Question 1
Combine averages 72 for 30 students and 76 for 25 students.
Steps only
Total score = 72×30 + 76×25 = 4060.
Students = 30+25 = 55.
Combined average = 4060÷55 ≈ 73.82.
Answer
Approximately 73.82.
CLASS 9 · GANITA MANJARI PART II · CHAPTER 10
How Quantities Combine: Understanding Data
Combine unequal groups with weighted means and interpret stacked and 100% stacked charts carefully.
60 worked answersAll 6 exercise setsAll 16 end-of-chapter questions
What you will learn
- Calculate weighted means and mixture concentrations.
- Update an average when data are added or removed.
- Distinguish absolute counts from percentage shares in charts.
60 worked answers · short steps first, explanations when you need them
Exercise Set 10.1
Textbook pages 13
Question 2
Equal amounts of fertilisers contain 1/10, 9/50 and 3/60 nitrogen. Find the mixture fraction.
Steps only
Equal quantities give equal weights.
(1/10 + 9/50 + 3/60)÷3 = (10/100+18/100+5/100)÷3.
= 33/100÷3 = 11/100.
Answer
11/100, or 11%.
Question 3
Combine 9 units at 12 varṇa, 5 at 10 varṇa and 17 at 11 varṇa.
Steps only
Pure-gold units = 9×12 + 5×10 + 17×11 = 345 varṇa-units.
Total amount = 31 units.
Purity = 345/31 ≈ 11.13 varṇa.
Answer
345/31 varṇa, approximately 11.13 varṇa.
Question 4
Write the combined daily rainfall average for May, June and July: 3.5, 10 and 8.7 mm.
Steps only
Weight by days: May 31, June 30, July 31.
Expression = (31×3.5 + 30×10 + 31×8.7)/(31+30+31).
Value ≈ 7.37 mm per day.
Answer
(31×3.5 + 30×10 + 31×8.7)/92 ≈ 7.37 mm per day.
Question 5(i)
Combine 100 mL at 5%, 200 mL at 10% and 300 mL at 15%.
Steps only
Spice = 5+20+45 = 70 mL-equivalent.
Total = 600 mL.
Concentration = 70/600×100 ≈ 11.67%.
Answer
Approximately 11.67%.
Question 5(ii)
Combine 300 mL at 5%, 200 mL at 10% and 100 mL at 15%.
Steps only
Spice = 15+20+15 = 50 mL-equivalent.
Total = 600 mL.
Concentration = 50/600×100 ≈ 8.33%.
Answer
Approximately 8.33%.
Exercise Set 10.2
Textbook pages 14–15
Question 1
Identify the correct expression for Savitri’s weighted percentage from 35/50, 44/60 and 80/100 with weights 3:4:5.
Steps only
Convert each mark to a percentage first: 70%, 73⅓%, 80%.
Weighted score = [70×3 + (73⅓)×4 + 80×5]/12.
= 75.28% approximately.
Answer
Expressions (iii) and (iv) are correct; the annual score is approximately 75.28%.
Exercise Set 10.3
Textbook pages 15–18
Question 1
A shop has ₹8000 book sales at 30% profit and ₹1000 cover sales at 50% profit. Find profit as a percentage of total sales.
Steps only
Profit amounts = 8000×0.30 + 1000×0.50 = ₹2900.
Total sales = ₹9000.
Percentage = 2900/9000×100 ≈ 32.22%.
Answer
Approximately 32.22%.
Question 2
A stork averaged 44.5 km for 20 days and flew 55 km on day 21. Find the new average.
Steps only
First 20 days total = 44.5×20 = 890 km.
New total = 945 km.
945÷21 = 45 km.
Answer
45 km per day.
Question 3
Mix 600 mL of 5% salt solution with 300 mL of 8% sugar solution.
Steps only
Salt = 600×0.05 = 30 mL-equivalent; sugar = 300×0.08 = 24.
Total mixture = 900 mL.
Salt = 30/900 = 3.33%; sugar = 24/900 = 2.67%.
Answer
Salt 3.33% and sugar 2.67%, option (v).
Question 4
How much water reduces 10 L of 8% spicy pani to three-fourths of its original concentration?
Steps only
Target concentration = (3/4)×8% = 6%.
Spice amount stays 0.8 L. Let final volume be V: 0.8/V = 0.06.
V = 13⅓ L, so water added = 3⅓ L.
Answer
Add 3⅓ litres of water.
Question 5(i)
Compare Rashi (60,65,70) and Keerthi (55,65,75) under weights 4:5:6 without full calculation.
Steps only
Their middle scores match.
Keerthi loses 5 in weight 4 but gains 5 in larger weight 6.
Net weighted advantage = 5(6−4) > 0.
Answer
Keerthi’s final score is higher.
Question 5(ii)
Calculate both final fitness scores.
Steps only
Rashi = (60×4+65×5+70×6)/15 = 985/15 ≈ 65.67.
Keerthi = (55×4+65×5+75×6)/15 = 995/15 ≈ 66.33.
Answer
Rashi ≈ 65.67; Keerthi ≈ 66.33.
Question 6
Find the restaurant rating when food, ambience and service have weights 6:5:4.
Steps only
Food mean = 43/10 = 4.3; ambience = 33/10 = 3.3; service = 28/10 = 2.8.
Combined = (4.3×6+3.3×5+2.8×4)/15.
= 53.5/15 ≈ 3.57.
Answer
Approximately 3.57 out of 5.
Question 7(i)
Choose the equation for male/female langurs when total count is 60 and combined mean weight is 14.925 kg.
Steps only
Let x and y be male and female counts.
x+y=60 and (16.5x+13.8y)/60 = 14.925.
Answer
Option (b), together with x+y=60.
Question 7(ii)
Find the number of male langurs.
Steps only
Use y=60−x.
16.5x+13.8(60−x)=14.925×60.
2.7x=67.5, so x=25.
Answer
25 male langurs and 35 female langurs.
Question 7(iii)
A 15.2 kg female joins the 35 females averaging 13.8 kg. Find the new mean.
Steps only
Old total = 35×13.8 = 483 kg.
New mean = (483+15.2)/36 ≈ 13.84 kg.
Answer
Approximately 13.84 kg.
Question 7(iv)
Two males weighing 16.9 kg and 16.1 kg leave the 25 males averaging 16.5 kg.
Steps only
Old total = 25×16.5 = 412.5 kg.
Remove 33 kg and two animals.
379.5÷23 = 16.5 kg.
Answer
16.5 kg.
Question 7(v)
After those releases, one male loses 1 kg. Find the group mean.
Steps only
Current total = 379.5 kg for 23 males.
New total = 378.5 kg.
378.5÷23 ≈ 16.46 kg.
Answer
Approximately 16.46 kg.
Question 8(i)
Mix 1 L Dead Sea water at 34% with 2 L purified water at 0.001%.
Steps only
Salt units = 1×34 + 2×0.001 = 34.002 percentage-litres.
Divide by 3 L: 34.002/3 = 11.334%.
Answer
Approximately 11.334% salinity.
Question 8(ii)
Can Dead Sea water and purified water make 0.01% groundwater salinity? Find purified water needed for 1 L Dead Sea water.
Steps only
0.01% lies between 34% and 0.001%, so it is possible.
Solve (34+0.001x)/(1+x)=0.01.
x = 33.99/0.009 ≈ 3776.67 L.
Answer
Yes; approximately 3776.67 litres of purified water.
Question 8(iii)
Can 34% Dead Sea water and 0.01% groundwater make 0.001% purified-water salinity?
Steps only
Every weighted mean of 34% and 0.01% is at least 0.01%.
The target 0.001% is below both starting values.
Answer
No quantity of groundwater can produce 0.001% salinity.
Exercise Set 10.4
Textbook pages 21–22
Question 1(i)
What does 14,234 above the 2019 IUCN column represent?
Steps only
It is the full height of all coloured sections in the 2019 bar.
Answer
A total of 14,234 listed animal species across all shown classes in 2019.
Question 1(ii)
Approximately how many reptile species appear in 2016?
Steps only
Read the thickness of the red reptile segment against the 3000-unit scale.
Its thickness is roughly one third of a 3000 interval.
Answer
About 1,000 reptile species; a nearby estimate is acceptable.
Question 1(iii)
Which classes show relatively small growth from 2007 to 2019?
Steps only
Compare the thickness of each coloured segment in the first and last columns.
The mammal and bird segments change comparatively little; reptiles also rise only modestly.
Answer
Mammals and birds show the smallest visible increases; reptiles are also relatively small compared with insects, molluscs, fish and others.
Question 2(i)
Give the segment lengths for home and overseas wicket bars.
Steps only
Home segments: Test 62, ODI 100, T20 32; total 194.
Overseas segments: Test 172, ODI 49, T20 71; total 292.
Answer
Draw stacked bars ending at 194 for home and 292 for overseas.
Question 2(ii)
Give home/overseas segments for each format.
Steps only
Test: 62 home + 172 overseas = 234.
ODI: 100+49 = 149; T20: 32+71 = 103.
Answer
Draw format bars with totals Test 234, ODI 149 and T20 103.
Exercise Set 10.5
Textbook pages 29
Question 1(i)
Give a reasonable 24-hour body-state row for a student.
Steps only
One possible estimate: lying down 8 h, sitting 8 h, standing/moving 8 h.
Check: 8+8+8=24.
Answer
Example: 8 h lying down, 8 h sitting and 8 h standing/moving.
Question 1(ii)
Suggest Julie’s work from 8 h lying, 10 h sitting and 6 h moving.
Steps only
The long sitting time suggests desk-based study or office work.
Movement and daily tasks can account for the remaining 6 hours.
Answer
A reasonable guess is an office worker or college student.
Question 1(iii)
Convert each person’s hours to a 100% stacked bar.
Steps only
Since each day is 24 h, percentage = hours÷24×100.
Pavani: 83.33%,12.5%,4.17%; Raghu:29.17%,16.67%,54.17%; Zakir:33.33%,25%,41.67%.
Example Sahana:33.33% each; Julie:33.33%,41.67%,25%.
Answer
Use the listed percentages as the three consecutive segments of each 100% bar.
End-of-Chapter Exercises
Textbook pages 30–36
Question 1(i)
A team scores 6 runs then 12 runs in its first two overs. Find the run rate.
Steps only
Total = 18 runs in 2 overs.
Run rate = 18÷2 = 9.
Answer
9 runs per over.
Question 1(ii)
After 19 overs the run rate is 6; the last over adds 12. Find the 20-over rate.
Steps only
Runs after 19 overs = 19×6 = 114.
Final total = 126.
126÷20 = 6.3.
Answer
6.3 runs per over.
Question 2(i)
Write the combined shuttlecock-price average using friends’ averages ay, as, ak, ap, ag.
Steps only
Their sample counts are 6,4,5,4,4.
Combined mean = (6ay+4as+5ak+4ap+4ag)/23.
Answer
(6ay + 4as + 5ak + 4ap + 4ag)/23.
Question 2(ii)
Write the combined mean using nylon mean an and feather mean af, and compare methods.
Steps only
There are 13 nylon and 10 feather prices.
Mean = (13an+10af)/23.
Both regroup the same 23 prices, so the results are equal.
Answer
(13an + 10af)/23; yes, it equals part (i).
Question 3(i)
Shreyas has 25 shares at ₹150 and buys 10 at ₹30. Find his new average.
Steps only
Old cost = 25×150 = ₹3750; new cost = ₹300.
Total cost = ₹4050 for 35 shares.
Average = 4050÷35 ≈ ₹115.71.
Answer
Approximately ₹115.71 per share.
Question 3(ii)
Vaishnavi has 5 shares at ₹150 and buys n at ₹30; new average is ₹70. Find n.
Steps only
(750+30n)/(5+n)=70.
750+30n=350+70n.
400=40n, so n=10.
Answer
10 shares.
Question 4
A 30-hasta pool has section lengths 4,5,6,7,8 and depths 9,7,7,3,2. Find mean depth.
Steps only
Weighted depth total = 4×9+5×7+6×7+7×3+8×2 = 150.
Total length = 30.
Mean depth = 150÷30 = 5 hastas.
Answer
5 hastas.
Question 5(i)
Starting with 1 g at ₹15k, purchase 1 g at ₹30k. Find the new average price.
Steps only
Average = (1×15 + 1×30)/(1+1).
= ₹22.5k per gram.
Answer
₹22.5k per gram.
Question 5(ii)
Starting with 1 g at ₹15k, purchase 2 g at ₹30k. Find the new average price.
Steps only
Average = (1×15 + 2×30)/(1+2).
= ₹25k per gram.
Answer
₹25k per gram.
Question 5(iii)
Starting with 1 g at ₹15k, purchase 1 g at ₹10k. Find the new average price.
Steps only
Average = (1×15 + 1×10)/(1+1).
= ₹12.5k per gram.
Answer
₹12.5k per gram.
Question 5(iv)
Starting with 1 g at ₹15k, purchase 10 g at ₹10k. Find the new average price.
Steps only
Average = (1×15 + 10×10)/(1+10).
= ₹10.45k per gram approximately.
Answer
₹10.45k per gram approximately.
Question 5(v)
Starting with 1 g at ₹15k, purchase 0.5 g at ₹30k. Find the new average price.
Steps only
Average = (1×15 + 0.5×30)/(1+0.5).
= ₹20k per gram.
Answer
₹20k per gram.
Question 5(vi)
Starting with 1 g at ₹15k, purchase 0.5 g at ₹15k. Find the new average price.
Steps only
Average = (1×15 + 0.5×15)/(1+0.5).
= ₹15k per gram.
Answer
₹15k per gram.
Question 6
What happens if every weight in a weighted mean is doubled?
Steps only
New mean = Σ(2wᵢxᵢ)/Σ(2wᵢ).
Cancel the common factor 2 to obtain Σwᵢxᵢ/Σwᵢ.
Answer
The weighted average does not change.
Question 7(i)
Estimate the orbiting-object total in 2003 and when it doubled.
Steps only
The 2003 bar is close to 10,000 objects.
The chart first reaches roughly 20,000 around 2020–2021.
Answer
About 10,000 in 2003; it doubled around 2020–2021.
Question 7(ii)
Estimate payload and other orbiting objects in 2025.
Steps only
The 2025 stack is about 33,000 in total.
Payload segment ≈ 17,500 and other segment ≈ 15,500.
Shares are about 53% payload and 47% other.
Answer
Approximately 17,500 payload objects (53%) and 15,500 other objects (47%).
Question 7(iii)
Describe payload and other-object trends and shares over time.
Steps only
Both cumulative counts rise, with much faster growth after about 2020.
Other objects dominate much of the earlier period.
Recent payload growth raises the payload share until it becomes slightly larger.
Answer
Both counts increase; recent payload growth is especially steep, so payloads form a growing and now slightly larger share.
Question 8(i)
Which playground-map claims follow from the printed percentages?
Steps only
Arunachal Pradesh is 68%, close to 2/3, so (b) follows.
Percentages alone do not prove (a) or (c), which compare numbers of schools.
Maharashtra has both more schools (given) and a higher rate, 93% versus Telangana’s 74%, so (d) follows.
Answer
Statements (b) and (d) are correct.
Question 8(ii)
Can the nationwide playground percentage be obtained from state percentages alone?
Steps only
No: states contain different numbers of schools.
Need each state’s total school count, then compute Σ(playground schools)/Σ(all schools)×100.
Answer
No; the total number of schools in every state is also needed.
Question 9(i)
Choose a play using only the 100% rating chart.
Steps only
Play C has the largest share of five-star ratings, about 40%.
If five-star share is the rule, choose Play C.
Answer
A reasonable choice is Play C based on its highest five-star proportion.
Question 9(ii)
Would the ordinary stacked chart change the choice?
Steps only
The ordinary chart shows Play B has far more ratings overall.
Its absolute number of high ratings is also large, while Play C has a smaller sample.
Answer
A reasonable revised choice is Play B because its strong rating pattern is supported by many more ratings; the decision rule should be stated.
Question 10(i)
Which chart suits triathlon segment times?
Steps only
The totals differ: Athlete 1 = 9:23, Athlete 2 = 9:50, Athlete 3 = 11:07.
A normal stacked bar preserves these total-time differences.
Answer
Use a stacked bar chart.
Question 10(ii)
Which listed triathlon questions can a 100% stacked chart answer?
Steps only
It cannot show who finished first or who ran longest because actual totals are hidden.
Athlete 1 cycled 300 of 563 minutes, about 53%, which its proportional bar can estimate.
Answer
Only (b): Athlete 1 spent a little over half, about 53%, of the race cycling.
Question 11
Write supported observations from the age-group 100% chart.
Steps only
Seeing forms a larger share as age rises, especially for age 60+.
Movement difficulty also becomes more prominent in older groups.
Speech and multiple disability occupy relatively larger shares among ages 0–19 than in several older groups.
Answer
The distribution changes with age; the chart compares shares within each age group, not the total number of people in different age groups.
Question 12(i)
Plan a body-state stacked-chart project.
Steps only
Record lying, sitting and moving hours for each chosen person/day.
Check every row totals 24 hours.
Draw one stacked bar per person/day on a common 0–24 h scale and explain estimates.
Answer
A complete response contains the data table, 24-hour checks, stacked chart and observations.
Question 12(ii)
Plan a three-month 100% family-expenditure chart.
Steps only
Choose categories and record monthly amounts with family help.
For each month compute category amount÷monthly total×100.
Draw equal-length 100% bars and compare changing shares.
Answer
A complete response contains raw amounts, percentage calculations, three 100% bars and evidence-based observations.
Question 13
Design a weighted 1–5 experience rating scheme.
Steps only
Choose 4–6 justified aspects and positive weights.
For each of at least 10 responses compute Σ(weight×rating)/Σweights.
Average individual scores; chart rating shares and discuss strengths and limits.
Answer
The project is complete when the aspects, weights, response data, calculations, 100% chart and reflection are all shown.
Question 14
How should a class combine families’ average ages?
Steps only
For family i, multiply its mean age aᵢ by member count nᵢ to recover total age.
Add all total ages and divide by all members.
Answer
Class-wide family-member mean = Σ(nᵢaᵢ)/Σnᵢ.
Question 15
What happens if 1 is added to every weight?
Steps only
Old mean = Σwᵢxᵢ/Σwᵢ. New mean = [Σwᵢxᵢ+Σxᵢ]/[Σwᵢ+n].
The new mean lies between the old weighted mean and the ordinary mean Σxᵢ/n.
Answer
It generally changes, moving toward the ordinary mean; it stays unchanged only when those two means are equal.
Question 16
All cow, sheep and chicken counts fall. Which listed share-change patterns are possible?
Steps only
Shares must still add to 100%, so all three cannot rise or all three fall.
Equal proportional declines keep all shares unchanged.
Different decline rates can make some shares rise while others fall or stay fixed.
Answer
Possibilities (iii), (iv) and (v) can occur; (i) and (ii) cannot.