How Quantities Combine: Understanding Data

Each weekly mean represents the same number of observations—5 days. Therefore, the weeks have equal weight.

If the numbers of days differed, the correct expression would be (n₁a₁ + n₂a₂ + n₃a₃) ÷ (n₁ + n₂ + n₃).

Multiply each mean by its number of students, add the total marks, then divide by the total number of students.

Since the amounts mixed are equal, take the ordinary average of the three fractions.

Varṇa purity measures the amount of gold on a 16-part scale, so the same weighted-mean rule applies.

Combined purity = 11.13 varṇa (approximately).

May has 31 days, June has 30 and July has 31. We must weight the monthly averages by days, rather than treating the months as equal.

Required expression: (31×3.5 + 30×10 + 31×8.7)/92.

(i) Spice = 100×0.05 + 200×0.10 + 300×0.15 = 5 + 20 + 45 = 70 mL-equivalent.

(ii) Spice = 300×0.05 + 200×0.10 + 100×0.15 = 15 + 20 + 15 = 50.

The same three percentages give different results because the quantities attached to each percentage change.

First convert the marks to proportions: 35/50 = 70%; 44/60 = 73⅓%; 80/100 = 80%.

Correct choices: (iii) and (iv). Both properly convert the different maximum marks to a common 100-point scale. (i) treats raw marks as comparable; (ii) incorrectly divides raw marks by 100.

Profit on books = 8,000×30% = ₹2,400. Profit on covers = 1,000×50% = ₹500.

This uses the question’s stated percentages of sales, not the conventional profit-on-cost formula.

The earlier distance total is 44.5×20 = 890 km. Including day 21, total = 945 km.

A quick prediction: the new average must be slightly above 44.5, because day 21 is higher than the earlier mean.

Salt = 600×5% = 30 mL-equivalent; sugar = 300×8% = 24 mL-equivalent. Total volume = 900 mL.

Correct choice: (v). Do not add 5% and 8%: they describe two different substances in different volumes.

Target = (3/4)×8% = 6%. The quantity of spice stays constant: 10×8% = 0.8 L-equivalent.

Let x litres of water be added.

(i) Keerthi loses 5 points in strength (weight 4) but gains 5 in agility (weight 6). Net weighted gain = 5×6 − 5×4 = 10; flexibility is the same. Hence Keerthi wins.

(ii) Rashi = (60×4+65×5+70×6)/15 = 985/15 ≈ 65.67.

Keerthi = (55×4+65×5+75×6)/15 = 995/15 ≈ 66.33.

Mean food rating = (5×5+4×3+3×2)/10 = 4.3.

Mean ambience rating = (4×4+3×5+2×1)/10 = 3.3.

Mean service rating = (5×1+4×2+3×2+2×4+1×1)/10 = 2.8.

Let x be males and y be females. Then x+y=60 and

Using y=60−x:

Without calculating: the overall average, 14.925 kg, is closer to the female average (13.8 kg) than to the male average (16.5 kg), so there must be more female langurs.

(i) As printed, option (a) is correct. The other options, as printed in the textbook, show a negative sign before 14.925 or an incorrect denominator. If option (b) was intended to show positive 14.925, it would also be equivalent because x+y=60; the printed sign matters.

(ii) 25 males and 35 females, so there are more females.

(iii) New female mean: 35×13.8 = 483 kg before admission. (483+15.2)/36 ≈ 13.84 kg.

(iv) New male mean: 25×16.5=412.5 kg before release; mass removed=16.9+16.1=33 kg, giving 379.5 kg for 23 males. Thus 379.5/23 = 16.5 kg (unchanged).

(v) After a further 1 kg loss: 378.5/23 ≈ 16.46 kg (approximately). Here “now” is taken to refer to the 23 males remaining after part (iv).

(i) Salinity = [1×34% + 2×0.001%]/3 = 34.002%/3 = 11.334%.

(ii) Yes, mathematically, because 0.01% lies between 34% and 0.001%. Let x L of purified water be added:

This is an extremely large amount of water because the target salinity is so low.

(iii) Impossible using only these two sources. Both Dead Sea water (34%) and groundwater (0.01%) are saltier than purified water (0.001%). A mixture of two concentrations cannot be below both of them.

1. Family C spent least on housing: ₹950.
2. Family A spent most on healthcare: ₹1,770.
3. Family A spent ₹1,280 on education (approximately ₹1,300 from the graph).
4. Family B’s largest spending category is housing: ₹1,750.
5. Family A spent more on healthcare (₹1,770) than food (₹1,700), a difference of ₹70.

Grouping bars by category is best for comparing families; grouping them by family is better for comparing spending categories within a family.

(6) Family C spent the least overall. (7) Its total is ₹7,040 (about ₹7,000). (8) Housing share = 950/7040×100 ≈ 13.49% (about 13.5%).

A stacked bar makes totals easy to compare; a 100% stacked bar is ideal for comparing percentage shares.

Compute each family’s healthcare expense as a fraction of its own total:

• A: 1770/8425×100 ≈ 21.0%.
• B: 1210/7286×100 ≈ 16.6%.
• C: 1300/7040×100 ≈ 18.5%.

Answer: Family B (approximately 16.6%).

(i) 14,234 is the total number of animal species across all the displayed classes represented in the 2019 stacked column, not a count of one class.

(ii) The reptile segment in 2016 represents about 1,100 species, estimated from the thickness of the reptile band against the vertical scale.

(iii) Mammals and birds show comparatively small rises compared with groups such as fish, molluscs and reptiles. Amphibians also increase more moderately than some fast-growing categories.

Home = 62+100+32 = 194 wickets. Overseas = 172+49+71 = 292 wickets.

Both bars use the same scale, with 300 wickets represented by the full track length.

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

Only (i) can be concluded. Within Fatima’s own garden, summer’s 50% is greater than monsoon’s 30%. The chart gives Naveen 40% in summer, 25% in monsoon and 35% in winter. A later sentence in the book swaps his monsoon and winter shares. The figure and both worked cases use 40%, 25% and 35%. Case 1 is Fatima 105, 63 and 42 out of 210, and Naveen 48, 30 and 42 out of 120. Case 2 is Fatima 90, 54 and 36 out of 180, and Naveen 96, 60 and 84 out of 240.

The 100% bars do not provide the total number of flowers in either garden. Hence no direct comparison of absolute summer or winter counts between gardens is justified; nor can their overall totals be declared equal.

The overall bar divides 24 hours among sleep, personal care, paid work, unpaid work/care, learning and leisure/travel. But a single national average mixes children, students, workers and retired people, whose daily patterns differ considerably.

The age-group graph is more informative: children allocate much more time to learning than working-age adults; paid work and unpaid care become more prominent in adulthood; sleep and leisure shares vary with age.

Lesson: an overall average can hide important differences between subgroups. For comparison, all these bars total 24 hours, so they are both actual stacked bars and 100% stacked bars.

The chart is an average over many people. For example, 34% sleeping of 24 hours is 0.34×24 = 8.16 hours. A 13% category corresponds to 3.12 hours; a 10% category corresponds to 2.4 hours.

The fractions add to roughly 100% (subject to rounding). A child, student, employed adult and retiree can have quite different activity patterns; their combined average is not an instruction for anyone’s day.

One possible answer is lying down 8 h, sitting 9 h, standing/moving 7 h. Check: 8+9+7 = 24 hours.

This is a sample estimate; many other reasonable combinations are acceptable if the three categories total 24 hours.

One reasonable guess is office worker, software developer, student or desk-based professional, because a large part of her time is spent sitting. The table alone cannot identify her occupation uniquely.

Each percentage = hours ÷ 24 × 100.

(i) Total runs = 6+12 = 18; overs = 2.

(ii) Runs after 19 overs = 19×6 = 114. New total = 114+12 = 126.

There are 6+4+5+4+4 = 23 observations.

(i) Combined mean by friend:

(ii) There are 13 nylon prices (sum ₹1,429) and 10 feather prices (sum ₹2,634):

Here a_n = 1429/13 ≈ ₹109.92 and a_f = 2634/10 = ₹263.40.

Yes, both expressions are equal, because they add the same 23 prices. Grouping by friend or by product type must not change the grand total.

(i) Existing amount = 25×150=₹3,750. New amount = 10×30=₹300.

(ii) Let x be shares purchased:

These are averages of purchase prices; they do not show current market values.

Use section lengths as the weights.

The textbook describes one hasta as approximately 18 inches, so this is about 90 inches (7.5 feet).

Always start afresh with the original 1 g at ₹15k; the six purchases are separate scenarios, not a sequence of purchases.

On the textbook number line (in thousands), mark the results at 22.5, 25, 12.5, about 10.45, 20 and 15.

Original mean = (w₁x₁ + w₂x₂ + ⋯ + wₙxₙ)/(w₁+w₂+⋯+wₙ).

Doubling all weights gives

No change. Only the relative sizes of weights affect the mean; multiplying all by the same positive factor leaves every relative weight unchanged.

(i) A reasonable graph reading is about 10,000 objects at the 2003 point, with roughly 20,000 around 2021. The exact year is approximate because the bars are not individually labelled.

(ii) For 2025 the stacked total is roughly 33,000. Approximate composition: 17,000 payload objects (about 52%) and 16,000 other objects (about 48%). Figures read from the image can vary modestly by eye.

(iii) Both the cumulative payload count and the other-object count rise, with payloads expanding especially rapidly in recent years. Thus the share of payloads can increase even while the number of non-payload objects also increases. Pay attention to the chart title: its bars count objects still in orbit as of the stated 2026 date, not necessarily the number that was orbiting at the end of each plotted historical year.

(i) Correct statements: (b) and (d).

• (a) Cannot be concluded: Haryana’s 90% exceeds Uttarakhand’s 78%, but higher percentage need not imply more schools in absolute count.
• (b) True: Arunachal Pradesh is about 68%, close to two of every three schools.
• (c) Cannot be concluded: A high share for Punjab does not prove it has India’s highest number of playground-equipped schools.
• (d) True: Maharashtra has about 93% versus Telangana’s 74%. If Maharashtra also has more schools in total, multiplying a greater count by a greater positive fraction ensures a greater count of schools with playgrounds.

(ii) No—not from the state percentages alone. We must also know how many schools each state/UT contains (and ensure the same reporting coverage and period).

Using an unweighted average of state percentages would be invalid.

Approximate readings from the two charts:

(i) From the 100% chart, Plays A and B look almost the same: about 72% of their ratings are 4 or 5 stars. Play A has a slightly bigger 5-star share. Play C has the highest share of 5-star ratings (about 40%) but also about one-third 1-star ratings, so opinions on it are sharply divided. On this chart alone, Play A (or B, nearly a tie) is a sensible choice; Play C is a gamble.

(ii) The stacked chart shows how many people rated each play: about 210 for A, 925 for B and 380 for C. I would choose Play B. Its favourable share is based on more than four times as many ratings as A’s. B also has more five-star ratings than C in absolute numbers: about 220 for B (24% of 925), compared with 150 for C (40% of 380).

This is an open-ended question: choosing Play A or C with an evidence-based reason is also acceptable. The core lesson is that rating shares and numbers of raters tell different stories; a 100% stacked bar suppresses sample-size information.

Convert all durations to minutes before adding.

(i) An ordinary stacked bar chart is better because total finish time determines the winner. Athlete 1 is fastest.

(ii) From the 100% bars alone, only (b) can be answered:

Without total durations, a percentage-only chart cannot tell who finished first or whose absolute running time was longest.

Approximate readings from the chart (share of each age group's disabled persons):

Possible observations: The seeing share falls from 18% to 15% before rising to 25% in the oldest group. The movement share increases across the four age groups. The shares of speech and intellectual disabilities shrink with age. Hearing is about the same share (18–20%) in every age group. Multiple disability has its largest share in the 60+ group.

What cannot be concluded: Because every bar totals 100%, the graph does not tell how many people in each age group have disabilities or the overall disability prevalence in an age group. It also cannot establish the causes of age differences.

Questions to explore: How large is each age group? Would counts tell a different story? How do diagnosis, access to assistive devices, or reporting practices affect the patterns?

Illustrative invented data for one person across three days (replace with your own diary):

Record each day in non-overlapping 30-minute intervals, combine similar activities, and plot each day as a stacked bar with three segments. Example inference: sitting time falls on the weekend while movement time rises.

These data are examples, not actual observations.

Illustrative data only (₹):

Inference: Housing remains ₹10,000 each month but its share falls from 50% to 40% in month 2 because the total rises. Collect real data for the assigned project.

Illustrative scenario: bus travel. Four aspects—Comfort (weight 4), Cleanliness (3), Punctuality (2), Accessibility (1)—are each rated from 1 to 5.

Comfort carries the highest illustrative weight because seating and ride quality matter on a long journey. Cleanliness is next, followed by punctuality and accessibility. These weights are debatable; a group could reasonably weight accessibility much higher.

Overall mean = (4.1+4.4+3.2+4.3+2.5+4.8+3.7+3.3+4.0+4.6)/10 = 38.9/10 = 3.89/5.

For the required 100% stacked chart, construct one bar per aspect and show how many of the 10 raters gave 1, 2, 3, 4 or 5 stars; each rater contributes 10% of that bar. The dataset above produces:

Limitations: A single overall score hides which specific criteria scored poorly; the chosen weights are subjective and 10 respondents are too few for broad conclusions. The textbook allows either collected or imagined ratings. State which your group used.

Let the families have respective mean ages a₁, a₂, … and sizes n₁, n₂, … . The sum of ages in family 1 is n₁a₁, and so on.

Example: A family of 4 averages 30 years and a family of 6 averages 25 years; their combined mean = (4×30+6×25)/10 = 27 years. Simply averaging 30 and 25 would be wrong.

Let W = Σwᵢ be the total original weight, M = Σ(wᵢxᵢ)/W the old weighted mean, and A = (Σxᵢ)/n the ordinary mean of n numbers.

After adding 1 to all n weights:

Thus M′ is a weighted average of the old weighted mean M and the unweighted mean A. When weights are nonnegative and W>0, it moves toward the ordinary mean. More generally, adding the same positive constant c to each weight gives M′ = (WM + ncA)/(W + nc).

• If A>M, the new weighted mean increases.
• If A<M, it decreases.
• If A=M, it does not change.

Example: values 10 and 20 have weights 1 and 3. Old mean = (10+60)/4 = 17.5. Add 1 to each weight: (2×10+4×20)/6 ≈ 16.67. The new mean moves toward the ordinary mean 15. (It does not always increase.)

Correct choices: (iii), (iv) and (v). All three percentages must always sum to 100%, so (i) all decreasing and (ii) all increasing are impossible.

To verify the possible options, start with a hypothetical farm of 60 cows, 25 sheep and 15 chickens (100 animals):

In each example, every absolute animal count is smaller than initially, yet some percentages rise because the denominator (the farm’s total population) has also fallen. This is why counts and shares are different concepts.