A sum of money invested at 8% per annum simple interest grows to Rs 180 in a certain number of years. The same sum of money invested for the same number of years at 4% per annum simple interest grows to Rs 120. For how many years is the money invested in each case?

Introduction / Context:
This question gives two different amounts obtained from the same principal over the same time but at two different simple interest rates. Using these two scenarios, you can form simultaneous equations and solve for both the principal and the time period. It highlights how changing the rate affects the final amount under simple interest and how to use the resulting equations to solve for unknowns.

Given Data / Assumptions:

Principal P is the same in both cases.
Time T (in years) is the same in both cases.
In the first case, rate R1 = 8% per annum and amount A1 = Rs 180.
In the second case, rate R2 = 4% per annum and amount A2 = Rs 120.
Simple interest formula: SI = (P * R * T) / 100, and Amount A = P + SI.

Concept / Approach:
We write down expressions for the amounts at each rate in terms of P and T. This gives two equations: one for A1 and one for A2. Solving these equations simultaneously allows us to determine P and T. Once T is known, we choose the option that matches this value. This is a standard two-variable simple interest system.

Step-by-Step Solution:
Step 1: For rate 8%, amount after T years is A1 = P + (P * 8 * T) / 100. Step 2: This can be written as A1 = P * (1 + 8T / 100). Step 3: Given A1 = 180, so P * (1 + 8T / 100) = 180. Step 4: For rate 4%, amount after T years is A2 = P + (P * 4 * T) / 100 = P * (1 + 4T / 100). Step 5: Given A2 = 120, so P * (1 + 4T / 100) = 120. Step 6: We now have two equations: P(1 + 8T / 100) = 180 and P(1 + 4T / 100) = 120. Step 7: Divide the first equation by the second to eliminate P: (1 + 8T / 100) / (1 + 4T / 100) = 180 / 120. Step 8: Simplify 180 / 120 to 3 / 2. Step 9: Let a = 8T / 100 and b = 4T / 100 for clarity. Then (1 + a) / (1 + b) = 3 / 2. Step 10: Substitute a and b: (1 + 0.08T) / (1 + 0.04T) = 3 / 2. Step 11: Cross-multiply: 2(1 + 0.08T) = 3(1 + 0.04T). Step 12: Expand: 2 + 0.16T = 3 + 0.12T. Step 13: Rearrange: 0.16T - 0.12T = 3 - 2, giving 0.04T = 1. Step 14: Therefore, T = 1 / 0.04 = 25 years.

Verification / Alternative check:
Once T = 25 is known, find P from the second equation. A2 = 120 = P * (1 + 4 * 25 / 100) = P * (1 + 1) = 2P. Thus, P = 60. Check the first scenario: SI at 8% for 25 years = (60 * 8 * 25) / 100 = 60 * 2 = 120, so A1 = 60 + 120 = 180, which matches the given amount. This confirms that T = 25 years and P = 60 are correct.

Why Other Options Are Wrong:
If T were 15, 20, 22, or 18 years, substituting these values into the equations would not yield consistent values for P that satisfy both amounts 180 and 120 simultaneously. Only T = 25 years works for both scenarios.

Common Pitfalls:
A common mistake is to subtract the equations incorrectly or to attempt to solve them without eliminating one variable first, leading to algebraic errors. Some students also confuse amount with interest and set up formulas incorrectly. Carefully define amount and interest for each rate, form two equations, and then use either substitution or division to eliminate P and solve for T.

Final Answer:
The money is invested for 25 years in both cases.