A man saves Rs. 200 in each of the first three months of his service. In each subsequent month his saving increases by Rs. 40 more than the saving of the immediately previous month. After how many months will his total saving from the start of service become Rs. 11040?

Introduction / Context:
This question involves arithmetic progressions and cumulative sums. The man saves a fixed amount for the first few months, and then his monthly savings follow a pattern where they increase by a constant amount. You are asked to find the total number of months required to reach a given total saving. This is a classic application of arithmetic progression formulas in aptitude exams.

Given Data / Assumptions:

• For the first three months he saves Rs. 200 per month.
• From the fourth month onward, each month he saves Rs. 40 more than the previous month.
• Total saving at some month is Rs. 11040.
• We must find the number of months from the start of service to reach this total.

Concept / Approach:
We split the problem into two parts: the first three months with constant savings, and the remaining months which form an arithmetic progression. The arithmetic progression has first term 240 and common difference 40. We then write the total saving as the sum of Rs. 600 from the first three months plus the sum of an arithmetic progression from month four onward. This leads to a quadratic equation in the number of additional months, which we solve to find the total time.

Step-by-Step Solution:
Step 1: Savings for first three months: 3 * 200 = 600.Step 2: Let the total number of months be n. Then the number of months after the first three is m = n - 3.Step 3: Savings from the fourth month form an arithmetic progression with first term a = 240 and common difference d = 40.Step 4: Sum of this arithmetic progression for m months is S = m / 2 * [2a + (m - 1) * d].Step 5: Substitute a = 240 and d = 40 to get S = m / 2 * [480 + 40m - 40] = m / 2 * (40m + 440) = 20m(m + 11).Step 6: Total saving is 600 + 20m(m + 11) = 11040.Step 7: So 20m(m + 11) = 11040 - 600 = 10440.Step 8: Divide both sides by 20 to get m(m + 11) = 522.Step 9: Solve m^2 + 11m - 522 = 0. The positive integer solution is m = 18 since 18 * 29 = 522.Step 10: Therefore n = m + 3 = 18 + 3 = 21 months.

Verification / Alternative check:
We can verify by calculating the sums directly. First three months contribute 600. The next 18 months form an arithmetic progression from 240 to 240 + 17 * 40 = 920. The sum is 18 / 2 * (240 + 920) = 9 * 1160 = 10440. Adding 600 gives 11040, which matches the required total. This confirms that 21 months is correct.

Why Other Options Are Wrong:
Using 19, 20 or 18 months leads to m values that do not satisfy the equation m(m + 11) = 522, and the total savings are either less than or greater than 11040. Similarly, 22 months would produce a total exceeding 11040. Only 21 months produces exactly the required total saving.

Common Pitfalls:
Some students forget to separate the constant saving phase and the arithmetic progression phase, and instead treat the entire period as one progression, which gives incorrect results. Others may make algebraic mistakes when forming or solving the quadratic equation. Carefully setting up the variables and using the standard sum formula for arithmetic progression avoids these issues.

Final Answer:
The man will have saved Rs. 11040 after 21 months from the start of his service.