Verified step-by-step solutions for your class notes.
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?
Answer: The combined average is calculated using the weighted mean formula:
Combined Average = $\frac{(72 \times 30) + (76 \times 25)}{30 + 25}$
= $\frac{2160 + 1900}{55} = \frac{4060}{55} \approx 73.81$
A farmer mixes three equal quantities of fertilisers. The first one contains $\frac{1}{10}$ nitrogen, the second contains $\frac{9}{50}$ nitrogen, and the third contains $\frac{3}{60}$ nitrogen. What is the fraction of nitrogen in the mixture?
Answer: Since the quantities are equal, we find the simple average of their concentrations:
Average = $\frac{\frac{1}{10} + \frac{9}{50} + \frac{3}{60}}{3} = \frac{\frac{10}{100} + \frac{18}{100} + \frac{5}{100}}{3} = \frac{\frac{33}{100}}{3} = \frac{11}{100}$ (or 11%).
Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varṇa, 5 units at 10 varṇa, and 17 units at 11 varṇa. Find the purity in varṇa of the combined gold.
Answer: Using the weighted mean:
Purity = $\frac{(9 \times 12) + (5 \times 10) + (17 \times 11)}{9 + 5 + 17}$
= $\frac{108 + 50 + 187}{31} = \frac{345}{31} \approx 11.13$ varṇa.
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.
Answer: The months of May, June, and July have 31, 30, and 31 days respectively.
Expression: $\frac{(3.5 \times 31) + (10 \times 30) + (8.7 \times 31)}{31 + 30 + 31}$
Calculate the concentration of spice mix in these two scenarios:
(i) A 100 mL kashayam with 5% spice, 200 mL with 10% spice, and 300 mL with 15% spice.
(ii) A 300 mL kashayam with 5% spice, 200 mL with 10% spice, and 100 mL with 15% spice.
Answer:
(i) Concentration = $\frac{(100 \times 5) + (200 \times 10) + (300 \times 15)}{100 + 200 + 300} = \frac{500 + 2000 + 4500}{600} = \frac{7000}{600} \approx 11.67\%$
(ii) Concentration = $\frac{(300 \times 5) + (200 \times 10) + (100 \times 15)}{300 + 200 + 100} = \frac{1500 + 2000 + 1500}{600} = \frac{5000}{600} \approx 8.33\%$
Savitri's marks in Kashmiri are 35/50 in internals, 44/60 in project, and 80/100 in the final exam. The ratio is 3:4:5. Which of the following expression(s) gives her annual score (as a percentage)?
Answer: Expressions (iii) and (iv) are correct. Both correctly convert the individual scores into fractions/percentages of their respective maximum marks before applying the 3:4:5 weights.
A stationery shop owner made ₹8000 selling books (30% profit) and ₹1000 selling book covers (50% profit). What is the percentage of profit on the total sales?
Answer: Total sales = ₹8000 + ₹1000 = ₹9000.
Total profit = (0.30 × 8000) + (0.50 × 1000) = 2400 + 500 = ₹2900.
Percentage profit = $\frac{2900}{9000} \times 100 \approx 32.22\%$
A white stork's average daily distance travelled 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?
Answer: Total distance for 20 days = $44.5 \times 20 = 890$ km.
Total distance for 21 days = 890 + 55 = 945 km.
New average = $\frac{945}{21} = 45$ km/day.
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?
Answer: Total volume = 600 + 300 = 900 mL.
Salt concentration = $\frac{600 \times 5\%}{900} = \frac{3000}{900} = 3.33\%$
Sugar concentration = $\frac{300 \times 8\%}{900} = \frac{2400}{900} = 2.67\%$
Correct Option: (v) Salt: 3.33%, Sugar: 2.67%
At a panipuri stall, the spice concentration is 8%. What quantity of regular water should be mixed into 10 litres of pani so that the spice level is reduced to (3/4)th of the original concentration?
Answer: Target concentration = $\frac{3}{4} \times 8\% = 6\%$.
Let $x$ be the amount of water added. The amount of spice remains constant: $10 \times 0.08 = 0.8$ litres.
$\frac{0.8}{10 + x} = 0.06 \implies 10 + x = \frac{0.8}{0.06} \approx 13.33$ litres.
Water added ($x$) = $13.33 - 10 = 3.33$ litres.
Physical fitness marks for strength, flexibility, and agility are weighted 4:5:6. Rashi scored 60, 65, 70. Keerthi scored 55, 65, 75.
(i) Find out whose total is more without calculating.
(ii) What are their final marks?
Answer:
(i) Keerthi's total is more because she scored higher (75) in agility, which carries the highest weight (6), compensating for her lower score in strength (weight 4).
(ii) Rashi's marks = $\frac{60(4) + 65(5) + 70(6)}{15} = \frac{985}{15} \approx 65.67$
Keerthi's marks = $\frac{55(4) + 65(5) + 75(6)}{15} = \frac{995}{15} \approx 66.33$
A restaurant collected ratings from 10 customers... What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?
| Metric | 5★ | 4★ | 3★ | 2★ | 1★ |
|---|---|---|---|---|---|
| Food | 5 | 3 | 2 | 0 | 0 |
| Ambience | 0 | 4 | 5 | 1 | 0 |
| Service | 1 | 2 | 2 | 4 | 1 |
Answer: First, find the average rating for each category from the 10 customers:
Food avg = $\frac{(5 \times 5) + (4 \times 0) + (3 \times 3) + (2 \times 2) + (1 \times 0)}{10} = 3.8$
Ambience avg = $\frac{(5 \times 0) + (4 \times 4) + (3 \times 5) + (2 \times 1) + (1 \times 0)}{10} = 3.3$
Service avg = $\frac{(5 \times 4) + (4 \times 1) + (3 \times 2) + (2 \times 2) + (1 \times 1)}{10} = 3.5$
Weighted Overall Average = $\frac{(3.8 \times 6) + (3.3 \times 5) + (3.5 \times 4)}{6 + 5 + 4} = \frac{53.3}{15} \approx 3.55$
Using the table of langur weights (Male avg=16.5, Female avg=13.8, Total avg=14.925, Total count=60):
(i) Which expression(s) describes the scenario?
(ii) Find out how many male langurs are present.
(iii) A female langur weighing 15.2 kg is admitted. New average weight of female langurs?
(iv) Two male langurs (16.9 kg and 16.1 kg) are released. New average weight of male langurs?
(v) One male langur loses 1 kg. New average weight of male langurs?
| Category | Male langurs | Female langurs | All langurs |
|---|---|---|---|
| Average weight | 16.5 kg | 13.8 kg | 14.925 kg |
| Number of langurs | $x$ | $y$ | 60 |
Answer:
(i) Expressions (a) $\frac{16.5x + 13.8y}{x + y} = 14.925$ and (b) $\frac{16.5x + 13.8y}{60} = 14.925$ (ignoring the typo negative sign in the textbook) correctly describe the scenario.
(ii) $16.5x + 13.8(60 - x) = 14.925 \times 60 \implies 2.7x = 67.5 \implies x = 25$. There are 25 male langurs (and 35 female langurs).
(iii) New female avg = $\frac{(13.8 \times 35) + 15.2}{36} = \frac{498.2}{36} \approx 13.84$ kg.
(iv) New male avg = $\frac{(16.5 \times 25) - (16.9 + 16.1)}{23} = \frac{412.5 - 33}{23} = \frac{379.5}{23} = 16.5$ kg.
(v) New male avg = $\frac{379.5 - 1}{23} = \frac{378.5}{23} \approx 16.46$ kg.
Using the salinity table (Dead Sea ~34%, Purified ~0.001%, Ground ~0.01%):
(i) Mix 1L Dead Sea with 2L Purified. Salinity?
(ii) Mix Dead Sea and Purified to get Groundwater? Quantity?
(iii) Mix Dead Sea and Ground to get Purified?
| Water Source | Salinity |
|---|---|
| Dead Sea | $\approx 34\%$ |
| Other seas and oceans | $\approx 3.5\%$ |
| Ground water | $\approx 0.01\%$ |
| Purified drinking water | $\approx 0.001\%$ |
Answer:
(i) Salinity = $\frac{(1 \times 34) + (2 \times 0.001)}{3} \approx 11.334\%$
(ii) Yes, it is possible because the target (0.01%) lies between 34% and 0.001%.
$0.01 = \frac{34(1) + 0.001(x)}{1 + x} \implies 0.01 + 0.01x = 34 + 0.001x \implies 0.009x = 33.99 \implies x = 3776.67$ Litres of purified water needed.
(iii) No, it is impossible because both Dead Sea (34%) and Groundwater (0.01%) have a higher salinity than the target Purified water (0.001%). You cannot lower concentration below your least concentrated ingredient.
Based on the IUCN Red List stacked column chart:
(i) What does 14,234 represent?
(ii) Approx how many reptile species in 2016?
(iii) Which class(es) had a small increase between 2007 and 2019?
Answer:
(i) It represents the total cumulative number of endangered animal species (across all listed classes) in the year 2019.
(ii) Looking at the maroon bar segment for 2016, it spans roughly from 1,000 to 2,000 on the y-axis, indicating approximately 1,000 reptile species.
(iii) 'Mammals' (bottom dark blue bar) and 'Birds' (light blue bar) show relatively small increases over that time period.
Wickets taken in International Cricket. Complete the given stacked bar charts (the bar lengths can be approximate).
Answer (i) Comparing total wickets taken at home vs. overseas:
Answer (ii) Total wickets taken in each format:
The smart watch data of 5 people...
(i) How much time does Sahana spend per day? (Guess)
(ii) Guess what activity Julie could be engaged in.
(iii) Complete the chart.
Answer:
(i) Total hours must be 24. A student might spend: Lying down = 8h, Sitting = 10h, Standing = 6h.
(ii) The bar for Julie shows ~33% lying down (8h), ~42% sitting (10h), and ~25% standing (6h). This pattern fits an office worker, a student, or a software engineer.
(iii) Chart completion representing the body states:
In cricket, run rate = average runs per over. Over 1: 6 runs.
(i) Over 2: scored 12 runs. What is run rate now?
(ii) After Over 19, run rate was 6. Over 20: 12 runs. What is the run rate after 20 overs?
Answer:
(i) Total runs = $6 + 12 = 18$. Overs = 2. Run rate = $\frac{18}{2} = 9$.
(ii) Runs after 19 overs = $19 \times 6 = 114$. Total runs = $114 + 12 = 126$. Run rate = $\frac{126}{20} = 6.3$.
Five friends collected shuttlecock prices across types (Nylon and Feather).
(i) If $a_y, a_s, a_k, a_p, a_g$ are their individual averages, write an expression for the combined average.
(ii) Will the average of nylon ($a_n$) and feather ($a_f$) prices give the same combined average?
Answer:
(i) Using weighted mean based on counts ($c$): $\frac{(a_y c_y + a_s c_s + a_k c_k + a_p c_p + a_g c_g)}{c_y + c_s + c_k + c_p + c_g}$
(ii) The simple average of averages $\frac{a_n + a_f}{2}$ will not be equal to the true average price. The true average must be weighted by the number of nylon vs. feather shuttlecocks recorded overall.
Shreyas holds 25 shares at ₹150 avg. Vaishnavi holds 5 shares at ₹150 avg.
(i) Shreyas buys 10 shares at ₹30. New average?
(ii) Vaishnavi buys $x$ shares at ₹30, new average is ₹70. Find $x$.
Answer:
(i) Shreyas's new avg = $\frac{(25 \times 150) + (10 \times 30)}{25 + 10} = \frac{3750 + 300}{35} = \frac{4050}{35} \approx 115.71$ rupees.
(ii) $\frac{(5 \times 150) + (x \times 30)}{5 + x} = 70 \implies 750 + 30x = 350 + 70x \implies 400 = 40x \implies x = 10$. She bought 10 shares.
A pool 30 hastas long is dug. Length sections: 4, 5, 6, 7, 8. Depths: 9, 7, 7, 3, 2. Find the mean depth.
Answer: Mean depth = $\frac{(4 \times 9) + (5 \times 7) + (6 \times 7) + (7 \times 3) + (8 \times 2)}{30} = \frac{36 + 35 + 42 + 21 + 16}{30} = \frac{150}{30} = 5$ hastas.
Suvarna has 1g gold at ₹15k. Find her new average price if she:
(i) Buys 1g at ₹30k
(ii) Buys 2g at ₹30k
(iii) Buys 1g at ₹10k
(iv) Buys 10g at ₹10k
(v) Buys 0.5g at ₹30k
(vi) Buys 0.5g at ₹15k
Answer:
(i) $\frac{15 + 30}{2} = 22.5$k
(ii) $\frac{15 + 60}{3} = 25$k
(iii) $\frac{15 + 10}{2} = 12.5$k
(iv) $\frac{15 + 100}{11} = 10.45$k
(v) $\frac{15 + 15}{1.5} = 20$k
(vi) $\frac{15 + 7.5}{1.5} = 15$k
Given some data with weights, how would the weighted average change if all weights are doubled?
Answer: It does not change. Algebraically, $\frac{2w_1x_1 + 2w_2x_2 + ...}{2w_1 + 2w_2 + ...} = \frac{2(w_1x_1 + ...)}{2(w_1 + ...)}$. The 2s cancel out, leaving the original weighted average.
Based on the Space Objects graph:
(i) Objects in 2003? Year it doubled?
(ii) Payload vs Other objects in 2025?
(iii) Trends of Payload vs Other objects?
Answer:
(i) In 2003, there were roughly 10,000 objects. It doubled (to 20,000) around the year 2019.
(ii) In 2025, total objects ~35,000. Payload ~18,000; Other objects ~17,000.
(iii) Payload objects show a recent, massive exponential spike (due to satellite constellations), whereas 'other objects' (debris) have grown steadily over time.
Based on the 'Schools with a Playground' map:
(i) Identify correct inferences: (a) Haryana > Uttarakhand, (b) Arunachal ~2/3, (c) Punjab highest number, (d) If Maharashtra has more schools than Telangana, it has more playground schools.
(ii) Can we find the nation-wide percentage from this?
Answer:
(i) Statements (a), (b), and (d) are correct inferences. Statement (c) is false because Punjab has the highest percentage (98%), not necessarily the highest absolute number of schools.
(ii) No. To find the national average, we require the total number of schools in each state to calculate the properly weighted mean.
Decision Dilemma: Which play to watch based on the 100% stacked bar vs. the standard stacked bar?
Answer: Looking only at the 100% stacked bar chart, Play A seems best because it has the largest proportion of 5-star ratings. However, looking at the standard stacked bar chart reveals that Play B has a massively larger total volume of reviews compared to Play A. This large sample size makes Play B's high rating far more reliable, which could change one's decision.
Triathlon finish times (Swimming, Cycling, Running):
(i) Suitable representation: stacked or 100% stacked?
(ii) If a 100% stacked bar is drawn, which questions can be answered?
| Athlete | Swimming | Cycling | Running | Total Time |
|---|---|---|---|---|
| Athlete 1 | 01:08 | 05:00 | 03:15 | 09:23 |
| Athlete 2 | 01:05 | 05:10 | 03:35 | 09:50 |
| Athlete 3 | 01:22 | 05:55 | 03:50 | 11:07 |
Answer:
(i) A standard stacked bar chart is much more suitable because the total length (overall finish time) dictates the winner of the race.
(ii) A 100% stacked bar chart only shows proportions, so it can ONLY answer question (b): "Approximately what fraction of their race time did Athlete 1 spend cycling?"
What would happen to the weighted average if all the weights are increased by a constant value, say 1?
Answer: The weighted average will change. Adding a constant flattens the relative differences between the heavy and light weights, pulling the final weighted average closer to the simple arithmetic mean of the values.
A farm has 60% cows, 25% sheep, 15% chickens. All animal populations decreased. Choose the possibilities for the change in their respective shares (%) of the population.
Answer: Because the percentages must always sum to 100%, it is impossible for all shares to increase or all shares to decrease simultaneously. However, depending on the varying rates of decline, some shares can increase while others decrease, or they can all remain identical.
Therefore, possibilities are:
(iii) % of all remain the same (if all decrease at the exact same rate)
(iv) % of cows decreased, % of sheep increased, % of chickens remained the same
(v) % of cows increased, % of sheep increased, % of chickens decreased.
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