---|
| Doctor consultation | The quality of diagnosis, explanation and treatment is the most important part of the visit. |
| Staff behaviour | Polite and helpful staff improve the patient's overall experience. |
| Waiting time | Long waiting times can make the visit inconvenient and stressful. |
| Cleanliness | Hygiene is particularly important in a healthcare facility. |
| Facilities | Availability and quality of seating, washrooms, equipment and other facilities affect comfort and convenience. |
(ii) Relative weights
After discussion, we assigned the following weights:
| Aspect | Weight |
|---|---|
| Doctor consultation | 30% |
| Staff behaviour | 20% |
| Waiting time | 20% |
| Cleanliness | 15% |
| Facilities | 15% |
| Total | 100% |
We gave the highest weight to doctor consultation because medical care is the main purpose of visiting a clinic or hospital. Staff behaviour and waiting time were given the next-highest weights, while cleanliness and facilities were also considered important.
(iii) Ratings collected from 10 people
Ratings are on a scale of 1 to 5, where 5 means very good.
| Person | Doctor | Staff | Waiting time | Cleanliness | Facilities |
|---|---|---|---|---|---|
| 1 | 5 | 4 | 3 | 5 | 4 |
| 2 | 4 | 4 | 4 | 4 | 4 |
| 3 | 5 | 5 | 4 | 4 | 5 |
| 4 | 3 | 4 | 2 | 4 | 3 |
| 5 | 4 | 3 | 3 | 5 | 4 |
| 6 | 5 | 4 | 4 | 5 | 5 |
| 7 | 4 | 5 | 3 | 4 | 4 |
| 8 | 3 | 3 | 2 | 3 | 3 |
| 9 | 4 | 4 | 5 | 4 | 4 |
| 10 | 5 | 4 | 3 | 4 | 5 |
(iv) Individual ratings and overall average
The weighted rating is calculated as:
Individual rating = (30 × Doctor + 20 × Staff + 20 × Waiting + 15 × Cleanliness + 15 × Facilities) / 100
For Person 1:
= (30 × 5 + 20 × 4 + 20 × 3 + 15 × 5 + 15 × 4) / 100
= 425 / 100
= 4.25
Similarly:
| Person | Final weighted rating |
|---|---|
| 1 | 4.25 |
| 2 | 4.00 |
| 3 | 4.65 |
| 4 | 3.15 |
| 5 | 3.75 |
| 6 | 4.60 |
| 7 | 4.00 |
| 8 | 2.80 |
| 9 | 4.20 |
| 10 | 4.25 |
Overall average:
= (4.25 + 4.00 + 4.65 + 3.15 + 3.75 + 4.60 + 4.00 + 2.80 + 4.20 + 4.25) / 10
= 39.65 / 10
= 3.965 ≈ 3.97 out of 5
Therefore, the overall rating of the clinic/hospital is approximately 3.97 out of 5.
(v) 100% stacked bar chart
For the chart, we can show the percentage of respondents who gave 1-star, 2-star, 3-star, 4-star and 5-star ratings to each aspect.
| Aspect | 1★ | 2★ | 3★ | 4★ | 5★ |
|---|---|---|---|---|---|
| Doctor consultation | 0% | 0% | 20% | 40% | 40% |
| Staff behaviour | 0% | 0% | 20% | 60% | 20% |
| Waiting time | 0% | 20% | 40% | 30% | 10% |
| Cleanliness | 0% | 0% | 10% | 60% | 30% |
| Facilities | 0% | 0% | 20% | 50% | 30% |
(vi) Observations and inferences
The overall rating of about 3.97 out of 5 indicates that the respondents generally had a positive experience. Doctor consultation received particularly strong ratings, with 80% of respondents giving it either 4 or 5. Cleanliness and facilities were also rated positively.
Waiting time appears to be the main area for improvement, as it received the largest share of 2- and 3-star ratings.
The rating system captures important factors such as medical care, staff behaviour, waiting time, cleanliness and facilities. However, it does not capture everything—for example, treatment cost, availability of medicines, emergency services, accessibility, or individual medical outcomes. The results may also change if a larger group of people is surveyed.
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:
To find the average age of all family members represented in the class, use the number of members in each family as the weight.
- Record each family’s average age and number of members.
- Multiply the average age by the number of members to find the total age of that family.
- Add the total ages of all families.
- Divide by the total number of family members.
If the family averages are a₁, a₂, …, aₙ and the family sizes are n₁, n₂, …, nₙ, then:
Combined average age = (n₁a₁ + n₂a₂ + … + nₙaₙ) / (n₁ + n₂ + … + nₙ)
For example:
- Family A: 4 members, average age = 25 years
Total age = 4 × 25 = 100 years - Family B: 6 members, average age = 30 years
Total age = 6 × 30 = 180 years - Family C: 5 members, average age = 28 years
Total age = 5 × 28 = 140 years
Therefore,
Combined average = (100 + 180 + 140) / (4 + 6 + 5)
= 420 / 15
= 28 years
Therefore, the average age of all the family members is 28 years.
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:
Unlike multiplying all the weights by the same number, adding 1 to every weight will generally change the weighted average.
For example, take the data values 10 and 20 with weights 1 and 3.
Original weighted average:
= (10 × 1 + 20 × 3) / (1 + 3)
= 70 / 4
= 17.5
Now increase each weight by 1. The new weights are 2 and 4.
New weighted average:
= (10 × 2 + 20 × 4) / (2 + 4)
= 100 / 6
≈ 16.67
So, the weighted average has changed.
Algebraic justification
Suppose the data values are x₁, x₂, …, xₙ with weights w₁, w₂, …, wₙ.
Original weighted average:
A = (w₁x₁ + w₂x₂ + … + wₙxₙ) / (w₁ + w₂ + … + wₙ)
If 1 is added to every weight, the new weighted average becomes:
A' = [(w₁ + 1)x₁ + (w₂ + 1)x₂ + … + (wₙ + 1)xₙ] / [(w₁ + 1) + (w₂ + 1) + … + (wₙ + 1)]
Simplifying:
A' = (w₁x₁ + w₂x₂ + … + wₙxₙ + x₁ + x₂ + … + xₙ) / (w₁ + w₂ + … + wₙ + n)
This is generally different from the original weighted average.
Observation: Adding the same positive number to all the weights makes the weights more similar to one another. Therefore, the new weighted average moves closer to the ordinary arithmetic mean of the data.
Hence, increasing all weights by 1 generally changes the weighted average; it does not remain unchanged.
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 percentages of cows, sheep, and chickens must always add up to 100%.
Even though the number of all three animals decreased, their percentage shares may increase, decrease, or stay the same depending on how much each population decreased.
- (i) All three percentages decreased — Not possible, because their shares must still total 100%.
- (ii) All three percentages increased — Not possible, for the same reason.
- (iii) All three percentages remained the same — Possible, if cows, sheep, and chickens all decreased by the same proportion.
- (iv) Cows decreased, sheep increased, chickens remained the same — Possible, if cows decreased relatively more, sheep relatively less, and chickens changed in a way that keeps their share unchanged.
- (v) Cows increased, sheep increased, chickens decreased — Possible, if chickens decreased by a larger proportion than cows and sheep.
Therefore, the possible cases are (iii), (iv), and (v).
Related Links: Chapter 10 Exercise-wise Solutions
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.1
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.2
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.3
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.4
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.5
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 End-of-Chapter Exercises