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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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