CBSE Class 9 Maths Ganita Manjari Part 2 Ch 10 How Quantities Combine Understanding Data Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. Three sections of a class took the same test. Section P has 25 students with a mean score of 68, Section Q has 30 students with a mean score of 72, and Section R has 45 students with a mean score of 80.
(i) Find the mean score of all 100 students.
(ii) Find the simple average of the three section means.
(iii) Explain why the two answers differ, and state which one answers the question 'what is the mean score of the class?'
(iv) State what the mean score of the class would have been if each section had contained the same number of students.
Answer:
(i) \( \frac{68 \times 25 + 72 \times 30 + 80 \times 45}{100} = \frac{1700 + 2160 + 3600}{100} = \) 74.6.
(ii) (68 + 72 + 80) ÷ 3 = 220 ÷ 3 ≈ 73.33.
(iii) The sections differ in size. R (45 students) should count more than P (25). 74.6 is the class mean; 73.33 is just the average of three numbers.
(iv) With equal sizes, the simple average would be correct: ≈ 73.33.
In simple words: The largest section scored highest, so the true class mean is above the plain average of the section means.

Teacher's Note:
1. Section sizes are the weights.
2. Equal sizes make both methods agree.

Question 2. One collection has n values with average a, and a second collection has m values with average b.
(i) Write down the sum of the values in each collection.
(ii) Hence derive a formula for the average of the combined collection.
(iii) Show that when n = m the formula reduces to \( \frac{a + b}{2} \).
(iv) Give one numerical example in which the simple average of a and b is correct, and one in which it is wrong, showing the correct value in the second case.
Answer:
(i) Sums: an and bm.
(ii) Combined average = \( \frac{an + bm}{n + m} \).
(iii) With m = n: \( \frac{an + bn}{2n} = \frac{a + b}{2} \).
(iv) Correct: 20 students with mean 50 and 20 with mean 70 give 2400 ÷ 40 = 60 = (50 + 70) ÷ 2.
Wrong: 20 students with mean 50 and 5 with mean 70. Simple average says 60, but the true value is (1000 + 350) ÷ 25 = 54.
In simple words: The general formula weights each average by its size; with equal sizes it becomes the simple average.

Teacher's Note:
1. Derive, specialise, then illustrate: a model long answer.
2. Use numbers that make the gap obvious.

Question 3. A vessel holds 40 litres of a solution containing 15% acid.
(i) Find the quantity of acid in the vessel.
(ii) Find the quantity of water that must be added to bring the concentration down to 10%.
(iii) Find the quantity of pure acid that must instead be added to the original 40 litres to raise the concentration to 20%.
(iv) Explain why the calculation in (ii) uses an added liquid of concentration 0% while the calculation in (iii) uses one of concentration 100%.
Answer:
(i) 40 × 0.15 = 6 litres of acid.
(ii) \( \frac{6}{40 + w} = 0.10 \Rightarrow 40 + w = 60 \Rightarrow w = \) 20 litres.
(iii) \( \frac{6 + a}{40 + a} = 0.20 \Rightarrow 6 + a = 8 + 0.2a \Rightarrow 0.8a = 2 \Rightarrow a = \) 2.5 litres.
(iv) Water contains no acid (0%), so it adds volume only. Pure acid is all acid (100%), so each litre adds to both the acid and the volume. These are the two extreme cases of the same mixture formula.
In simple words: Water dilutes without adding acid; pure acid adds acid and volume together.

Teacher's Note:
1. Hold the acid fixed when adding water.
2. Add to both top and bottom when adding pure acid.

Question 4. Brass is an alloy of copper and zinc. Batch A weighs 250 kg and is 72% copper, Batch B weighs 150 kg and is 52% copper, and Batch C is 40% copper. When all three batches are melted together, the alloy obtained is 60% copper.
(i) Write an equation in y, the weight of Batch C in kg.
(ii) Solve it to find the weight of Batch C.
(iii) Verify your answer by computing the copper content of the combined alloy.
(iv) Batches A and B together are 64.5% copper. Explain why the weight of Batch C had to come out smaller than the combined weight of Batches A and B.
Answer:
(i) Copper = 180 + 78 + 0.4y; total = 400 + y. \( \frac{258 + 0.4y}{400 + y} = 0.60 \)
(ii) 258 + 0.4y = 240 + 0.6y ⇒ 0.2y = 18 ⇒ y = 90 kg.
(iii) Copper = 180 + 78 + 36 = 294 kg in 490 kg; 294 ÷ 490 = 60% ✓
(iv) A and B together (64.5%) are only 4.5 points above 60%, while C is 20 points below. Weights go in the inverse ratio 20 : 4.5, so C needs only 4.5/20 of 400 kg = 90 kg.
In simple words: Batch C is far from the target, so only a little of it is needed to pull the mix down to 60%.

Teacher's Note:
1. A far-off ingredient has a strong pull.
2. Inverse-ratio reasoning confirms the algebra.

Question 5. Savitri's marks in a subject are 36 out of 50 in the internal tests, 39 out of 60 in the project and 78 out of 100 in the final exam. The three are combined in the ratio 2 : 3 : 5.
(i) Express each of the three marks as a percentage.
(ii) Find Savitri's annual percentage score.
(iii) Explain why the raw marks 36, 39 and 78 must not be combined directly with the weights 2, 3 and 5.
(iv) Find what her annual percentage score would have been if the three assessments had carried equal weight.
Answer:
(i) 36/50 = 72%; 39/60 = 65%; 78/100 = 78%.
(ii) \( \frac{72 \times 2 + 65 \times 3 + 78 \times 5}{10} = \frac{144 + 195 + 390}{10} = \) 72.9%.
(iii) The raw marks are out of different maximums (50, 60, 100), so they are not comparable. Converting to percentages puts them on one scale first.
(iv) (72 + 65 + 78) ÷ 3 = 215 ÷ 3 ≈ 71.67%.
In simple words: Put all marks on the same 100-point scale first, then apply the weights.

Teacher's Note:
1. Common scale first, weights second.
2. Weighting helps her because the final exam is her best.

Question 6. Consider the values 6, 10 and 15 with weights 2, 3 and 5 respectively.
(i) Find the weighted mean.
(ii) Find the weighted mean again after every weight is doubled, and comment on the result.
(iii) Find the weighted mean again after 1 is added to every original weight, giving your answer correct to two decimal places, and comment on the result.
(iv) Explain algebraically why one of these two changes leaves the weighted mean unaltered while the other does not.
Answer:
(i) (12 + 30 + 75) ÷ 10 = 11.7.
(ii) Weights 4, 6, 10: (24 + 60 + 150) ÷ 20 = 11.7, which is unchanged.
(iii) Weights 3, 4, 6: (18 + 40 + 90) ÷ 13 = 148 ÷ 13 ≈ 11.38, which is changed.
(iv) Multiplying by k gives k in both numerator and denominator, which cancels; the ratio 2 : 3 : 5 stays the same. Adding 1 gives 3 : 4 : 6, a different ratio, with nothing to cancel. The weights become more equal, so the mean moves towards the plain mean 10.33.
In simple words: Scaling weights keeps their ratio; adding to them changes it.

Teacher's Note:
1. Ratio preserved ⇒ mean preserved.
2. Adding flattens weights towards equality.

Question 7. A bowler's wickets in international cricket are: Test 70 at home and 130 overseas; ODI 90 at home and 60 overseas; T20 40 at home and 60 overseas.
(i) Find the total number of wickets taken in each format, and the overall total.
(ii) Find the percentage of the Test wickets that were taken overseas.
(iii) Find the format in which the largest share of the wickets was taken at home.
(iv) State which kind of chart you would draw to answer (i) and which to answer (iii), giving a reason in each case.
Answer:
(i) Test 200, ODI 150, T20 100; overall 450.
(ii) 130 ÷ 200 × 100 = 65%.
(iii) Home shares: Test 35%, ODI 60%, T20 40%. Largest: ODI.
(iv) For (i), a stacked bar chart: bar lengths give totals. For (iii), a 100% stacked bar chart: all bars equal, so home segments compare as shares. On the stacked chart ODI's bar is shorter, which hides its large home share.
In simple words: Totals need a stacked bar; shares need a 100% stacked bar.

Teacher's Note:
1. Match chart type to question type.
2. Shorter bars can hide large shares.

Question 8. The average daily electricity use of a house, in units, is: 2005 — Lighting 400, Cooling 300, Kitchen 100, Other 200; 2025 — Lighting 240, Cooling 960, Kitchen 480, Other 320.
(i) Find the total for each year and the percentage share of each category in each year.
(ii) The share of lighting fell from 40% to 12%. State whether the actual number of units used for lighting also fell, and justify your answer with the figures.
(iii) Find the factor by which the actual cooling use rose.
(iv) The share of the 'Other' category fell between the two years. Show that its actual use nevertheless rose, and explain what this teaches about reading a 100% stacked bar chart.
Answer:
(i) 2005 total 1000: 40%, 30%, 10%, 20%. 2025 total 2000: 12%, 48%, 24%, 16%.
(ii) Yes: 400 → 240 units, a fall of 160. Here share and amount agree, but only by chance.
(iii) 960 ÷ 300 = 3.2 times.
(iv) 'Other' share fell 20% → 16%, but units rose 200 → 320. The total doubled while 'Other' grew only 1.6 times. A shrinking segment does not mean a shrinking quantity; totals are needed to judge amounts.
In simple words: A category can grow in amount yet shrink in share if the total grows faster.

Teacher's Note:
1. The key lesson of 100% charts.
2. Compare growth factors: 1.6 vs 2.

Question 9. The 100% stacked bar chart shows the share of the year's blooms in two gardens across three seasons.
(i) State which of the following can be inferred from the chart alone, giving a reason in each case.
a) In Fatima's garden there were more blooms in summer than in winter.
b) Fatima's garden had more summer blooms than Naveen's garden.
c) The two gardens had the same number of blooms over the year.
(ii) Suppose Fatima's garden had 200 blooms in the year and Naveen's had 100. Find the number of summer blooms in each garden.
(iii) Suppose instead that Fatima's garden had 200 blooms and Naveen's had 500. Find the number of summer blooms in each garden.
(iv) Using your answers to (ii) and (iii), explain what a 100% stacked bar chart does and does not tell us.
Answer:
(i) a) Can be inferred: within one bar, summer 50% > winter 20%.
b) Cannot: 50% and 40% are shares of different, unknown totals.
c) Cannot: every bar is drawn the same length regardless of its total.
(ii) Fatima 0.50 × 200 = 100; Naveen 0.40 × 100 = 40.
(iii) Fatima 100; Naveen 0.40 × 500 = 200.
(iv) The same chart fits both cases, which give opposite answers. So the chart tells us shares within each bar, but not amounts or comparisons across bars.
In simple words: Within a bar the chart is reliable; across bars it needs the totals.

Teacher's Note:
1. Use two scenarios to show the chart cannot decide.
2. Within-bar = safe; across-bar = unsafe.

Question 10. In a 20-over cricket match a team scores at a run rate of 9 in the first 6 overs, at a run rate of 6 in the next 10 overs, and at a run rate of 12 in the last 4 overs.
(i) Find the runs scored in each of the three phases, and the team's total score.
(ii) Find the team's run rate over the full 20 overs.
(iii) Explain why this run rate is not the simple average of 9, 6 and 12.
(iv) Find the run rate the team would have needed in the last 4 overs for the run rate over the full 20 overs to be exactly 9.
Answer:
(i) 6 × 9 = 54; 10 × 6 = 60; 4 × 12 = 48; total 162.
(ii) 162 ÷ 20 = 8.1.
(iii) The phases have different lengths (6, 10, 4 overs). The longest phase is the slowest, so it pulls the overall rate below the simple average of 9.
(iv) Target 20 × 9 = 180; first 16 overs gave 114; need 66 in 4 overs: rate 16.5.
In simple words: Overs are the weights; the long slow phase drags the rate down to 8.1.

Teacher's Note:
1. Run rate = weighted mean of phase rates.
2. Target total − runs so far = runs needed.

Question 11. A hostel has two wings. Wing A has 30 students whose mean weight is 48 kg. Wing B has m students whose mean weight is 56 kg. The mean weight of all the students in the hostel is 52.8 kg.
(i) Write an equation in m.
(ii) Solve it to find the number of students in Wing B.
(iii) Find the total number of students in the hostel.
(iv) Explain why the combined mean had to lie between 48 kg and 56 kg, and why it lies nearer to 56 kg than to 48 kg.
Answer:
(i) \( \frac{1440 + 56m}{30 + m} = 52.8 \)
(ii) 1440 + 56m = 1584 + 52.8m ⇒ 3.2m = 144 ⇒ m = 45.
(iii) 30 + 45 = 75 students.
(iv) A weighted mean always lies between the values combined. It is nearer 56 because Wing B is larger (45 vs 30). Check: distances 4.8 and 3.2 are in the ratio 3 : 2, the inverse of 30 : 45.
In simple words: The bigger wing is heavier on average, so the overall mean leans towards 56 kg.

Teacher's Note:
1. Solve for an unknown group size from the combined mean.
2. Inverse-ratio check confirms it.

Question 12. A wildlife facility houses 100 deer. The mean weight of the male deer is 80 kg, the mean weight of the female deer is 62 kg, and the mean weight of all 100 deer is 69.2 kg.
(i) Taking the number of male deer to be x, write an equation in x and solve it.
(ii) Five female deer with a total weight of 365 kg are moved to another facility. Find the new mean weight of the female deer.
(iii) Returning to the original data, one male deer loses 4 kg. Find the new mean weight of the male deer.
(iv) Explain why the mean weight of all the deer, 69.2 kg, lies nearer to 62 kg than to 80 kg.
Answer:
(i) 80x + 62(100 − x) = 6920 ⇒ 18x = 720 ⇒ x = 40 males, 60 females.
(ii) Females total 3720 kg; remove 365 kg: 3355 ÷ 55 = 61 kg.
(iii) Males total 3200 kg; minus 4 = 3196 ÷ 40 = 79.9 kg.
(iv) There are more females (60) than males (40), so their mean pulls harder. Distances 7.2 and 10.8 are in the ratio 2 : 3, the inverse of 40 : 60.
In simple words: There are more females, so the overall mean sits closer to the female mean.

Teacher's Note:
1. A 4 kg loss spread over 40 deer lowers the mean by only 0.1 kg.
2. Removing heavier-than-average animals lowers the mean.

Question 13. In the old Indian measure described by Śrīdharācārya, a purity of k varna meant that the alloy was \( \frac{k}{16} \) gold, with 16 varna standing for pure gold. Today the same idea is expressed in karat, with 16 varna = 24 karat. A goldsmith melts together 8 units of gold of 15 varna, 6 units of 12 varna and 6 units of 9 varna.
(i) Find the purity of the combined gold in varna.
(ii) Express this purity in karat.
(iii) Find the fraction of the combined metal that is gold.
(iv) Explain why this calculation is a weighted mean, naming the quantity that acts as the weight.
Answer:
(i) \( \frac{15 \times 8 + 12 \times 6 + 9 \times 6}{20} = \frac{246}{20} = \) 12.3 varna.
(ii) 1 varna = 1.5 karat, so 12.3 × 1.5 = 18.45 karat.
(iii) 12.3 ÷ 16 = 0.76875 (76.875% gold).
(iv) Each piece influences the result in proportion to its amount; the quantity of gold is the weight. The 8 units at 15 varna pull the answer above the plain average of 12.
In simple words: Bigger lumps of purer gold count for more, so the mix is 12.3 varna.

Teacher's Note:
1. Historical link: Śrīdharācārya (c. 750 CE).
2. Convert units only after finding the mean.

Question 14. A clinic asks five patients to rate four aspects of their visit on a scale of 1 to 5. Doctor's care: 5, 4, 5, 4, 5. Waiting time: 2, 3, 2, 3, 5. Cleanliness: 4, 4, 5, 3, 4. Staff behaviour: 3, 4, 3, 5, 5. The four aspects are combined with weights in the ratio 5 : 3 : 4 : 2.
(i) Find the mean rating for each aspect.
(ii) Find the clinic's overall weighted rating.
(iii) Find the simple average of the four aspect means.
(iv) Explain why the weighted rating is higher than the simple average, and what the clinic's choice of weights says about its priorities.
Answer:
(i) Doctor's care 4.6; waiting time 3.0; cleanliness 4.0; staff behaviour 4.0.
(ii) \( \frac{4.6 \times 5 + 3.0 \times 3 + 4.0 \times 4 + 4.0 \times 2}{14} = \frac{56}{14} = \) 4.0.
(iii) 15.6 ÷ 4 = 3.9.
(iv) The heaviest weight is on doctor's care (best score), and the worst aspect (waiting time) has a small weight. The clinic values medical care most; other groups might choose different weights, which is why weights should be published.
In simple words: Giving most weight to the best-scoring aspect lifts the overall rating.

Teacher's Note:
1. Weights express priorities.
2. Discuss fairness of weight choices.

Question 15. The stacked bar chart shows the monthly expenditure of three families across four categories.
(i) Find the total monthly expenditure of each family.
(ii) Find which family spent the largest amount on healthcare.
(iii) Find the percentage share of each category for Family B and for Family C.
(iv) Families B and C would have identical 100% stacked bar charts. Explain how this can happen although the two families spend different amounts, and state which of the two charts should be used to compare the amounts spent on healthcare.
Answer:
(i) A: ₹15 000; B: ₹12 000; C: ₹10 000.
(ii) Healthcare: A ₹1500, B ₹3000, C ₹2500. Largest: Family B.
(iii) B and C both: Housing 30%, Food 25%, Education 20%, Healthcare 25%.
(iv) C spends exactly 5/6 of B in every category, so every share is the same. The 100% chart divides out the total. To compare healthcare amounts, use the stacked bar chart (B ₹3000 vs C ₹2500).
In simple words: Scaling every category by the same factor keeps shares the same, so only the stacked chart shows the difference.

Teacher's Note:
1. Proportional spending ⇒ identical 100% bars.
2. Amounts need the stacked chart.

Question 16. Two plays are rated by the audience on a scale of 1 to 5 stars. Play X (100 ratings): 5★ 40%, 4★ 30%, 3★ 20%, 2★ 7%, 1★ 3%. Play Y (400 ratings): 5★ 30%, 4★ 35%, 3★ 25%, 2★ 7%, 1★ 3%.
(i) Find the mean rating of each play.
(ii) Find the actual number of 5★ ratings received by each play.
(iii) State which play appears better from the shares alone, and which play's mean rating you would trust more, giving a reason.
(iv) Explain what a 100% stacked bar chart of this data would show clearly, and what it would hide.
Answer:
(i) X: 2.00 + 1.20 + 0.60 + 0.14 + 0.03 = 3.97. Y: 1.50 + 1.40 + 0.75 + 0.14 + 0.03 = 3.82.
(ii) X: 40; Y: 120.
(iii) X looks better from shares, but Y's mean is more trustworthy: it rests on 400 ratings, not 100, and Y has three times as many 5★ ratings.
(iv) It would show the rating profile clearly (both mostly 4–5★, X's 5★ segment longer), but hide that Y was rated four times as often.
In simple words: Percentages show the pattern, but the number of ratings tells you how reliable it is.

Teacher's Note:
1. Discuss sample size and reliability.
2. Shares as weights give a mean rating.

Question 17. A shopkeeper mixes 25 kg of rice costing ₹48 per kg, 25 kg costing ₹60 per kg and 50 kg costing ₹66 per kg.
(i) Find the cost price per kg of the mixture.
(ii) He sells the mixture at ₹69 per kg. Find his percentage profit.
(iii) Find the quantity of a variety costing ₹90 per kg that must be added to the 100 kg mixture to raise its cost price to ₹65 per kg.
(iv) Explain why the calculation in (iii) is a weighted-mean calculation worked backwards.
Answer:
(i) (1200 + 1500 + 3300) ÷ 100 = ₹60 per kg.
(ii) Profit ₹9 on ₹60 = 15%.
(iii) \( \frac{6000 + 90x}{100 + x} = 65 \Rightarrow 25x = 500 \Rightarrow x = \) 20 kg.
(iv) In (i) the weights were known and the mean found. In (iii) the mean is fixed (₹65) and a weight is unknown: same equation, solved for a weight. Check: ₹65 is 5 from ₹60 and 25 from ₹90, ratio 5 : 1 = 100 : 20.
In simple words: Fix the target price and solve for how much of the costly rice is needed.

Teacher's Note:
1. Forward: find the mean. Backward: find a weight.
2. Balance check gives the answer mentally.

Question 18. The mean daily rainfall at a station was 3.5 mm over the 30 days of April, 8 mm over the 31 days of May and 15 mm over the 30 days of June.
(i) Write an expression that gives the mean daily rainfall over the three months together.
(ii) Evaluate it, correct to two decimal places.
(iii) Find the simple average of the three monthly means, correct to two decimal places, and compare it with your answer to (ii).
(iv) Explain why the two answers are so close in this case, and describe a situation in which the simple average would be badly wrong.
Answer:
(i) \( \frac{30 \times 3.5 + 31 \times 8 + 30 \times 15}{91} \)
(ii) (105 + 248 + 450) ÷ 91 = 803 ÷ 91 ≈ 8.82 mm.
(iii) 26.5 ÷ 3 ≈ 8.83 mm, just 0.01 mm more.
(iv) The months are almost equal in length (30, 31, 30), so the weights are nearly equal. It would go badly wrong for, say, 90 dry days at 2 mm and 2 cloudburst days at 60 mm: simple average 31 mm, true mean (180 + 120) ÷ 92 ≈ 3.26 mm.
In simple words: Nearly equal weights make the two methods agree; very unequal periods make them disagree badly.

Teacher's Note:
1. Great example of when shortcuts are safe.
2. Construct extreme cases to show the danger.

Question 19. A smart watch records how many hours a day four people spend lying down, sitting, and standing or moving around. Pavani (patient): 18, 3, 3. Raghu (nurse): 6, 6, 12. Zakir (teacher): 9, 9, 6. Sahana (student): 9, 12, 3.
(i) Show that each row accounts for a full day.
(ii) Find the percentage share of each body-state for Raghu and for Sahana.
(iii) State whose bar would have the longest 'standing or moving' segment in a 100% stacked bar chart of this data, and give its share.
(iv) Explain why a single bar of this chart can be read both as a stacked bar and as a 100% stacked bar.
(v) Suggest one other occupation whose bar might look similar to Raghu's, and give a reason.
Answer:
(i) Each row adds to 24 hours.
(ii) Raghu: 25%, 25%, 50%. Sahana: 37.5%, 50%, 12.5%.
(iii) Raghu, with 50%.
(iv) Segments are actual hours (stacked bar), and every bar totals 24 hours, so lengths are also shares of the day (100% stacked bar).
(v) A security guard or shop assistant: they spend most of the day on their feet.
In simple words: Every bar is exactly one day, so hours and shares tell the same story.

Teacher's Note:
1. Equal totals make both readings coincide.
2. Invite students to suggest other jobs.

Question 20. Write an account of how to choose between the three kinds of chart studied in this chapter. Your account must include
(i) what a cluster-column chart shows well;
(ii) what a stacked bar chart shows well;
(iii) what a 100% stacked bar chart shows well;
(iv) one question that each of the three charts answers badly;
(v) an explanation of why a 100% stacked bar chart can always be drawn from a stacked bar chart, but never the other way round.
Answer:
(i) Cluster-column: individual values side by side from a common zero, which is good for comparing any one value with another. The choice of cluster (by category or by group) decides which comparison is easiest.
(ii) Stacked bar: parts and totals together; bar length = total, segments = parts, with actual amounts kept.
(iii) 100% stacked bar: proportions within each group; a clearer alternative to several pie charts.
(iv) Cluster-column is poor at totals (they must be added by eye). Stacked bar is poor at comparing inner segments across bars (they start at different points). 100% stacked bar is poor at anything absolute: it cannot say 'how much'.
(v) Stacked → 100%: divide each segment by its bar total, which is always possible. 100% → stacked: impossible, because totals were divided out; e.g. 30%, 25%, 20%, 25% could come from ₹12 000 or ₹10 000 or any total.
In simple words: Pick the chart by the question: single values, totals with parts, or shares.

Teacher's Note:
1. Organise under the five headings.
2. Use one data set to illustrate all three charts.

Case Study Based Questions

Question 21. Sports Day Points: At a school sports day, Red House has 40 students who scored a mean of 62 points each, and Blue House has 60 students who scored a mean of 72 points each.
(a) Find the total number of points scored by Red House.
(b) Find the mean number of points scored by all 100 students.
(c) The simple average of 62 and 72 is 67. Explain why the answer to (b) is not 67.
(d) Find what the mean for all the students would have been if each house had 50 students.
Answer:
(a) 40 × 62 = 2480.
(b) (2480 + 4320) ÷ 100 = 68.
(c) Blue House is larger (60 vs 40), so its higher mean pulls the result above 67.
(d) Equal sizes: (62 + 72) ÷ 2 = 67.
In simple words: The bigger, higher-scoring house lifts the overall mean to 68.

Teacher's Note:
1. Totals first, then combine.
2. Equal sizes ⇒ simple average.

Question 22. The Lemonade Stall: A stall has 3 litres of lemonade containing 12% sugar and 2 litres of lemonade containing 7% sugar. The two are poured into one large jar.
(a) Find the quantity of sugar in the 3 litres.
(b) Find the quantity of sugar in the 2 litres.
(c) Find the percentage of sugar in the mixture.
(d) The simple average of 12% and 7% is 9.5%. State whether your answer to (c) is above or below this, and explain why.
Answer:
(a) 3 × 0.12 = 0.36 L.
(b) 2 × 0.07 = 0.14 L.
(c) 0.50 ÷ 5 = 10%.
(d) Above: there is more of the sweeter 12% lemonade, so it pulls the mixture above the midpoint.
In simple words: More sweet lemonade than weak lemonade makes the mix 10%, above halfway.

Teacher's Note:
1. Quantity of the dissolved substance first.
2. Predict the side of the midpoint.

Question 23. Meera's Report Card: Meera's marks in Science are 18 out of 25 in the internal tests, 27 out of 50 in the project and 78 out of 100 in the final exam. The school combines the three in the ratio 3 : 2 : 5.
(a) Express each of the three marks as a percentage.
(b) Find Meera's weighted percentage score in Science.
(c) Find the simple average of the three percentages.
(d) Explain why the weighted score is higher than the simple average.
Answer:
(a) 72%, 54%, 78%.
(b) (216 + 108 + 390) ÷ 10 = 71.4%.
(c) 204 ÷ 3 = 68%.
(d) Half the weight is on the final exam (her best, 78%), and her weakest score (54%) has the smallest weight.
In simple words: Her strongest paper counts most, so the weighted score beats the plain average.

Teacher's Note:
1. Convert to percentages before weighting.
2. Explain using where the weights fall.

Question 24. Chasing a Target: A team batting in a 20-over match scores at a run rate of 7 in the first 10 overs.
(a) Find the number of runs scored in the first 10 overs.
(b) In the next 5 overs the team scores 50 runs. Find the run rate after 15 overs.
(c) The team needs 180 runs in all. Find the number of runs still needed in the last 5 overs.
(d) Find the run rate required in the last 5 overs.
Answer:
(a) 10 × 7 = 70.
(b) 120 ÷ 15 = 8.
(c) 180 − 120 = 60.
(d) 60 ÷ 5 = 12.
In simple words: Keep track of runs and overs; the required rate is runs needed ÷ overs left.

Teacher's Note:
1. Run rate problems are mean problems.
2. Real-life context engages students.

Question 25. The Electricity Bill: A house used the following units of electricity during a year: Lighting 600, Cooling 1200, Kitchen 900, Other 300.
(a) Find the total number of units used.
(b) Find the percentage share of cooling.
(c) Name the category that would have the shortest segment in a 100% stacked bar chart of this data, and give its share.
(d) Next year the total use rises to 4000 units while the cooling use stays at 1200 units. State whether the cooling segment of the 100% stacked bar chart becomes longer or shorter, and give its new share.
Answer:
(a) 3000 units.
(b) 1200 ÷ 3000 = 40%.
(c) Other, 300 ÷ 3000 = 10%.
(d) Shorter: 1200 ÷ 4000 = 30%, although cooling use has not changed.
In simple words: A share can shrink even when the amount stays the same, if the total grows.

Teacher's Note:
1. Shares depend on the total.
2. Never read a shorter segment as 'less used' without totals.

Question 26. The Brass Workshop: A workshop melts together 80 kg of brass containing 65% copper and 120 kg of brass containing 45% copper.
(a) Find the mass of copper in the first batch.
(b) Find the mass of copper in the second batch.
(c) Find the percentage of copper in the combined alloy.
(d) Find the mass of pure copper that must be added to the combined alloy to make it 60% copper.
Answer:
(a) 80 × 0.65 = 52 kg.
(b) 120 × 0.45 = 54 kg.
(c) 106 ÷ 200 = 53%.
(d) \( \frac{106 + c}{200 + c} = 0.6 \Rightarrow 0.4c = 14 \Rightarrow c = \) 35 kg.
In simple words: Adding pure copper raises both the copper and the total, so solve the equation.

Teacher's Note:
1. Pure copper = 100% concentration.
2. Check: 141 ÷ 235 = 0.6 ✓

Question 27. Two School Libraries: The chart shows how the books in two school libraries are divided between three categories.
(a) Name the category with the largest share in School P's library.
(b) State whether the chart alone tells you which school has more fiction books.
(c) School P has 3000 books in all and School Q has 5000. Find which school has more fiction books.
(d) State the one piece of information the chart does not give, which was needed for part (c).
Answer:
(a) Fiction (45%).
(b) No: the shares are of different, unknown totals.
(c) P: 45% of 3000 = 1350. Q: 30% of 5000 = 1500. School Q has more.
(d) The total number of books in each library.
In simple words: A bigger share of a smaller library can still mean fewer books.

Teacher's Note:
1. Classic 100% chart trap.
2. Share × total = number.

Question 28. Blending Tea: A shopkeeper blends 4 kg of tea costing ₹250 per kg with 6 kg of tea costing ₹400 per kg.
(a) Find the total cost of the 10 kg blend.
(b) Find the cost price per kg of the blend.
(c) State whether your answer to (b) is nearer ₹250 or ₹400, and explain why.
(d) He sells the blend at ₹391 per kg. Find his percentage profit.
Answer:
(a) 1000 + 2400 = ₹3400.
(b) 3400 ÷ 10 = ₹340.
(c) Nearer ₹400: there is more of the costlier tea (6 kg vs 4 kg).
(d) Profit ₹51 on ₹340 = 15%.
In simple words: More of the expensive tea makes the blend cost closer to ₹400.

Teacher's Note:
1. Weighted mean = cost per kg of blend.
2. Profit % is on cost price.

Question 29. Water from the Dead Sea: Dorjee has 1 litre of water taken from the Dead Sea, in which the salinity is 32%. Purified drinking water may be taken to have a salinity of 0%.
(a) Find the quantity of salt in his 1 litre of Dead Sea water.
(b) Find the salinity of the mixture if he adds 3 litres of purified water.
(c) Find the quantity of purified water he must add to his 1 litre to bring the salinity down to 2%.
(d) Explain whether he can ever reach a salinity of exactly 0% by adding purified water.
Answer:
(a) 0.32 L.
(b) 0.32 ÷ 4 = 8%.
(c) 0.32 ÷ (1 + w) = 0.02 ⇒ 1 + w = 16 ⇒ w = 15 L.
(d) No. The 0.32 L of salt never disappears; salinity 0.32 ÷ (1 + w) stays positive, however large w is.
In simple words: Adding water can make the salt very dilute but never remove it completely.

Teacher's Note:
1. Good link to the idea of a limit.
2. The salt amount is fixed.

Question 30. The Class Project: For a class project, three families report: Family P has 4 members with mean age 30 years; Family Q has 6 members with mean age 25 years; Family R has 5 members with mean age 36 years.
(a) Find the sum of the ages of the members of Family R.
(b) Find the mean age of all the people in the three families.
(c) Find the simple average of the three family mean ages, correct to two decimal places.
(d) Explain why the two answers differ, and state which one answers the question 'what is the average age of all these people?'
Answer:
(a) 5 × 36 = 180 years.
(b) (120 + 150 + 180) ÷ 15 = 30 years.
(c) 91 ÷ 3 ≈ 30.33 years.
(d) Families differ in size, so each person must count once. 30 years answers the question; 30.33 is just the average of three family averages.
In simple words: Counting every person once gives 30 years; averaging the families gives a slightly different number.

Teacher's Note:
1. Ask: is the question about people or about families?
2. Group sizes are the weights.