Two Awareness Teams: A Tripling Model against a Steady Addition
During a dengue-awareness drive in a block, two teams work for six days. Team A's volunteers inform 5 households on day 1, and the block officer's model says the number of newly informed households triples every day after that, because each informed household is expected to tell two more. Team B, working from a prepared list, informs 40 households on day 1 and manages exactly 30 more households each day than on the day before. The block officer must send daily figures to the district and recommend which team should receive the second week's funds. On day 4 she notices that Team A's daily figure has just overtaken Team B's for the first time, although Team B's running total is still far ahead and its volunteers are exhausted. Funds are released at ₹20 per household informed in total over the six days, and no team may be credited with more than 1000 households over the six days without a written audit. Team B asks what daily increment would let it match Team A on day 6.
1. On which day does Team A's daily figure first exceed Team B's, and by how much? Explain why this cannot be settled by simply solving an equation.
Read the worked answer
Model the daily figures: Team A is a GP with a = 5, r = 3, so day n gives 5·3^(n−1). Team B is an AP with a = 40, d = 30, so day n gives 40 + 30(n − 1).
Test the days in order, because 5·3^(n−1) = 40 + 30(n − 1) is not an equation that school algebra can solve for n:
- n = 1: 5 vs 40
- n = 2: 15 vs 70
- n = 3: 45 vs 100 (gap 55)
- n = 4: 135 vs 130 (gap −5)
So the first crossing is on day 4, with Team A ahead by 5 households. The judgement point is that the GP's nth term grows geometrically, so the comparison has to be checked on successive integers; the crossover is found by testing, and the gap then widens extremely fast (day 5: 405 vs 160).
2. Compute both six-day totals, the funds due to each team, and decide whether the audit rule is triggered.
Read the worked answer
Team A, GP with a = 5, r = 3, n = 6: S = 5(3⁶ − 1)/(3 − 1) = 5(729 − 1)/2 = 5 × 364 = 1820 households.
Team B, AP with a = 40, d = 30, n = 6: S = (6/2)[2(40) + 5(30)] = 3(80 + 150) = 3 × 230 = 690 households.
Funds at ₹20 per household: Team A gets 1820 × 20 = ₹36,400; Team B gets 690 × 20 = ₹13,800.
Audit: 1820 > 1000, so Team A's figure cannot be credited without a written audit, whereas 690 is within the limit. A second judgement matters here: the tripling model is a projection, not a count. If Team A's register shows actual households rather than the assumed tripling, the district figure must be built from the register; crediting 1820 on the strength of a model would be exactly what the audit rule exists to prevent.
3. Team B wants to increase its daily increment so that its day-6 figure at least matches Team A's day-6 figure. Find the minimum whole-number increment and comment on whether it is realistic.
Read the worked answer
Team A's day-6 figure is 5·3⁵ = 5 × 243 = 1215.
Team B's day-6 figure with increment d is 40 + 5d. Require 40 + 5d ≥ 1215, so 5d ≥ 1175 and d ≥ 235. Since d must be a whole number of households, the minimum is d = 236, giving 40 + 1180 = 1220.
Check: with d = 235, Team B reaches exactly 1215, which matches but does not exceed; the question asks for at least a match, so 235 is also defensible if "match" is read strictly. State the reading chosen.
Realism: Team B currently manages 30 extra households a day; 236 is nearly eight times that, and the six-day total would become 3[80 + 5(236)] = 3 × 1260 = 3780 households, far beyond a block's household count and well past the audit threshold. The arithmetic answer is 236, but the practical judgement is that no list-based team can do this, which is why the geometric model of Team A should itself be audited.