Ganithub

Answers to all your questions

Delhi, India

|

Chapter-10 How Quantities Combine: Understanding Data

EXERCISE SET 10.1

Question 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?

Solution:

  • Number of students in Section A (n1) = 30
  • Average score of Section A (A1) = 72
  • Total score of Section A = 30 × 72 = 2160
  • Number of students in Section B (n2) = 25
  • Average score of Section B (A2) = 76
  • Total score of Section B = 25 × 76 = 1900
  • Total score of both sections = 2160 + 1900 = 4060
  • Total number of students = 30 + 25 = 55

Combined Average = Total Score / Total Students

Combined Average = 4060 / 55 = 73.82

Answer:

The combined average score of both sections is 73.82 (or 73 and 9/11).

Question 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?

Solution:

Let the equal quantity taken from each fertilizer be 1 unit.

  • Nitrogen from 1st fertilizer = 1/10
  • Nitrogen from 2nd fertilizer = 9/50
  • Nitrogen from 3rd fertilizer = 3/60 = 1/20

Total Nitrogen = 1/10 + 9/50 + 1/20

Taking LCM of 10, 50, and 20 (which is 100):

Total Nitrogen = (10 + 18 + 5) / 100 = 33/100

Total Quantity of Mixture = 1 + 1 + 1 = 3 units

Fraction of Nitrogen in Mixture = (33/100) / 3 = 11/100

Answer:

The fraction of nitrogen in the mixture is 11/100 (or 11%).

Question 3

(Sridharacharya, Patiganita, c. 750 CE) 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.

Solution:

Weighted Average Purity = Total Pure Gold Content / Total Weight

  • Piece 1: Weight = 9, Purity = 12 varna
  • Piece 2: Weight = 5, Purity = 10 varna
  • Piece 3: Weight = 17, Purity = 11 varna

Total Weight = 9 + 5 + 17 = 31 units

Total Pure Gold Content = (9 × 12) + (5 × 10) + (17 × 11)

= 108 + 50 + 187 = 345

Combined Purity = 345 / 31 = 11.13 varna

Answer:

The purity of the combined gold is 11.13 varna (or 11 and 4/31 varna).

Question 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.

Solution:

  • Days in May = 31
  • Days in June = 30
  • Days in July = 31
  • Total Days = 31 + 30 + 31 = 92

Answer:

Required Expression:

(31 × 3.5 + 30 × 10 + 31 × 8.7) / 92

Question 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.

Solution:

(i)

  • Total volume = 100 + 200 + 300 = 600 mL
  • Total spice content = (100 × 5%) + (200 × 10%) + (300 × 15%)

  • = 5 + 20 + 45 = 70 mL

Concentration = (70 / 600) × 100% = 11.67%

(ii)

  • Total volume = 300 + 200 + 100 = 600 mL
  • Total spice content = (300 × 5%) + (200 × 10%) + (100 × 15%)

  • = 15 + 20 + 15 = 50 mL

Concentration = (50 / 600) × 100% = 8.33%

Answer:

(i) 11.67% (or 11 and 2/3 %)

(ii) 8.33% (or 8 and 1/3 %)

EXERCISE SET 10.2

Question 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?

Solution:

Score ratios as fractions:

  • Internal score fraction = 35/50
  • Project score fraction = 44/60
  • Final exam score fraction = 80/100

Weights for the components are 3, 4, 5 respectively. Total weight = 3 + 4 + 5 = 12.

Combining fractional scores:

[ (35/50) × 3 + (44/60) × 4 + (80/100) × 5 ] / (3 + 4 + 5) × 100

Alternatively, combining percentage scores directly:

[ (35/50 × 100) × 3 + (44/60 × 100) × 4 + (80/100 × 100) × 5 ] / (3 + 4 + 5)

Both expressions (iii) and (iv) correctly calculate her percentage score.

Answer:

Options (iii) and (iv).

EXERCISE SET 10.3

Question 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?

Solution:

  • Profit from books = 30% of 8000 = ₹2400
  • Profit from book covers = 50% of 1000 = ₹500

Total Profit = 2400 + 500 = ₹2900

Total Sales = 8000 + 1000 = ₹9000

Profit Percentage = (Total Profit / Total Sales) × 100

Profit Percentage = (2900 / 9000) × 100 = 32.22%

Answer:

The profit percentage on total sales is 32.22% (or 32 and 2/9 %).

Question 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.

Solution:

  • Guess: Since 55 km is higher than the previous average of 44.5 km, the overall average will rise slightly to roughly 45 km.
  • Calculation:

  •
    • Total distance over 20 days = 20 × 44.5 = 890 km
    • Distance on 21st day = 55 km
    • Total distance over 21 days = 890 + 55 = 945 km

New Average = 945 / 21 = 45 km

Answer:

The average daily distance travelled over 21 days is 45 km.

Question 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?

Solution:

  • Total volume of mixture = 600 + 300 = 900 mL
  • Volume of salt = 5% of 600 = 30 mL
  • Volume of sugar = 8% of 300 = 24 mL

Concentration of Salt = (30 / 900) × 100% = 3.33%

Concentration of Sugar = (24 / 900) × 100% = 2.67%

Answer:

(iv) Salt: 3.33%, Sugar: 2.67%.

Question 4

At a panipuri stall, the concentration of spice in the pani was 8%. 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?

Solution:

  • Initial volume of pani = 10 litres
  • Initial concentration = 8%
  • Target concentration = (3/4) × 8% = 6%

Amount of spice in original pani = 8% of 10 = 0.8 litres

Let x be the amount of regular water added.

New Total Volume = 10 + x

Setting up the concentration equation:

0.8 / (10 + x) = 6 / 100

0.8 × 100 = 6 × (10 + x)

80 = 60 + 6x

6x = 20

x = 20 / 6 = 3.33 litres

Answer:

3.33 litres (or 3 and 1/3 litres) of regular water should be added.

Question 5

Physical fitness evaluation: Strength, flexibility, and agility in ratio 4:5:6. Rashi scores: 60, 65, 70. Keerthi scores: 55, 65, 75 respectively.

(i) Find out whose total is more without calculating.

(ii) What are their final marks?

Solution:

(i) Without Calculation:

Both Rashi and Keerthi scored identically in flexibility (65 marks).

Comparing the other components, Keerthi scored 5 marks lower in strength (weight 4) but 5 marks higher in agility (weight 6). Because agility carries a higher weight (6 > 4), Keerthi’s score gain outweighs her score loss.

Therefore, Keerthi’s final total marks will be higher.

(ii) Calculations:

Total Weight = 4 + 5 + 6 = 15

Rashi’s Final Mark = [ (60 × 4) + (65 × 5) + (70 × 6) ] / 15

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

Keerthi’s Final Mark = [ (55 × 4) + (65 × 5) + (75 × 6) ] / 15

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

Answer:

(i) Keerthi

(ii) Rashi = 65.67, Keerthi = 66.33

Question 6

A restaurant collected ratings from 10 customers. What is the average rating if metrics are combined with weights food: ambience: service = 6:5:4?

Data from table:

  • Food ratings sum = 44 -> Average = 4.4
  • Ambience ratings sum = 33 -> Average = 3.3
  • Service ratings sum = 29 -> Average = 2.9

Solution:

Weighted Average = [ (4.4 × 6) + (3.3 × 5) + (2.9 × 4) ] / (6 + 5 + 4)

= (26.4 + 16.5 + 11.6) / 15

= 54.5 / 15 = 3.63

Answer:

The weighted average rating is 3.63.

Question 7

Langurs Weight Data:

  • Male average = 16.5 kg, Female average = 13.8 kg, Combined average = 14.925 kg
  • Total langurs = 60

(i) Expression describing scenario:

(16.5x + 13.8y) / (x + y) = 14.925 where x + y = 60.

Answer: (a)

(ii) Number of male langurs:

Let x = males, y = females = 60 – x.

16.5x + 13.8(60 – x) = 14.925 × 60

16.5x + 828 – 13.8x = 895.5

2.7x = 67.5

x = 67.5 / 2.7 = 25

Answer: 25 male langurs (and 35 female langurs).

(iii) A female langur weighing 15.2 kg is admitted:

Initial sum of female weights = 35 × 13.8 = 483 kg

New total weight = 483 + 15.2 = 498.2 kg

New female count = 36

New Female Average = 498.2 / 36 = 13.84 kg

(iv) Two male langurs (16.9 kg, 16.1 kg) are released:

Initial sum of male weights = 25 × 16.5 = 412.5 kg

Weight removed = 16.9 + 16.1 = 33 kg

New total weight = 412.5 – 33 = 379.5 kg

New male count = 23

New Male Average = 379.5 / 23 = 16.5 kg

(v) One male langur lost 1 kg of weight:

New total male weight = 412.5 – 1 = 411.5 kg

Total male count = 25

New Male Average = 411.5 / 25 = 16.46 kg

Question 8

Dead Sea Water Questions:

  • Salinities: Dead Sea = 34%, Purified Water = 0.001%, Groundwater = 0.01%

(i) 1 L Dead Sea water + 2 L purified water:

Salinity = [ (1 × 34%) + (2 × 0.001%) ] / (1 + 2)

= (34 + 0.002) / 3 = 11.33%

(ii) Mix to get groundwater salinity (0.01%):

  • Possibility: Yes, because 0.01% falls between 34% and 0.001%.
  • Calculation: Let V be the litres of purified water.

  • (1 × 34 + V × 0.001) / (1 + V) = 0.01

  • 34 + 0.001V = 0.01 + 0.01V

  • 33.99 = 0.009V

  • V = 33.99 / 0.009 = 3776.67 litres

(iii) Mix Dead Sea and groundwater (0.01%) to get purified water salinity (0.001%):

  • Possibility: No. A mixture’s concentration can never be lower than the concentration of its lowest ingredient (0.01%).

EXERCISE SET 10.4

Question 1

IUCN Red List Chart:

(i) What does 14,234 represent?

Answer: Total number of endangered species listed in 2019.

(ii) Approximate reptile species in 2016:

Answer: Approximately 1,200 to 1,500 species.

(iii) Class(es) with small increase:

Answer: Mammals and Birds.

Question 2

Cricket Wickets Stacked Bar Charts:

  • Home Wickets: Test = 62, ODI = 100, T20 = 32 -> Total = 194

  • Overseas Wickets: Test = 172, ODI = 49, T20 = 71 -> Total = 292

Format Totals:

  • Test = 62 + 172 = 234
  • ODI = 100 + 49 = 149
  • T20 = 32 + 71 = 103

(i) Chart 1 Values (Home vs Overseas):

  • Home Bar: Total 194 (Test: 62, ODI: 100, T20: 32)
  • Overseas Bar: Total 292 (Test: 172, ODI: 49, T20: 71)

(ii) Chart 2 Values (By Format):

  • Test Bar: Length = 234 (Home: 62, Overseas: 172)
  • ODI Bar: Length = 149 (Home: 100, Overseas: 49)
  • T20 Bar: Length = 103 (Home: 32, Overseas: 71)

EXERCISE SET 10.5

Question 1

Smart Watch Data:

(i) Sahana’s Row (Student Routine Guess):

  • Lying down: 8 hours
  • Sitting: 10 hours
  • Standing/Moving: 6 hours

  • (Total = 24 hours)

(ii) Julie’s Activity:

  • Julie spends 8 hrs lying down, 10 hrs sitting, and 6 hrs standing/moving. She is likely a desk-job worker, software programmer, or student.

(iii) Chart Percentages:

  • Pavani: Lying = 83.3%, Sitting = 12.5%, Standing = 4.2%
  • Raghu: Lying = 29.2%, Sitting = 16.7%, Standing = 54.1%
  • Zakir: Lying = 33.3%, Sitting = 25.0%, Standing = 41.7%

END-OF-CHAPTER EXERCISES

Question 1

Cricket Run Rate:

(i) 6 runs in 1st over, 12 in 2nd over:

Run Rate = (6 + 12) / 2 = 18 / 2 = 9 runs per over

(ii) Run rate 6 after 19 overs, 12 in 20th over:

Runs in 19 overs = 19 × 6 = 114 runs

Runs in 20 overs = 114 + 12 = 126 runs

New Run Rate = 126 / 20 = 6.3 runs per over

Question 2

Shuttlecock Average Prices:

(i) Expression using group averages (ay, as, ak, ap, ag) with counts (6, 4, 5, 4, 4):

(6ay + 4as + 5ak + 4ap + 4ag) / 23

(ii) Expression using Nylon (an) and Feather (af) averages:

Nylon count = 13, Feather count = 10. Total = 23.

(13an + 10af) / 23

Yes, this will give the exact same overall average value.

Question 3

Share Prices:

(i) Shreyas:

  • 25 shares at ₹150 = ₹3750
  • 10 shares at ₹30 = ₹300

  • New Average = (3750 + 300) / (25 + 10) = 4050 / 35 = ₹115.71

(ii) Vaishnavi:

  • 5 shares at ₹150 = ₹750
  • Buys x shares at ₹30 to get average ₹70.

  • (750 + 30x) / (5 + x) = 70

  • 750 + 30x = 350 + 70x

  • 400 = 40x

  • x = 10 shares

Answer:

Vaishnavi bought 10 shares.

Question 4

Mean Depth of Pool:

Mean Depth = Total Volume / Total Length

Mean Depth = [ (4 × 9) + (5 × 7) + (6 × 7) + (7 × 3) + (8 × 2) ] / (4 + 5 + 6 + 7 + 8)

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

= 150 / 30 = 5 hastas

Answer:

The mean depth is 5 hastas.

Question 5

Gold Purchase Transactions (Initial: 1g at ₹15k):

(i) Purchase 1g at ₹30k -> Average = (15 + 30) / 2 = 22.5k

(ii) Purchase 2g at ₹30k -> Average = (15 + 60) / 3 = 25k

(iii) Purchase 1g at ₹10k -> Average = (15 + 10) / 2 = 12.5k

(iv) Purchase 10g at ₹10k -> Average = (15 + 100) / 11 = 10.45k

(v) Purchase 0.5g at ₹30k -> Average = (15 + 15) / 1.5 = 20k

(vi) Purchase 0.5g at ₹15k -> Average = (15 + 7.5) / 1.5 = 15k

Question 6

Effect of doubling all weights:

If all weights are doubled, both the numerator and the denominator are multiplied by 2. The factor of 2 cancels out completely.

Observation:

The weighted average remains unchanged.

Question 7

Orbiting Objects Chart:

(i) Total objects in 2003 = approx. 10,000. Doubled to approx. 20,000 around 2018–2020.

(ii) In 2025: Total objects approx. 30,000; Payloads approx. 15,000 (50%), Other objects approx. 15,000 (50%).

(iii) Both payload and other objects have grown over time, with payload growth expanding rapidly after 2015 due to low-Earth orbit satellite constellations.

Question 8

Playground Infographic:

(i) Correct Inference: (b) Approximately every 2 out of 3 (67%) schools in Arunachal Pradesh have a playground.

(ii) Nation-wide percentage: No, we cannot calculate it without knowing the total number of schools in each state.

Question 9

Play Ratings Dilemma:

(i) 100% Stacked Chart shows Play C has the highest percentage of 5-star ratings.

(ii) Absolute Stacked Chart reveals Play B has a significantly larger total number of ratings. One might choose Play B because a larger sample size provides more reliable feedback.

Question 10

Triathlon Race Analysis:

(i) A stacked bar chart is suitable because the total time spent across all three segments determines the overall winner.

(ii) A 100% stacked bar chart can answer (b) Approximately what fraction of total time Athlete 1 spent cycling.

Question 11

Disabled Persons Distribution:

  • Inferences: Hearing and seeing disabilities make up a large share in lower age groups (0–19). Movement disability increases substantially in older age categories (60+).

Questions 12–14

(Hands-on classroom survey and data visualization activities)

Question 15*

Effect of increasing all weights by a constant (c = 1):

Adding 1 to each weight increases smaller weights proportionally more than larger weights.

Observation:

The new weighted average shifts closer to the simple unweighted average of the values.

Question 16*

Farm Animal Shares Reduction:

Since all animal populations decreased, their relative percentages must still sum to 100%.

Possible valid options depending on the number of animals lost from each group: (iii), (iv), and (v).