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NCERT Solutions Class 9 Maths Ch 10: Understanding Data

Class IX Maths 📘 NCERT Solutions Ch 10: How Quantities Combine: Understanding Data NCERT book: Ganita Manjari
Updated 09 Nov 2026

NCERT Solutions · Chapter 10

How Quantities Combine: Understanding Data

  • 32Questions solved
  • 16Charts and figures
  • 5Sections

Prepared by Sumeet Sahu, Unique Study Point, Indore · Every calculation re-done

These are the solutions for Class 9 Maths Chapter 10, How Quantities Combine: Understanding Data, from the new NCERT textbook Ganita Manjari (Part 1, session 2026-27). All 32 questions are solved step by step: combined and weighted averages, mixtures, and reading stacked bar charts. Chart readings are approximate.

★ Key takeaways
  1. When groups of different sizes are combined, the combined average is a weighted average: (sum of weight × value) ÷ (sum of weights).
  2. Mixtures work the same way: the concentration of a mixture is the quantity-weighted average of the concentrations. Adding plain water adds quantity but no substance.
  3. Multiplying all weights by the same number does not change a weighted average. Adding the same number to every weight can change it.
  4. A stacked bar chart shows actual amounts; a 100% stacked bar chart shows shares and hides the totals.
🎯 Jump to a question (32 questions)

Combined and weighted averages (Q1 to Q9)

Question 1: Combined average of two sections

Q1

The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has 30 students and Section B has 25 students?

Answer

The combined average is 81211 ≈ 73.82.

  1. Total score of each sectionA: 72 × 30 = 2160   and   B: 76 × 25 = 1900
  2. CombineTotal score = 2160 + 1900 = 4060   and   total students = 30 + 25 = 55
  3. Average406055 = 81211 ≈ 73.82

The result is nearer to 72 than to 76 because Section A has more students (a larger weight).

Question 2: Mixing three fertilisers

Q2

A farmer mixes three equal quantities of fertilisers. The first one contains 110 nitrogen, the second contains 950 nitrogen, and the third contains 360 nitrogen. What is the fraction of nitrogen in the mixture?

Answer

The mixture is 11100 nitrogen.

  1. Equal quantitiesThe quantities are equal, so each fraction has the same weight. Take the ordinary average.
  2. Add the fractions110 + 950 + 360 = 0.10 + 0.18 + 0.05 = 0.33
  3. Divide by 30.33 ÷ 3 = 0.11 = 11100

Question 3: Purity of gold in varna

Q3

(Śrīdharācārya, Pāṭīgaṇita, c. 750 CE) In ancient India, Varṇa was the measure of gold purity. A purity of 16 varna meant pure gold; in general, a purity of k varna meant that the gold-alloy was k/16 gold and the rest impurities. Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varna, 5 units at 10 varna, and 17 units at 11 varna. Find the purity in varna of the combined gold.

Answer

The combined purity is 34531 ≈ 11.13 varna.

  1. Total quantity9 + 5 + 17 = 31 units.
  2. Weighted average of purity9 × 12 + 5 × 10 + 17 × 1131 = 108 + 50 + 18731 = 34531 ≈ 11.13 varna

Question 4: Average rainfall over three months

Q4

The average rainfall per day in the months of May, June, and July in a certain location are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression that gives their combined average.

Answer

Combined average = 3.5 × 31 + 10 × 30 + 8.7 × 3131 + 30 + 31 ≈ 7.37 mm per day.

  1. WeightsThe weights are the numbers of days: May has 31, June has 30 and July has 31.
  2. Expression3.5 × 31 + 10 × 30 + 8.7 × 3131 + 30 + 31
  3. Value108.5 + 300 + 269.792 = 678.292 ≈ 7.37 mm per day

Question 5: Concentration of spice mix

Q5

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

Answer

(i) 353% ≈ 11.67%. (ii) 253% ≈ 8.33%.

  1. (i)Total = 100 + 200 + 300 = 600 mL.100 × 5 + 200 × 10 + 300 × 15600 = 7000600 = 353% ≈ 11.67%
  2. (ii)Total = 300 + 200 + 100 = 600 mL.300 × 5 + 200 × 10 + 100 × 15600 = 5000600 = 253% ≈ 8.33%

The first mixture is stronger because more of it comes from the 15% kadha.

Question 6: Savitri’s annual score

Q6

Savitri’s marks in Kashmiri are as follows: 35 out of 50 in internal tests, 44 out of 60 in the project, and 80 out of 100 in the final exam. The annual percentage score is calculated by combining the internals, project, and final exam in the ratio 3 : 4 : 5. Which of the following expression(s) gives her annual score (as a percentage) in Kashmiri? i. 35 × 3 + 44 × 4 + 80 × 53 + 4 + 5   ii. (35/100) × 3 + (44/100) × 4 + (80/100) × 53 + 4 + 5   iii. (35/50) × 3 + (44/60) × 4 + (80/100) × 53 + 4 + 5 × 100   iv. (35/50 × 100) × 3 + (44/60 × 100) × 4 + (80/100 × 100) × 53 + 4 + 5

Answer

Expressions (iii) and (iv) are correct. The annual score is 135518% ≈ 75.28%.

  1. Put every score on the same scale (percentages)35/50 × 100 = 70%,   44/60 × 100 = 2203%,   80/100 × 100 = 80%
  2. Weighted average with weights 3, 4, 570 × 3 + 2203 × 4 + 80 × 512 = 210 + 293.33 + 40012 ≈ 75.28%
  3. Check the expressions(i) adds marks with different maximum marks (50, 60, 100) without converting them. (ii) divides every score by 100, but the maximum marks are 50, 60 and 100, and it never multiplies by 100 to get a percentage. (iii) and (iv) convert each score to a fraction of its own maximum, so they are correct.

Question 7: Profit on total sales

Q7

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

The profit is 2909% ≈ 32.22% of the total sales.

  1. Profit on books30% of 8000 = ₹2400
  2. Profit on covers50% of 1000 = ₹500
  3. Total profit and total sales₹2400 + ₹500 = ₹2900   and   ₹8000 + ₹1000 = ₹9000
  4. Percentage29009000 × 100 = 2909% ≈ 32.22%

Question 8: The white stork’s migration

Q8

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

The new average is 45 km per day.

  1. GuessOnly one extra day is added, and it is above the old average, so the new average will be a little above 44.5 km, about 45 km.
  2. Total distance44.5 × 20 = 890 km   and   890 + 55 = 945 km
  3. New average94521 = 45 km per day

Question 9: Salt and sugar solutions

Q9

A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture? i. Salt: 5%, Sugar: 8%   ii. Salt: 13%, Sugar: 3%   iii. Salt: 6%, Sugar: 6%   iv. Salt: 5.55%, Sugar: 8.88%   v. Salt: 3.33%, Sugar: 2.67%   vi. Salt: 4.1%, Sugar: 7.08%

Answer

Option (v): salt 3.33% and sugar 2.67%.

  1. Total volume600 + 300 = 900 mL.
  2. SaltOnly the first solution has salt (the second has 0%).600 × 5 + 300 × 0900 = 103% ≈ 3.33%
  3. SugarOnly the second solution has sugar.600 × 0 + 300 × 8900 = 83% ≈ 2.67%

Both concentrations fall because each solution is diluted by the other.

Mixtures and weights (Q10 to Q14)

Question 10: Reducing the spice in pani

Q10

At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into the 10 litres of pani so that the spice level is reduced to 34 of the original concentration?

Answer

Add 103 = 313 litres (about 3.33 litres) of regular water.

  1. Target concentration34 × 8% = 6%
  2. Set upLet x litres of water (0% spice) be added. The amount of spice does not change, but the total becomes 10 + x litres.10 × 8 + x × 010 + x = 6
  3. Solve80 = 60 + 6x ⇒ 6x = 20 ⇒ x = 103 ≈ 3.33 litres

The PDF wrote x = 3⅓ and then said “add 3 litres”. The correct amount is 3⅓ litres, about 3.33 litres.

Question 11: Fitness evaluation: Rashi and Keerthi

Q11

A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio 4 : 5 : 6. Rashi has scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively. i. Find out whose total is more without calculating. ii. What are their final marks?

Answer

(i) Keerthi’s final marks are higher. (ii) Rashi: 1973 ≈ 65.67. Keerthi: 1993 ≈ 66.33.

  1. (i) Without calculatingCompared with Rashi, Keerthi scores 5 less in strength (weight 4) and 5 more in agility (weight 6). Flexibility is equal. The gain of 5 × 6 is more than the loss of 5 × 4, so Keerthi’s weighted total is higher.
  2. (ii) Rashi60 × 4 + 65 × 5 + 70 × 615 = 240 + 325 + 42015 = 98515 ≈ 65.67
  3. (ii) Keerthi55 × 4 + 65 × 5 + 75 × 615 = 220 + 325 + 45015 = 99515 ≈ 66.33

Question 12: Restaurant ratings

Q12

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?

5★4★3★2★1★
Food53200
Ambience04510
Service12241
Answer

The combined average rating is 10730 ≈ 3.57 out of 5.

  1. Food5 × 5 + 3 × 4 + 2 × 310 = 4310 = 4.3
  2. Ambience4 × 4 + 5 × 3 + 1 × 210 = 3310 = 3.3
  3. Service1 × 5 + 2 × 4 + 2 × 3 + 4 × 2 + 1 × 110 = 2810 = 2.8
  4. Combine with weights 6 : 5 : 44.3 × 6 + 3.3 × 5 + 2.8 × 415 = 25.8 + 16.5 + 11.215 = 53.515 ≈ 3.57

Question 13: Langurs in an animal facility

Q13

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 langursFemale langursAll langurs
Average weight16.5 kg13.8 kg14.925 kg
Number of langurs60

i. Which of the following expression(s) describes the given scenario? a. 16.5x + 13.8yx + y = 14.925   b. 16.5x + 13.8y60 = −14.925   c. 16.5x + 13.8y2 = −14.925   d. 16.5x + 13.8y16.5 + 13.8 = −14.925t. 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

More female langurs (35 females and 25 males). (i) Option (a). (iii) ≈ 13.84 kg. (iv) 16.5 kg. (v) ≈ 16.46 kg.

  1. Without computingThe overall average 14.925 kg is closer to the female average (13.8) than to the male average (16.5), so the females have more weight in the combination. There are more females.
  2. (i) The expressionLet x = number of males and y = number of females, so x + y = 60. The weighted average is 16.5x + 13.8yx + y = 14.925, which is option (a).
  3. (ii) Number of males16.5x + 13.8(60 − x) = 14.925 × 60 ⇒ 2.7x + 828 = 895.5 ⇒ x = 25So there are 25 males and 60 − 25 = 35 females.
  4. (iii) Female average after admitting one(35 × 13.8 + 15.2) ÷ 36 = 498.2 ÷ 36 ≈ 13.84 kg
  5. (iv) Male average after two are released(25 × 16.5 − 16.9 − 16.1) ÷ 23 = 379.5 ÷ 23 = 16.5 kg
  6. (v) After one male loses 1 kg(379.5 − 1) ÷ 23 = 378.5 ÷ 23 ≈ 16.46 kg

Option (b) is “(16.5x + 13.8y) ÷ 60 = −14.925” in the PDF: with the minus sign it is wrong. Without the minus sign it would also be correct, because x + y = 60. Options (c) and (d) are wrong in any case. Please check the minus signs against your textbook.

Question 14: Salinity of water

Q14

Dorjee has collected 1 litre of water from the Dead Sea! Using the table (Dead Sea ≈ 34%, other seas and oceans ≈ 3.5%, ground water ≈ 0.01%, purified drinking water ≈ 0.001%), answer the following questions. 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

(i) 11.334%. (ii) Yes, about 3777 litres of purified water. (iii) No.

  1. (i) 1 L Dead Sea + 2 L purified1 × 34 + 2 × 0.0013 = 34.0023 = 11.334%
  2. (ii) Possible?Yes. 0.01% lies between 0.001% (purified) and 34% (Dead Sea), and a mixture’s salinity always lies between the salinities of what is mixed.
  3. (ii) How much purified water?Let x litres be added.34 + 0.001x1 + x = 0.01 ⇒ 34 + 0.001x = 0.01 + 0.01x ⇒ 33.99 = 0.009x ⇒ x ≈ 3776.67 litres
  4. (iii) Possible?No. Both Dead Sea water (34%) and groundwater (0.01%) are saltier than 0.001%. A mixture’s salinity cannot fall below the smaller of the two, 0.01%.
  5. (iii) Check with algebra34 + 0.01x1 + x = 0.001 ⇒ 0.009x = −33.999 ⇒ x ≈ −3777.67A negative quantity cannot be added, so it is impossible.

The salinities in the table are approximate, so the answers are approximate too.

Reading charts and more averages (Q15 to Q20)

Question 15: IUCN Red List chart

Q15

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?

Stacked column chart of the number of animal species on the IUCN Red List by class for 2007, 2010, 2013, 2016 and 2019, with totals 7,851, 9,618, 11,212, 12,630 and 14,234
Answer

(i) The total number of species on the list in 2019. (ii) About 1,100 reptile species. (iii) Mammals, birds, reptiles and amphibians.

  1. (i)The number at the top of a column is the total of all the stacked classes. So 14,234 is the total number of animal species on the Red List in 2019.
  2. (ii) Reptiles in 2016The reptile (dark red) segment in the 2016 column starts at about 1,700 and ends at about 2,900, so its height is about 1,200. A reading of about 1,000 to 1,200 is reasonable.
  3. (iii) Small increaseCompare the height of each segment in 2007 and in 2019. Mammals (about 1,200 to 1,900), birds, reptiles and amphibians grew by about one and a half times or less, while insects, molluscs, fish and the “others” group roughly doubled or more.

These are readings from the chart, so your answers can differ by a small amount.

Question 16: Wickets taken by a bowler

Q16

The wickets taken by a bowler in International Cricket matches till 2025 are shown in the table (Home: Test 62, ODI 100, T20 32; Overseas: Test 172, ODI 49, T20 71). 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.

TestODIT20
Home6210032
Overseas1724971
Empty stacked bar chart template for wickets at home and overseas, axis from 0 to 300, with Test, ODI and T20 in the legend
Empty stacked bar chart template for wickets in the Test, ODI and T20 formats, axis from 0 to 300, with Home and Overseas in the legend
Answer

Home total 194, overseas total 292. Test 234, ODI 149, T20 103.

  1. (i) Home and overseasStack the segments in the order Test, ODI, T20.Home: 62 + 100 + 32 = 194, boundaries 0, 62, 162, 194Overseas: 172 + 49 + 71 = 292, boundaries 0, 172, 221, 292More wickets were taken overseas.
  2. (ii) By formatStack Home first, then Overseas.Test: 62 + 172 = 234, boundaries 0, 62, 234ODI: 100 + 49 = 149, boundaries 0, 100, 149T20: 32 + 71 = 103, boundaries 0, 32, 103The most wickets were taken in Test matches.
Completed stacked bar chart of wickets at home (62 Test, 100 ODI, 32 T20, total 194) and overseas (172 Test, 49 ODI, 71 T20, total 292)
Completed stacked bar chart of wickets by format: Test 62 home and 172 overseas (234), ODI 100 and 49 (149), T20 32 and 71 (103)

Question 17: Run rate in cricket

Q17

In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6. i. In the second over they scored 12 runs. What is the run rate now? ii. After Over 19, their run rate was 6. What is the run rate after 20 overs given the team made 12 runs in the last over?

Answer

(i) 9 runs per over. (ii) 6.3 runs per over.

  1. (i)Runs after 2 overs = 6 + 12 = 18   ⇒   run rate = 18 ÷ 2 = 9
  2. (ii)Runs after 19 overs = 19 × 6 = 114Runs after 20 overs = 114 + 12 = 126   ⇒   run rate = 126 ÷ 20 = 6.3

Question 18: Shuttlecock prices

Q18

Five friends who play badminton surveyed the city and collected the price of shuttlecocks across brands and types (Nylon and Feather). 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?

A green nylon shuttlecock labelled N and a white feather shuttlecock labelled F
Answer

Combined average = ₹4063 ÷ 23 ≈ ₹176.65. Both methods give the same value.

  1. How they can work togetherEach friend reports the sum of the prices and the number of prices. Add all the sums and divide by the total number of prices. Equivalently, combine the averages using the counts as weights.
  2. (i) Using each friend’s averageCounts: Yusuf 6, Srikanth 4, Kashvi 5, Prasanna 4, Gracy 4; total 23.Combined average = 6ay + 4as + 5ak + 4ap + 4ag23Price totals: 917, 493, 766, 897, 990 ⇒ 917 + 493 + 766 + 897 + 99023 = 406323 ≈ ₹176.65
  3. (ii) By typeThere are 13 nylon and 10 feather prices.Combined average = 13an + 10af23Nylon total ₹1429, feather total ₹2634 ⇒ 1429 + 263423 = 406323 ≈ ₹176.65
  4. Are they equal?Yes. Both methods combine the same 23 prices, only grouped in different ways.

Question 19: Shares bought at different prices

Q19

Shreyas holds 25 shares of a company at an average price of ₹150 and Vaishnavi holds 5 shares of the same company at an average price of ₹150. i. Shreyas buys 10 shares of this company at a price of ₹30 each. What is the average price per share for him after the purchase? ii. Vaishnavi buys some shares of this company at a price of ₹30 each and the average price per share for her after the purchase is ₹70. How many shares did she buy?

Answer

(i) ₹8107 ≈ ₹115.71. (ii) 10 shares.

  1. (i) ShreyasTotal cost = 25 × 150 + 10 × 30 = 3750 + 300 = ₹4050   for 35 sharesAverage = 405035 = 8107 ≈ ₹115.71
  2. (ii) VaishnaviLet x be the number of shares bought.5 × 150 + 30x5 + x = 70 ⇒ 750 + 30x = 350 + 70x ⇒ 400 = 40x ⇒ x = 10

Question 20: Mean depth of a pool

Q20

(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

The mean depth is 5 hastas.

  1. Weight each depth by the lengthEach depth applies over a different length, so the lengths are the weights.
  2. Calculate4 × 9 + 5 × 7 + 6 × 7 + 7 × 3 + 8 × 230 = 36 + 35 + 42 + 21 + 1630 = 15030 = 5 hastas

Reading graphs and weights (Q21 to Q27)

Question 21: Suvarna’s gold

Q21

Suvarna had purchased 1 g gold at ₹15k (short for ₹15,000). This is shown by point O denoting the average price of gold that she possesses. Six different scenarios are given below for her next transaction. For each scenario estimate and mark the average price of gold she will have after the transaction. i. Purchase 1 g gold at ₹30k   ii. Purchase 2 g gold at ₹30k   iii. Purchase 1 g gold at ₹10k   iv. Purchase 10 g gold at ₹10k   v. Purchase 0.5 g gold at ₹30k   vi. Purchase 0.5 g gold at ₹15k

A number line from 0 to 30 with the point O marked at 15
Answer

(i) ₹22.5k (ii) ₹25k (iii) ₹12.5k (iv) about ₹10.45k (v) ₹20k (vi) ₹15k

  1. The ruleAverage price per gram = total cost ÷ total grams. The new average always lies between the old average and the new price, and it is nearer to the price of the larger quantity.
ScenarioCalculationNew averageWhy
i. 1 g at ₹30k(15 + 30) ÷ 2₹22.5kEqual quantities give the midpoint.
ii. 2 g at ₹30k(15 + 60) ÷ 3₹25kMore is bought at 30k, so it is nearer 30k.
iii. 1 g at ₹10k(15 + 10) ÷ 2₹12.5kEqual quantities give the midpoint.
iv. 10 g at ₹10k(15 + 100) ÷ 11 ≈ 10.45about ₹10.45kMost of the gold is at 10k, so just above 10k.
v. 0.5 g at ₹30k(15 + 15) ÷ 1.5₹20kOnly half as much is bought as already held.
vi. 0.5 g at ₹15k(15 + 7.5) ÷ 1.5₹15kSame price, so no change.
Six number lines from 0 to 30 showing the new average price after each of the six purchases: 22.5, 25, 12.5, 10.45, 20 and 15

Question 22: Doubling all the weights

Q22

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.

  1. An experimentValues 10 and 20 with weights 1 and 3: (1 × 10 + 3 × 20) ÷ 4 = 70 ÷ 4 = 17.5Double the weights to 2 and 6: (2 × 10 + 6 × 20) ÷ 8 = 140 ÷ 8 = 17.5
  2. AlgebraThe weighted average is M = w₁x₁ + w₂x₂ + ⋯ + wₙxₙw₁ + w₂ + ⋯ + wₙ. With every weight doubled:M′ = 2w₁x₁ + 2w₂x₂ + ⋯ + 2wₙxₙ2w₁ + 2w₂ + ⋯ + 2wₙ = 2(w₁x₁ + w₂x₂ + ⋯ + wₙxₙ)2(w₁ + w₂ + ⋯ + wₙ) = MThe factor 2 cancels, because the ratios of the weights stay the same.

Question 23: Objects orbiting Earth

Q23

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?

Stacked column chart of the cumulative number of objects orbiting Earth by year from 1960, showing payload objects and other objects, rising to about 35,000
Answer

(i) About 10,000 objects in 2003, which doubled to about 20,000 around 2021. (ii) In 2025 about 33,000 to 34,000 in total, with payload objects about 46% and other objects about 54%. (iii) Both counts keep rising.

  1. (i)The 2003 column reaches about 10,000. Twice this is 20,000, which the columns reach around 2021.
  2. (ii) The year 2025The column for 2025 is about 33,500 in total, of which payload objects (green) are about 15,500 and other objects (blue) about 18,000.Payload share ≈ 15,50033,500 ≈ 46%   and   other share ≈ 54%
  3. (iii) TrendsThe number of both payload and other objects has increased over time. Payload objects rise especially sharply in recent years, and the other objects rise sharply around 2007 and again recently. The share of payload objects was small in the early period and has grown to about half of the total recently, so the share of other objects has fallen correspondingly, even though their number is still large.

Chart readings are approximate. The last column of the chart (April 2026) is about 35,000, with about 17,500 payload objects.

Question 24: Schools with a playground

Q24

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) Only (b) and (d) are correct inferences. (ii) No: we also need the number of schools in each state.

  1. (i)(a) Not an inferenceHaryana has a higher percentage (90% against 78% for Uttarakhand), but we do not know how many schools each state has.
  2. (i)(b) CorrectArunachal Pradesh is at about 68%, which is close to 2/3 (about 66.7%).
  3. (i)(c) Not an inferenceA high percentage does not mean a high number. Punjab’s 98% says nothing about how many schools it has (and Dadra and Nagar Haveli and Daman and Diu is even higher at 99.8%).
  4. (i)(d) CorrectMaharashtra has a higher percentage (93%) than Telangana (74%) and more schools. If their totals are M and T with M > T, then 0.93M > 0.93T > 0.74T, so Maharashtra has more schools with a playground.
  5. (ii) National percentageNo. We need the total number of schools in each state and union territory (or the actual number with playgrounds). The percentages must be weighted by the number of schools:National % = N₁p₁ + N₂p₂ + ⋯N₁ + N₂ + ⋯
Map of India showing the percentage of schools with a playground in each state, for example Punjab 98 percent, Haryana 90 percent, Maharashtra 93 percent, Telangana 74 percent, Uttarakhand 78 percent and Arunachal Pradesh 68 percent

Question 25: Decision dilemma: which play to watch?

Q25

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.

100 percent stacked bar chart of the star ratings of Play A, Play B and Play C
Stacked bar chart of the number of star ratings of Play A, Play B and Play C, with axis up to about 900
Answer

There is no single right answer. A reasonable choice is Play A from the percentage chart, and Play B once the number of ratings is known.

  1. (i) From the 100% chartPlay A has a high combined share of 4-star and 5-star ratings (about 72%), and a 5-star share a little larger than Play B’s. Play C has the largest 5-star share (about 39%) but also a big share of 1-star and 2-star ratings, so opinions about it are divided.
  2. (ii) From the stacked chartThe second chart shows how many ratings each play has: about 900 or more for Play B, about 400 for Play C and only about 230 for Play A. Play B’s high share of good ratings is supported by far more people, so it is more reliable. Many people would now choose Play B.
  3. LessonA percentage chart hides how many responses there are. A share based on a few ratings can be misleading.

Question 26: A triathlon

Q26

A triathlon is an endurance race consisting of swimming 3.8 km, cycling 180 km, and running 42.2 km. The finish time of 3 athletes in each segment is given (hh : mm): 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?

Answer

(i) A stacked bar chart with actual times. (ii) Only (b) can be answered from a 100% chart.

  1. Total times
    AthleteSwimCycleRunTotal
    11 h 08 min5 h3 h 15 min9 h 23 min = 563 min
    21 h 05 min5 h 10 min3 h 35 min9 h 50 min = 590 min
    31 h 22 min5 h 55 min3 h 50 min11 h 07 min = 667 min
  2. (i) Which chart?A stacked bar chart of actual times is better: it shows each segment’s time and also each athlete’s total time, so we can see who finished first. Athlete 1 has the shortest bar.
  3. (ii) What a 100% chart can tell(a) No: all the bars have the same length, so the totals are hidden. (b) Yes: Athlete 1’s cycling is 300563 ≈ 0.533, about 53.3% of the time. (c) No: the biggest running share need not be the longest running time. (The table shows Athlete 3 took longest to run, but the percentage chart alone cannot show it.)

Question 27: Disability by age group

Q27

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

100 percent stacked bar chart of the distribution of disabled persons by age group and type of disability in India, Census 2011
Answer

The type of disability changes with age. Each bar shows shares, not numbers.

  1. What I noticeEach bar adds up to 100%, so the chart compares the types of disability within each age group. The shares of seeing difficulties (about 19% to 27%) and movement difficulties (about 15% to 23%) are larger in the oldest group (60 and above) than in the youngest (0 to 19). The share of speech difficulties is smaller in the oldest group.
  2. What I wonderHow many people does each bar stand for? How does support differ across age groups? How has the distribution changed since the 2011 Census?
  3. InferencesThe distribution of disability types varies with age. The chart does not show the number of persons in each age group, or what share of all people in that age group have a disability, and it cannot tell us the causes.

Answers will differ. Any careful observation and a sensible question is acceptable.

Projects and reasoning (Q28 to Q32)

Question 28: Individual project: body states and family budget

Q28

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. 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. identify the expense categories; b. decide how to collect and organise the data; c. collect expense data for at least 3 months; d. represent the data in a 100% stacked bar chart; e. write your observations and inferences.

Answer

This is a project, so your answers will be your own. Below is a worked example with sample numbers.

  1. (i) Body-state project (example data)
    PersonLying down (h)Sitting (h)Standing / moving (h)Total
    Student810624
    Family member A88824
    Family member B751224

    How to estimate: make a time diary from waking to bedtime, put each interval into one body state (seated travel counts as sitting), add up each state, and check that every day totals 24 hours. Ask the family members to check their own estimates. Draw each person as one horizontal bar from 0 to 24 hours.

    Stacked bar chart of hours spent lying down, sitting and standing or moving for the student and two family members, each totalling 24 hours
  2. (ii) Monthly expenditure project (example data)

    a. Categories: housing, food, education, transport, healthcare, other. b. Note each expense in a notebook or spreadsheet, using bills and payment records, with each expense in one category only. c. Example data for 3 months (₹):

    CategoryMonth 1Month 2Month 3
    Housing100001000010000
    Food750080008400
    Education500050005600
    Transport375040004200
    Healthcare250040005600
    Other125010001200
    Total300003200035000

    d. Convert each amount into a percentage of its own month’s total (for example, Month 1 housing = 10000 ÷ 30000 × 100 ≈ 33.33%):

    CategoryMonth 1Month 2Month 3
    Housing33.33%31.25%28.57%
    Food25.00%25.00%24.00%
    Education16.67%15.63%16.00%
    Transport12.50%12.50%12.00%
    Healthcare8.33%12.50%16.00%
    Other4.17%3.13%3.43%
    100 percent stacked bar chart of the monthly family expenditure for three months, split into housing, food, education, transport, healthcare and other
  3. e. ObservationsTotal spending rises from ₹30,000 to ₹35,000. Housing stays at ₹10,000, but its share falls from 33.33% to 28.57% because the total rises. Healthcare rises in both rupees and share. Food rises from ₹7,500 to ₹8,400, but its share falls from 25% to 24%. A percentage chart does not show the rise in the total, so the totals must be given separately.

All the numbers in this answer are sample data for illustration. Use your own family’s data for your project.

Question 29: Small-group project: a custom rating scheme

Q29

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, and justify why each matters. 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. 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

Worked example for the scenario “travel experience in a bus”, with imagined ratings. The overall average is 4.01 out of 5.

  1. (i), (ii) Aspects and weights
    AspectWeightWhy it matters
    Safety4Careful driving and safe boarding are essential.
    Punctuality3Passengers need to reach their destinations on time.
    Comfort2Seating, space and ventilation affect the journey.
    Cleanliness1A clean bus improves the travelling experience.

    The weights add up to 10.

  2. (iii) Imagined ratings of 10 passengers
    PassengerSafetyPunctualityComfortCleanliness
    15443
    24434
    35544
    43433
    54343
    65454
    74545
    83334
    94444
    105435
  3. (iv) Individual and overall ratingsRating = (4 × Safety + 3 × Punctuality + 2 × Comfort + 1 × Cleanliness) ÷ 10

    For example, passenger 1: (20 + 12 + 8 + 3) ÷ 10 = 4.3.

    Passenger12345678910
    Rating4.33.84.73.33.64.64.43.14.04.3
    Overall average = 40.1 ÷ 10 = 4.01 out of 5
  4. (v) 100% stacked bar chartEach passenger is 10% of a bar.
    Aspect1★2★3★4★5★
    Safety0%0%20%40%40%
    Punctuality0%0%20%60%20%
    Comfort0%0%40%50%10%
    Cleanliness0%0%30%50%20%
    100 percent stacked bar chart of the star ratings of 10 passengers for safety, punctuality, comfort and cleanliness
  5. (vi) Observations and inferencesSafety has the highest share of 5-star ratings (40%). Comfort has the highest share of 3-star ratings (40%), which shows room for improvement. No passenger rated any aspect below 3. The weighted rating of 4.01 out of 5 shows a generally favourable experience in this imagined data. The scheme captures four important parts of the journey and gives safety the most importance. It misses fare, accessibility and staff behaviour, and different weights would give different overall ratings.

The ratings in this example are imagined. Use your own group’s data for your project.

Question 30: Whole class project: average age of all families

Q30

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

Use the number of members in each family as its weight.

  1. Method1. Record each family’s average age and number of members. 2. Multiply them to get the sum of the ages in that family. 3. Add these sums for all the families. 4. Divide by the total number of family members.
  2. FormulaIf the family averages are a₁, a₂, …, aₙ and the family sizes are n₁, n₂, …, nₙ:Combined average age = n₁a₁ + n₂a₂ + ⋯ + nₙaₙn₁ + n₂ + ⋯ + nₙ
  3. Example (sample data)Family A: 4 members, average 25 years, so the sum of ages is 100. Family B: 6 members, average 30, sum 180. Family C: 5 members, average 28, sum 140.100 + 180 + 1404 + 6 + 5 = 42015 = 28 years

Question 31: Adding 1 to all the weights

Q31

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

The weighted average can change. It moves towards the ordinary average of the values.

  1. ExperimentsValues 10 and 20.
    WeightsWeighted averageAfter adding 1 to each weightNew average
    3 and 130 + 204 = 12.54 and 240 + 406 = 1313
    1 and 310 + 604 = 17.52 and 420 + 806 = 1623
    1 and 1152 and 215 (no change)
  2. ObservationUnlike multiplying all weights by the same number, adding 1 does not keep the ratios of the weights. The mean can go up, go down, or stay the same.
  3. AlgebraLet there be n values with total weight W and weighted mean M, so w₁x₁ + ⋯ + wₙxₙ = WM. Let A be the ordinary average of the values, so x₁ + ⋯ + xₙ = nA. After adding 1 to every weight:M′ = (w₁ + 1)x₁ + ⋯ + (wₙ + 1)xₙW + n = WM + nAW + nM′ − M = n(A − M)W + nSo M′ > M if A > M, M′ < M if A < M, and M′ = M if A = M.

For a positive constant c added to every weight, the same argument gives M′ = (WM + cnA) ÷ (W + cn).

Question 32: Changing shares of cows, sheep and chickens

Q32

A farm has some cows, sheep, and chickens. Last year the cows made up 60%, the sheep 25%, and the chickens 15%. There was a decrease in the number of all three animals’ population over the year. Choose the possibilities for the change in their respective shares of the population: i. % of cows decreased, % of sheep decreased, % of chickens decreased. ii. % of cows increased, % of sheep increased, % of chickens increased. iii. % of cows remained the same, % of sheep remained the same, % of chickens remained the same. iv. % of cows decreased, % of sheep increased, % of chickens remained the same. v. % of cows increased, % of sheep increased, % of chickens decreased.

Answer

(iii), (iv) and (v) are possible. (i) and (ii) are impossible.

  1. Shares add up to 100%The three shares always add up to 100%, so all three cannot decrease together and all three cannot increase together. Options (i) and (ii) are impossible.
  2. Example with 200 animalsLast year: 120 cows, 50 sheep, 30 chickens (60%, 25%, 15%). Suppose every number goes down:
  3. (iii) Same shares60 cows, 25 sheep, 15 chickens (total 100): shares 60%, 25%, 15%. Every number fell, and every share is unchanged.
  4. (iv)55 cows, 30 sheep, 15 chickens (total 100): shares 55%, 30%, 15%. The cows’ share fell, the sheep’s share rose, and the chickens’ share is the same. All three numbers are below the old numbers.
  5. (v)65 cows, 26 sheep, 9 chickens (total 100): shares 65%, 26%, 9%. The cows’ and sheep’s shares rose and the chickens’ share fell, while every number fell.

Important formulas and ideas of Chapter 10

IdeaRule
Weighted average(w₁x₁ + w₂x₂ + ⋯ + wₙxₙ) ÷ (w₁ + w₂ + ⋯ + wₙ)
Combined average of groupsUse the size of each group as its weight
Mixture concentration(quantity₁ × concentration₁ + quantity₂ × concentration₂ + ⋯) ÷ total quantity
Adding plain waterAdds to the total quantity but adds 0% of the substance
Multiplying all weights by the same numberThe weighted average does not change
Adding the same number to every weightThe weighted average moves towards the ordinary average
Run rateTotal runs ÷ number of overs
Stacked bar chartShows actual amounts and totals
100% stacked bar chartShows shares only; totals are hidden

Practice: multiple choice questions

Tap an option to check it at once. Your score updates as you answer.

Your score0 / 12
1

The weighted average of x₁, x₂ with weights w₁, w₂ is:

Multiply each value by its weight, add, and divide by the total weight.
2

Section A (30 students) averages 72 and Section B (25 students) averages 76. The combined average is about:

(2160 + 1900) ÷ 55 = 4060 ÷ 55 ≈ 73.82.
3

A 100% stacked bar chart shows:

Every bar has the same length (100%), so it shows shares and hides totals.
4

If every weight is doubled, the weighted average:

The factor 2 cancels from the numerator and the denominator.
5

Mixing plain water into a salt solution makes the concentration:

The amount of salt stays the same but the total volume increases.
6

The mean depth of the pool in Q20 (lengths 4, 5, 6, 7, 8; depths 9, 7, 7, 3, 2) is:

150 ÷ 30 = 5 hastas.
7

A stork averaged 44.5 km a day for 20 days and flew 55 km on day 21. The 21-day average is:

(890 + 55) ÷ 21 = 45.
8

600 mL of 5% salt solution is mixed with 300 mL of 8% sugar solution. The salt concentration is about:

600 × 5 ÷ 900 = 3.33%.
9

To find the average of several groups’ averages, use as weights the:

The number of members in each group is its weight.
10

A team scores 6 runs in over 1 and 12 runs in over 2. The run rate after 2 overs is:

18 runs ÷ 2 overs = 9.
11

Assertion (A): Doubling every weight does not change a weighted average.
Reason (R): The factor 2 cancels from the numerator and the denominator.

Both are true, and the cancelling factor is exactly the reason.
12

Assertion (A): Adding the same number to every weight never changes the weighted average.
Reason (R): Adding a constant to every weight changes the ratios of the weights.

A is false: the weighted average can change. R is true, and it is why.

Frequently asked questions

How many questions are solved in Class 9 Maths Chapter 10?

All 32 questions of How Quantities Combine: Understanding Data from the new NCERT Ganita Manjari textbook (2026-27).

What is a weighted average?

An average in which some values count more than others. Each value is multiplied by its weight, the products are added, and the sum is divided by the total weight.

How do I find the concentration of a mixture?

Multiply each quantity by its concentration, add them, and divide by the total quantity. Plain water has concentration 0%.

What is the difference between a stacked bar chart and a 100% stacked bar chart?

A stacked bar chart shows actual amounts (so the bar lengths show totals). A 100% stacked bar chart shows each category as a share of the total, so all bars have the same length.

Why does doubling all the weights not change the weighted average?

The factor 2 appears in both the numerator and the denominator, so it cancels.

Where can I get the PPT and worksheet for this chapter?

Use the links below, where available, for the free PPT and worksheet on this site.

More free CBSE study material

Notes, worksheets, PPTs and practice papers for Classes 6 to 10 are available at uniquestudyonline.com.

Solutions prepared by Unique Study Point by Sumeet Sahu. The questions are from the NCERT Mathematics textbook Ganita Manjari (Class 9, Part 1).

More study material: Class IX Maths

Sumeet Sahu (14 years of teaching experience), Unique Study Point, Amitesh Nagar, Indore. Free NCERT/CBSE study material for Class VI to X.