“Ch 10 How Quantities Combine: Understanding Data Solutions Class 9“ is prepared on the basis on Ganita Manjari Part 2 class 9 NCERT. And also we have prepared it after teaching and solving the questions in the actual class.
“Ch 10 How Quantities Combine: Understanding Data Solutions Class 9“ is strictly based on Ganita Manjari Grade 9 NCERT textbook.
We have tried to strike a balance between the solution and explanations. So that you do not get confused or find difficulty understanding the solutions.
For any further assistance on any question please contact us.
For more chapter solutions | Class 9 Maths Solutions
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?
Q1. Solution
Given,
Section A
Av. Score = 72
No. of Students = 30
Section B
Av. Score = 76
No. of Students = 25
∴ using,
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?
Q2. Solution
Given,
Let us assume that Quantity of each fertilizer = y.
Fertiliser 1
Nitrogen = 110
Fertiliser 2
Nitrogen = 950
Fertiliser 3
Nitrogen = 360
∴ using,
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 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.
Q3. Solution
Given,
Piece 1
Quantity = 9 units
Purity = 12 varna
Piece 2
Quantity = 5 units
Purity = 10 varna
Piece 3
Quantity = 17 units
Purity = 11 varna
∴ using,
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.
Q4. Solution
Given,
May
Av. Rainfall per day = 3.5 mm
No. of days = 31
June
Av. Rainfall per day = 10 mm
No. of days = 30
July
Av. Rainfall per day = 8.7 mm
No. of days = 31
∴ using,
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
Q5. Solution
Part (i)
Given,
Kadha 1: Volume = 100 mL, Concentration = 5%
Kadha 2: Volume = 200 mL, Concentration = 10%
Kadha 3: Volume = 300 mL, Concentration = 15%
∴ using,
Part (ii)
Given,
Kadha 1: Volume = 300 mL, Concentration = 5%
Kadha 2: Volume = 200 mL, Concentration = 10%
Kadha 3: Volume = 100 mL, Concentration = 15%
∴ using the same formula,
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 x 3 + 44 x 4 + 80 x 5) / (3 + 4 + 5)
(ii) ((35/100) x 3 + (44/100) x 4 + (80/100) x 5) / (3 + 4 + 5)
(iii) (((35/50) x 3 + (44/60) x 4 + (80/100) x 5) / (3 + 4 + 5)) x 100
(iv) ((35/50 x 100) x 3 + (44/60 x 100) x 4 + (80/100 x 100) x 5) / (3 + 4 + 5)
Q1. Solution
Given,
Internals: 35 out of 50
Project: 44 out of 60
Final exam: 80 out of 100
Ratio of weights = 3 : 4 : 5
∴ using,
Percentage in each component:
Annual Score
Checking options (i) and (ii):
Both use the raw marks as if every component was out of 100.
(i) Not correct
(ii) Not correct
(iii) Correct
(iv) 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?
Q1. Solution
Given,
Books
Sales = ₹8000
Profit = 30% of sales
Book covers
Sales = ₹1000
Profit = 50% of sales
∴ using,
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.
Q2. Solution
Given,
First 20 days
Av. distance per day = 44.5 km
No. of days = 20
21st day
Distance = 55 km
No. of days = 1
Guess
55 km is more than 44.5 km, so the average will go up.
But it is only 1 day out of 21, so it will go up by a little.
∴ using,
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%
Q3. Solution
Given,
Solution 1
Volume = 600 mL
Salt = 5%, Sugar = 0%
Solution 2
Volume = 300 mL
Salt = 0%, Sugar = 8%
∴ using,
Salt
Sugar
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?
Q4. Solution
Given,
Pani
Volume = 10 litres
Spice = 8%
Regular water
Volume = x litres (to be found)
Spice = 0%
Required
∴ using,
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?
Q5. Solution
Given,
Weights (Strength : Flexibility : Agility) = 4 : 5 : 6
Rashi
Strength = 60, Flexibility = 65, Agility = 70
Keerthi
Strength = 55, Flexibility = 65, Agility = 75
Part (i)
Compare Keerthi with Rashi, test by test:
Strength (weight 4): Keerthi has 5 less
Flexibility (weight 5): equal
Agility (weight 6): Keerthi has 5 more
Keerthi loses 5 marks in a test with a smaller weight.
She gains 5 marks in a test with a bigger weight.
Part (ii)
∴ using,
Rashi
Keerthi
Rashi = 65.67
Keerthi = 66.33
Keerthi’s final marks are more, as we found in part (i).
Q6.
- A restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting data is shown in the table below.
| 5★ | 4★ | 3★ | 2★ | 1★ | |
|---|---|---|---|---|---|
| Food | 5 | 3 | 2 | 0 | 0 |
| Ambience | 0 | 4 | 5 | 1 | 0 |
| Service | 1 | 2 | 2 | 4 | 1 |
What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?
Q6. Solution
Given,
| Metric | 5★ | 4★ | 3★ | 2★ | 1★ |
|---|---|---|---|---|---|
| Food | 5 | 3 | 2 | 0 | 0 |
| Ambience | 0 | 4 | 5 | 1 | 0 |
| Service | 1 | 2 | 2 | 4 | 1 |
Customers = 10
Weights (Food : Ambience : Service) = 6 : 5 : 4
∴ using,
Food
Ambience
Service
Now combine the three metrics with their weights:
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?
| Male langurs | Female langurs | All langurs | |
|---|---|---|---|
| Average weight | 16.5 kg | 13.8 kg | 14.925 kg |
| Number of langurs | 60 |
(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?
Q7. Solution
Given,
| Data | Male langurs | Female langurs | All langurs |
|---|---|---|---|
| Average weight | 16.5 kg | 13.8 kg | 14.925 kg |
| Number of langurs | x | y | 60 |
Without computing
Overall average = 14.925 kg
It lies between 13.8 kg and 16.5 kg.
It is closer to the female average, 13.8 kg.
Part (i)
∴ using,
(a) Correct
(b) Correct
(c) Not correct
(d) Not correct
Part (ii)
Total number: x + y = 60 …(1)
Total weight: 16.5x + 13.8y = 14.925 × 60 = 895.5 …(2)
From (1), y = 60 − x
16.5x + 13.8 (60 − x) = 895.5
16.5x + 828 − 13.8x = 895.5
2.7x = 67.5
Then y = 60 − 25 = 35
Part (iii)
Total weight of females = 13.8 × 35 = 483 kg
Part (iv)
Total weight of males = 16.5 × 25 = 412.5 kg
Weight released = 16.9 + 16.1 = 33 kg
Part (v)
After part (iv), there are 23 male langurs.
Total weight = 379.5 kg
One loses 1 kg: total weight = 379.5 − 1 = 378.5 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
Q8. Solution
Given,
| Water source | Salinity |
|---|---|
| Dead Sea | ≈ 34% |
| Other seas and oceans | ≈ 3.5% |
| Ground water | ≈ 0.01% |
| Purified drinking water | ≈ 0.001% |
Dead Sea water collected = 1 litre
Part (i)
Dead Sea: Volume = 1 litre, Salinity = 34%
Purified water: Volume = 2 litres, Salinity = 0.001%
∴ using,
Part (ii)
Ground water = 0.01%
Lowest salinity = 0.001% (purified water)
Highest salinity = 34% (Dead Sea)
0.01% lies between 0.001% and 34%, so it is possible.
Let the purified water to be added be x litres.
34 + 0.001x = 0.01 (1 + x)
34 + 0.001x = 0.01 + 0.01x
34 − 0.01 = 0.01x − 0.001x
33.99 = 0.009x
Possible: Yes
Purified water ≈ 3776.67 litres
Part (iii)
Needed salinity = 0.001%
Salinity of Dead Sea = 34%, ground water = 0.01%
0.001% is less than both, so it is not possible.
Check: let the ground water be x litres.
34 + 0.01x = 0.001 + 0.001x
34 − 0.001 = 0.001x − 0.01x
33.999 = −0.009x
A negative quantity cannot be mixed in.
Possible: No
No quantity of ground water works
Exercise Set 10.4
Q1.
Q1. Solution
Part (i)
The number 14,234 is the height of the whole stacked column of 2019.
It is the sum of all the small bars in that column:
Mammals + Reptiles + Birds + Insects + Amphibians + Molluscs + Fish + Others
Part (ii)
Look at the column of 2016.
Reptiles are the second small bar from the bottom (dark red).
Bottom of the reptile bar ≈ 1,100
Top of the reptile bar ≈ 2,300
Part (iii)
Compare the thickness of each small bar in 2007 and in 2019.
| Class | 2007 (approx.) | 2019 (approx.) | Increase (approx.) |
|---|---|---|---|
| Mammals | 1,100 | 1,250 | 150 |
| Reptiles | 450 | 1,350 | 900 |
| Birds | 1,200 | 1,500 | 300 |
| Insects | 600 | 1,800 | 1,200 |
| Amphibians | 1,800 | 2,200 | 400 |
| Molluscs | 1,050 | 2,200 | 1,150 |
| Fish | 1,200 | 2,650 | 1,450 |
| Others | 550 | 1,300 | 750 |
The smallest increases are in Mammals (about 150) and Birds (about 300).
Amphibians also rose only a little (about 400).
Mammals
Birds
Q2.
Q2. Solution
Given,
| Test | ODI | T20 | |
|---|---|---|---|
| Home | 62 | 100 | 32 |
| Overseas | 172 | 49 | 71 |
Part (i)
Each bar is one place. Its small bars are Test, ODI and T20.
Home
Test: 0 to 62
ODI: 62 to 62 + 100 = 162
T20: 162 to 162 + 32 = 194
Overseas
Test: 0 to 172
ODI: 172 to 172 + 49 = 221
T20: 221 to 221 + 71 = 292
Home = 194
Overseas = 292
Part (ii)
Now each bar is a format. Its small bars are Home and Overseas.
Test
Home: 0 to 62
Overseas: 62 to 62 + 172 = 234
ODI
Home: 0 to 100
Overseas: 100 to 100 + 49 = 149
T20
Home: 0 to 32
Overseas: 32 to 32 + 71 = 103
Test = 234
ODI = 149
T20 = 103
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
Q1. Solution
Given,
| Lying down | Sitting | Standing or moving | |
|---|---|---|---|
| Pavani (Patient) | 20 | 3 | 1 |
| Raghu (Nurse) | 7 | 4 | 13 |
| Zakir (Teacher) | 8 | 6 | 10 |
| Sahana (Student) | ? | ? | ? |
| Julie (?) | 8 | 10 | 6 |
All values are hours per day.
Part (i)
A day has 24 hours, so every row must add up to 24.
Check: Pavani 20 + 3 + 1 = 24, and Julie 8 + 10 + 6 = 24.
A reasonable guess for a student’s day:
Lying down (sleep and rest) ≈ 9 hours
Sitting (classes, homework, meals, screens) ≈ 7 hours
Standing or moving (walking, games, chores) ≈ 8 hours
Check: 9 + 7 + 8 = 24
Part (ii)
Julie: Lying down = 8, Sitting = 10, Standing or moving = 6
She sits the most (10 hours) and stands or moves the least (6 hours).
Compare: Raghu (nurse) stands or moves for 13 hours, and Zakir (teacher) for 10 hours.
So Julie probably works at a desk.
Part (iii)
Each bar is one full day, 24 hours, which is 100%.
Example, Sahana: lying down 924 × 100 = 37.5 ≈ 38%
Sitting 724 × 100 = 29.2 ≈ 29%
Standing or moving 824 × 100 = 33.3 ≈ 33%
Doing the same for every row:
| Lying down | Sitting | Standing or moving | |
|---|---|---|---|
| Pavani | 83% | 13% | 4% |
| Raghu | 29% | 17% | 54% |
| Zakir | 33% | 25% | 42% |
| Sahana | 38% | 29% | 33% |
| Julie | 33% | 42% | 25% |
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?
Q1. Solution
Given,
Run rate = average number of runs scored per over
Over 1: 6 runs, so the run rate is 6
Part (i)
Over 2: 12 runs
∴ using,
Part (ii)
After 19 overs: run rate = 6
20th over: 12 runs
∴ using,
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?
Q2. Solution
Given,
Count the shuttlecocks collected by each friend, and of each type:
| Friend | Total | Nylon (N) | Feather (F) |
|---|---|---|---|
| Yusuf | 6 | 3 | 3 |
| Srikanth | 4 | 4 | 0 |
| Kashvi | 5 | 3 | 2 |
| Prasanna | 4 | 2 | 2 |
| Gracy | 4 | 1 | 3 |
| All | 23 | 13 | 10 |
Way to work together: each friend shares the average and the number of prices collected.
Part (i)
Averages: ay, as, ak, ap, ag
Number of prices: Yusuf 6, Srikanth 4, Kashvi 5, Prasanna 4, Gracy 4
∴ using,
Part (ii)
Averages: nylon an, feather af
Number of shuttlecocks: nylon 13, feather 10
Will this be equal to the answer in part (i)?
Use the actual prices to check.
Part (i): 6ay = 917, 4as = 493, 5ak = 766, 4ap = 897, 4ag = 990
Total = 917 + 493 + 766 + 897 + 990 = 4063
Part (ii): 13an = 1429, 10af = 2634
Total = 1429 + 2634 = 4063
Both give 406323 ≈ 176.65
Yes, they are equal
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?
Q3. Solution
Given,
Shreyas
Shares held = 25
Average price = ₹150 per share
Vaishnavi
Shares held = 5
Average price = ₹150 per share
Part (i)
Shreyas buys 10 shares at ₹30 each.
∴ using,
Part (ii)
Vaishnavi buys x shares at ₹30 each.
New average price = ₹70 per share
750 + 30x = 70 (5 + x)
750 + 30x = 350 + 70x
750 − 350 = 70x − 30x
400 = 40x
Check: 750 + 30015 = 105015 = 70
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.
Q4. Solution
Given,
Length of the pool = 30 hastas, divided into 5 sections.
| Length (hastas) | Depth (hastas) | |
|---|---|---|
| Section 1 | 4 | 9 |
| Section 2 | 5 | 7 |
| Section 3 | 6 | 7 |
| Section 4 | 7 | 3 |
| Section 5 | 8 | 2 |
∴ using,
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
Q5. Solution
Given,
Suvarna has 1 g of gold at ₹15k per gram. This is the point O = 15.
Prices are in thousands of rupees (k).
∴ using,
(i) Purchase 1 g at ₹30k
Equal grams, so exactly midway between 15 and 30.
(ii) Purchase 2 g at ₹30k
More grams at 30 than at 15, so it is above the midway value 22.5.
(iii) Purchase 1 g at ₹10k
Equal grams, so exactly midway between 15 and 10.
(iv) Purchase 10 g at ₹10k
Far more grams at 10, so it is very close to 10.
(v) Purchase 0.5 g at ₹30k
Fewer grams at 30 than at 15, so it is between 15 and 22.5.
(vi) Purchase 0.5 g at ₹15k
Same price as before, so the average does not change.
Marking the averages on the number line:
| Scenario | Purchase | New average (k) |
|---|---|---|
| (i) | 1 g at 30 | 22.5 |
| (ii) | 2 g at 30 | 25 |
| (iii) | 1 g at 10 | 12.5 |
| (iv) | 10 g at 10 | 10.45 |
| (v) | 0.5 g at 30 | 20 |
| (vi) | 0.5 g at 15 | 15 |
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.
Q6. Solution
Experiment
Let’s take these numbers: 60, 64, 73
Weights: 3, 2, 5
Average with these weights:
Now double all the weights: 6, 4, 10
Observation: the average does not change.
Algebra
Double every weight: w1, w2, …, wn become 2w1, 2w2, …, 2wn.
New average:
The same happens if all weights are multiplied by any number other than 0.
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?
Q7. Solution
Given,
A stacked column chart of the objects orbiting Earth, from 1960 to 2026.
Each column shows two parts: payload objects and other objects.
Readings from the chart are approximate.
Part (i)
Find the column of 2003. The top of the full column is just below 10,000.
Total objects in 2003 ≈ 10,000
Double of this = 2 × 10,000 = 20,000
The top of the columns reaches 20,000 in about 2021.
Objects in 2003 ≈ 10,000
Doubled to 20,000 in ≈ 2021
Part (ii)
Look at the column of 2025.
Top of the full column ≈ 33,000
Top of the payload part ≈ 17,000
Other objects = 33,000 − 17,000 = 16,000
Payload objects ≈ 17,000 (52%)
Other objects ≈ 16,000 (48%)
Part (iii)
Reading the payload part and the whole column for some years:
| Year | Total | Payload | Other | Payload share | Other share |
|---|---|---|---|---|---|
| 2003 | 10,000 | 2,400 | 7,600 | 24% | 76% |
| 2007 | 13,200 | 2,500 | 10,700 | 19% | 81% |
| 2010 | 14,700 | 2,800 | 11,900 | 19% | 81% |
| 2015 | 16,400 | 3,300 | 13,100 | 20% | 80% |
| 2019 | 17,800 | 3,800 | 14,000 | 21% | 79% |
| 2022 | 22,500 | 7,300 | 15,200 | 32% | 68% |
| 2025 | 33,000 | 17,000 | 16,000 | 52% | 48% |
Number of payload objects: keeps increasing. Slowly up to about 2019 (2,400 to 3,800), and very fast after 2020 (to about 17,000 by 2025).
Number of other objects: also keeps increasing. There is a sudden jump around 2007 (about 7,600 to 10,700), and slower growth later (about 14,000 in 2019 to 16,000 in 2025).
Share of payload objects: stays near 20% for many years, then rises sharply after 2020 to about half by 2025.
Share of other objects: stays near 80% for many years, then falls to about half by 2025.
Both numbers increase
Payload share rises, other share falls
Q8.
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?
Q8. Solution
Given,
The map gives the percentage of schools with a playground in each state.
| State | Schools with a playground |
|---|---|
| Haryana | 90% |
| Uttarakhand | 78% |
| Arunachal Pradesh | 68% |
| Punjab | 98% |
| Maharashtra | 93% |
| Telangana | 74% |
A percentage tells the share within a state. It does not tell the number of schools.
Part (i)
(a)
Haryana 90% and Uttarakhand 78% compare shares only.
The number of schools in each state is not given.
So we cannot say Haryana has more schools with a playground.
(b)
68% is close to 66.7%, so about 2 out of 3 schools have a playground. Correct.
(c)
Punjab’s 98% is a percentage, not a number of schools.
Also, 99.8% (Dadra & Nagar Haveli and Daman & Diu) is higher.
We cannot infer this.
(d)
Let Maharashtra have M schools and Telangana T schools, with M > T.
Playgrounds: 0.93 M in Maharashtra, 0.74 T in Telangana.
So Maharashtra has more schools with a playground. Correct.
(a) Not correct
(b) Correct
(c) Not correct
(d) Correct
Part (ii)
The map gives the percentages, but not the number of schools in each state.
Without the weights, we cannot find the weighted average.
No, we cannot find it
Needed: the number of schools in each state or UT
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.
Q9. Solution
Given,
Chart 1 is a 100% stacked bar: it shows the share of each rating.
Chart 2 is a stacked bar: it shows the number of each rating.
Readings from the charts are approximate.
Part (i)
Customers who liked a play gave it 5★ or 4★. In Chart 1 this is the blue and red parts together.
Play A
5★ + 4★ reaches about 71%; 2★ + 1★ is about 14%
Play B
5★ + 4★ reaches about 71%; 2★ + 1★ is about 14%
Play C
5★ + 4★ reaches only about 53%; 2★ + 1★ is about 43%
Play A and Play B look almost the same, and both do better than Play C.
Choose Play A or Play B
Play C is the risky one
Part (ii)
Chart 2 shows how many people gave each rating (approximate):
| 5★ | 4★ | 3★ | 2★ | 1★ | Total | 5★ + 4★ | |
|---|---|---|---|---|---|---|---|
| Play A | 60 | 90 | 30 | 5 | 25 | 210 | 150 |
| Play B | 220 | 440 | 130 | 35 | 100 | 925 | 660 |
| Play C | 150 | 50 | 15 | 40 | 125 | 380 | 200 |
Play A: only about 210 people rated it.
Play B: about 925 people rated it, and about 660 of them gave 5★ or 4★.
Play C: about 380 people rated it.
Play B and Play A have the same share of good ratings (about 71%).
But Play B’s share comes from many more people, so it is more reliable.
Choose Play B
If I chose Play A in part (i), I would change.
If I chose Play B, my decision stays.
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)
Table 10.4: Finish time of three athletes in each segment of a triathlon race (values in hh:mm)
| Athlete | Swimming | Cycling | Running |
|---|---|---|---|
| Athlete 1 | 01:08 | 05:00 | 03:15 |
| Athlete 2 | 01:05 | 05:10 | 03:35 |
| Athlete 3 | 01:22 | 05:55 | 03:50 |
(i) What is a suitable representation of this data — a stacked
bar chart or a 100% stacked bar chart? Why do you think so?
(ii) Suppose a 100% stacked bar chart is drawn. Which of the
following question(s) can be answered by looking at just
that chart?
(a) Who finished the race first?
(b) Approximately what fraction of their race time did
Athlete 1 spend cycling?
(c) Which athlete took the longest for running?
Q10. Solution
Given,
Triathlon: swimming 3.8 km, cycling 180 km, running 42.2 km.
| Swimming | Cycling | Running | Total | |
|---|---|---|---|---|
| Athlete 1 | 01:08 | 05:00 | 03:15 | 09:23 |
| Athlete 2 | 01:05 | 05:10 | 03:35 | 09:50 |
| Athlete 3 | 01:22 | 05:55 | 03:50 | 11:07 |
Times are in hh : mm. The Total column is the sum of the three times.
Athlete 1: 68 + 300 + 195 = 563 minutes = 9 h 23 min
Athlete 2: 65 + 310 + 215 = 590 minutes = 9 h 50 min
Athlete 3: 82 + 355 + 230 = 667 minutes = 11 h 07 min
Part (i)
The total times are different: 9:23, 9:50 and 11:07.
The total time decides who wins, and we also want to see the parts.
A stacked bar chart is suitable
It keeps the totals and the parts
Part (ii)
In a 100% stacked bar chart, every bar is the same length.
We can read only the share of each part.
(a) Who finished first?
All bars are equally long, so we cannot compare total times. Cannot be answered.
(b) Fraction of time Athlete 1 spent cycling
The cycling part is a little more than half of the bar. Can be answered.
(c) Which athlete took the longest for running?
The chart shows shares, not times. Athlete 2 has the largest running share (36%).
But Athlete 3 actually ran the longest, 3:50. So we cannot tell from the chart.
Only (b) can be answered
(a) and (c) need the actual times
Q11.
Look at the following graph. What do you notice? What do you
wonder? Write your inferences
Q11. Solution
Given,
A 100% stacked bar chart: disabled persons by age group and type of disability in India (Census 2011).
Each bar is 100% of the disabled persons in that age group. Values below are approximate.
| Age group | Seeing | Hearing | Speech | Movement | Mental retardation | Mental illness | Any other | Multiple disability |
|---|---|---|---|---|---|---|---|---|
| 0-19 | 18 | 20 | 9 | 13 | 8 | 2 | 21 | 9 |
| 20-39 | 15 | 18 | 9 | 22 | 7 | 3 | 20 | 6 |
| 40-59 | 19 | 18 | 8 | 23 | 5 | 4 | 17 | 6 |
| 60 & above | 25 | 19 | 4 | 25 | 2 | 2 | 11 | 12 |
What do you notice?
Movement disability: 13%, 22%, 23%, 25%. Its share almost doubles from 0-19 to 60 and above.
Seeing disability: 18%, 15%, 19%, 25%. It is the highest in the 60 and above group, one in four.
Hearing disability: stays close to 18% to 20% in every age group.
Speech disability: 9%, 9%, 8%, 4%. Its share falls with age.
Mental retardation: 8% in 0-19, falling to 2% in 60 and above. Mental illness is small, 2% to 4%.
Multiple disability: 9%, 6%, 6%, 12%. It is highest in the 60 and above group.
Any other: 21%, 20%, 17%, 11%. Its share falls with age.
What do you wonder?
How many disabled persons are there in each age group? The chart does not say.
Why does the share of movement and seeing disability rise with age?
Why is the share of multiple disabilities higher among the elderly?
Why are mental retardation and speech disability a larger part among the young?
Would the shares look the same in rural and urban areas, or in other years?
Inferences
Among disabled persons, movement and seeing disabilities make up a larger part in older age groups.
Speech, mental retardation and “any other” make up a smaller part in older age groups.
We cannot say that older people have more persons with movement disability.
We can say that among disabled persons aged 60 and above, about 25% have it.
Movement and seeing rise with age
Speech and “any other” fall with age
Shares tell us the mix, not the numbers
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.
Q12. Solution
Given,
This is an individual project, so there is no single answer. We do option (i) and option (ii) with sample values.
Replace the sample values with your own family’s data.
Option (i): body-states in a day
Recall the previous day. A day has 24 hours, so the three states must add up to 24.
| Lying down | Sitting | Standing or moving | Total | |
|---|---|---|---|---|
| Me (student) | 9 | 7 | 8 | 24 |
| Mother (homemaker) | 7 | 4 | 13 | 24 |
| Father (office worker) | 7 | 10 | 7 | 24 |
How the estimates were made:
Me: slept 8 hours and rested 1 hour = 9. School 5, homework 1, meals and screens 1 = 7. The rest, 8, was assembly, games, walking and chores.
Mother: slept 7 hours. Sat for meals and sewing, 4. Cooking, cleaning and errands filled the rest, 13.
Father: slept 7 hours. Desk work, travel and meals gave 10 hours of sitting. The rest, 7, was walking and chores.
Observations: Father sits the most (10 hours). Mother moves the most (13 hours). Everyone sleeps or rests 7 to 9 hours.
Every bar has the same total, 24 hours, so this chart can also be read as a 100% stacked bar chart.
Option (ii): family’s monthly expenditure
(a) Categories
Rent, Food, Education, Transport, Health, Other.
(b) Collecting the data
Keep a notebook. Write each expense in its category. Check with bills and ask family members.
(c) Expenses for 3 months (in ₹)
| Rent | Food | Education | Transport | Health | Other | Total | |
|---|---|---|---|---|---|---|---|
| July | 8,000 | 6,000 | 2,000 | 1,600 | 800 | 1,600 | 20,000 |
| August | 8,000 | 7,500 | 3,000 | 2,500 | 1,250 | 2,750 | 25,000 |
| September | 8,000 | 5,000 | 2,400 | 1,600 | 1,000 | 2,000 | 20,000 |
(d) 100% stacked bar chart
| Rent | Food | Education | Transport | Health | Other | |
|---|---|---|---|---|---|---|
| July | 40% | 30% | 10% | 8% | 4% | 8% |
| August | 32% | 30% | 12% | 10% | 5% | 11% |
| September | 40% | 25% | 12% | 8% | 5% | 10% |
(e) Observations and inferences
Rent is the biggest part every month, 32% to 40%. Food is next, 25% to 30%.
Rent is ₹8,000 in all three months, but its share fell to 32% in August.
The reason is that the total rose to ₹25,000, so every category is a smaller part of it.
The chart shows shares only. The totals (₹20,000, ₹25,000, ₹20,000) are in the table.
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.
Q13. Solution
Given,
This is a group project, so there is no single answer. This is a model answer with sample (imagined) ratings.
Scenario chosen: (b) travel experience in a bus.
(i) Aspects to rate
Safety
Rash driving or sudden braking can cause injury.
Punctuality
Late buses make people miss school, work and trains.
Cleanliness
A dirty bus makes the journey unpleasant and unhealthy.
Seat comfort
Long journeys are tiring without a comfortable seat.
Staff behaviour
A polite driver and conductor make travel easier.
(ii) Weights
Safety : Punctuality : Cleanliness : Seat comfort : Staff behaviour = 5 : 4 : 3 : 2 : 2
Safety matters most, since it affects life. Punctuality comes next. Comfort and staff matter, but a little less.
Sum of weights = 5 + 4 + 3 + 2 + 2 = 16
(iii) Ratings from 10 people (scale 1 to 5)
| Person | Safety | Punctuality | Cleanliness | Seat comfort | Staff behaviour | Rating |
|---|---|---|---|---|---|---|
| 1 | 4 | 3 | 3 | 4 | 5 | 3.69 |
| 2 | 5 | 4 | 4 | 3 | 4 | 4.19 |
| 3 | 3 | 2 | 3 | 3 | 4 | 2.88 |
| 4 | 4 | 4 | 2 | 3 | 3 | 3.38 |
| 5 | 5 | 5 | 4 | 4 | 5 | 4.69 |
| 6 | 2 | 3 | 2 | 2 | 3 | 2.38 |
| 7 | 4 | 3 | 3 | 4 | 4 | 3.56 |
| 8 | 3 | 2 | 2 | 3 | 3 | 2.56 |
| 9 | 5 | 4 | 5 | 4 | 5 | 4.62 |
| 10 | 4 | 4 | 3 | 3 | 4 | 3.69 |
| Average | 3.9 | 3.4 | 3.1 | 3.3 | 4.0 | 3.56 |
(iv) Individual rating and overall average
Example, Person 1:
The other nine ratings are found in the same way. They are in the last column of the table.
Overall average = average of the 10 individual ratings:
(v) 100% stacked bar chart
For each aspect, count how many people gave 5, 4, 3, 2 and 1 stars. Then find the share of each.
| Aspect | 5★ | 4★ | 3★ | 2★ | 1★ |
|---|---|---|---|---|---|
| Safety | 3 | 4 | 2 | 1 | 0 |
| Punctuality | 1 | 4 | 3 | 2 | 0 |
| Cleanliness | 1 | 2 | 4 | 3 | 0 |
| Seat comfort | 0 | 4 | 5 | 1 | 0 |
| Staff behaviour | 3 | 4 | 3 | 0 | 0 |
(vi) Observations and inferences
Staff behaviour has the highest average (4.0) and Safety is next (3.9). Cleanliness has the lowest (3.1).
Nobody gave 1 star to any aspect. Cleanliness has the most 2 star ratings (3 people).
The overall rating is 3.56, close to the plain average of the five aspects (3.54), because the heavily weighted aspect, Safety, is rated well.
What it captures well: what matters most (safety) counts the most, and one number sums up each person’s experience.
What it misses: only 10 people were asked. Weights are our choice. Reasons behind low ratings, and differences between routes and times, are not shown.
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.
Q14. Solution
Given,
Each student shares the average age of their family and the number of family members.
This is a class activity. The table below uses sample values for 6 students.
| Family | Members | Average age (years) |
|---|---|---|
| Student 1 | 4 | 28.5 |
| Student 2 | 5 | 25.2 |
| Student 3 | 3 | 33 |
| Student 4 | 6 | 24.5 |
| Student 5 | 4 | 30 |
| Student 6 | 5 | 27.6 |
The method
Step 1: Each family finds its total age = average age × number of members.
Step 2: The class writes down the members and the total age of every family.
Step 3: Add all the total ages, and add all the members.
Step 4: Divide the sum of total ages by the sum of members.
Applying it to the sample
| Family | Members | Average age | Total age |
|---|---|---|---|
| Student 1 | 4 | 28.5 | 114 |
| Student 2 | 5 | 25.2 | 126 |
| Student 3 | 3 | 33 | 99 |
| Student 4 | 6 | 24.5 | 147 |
| Student 5 | 4 | 30 | 120 |
| Student 6 | 5 | 27.6 | 138 |
| All | 27 | 744 |
Why not average the six averages?
This is different from 27.56, so it is not correct.
Use the weighted mean
Weights = number of family members
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.
Q15. Solution
Experiment
Take the data 60, 64, 73 with weights 3, 2, 5:
Increase every weight by 1: 4, 3, 6
The plain average of 60, 64, 73 is 1973 ≈ 65.67.
Observation: the weighted average changes. It moves from 67.3 towards the plain average 65.67.
Try equal weights 1, 1, 1: the average is 65.67. With weights 2, 2, 2 it is still 65.67. Here it does not change.
Algebra
Increase every weight by 1: w1 + 1, w2 + 1, …, wn + 1.
Let W = w1 + w2 + … + wn and let x̄ = x1 + x2 + … + xnn be the plain average.
Then the top is W x + n x̄, and the bottom is W + n.
Check with numbers: W = 10, x = 67.3, n = 3, x̄ = 65.67:
The weighted average changes
It moves towards the plain average x̄
It stays the same only if x = x̄
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.
Q16. Solution
Given,
Last year: cows 60%, sheep 25%, chickens 15%. The number of all three animals decreased.
To test each case, suppose last year there were 1,000 animals: 600 cows, 250 sheep, 150 chickens.
(i) and (ii)
The three shares always add up to 100%.
(i) All three shares decreasing would make the total less than 100%. Not possible.
(ii) All three shares increasing would make the total more than 100%. Not possible.
(iii), (iv) and (v)
Each of these can be shown with an example where every population decreases:
| Case | Cows | Sheep | Chickens | Total | Shares |
|---|---|---|---|---|---|
| Last year | 600 | 250 | 150 | 1000 | 60%, 25%, 15% |
| (iii) | 480 | 200 | 120 | 800 | 60%, 25%, 15% |
| (iv) | 460 | 220 | 120 | 800 | 57.5%, 27.5%, 15% |
| (v) | 540 | 225 | 45 | 810 | 66.7%, 27.8%, 5.6% |
(iii): every population falls by 20%, so the shares stay 60%, 25%, 15%. Possible.
(iv): cows fall a lot, sheep fall a little, chickens fall by the same rate as the total. Cows↓, sheep↑, chickens same. Possible.
(v): chickens fall a lot, cows and sheep only a little. Chickens↓, cows↑, sheep↑. Possible.
(iii), (iv) and (v) are possible
(i) and (ii) are not possible
FAQ: Ch 10 How Quantities Combine: Understanding Data Solutions
Some solutions are lengthy, Should I write all of them in my exam or elsewhere?
As stated at the beginning, the solutions are also meant to explain the steps. Moreover, some lengthy questions will hardly appear in the exam.
Consult your teacher regarding the steps to be followed while solving the questions.
