How Quantities Combine: Understanding Data

The combined average is a weighted mean: each section’s average is weighted by its number of students.

Both weights are multiples of 5, so cancel 5 before multiplying:

The combined average is about 73.82. It is closer to 72 than to 76 because Section A has more students.

Let each quantity be \(q\). The nitrogen in the mixture is

and the whole mixture is \(3q\). So the fraction is

(the \(q\) cancels: with equal quantities the answer is just the simple average). Over the common denominator 300:

Divide by 3, cancelling before multiplying:

The mixture is \(\tfrac{11}{100}\) nitrogen, that is 11%.

Purity in varṇa mixes like a concentration, so the purity of the combined gold is the weighted mean of the three purities, weighted by the amount of each piece.

The combined gold is about 11.13 varṇa. Check: it lies between the lowest (10) and highest (12) purities, and close to 11 because the largest piece (17 units) is at 11 varṇa.

Each month’s average must be weighted by its number of days: May has 31, June 30 and July 31 days, 92 days in all.

Evaluating it:

Concentration of the mixture = (total spice) ÷ (total volume), i.e. the weighted mean of the concentrations with the volumes as weights. The total volume is 600 mL in both cases.

(i)

Cancel 100 from every term first:

(ii)

Same three drinks, same total volume, different answers: in (i) most of the liquid is the strong 15% one, in (ii) most is the weak 5% one. The weights decide where the average lands.

The three marks are out of different totals (50, 60, 100), so each must first be turned into a percentage of its own total; only then can the percentages be weighted 3 : 4 : 5.

(iii) and (iv) are equal, because multiplying every term by 100 inside or once outside gives the same result. Answer: (iii) and (iv). Her score, using (iv):

Profit on books and on covers:

Total profit ₹2900 on total sales ₹9000:

This is the weighted mean of 30% and 50% with weights 8000 and 1000, so it sits close to 30%: about 32.22% profit.

Guess: 55 km is more than the average, so the new average goes up, but only a little because it is one day out of 21 — perhaps about 45 km.

Total distance in the first 20 days and then over 21 days:

The salt comes only from the first solution and the sugar only from the second, but each is spread through the whole 900 mL.

Answer: (v). Both concentrations fall, because each substance is diluted by the other solution.

New concentration:

Spice in the 10 L of pani (this does not change when water is added):

Let \(w\) litres of water be added. Then

Add \(\tfrac{10}{3}\) litres (about 3 L 333 mL) of water. Check: \(0.8 \div 13.33 = 0.06 = 6\%\).

(i) Compare them part by part. Flexibility is equal (65). Keerthi is 5 marks lower in strength, which has weight 4, and 5 marks higher in agility, which has the bigger weight 6. The gain counts more than the loss, so Keerthi’s final mark is higher.

(ii) Total weight \(4 + 5 + 6 = 15\).

The difference is \(\tfrac{10}{15}\), exactly the \(5 \times (6 - 4) = 10\) found in (i), divided by 15.

Step 1: average rating of each metric (a weighted mean of the stars, weighted by the number of customers).

Step 2: combine with weights 6 : 5 : 4 (total 15).

The restaurant’s combined rating is about 3.57 stars.

Without computing: if the numbers were equal, the overall average would be midway, \((16.5 + 13.8) \div 2 = 15.15\) kg. The actual 14.925 kg is below that, closer to the female average, so there are more female langurs.

(i) Let \(x\) = number of males and \(y\) = number of females. The total weight divided by the total number is the overall average, so (a) is correct. Since \(x + y = 60\), the expression in (b) would also be correct if its right-hand side were 14.925. As printed, (b), (c) and (d) equal −14.925, which is impossible (an average of positive weights cannot be negative). (c) and (d) are wrong anyway: they divide by 2 and by \(16.5 + 13.8\) instead of by the number of langurs.

(ii) Put \(y = 60 - x\) into (a):

25 male langurs (and 35 female ones).

(iii) Total female weight \(35 \times 13.8 = 483\) kg. Add the new langur:

(iv) Total male weight \(25 \times 16.5 = 412.5\) kg. Remove the two langurs:

The average does not change, because the two released langurs average \((16.9 + 16.1) \div 2 = 16.5\) kg themselves.

(v) The total drops by 1 kg among the 23 males:

(i) Weighted mean of the salinities, weighted by volume:

(ii) Yes. The salinity of a mixture always lies between the salinities of the two parts. Groundwater (0.01%) lies between purified water (0.001%) and the Dead Sea (34%), so the right mixture can reach it.

Let \(w\) litres of purified water be mixed with 1 litre of Dead Sea water:

About 3777 litres (\(3776\tfrac{2}{3}\) L) of purified water — the Dead Sea is so salty that it takes a lot of fresh water to dilute it.

(iii) No. Both the Dead Sea (34%) and groundwater (0.01%) are saltier than purified water (0.001%). A mixture’s salinity lies between its two parts, so it is at least 0.01%; it can never come down to 0.001%. No quantity of groundwater works. (Trying to solve it gives \(34 + 0.01w = 0.001 + 0.001w\), so \(w\) would have to be negative.)

(i) It is the total number of animal species on the IUCN Red List in 2019, all classes together (mammals, reptiles, birds, insects, amphibians, molluscs, fish and others). In a stacked column, the height of the whole column is the sum of its parts.

(ii) The reptile part (dark red) of the 2016 column runs from about 2,100 to about 1,100 on the scale. Its length is

So there were about 1,000 reptile species (any answer close to 1,000 is a fair reading). Note that the reptile bar does not start at 0, so read its length, not where its top reaches.

(iii) Compare the length of each coloured part in 2007 and in 2019:

Mammals and birds have grown only a little, while reptiles, insects, molluscs and fish have grown a lot (reptiles more than tripled).

(i) One bar for Home and one for Overseas; each bar is split into Test, ODI and T20 in that order. Bar lengths are the totals:

The bowler has taken many more wickets overseas, mostly in Tests.

(ii) One bar per format, each split into Home and Overseas:

Test cricket gives the most wickets; in ODIs most wickets came at home, in Tests and T20s most came overseas. Both charts use the same six numbers, and both have the same grand total \(194 + 292 = 234 + 149 + 103 = 486\).

(i) Any reasonable guess is fine, as long as the three numbers add up to 24 hours. One guess for a school student:

(ii) Julie sits for 10 hours a day, more than anyone else in the table, and lies down for 8 hours. She is likely to have a desk job: an office worker, bank clerk, software engineer, call-centre worker or a driver. (Your classmates may guess differently; any job with long hours of sitting fits.)

(iii) Each bar is one person’s 24 hours, so each part is (hours ÷ 24) × 100%. For example:

Since every person’s day is 24 hours, this chart is both a stacked bar chart and a 100% stacked bar chart.

(i) Run rate = total runs ÷ overs.

(ii) After 19 overs the total is \(19 \times 6 = 114\) runs.

The run rate rises only from 6 to 6.3: the extra 6 runs of the last over are shared over 20 overs, \(6 \div 20 = 0.3\).

A way to work together: each friend reports two numbers — their average and how many prices they collected. The combined average is then the weighted mean, weighted by the counts. (Simply averaging the five averages would be wrong, because they collected different numbers of prices.)

(i) There are \(6 + 4 + 5 + 4 + 4 = 23\) prices.

With the data, each \(a\) times its count is just that friend’s total: 917, 493, 766, 897 and 990.

The average price is about ₹176.65.

(ii) There are 13 nylon and 10 feather prices.

Yes, it is equal. Both expressions are (sum of all 23 prices) ÷ 23, only grouped differently. Check: nylon total 1429, feather total 2634, and

(i)

About ₹115.71 per share.

(ii) Let her buy \(k\) shares.

She bought 10 shares. Check: \((750 + 300) \div 15 = 70\).

Brahmagupta’s rule: weight each depth by the length of its section.

The mean depth is 5 hastas (about 90 inches). A pool 30 hastas long and 5 hastas deep everywhere would hold the same amount of water (for the same width).

New average = (total cost) ÷ (total grams), in ₹k per gram. Before calculating: the new average lies between 15 and the new price, nearer to whichever has more grams.

Notice: (ii) moves further from O than (i) because more gold is bought at 30k; (iv) lands very close to 10 because 10 g at 10k swamps the 1 g at 15k; (vi) stays exactly at O, since buying at the current average price never changes the average.

Experiment: values 60, 80 with weights 1, 3:

Doubled weights 2, 6:

Observation: the weighted average does not change.

Algebra: with weights \(2w_1, 2w_2, \ldots, 2w_n\),

The 2 cancels, leaving the original weighted mean. The same is true for multiplying all weights by any positive number: only the ratio of the weights matters (which is why ratios like 3 : 2 : 5 are enough).

(i) The 2003 column reaches about 10,000 objects. Double that is about 20,000, which is first reached by the column for 2021.

(ii) The 2025 column is about 33,000 tall; its payload (light green) part reaches about 16,000.

So in 2025, roughly half the objects were payloads and half were other objects.

(iii) Numbers: both keep rising (the chart is cumulative). Payload objects grew slowly for decades (about 2,000 in 2000, 4,000 in 2019) and then shot up after about 2019. Other objects (debris, rocket stages) grew steadily, with a sudden jump around 2007, and are now growing more slowly.

Shares: for most of the history, other objects made up the large majority (about 75–80%). Since about 2020 the share of payload objects has risen sharply, from about one-fifth to about one-half, so the share of other objects has fallen to about one-half.

The map gives percentages only: Haryana 90%, Uttarakhand 78%, Arunachal Pradesh 68%, Punjab 98%, Dadra and Nagar Haveli and Daman and Diu 99.8%, Maharashtra 93%, Telangana 74%.

Answer: (b) and (d).

(ii) No. The nation-wide percentage is the weighted mean of the state percentages, weighted by the number of schools in each state:

So we also need the number of schools in each state/UT. A simple average of the percentages would be wrong, because a state with many schools should count for more.

Approximate readings from the two charts:

(i) Plays A and B look almost the same: about 72% of viewers gave each of them 4 or 5 stars. Play C has the biggest 5-star share but also about one-third 1-star ratings — people either loved it or hated it. A safe choice is A or B (their weighted average ratings are about 3.75 and 3.71 stars, against about 3.14 for C).

(ii) The stacked bar chart shows the actual number of ratings: B has about 925, A only about 210. A 72% share of 925 ratings is far more reliable than the same share of 210. So between A and B, B is the better choice; the 100% chart hid how many people had rated each play. (Any reasoned answer is acceptable — e.g. a viewer who likes daring plays might still pick C.)

Total times, in minutes:

(i) A stacked bar chart. The race is decided by the total time, which is the length of a stacked bar, and we can still see each segment’s time. A 100% chart would make all three bars equal and hide who was fastest.

(ii) A 100% chart shows only each athlete’s fractions of their own time.

Answer: only (b). (From the table, Athlete 1 finished first.)

Approximate shares read from the chart:

Things to notice: seeing and movement disabilities make up a much bigger share in older age groups (movement nearly doubles, about 13% to 25%); speech disability and mental retardation shrink from the young to the old groups; hearing stays at about one-fifth in every group; ‘any other’ is a large share for the under-60 groups.

Inferences: ageing affects eyesight and mobility, so care for the elderly should focus on these. Disabilities present from birth form a larger share among the young.

Things to wonder: this is a 100% chart, so it shows shares within each age group, not the number of people — we cannot tell which age group has the most disabled persons. A smaller share of mental retardation among the elderly could also mean fewer such people survive to old age, or that they are under-counted.

(i) Make a table with one row per person and three columns (lying, sitting, standing/moving). Check that each row adds to 24 hours. Draw one bar per person, each 24 units long, split into the three parts in the same order and colours, like Exercise Set 10.5. Because every bar totals 24 h, it is also a 100% chart.

(ii) Choose 5–7 categories (e.g. food, housing/rent, education, transport, health, others). For each month find the total, then each category’s share:

Draw one bar per month, each the same length (100%), split by these shares. In your inferences, compare shares across months (e.g. ‘education rose in the month fees were paid’), and remember the chart does not show whether the total went up or down.

Sample (bus travel): aspects punctuality, safety, cleanliness, crowding, fare, with weights 3 : 4 : 2 : 2 : 1 (safety matters most). One traveller rates them 4, 5, 3, 2, 4:

Compute this for all 10 people and take the ordinary average of the 10 results (each person counts equally). For the 100% chart, draw one bar per aspect showing the share of 5★, 4★, …, 1★ ratings it received. In your note, mention what the weights hide: e.g. a very unsafe bus can still score 3 if it is cheap and punctual.

Each family’s average must be weighted by its number of members (a family of 6 has more people than a family of 3). If family \(i\) has \(n_i\) members with average age \(a_i\):

Each \(n_ia_i\) is that family’s total age, so the class simply adds all the totals and divides by the total number of people. Example: families (4 members, 30 years), (6, 35), (3, 28):

(The simple average of 30, 35, 28 is 31, which is wrong.) If two students are siblings, count their family only once.

Experiment: values 60, 80 with weights 1, 3 give 75 (as in Q6). Add 1 to each weight, giving 2 and 4:

The average changes: it moves from 75 towards the simple mean \((60 + 80) \div 2 = 70\). Adding 1 makes the weights ‘more equal’ (1 : 3 became 2 : 4 = 1 : 2).

Algebra: let \(W = w_1 + \cdots + w_n\), \(S = w_1x_1 + \cdots + w_nx_n\), so the old mean is \(\bar x_w = S \div W\). Adding \(c\) to every weight:

Write \(x_1 + \cdots + x_n = n\bar x\), where \(\bar x\) is the simple mean:

This is a weighted mean of the old weighted mean \(\bar x_w\) and the simple mean \(\bar x\), with weights \(W\) and \(nc\). So the new average lies between the two, and moves closer to the simple mean as \(c\) grows. It stays the same only when \(\bar x_w = \bar x\) (for example when all the weights were already equal).

The three shares always add up to 100%. So they cannot all go down, and they cannot all go up: (i) and (ii) are impossible. Check the others with examples, starting from 60 cows, 25 sheep, 15 chickens (100 animals), with every number decreasing:

Every animal count went down in each example, and the shares changed as stated. Possible: (iii), (iv) and (v). Not possible: (i) and (ii).