Savings Plans with Compounding Deposits and Affordability Cap
A student wants to buy a laptop costing K250,000. Plan A: save K4,000 in month 1 and increase the monthly deposit by K800 each month. Plan B: save K4,000 in month 1 and increase the monthly deposit by 10% each month. Deposits are made at the start of each month and no interest is paid. The student can afford at most K20,000 in any month. The student asks whether Plan B overtakes Plan A by month 21, and whether that makes Plan B the better choice. The mathematics teacher warns that a geometric progression can look impressive in total while becoming unaffordable month by month. The student has irregular income from part-time work and cannot borrow to maintain a deposit schedule. The comparison must use the actual deposit cap, not just the uncapped progression formula.
1. Write an expression for the total saved after n months under each plan. Compute the totals after 12 months and state which is larger.
Read the worked answer
Plan A deposits form an AP with first term 4000 and common difference 800: A_n=n/2[2(4000)+(n-1)800]=n/2[8000+800(n-1)]. At n=12, A_12=6[8000+8800]=6×16800=100800. Plan B deposits form a GP with first term 4000 and ratio 1.1: B_n=4000(1.1^n-1)/(1.1-1)=40000(1.1^n-1). At n=12, 1.1^12≈3.13843, so B_12≈40000(2.13843)=85537. After 12 months Plan A is larger by about K15,263.
2. Find the first month in which each plan's monthly deposit exceeds K20,000. Explain what this means for a student who can afford at most K20,000 in any month.
Read the worked answer
Plan A deposit in month n is 4000+800(n-1). Solve 4000+800(n-1)>20000: 800(n-1)>16000, so n-1>20 and n>21. Month 22 is the first month to exceed the cap; month 21 is exactly 20,000. Plan B deposit in month n is 4000(1.1)^(n-1). Need 1.1^(n-1)>5, so (n-1)ln1.1>ln5 and n-1>16.887, giving n≥18. Check: month 17 gives about 18,380; month 18 gives about 20,216. Thus Plan B breaches the cap in month 18. The student cannot follow Plan B as written beyond month 17 without exceeding the cap. Plan A remains affordable through month 21 and first breaches in month 22. A comparison that ignores this is misleading.
3. Using the uncapped model, find the first month n for which Plan B's total exceeds Plan A's total. Then explain why this does not by itself justify choosing Plan B for this student.
Read the worked answer
At n=20: A=10[8000+19×800]=232000; B=40000(1.1^20-1)≈40000(6.7275-1)=229100, so B is lower. At n=21: A=10.5[8000+20×800]=252000; B=40000(1.1^21-1)≈40000(7.40025-1)=256010. So the uncapped crossover is at n=21. However, from month 18 the Plan B deposit exceeds K20,000. If the cap is enforced by depositing only K20,000 in months 18 to 21, the Plan B total is the sum of months 1 to 17 plus 4×20000. The sum of the first 17 deposits is 40000(1.1^17-1)≈40000(5.05447-1)=162179; adding 80000 gives 242179, still below Plan A's 252000. The uncapped crossover relies on unaffordable deposits, so Plan B does not actually beat Plan A under the student's constraint.