Reading Two Salary Offers and a Savings Plan
A small printing firm offers a school-leaver two payment options. Option A starts at ₹18,000 in the first month and rises by ₹1,200 every month. Option B starts at ₹15,000 in the first month and rises by ₹1,800 every month. The offer letter says the contract will run for 24 months. Ravi, the candidate, also plans a personal savings scheme: he will deposit ₹500 in month 1, ₹700 in month 2, ₹900 in month 3, and so on, increasing by ₹200 each month. His uncle reads the offer and says two things: 'Option B pays more than Option A in every single month from the first one, so it is certainly the better total over 24 months,' and 'the two 24-month totals differ by exactly ₹14,400.' Ravi wants to check both remarks with arithmetic before he signs, and he also wants to know in which month his monthly deposit first reaches ₹4,100.
1. Which option pays more in the 12th month, and by how much? Also find the month in which the two monthly pays become equal, and test the uncle's 'every single month' claim.
Read the worked answer
Both are APs: a = first month's pay, d = monthly increment, and the nth term is a + (n − 1)d.
Option A: a₁₂ = 18,000 + 11 × 1,200 = 18,000 + 13,200 = ₹31,200.
Option B: a₁₂ = 15,000 + 11 × 1,800 = 15,000 + 19,800 = ₹34,800.
B pays more in month 12 by ₹3,600.
Crossover: 18,000 + 1,200(n − 1) = 15,000 + 1,800(n − 1) → 3,000 = 600(n − 1) → n = 6, when both pay ₹24,000.
So the uncle is wrong: A pays more in months 1–5, the two are equal in month 6, and B leads only from month 7.
2. Compute both 24-month totals and judge the claim that they differ by exactly ₹14,400.
Read the worked answer
S₂₄ = 24/2 × [2a + 23d] = 12(2a + 23d).
Option A: 12[36,000 + 27,600] = 12 × 63,600 = ₹7,63,200.
Option B: 12[30,000 + 41,400] = 12 × 71,400 = ₹8,56,800.
Difference = ₹93,600, in favour of B.
Judgement: the ₹14,400 figure is wrong - it is 6.5 times too small (14,400 × 6.5 = 93,600).
Independent check by differencing the monthly pays: A − B in month n = 3,000 − 600(n − 1). Summing for n = 1 to 24: 24 × 3,000 − 600 × (23 × 24/2) = 72,000 − 1,65,600 = −93,600, i.e. B is ahead by ₹93,600. The two methods agree.
3. In which month does Ravi's deposit first reach ₹4,100, and what is his total saving up to that month?
Read the worked answer
Deposits form an AP with a = 500, d = 200, so month n deposit = 500 + 200(n − 1).
Set 500 + 200(n − 1) = 4,100 → 200(n − 1) = 3,600 → n − 1 = 18 → n = 19.
Check: month 19 deposit = 500 + 18 × 200 = ₹4,100 ✔
Total saved up to month 19: S₁₉ = 19/2 [2(500) + 18(200)] = 19/2 (1,000 + 3,600) = 19/2 × 4,600 = ₹43,700.