In aptitude (Simple Interest), for what principal sum (in ₹) will the simple interest at R% per annum for 2 years be exactly ₹R? (Here R is both the numerical value of the rate in percent and the simple interest amount in rupees.)

Introduction / Context:
This question is a classic algebra based simple interest puzzle where a symbol R is used both as the annual rate of interest in percent and as the simple interest amount in rupees. The goal is to express the principal in terms of R and to understand how the formula behaves when the same symbol appears in different roles. Such questions test algebraic manipulation and conceptual clarity rather than heavy calculation.

Given Data / Assumptions:

• Simple interest rate per annum = R%.
• Time period t = 2 years.
• Simple interest amount SI = ₹R.
• We must find the principal sum P in rupees.
• Interest is simple, not compound.

Concept / Approach:
The simple interest formula is: SI = (P * r * t) / 100 Here SI is the interest, P is the principal, r is the rate in percent, and t is the time in years. In this question, SI is given as R and r is also R. Therefore, the formula becomes an equation with P and R. We will substitute SI = R, r = R, t = 2 and solve for P algebraically. The presence of the same symbol R in two places should not confuse us; we treat it simply as a number that remains unspecified.

Step-by-Step Solution:
Step 1: Start from the standard formula SI = (P * r * t) / 100. Step 2: Substitute SI = R, r = R, and t = 2. Step 3: This gives R = (P * R * 2) / 100. Step 4: Simplify the right side: R = (2 * P * R) / 100. Step 5: Since R is nonzero, cancel R from both sides of the equation. Step 6: This yields 1 = (2 * P) / 100. Step 7: Rearrange for P: P = 100 / 2. Step 8: Therefore, P = 50. Step 9: So the principal sum is ₹50, which does not depend on the specific value of R as long as R is nonzero.

Verification / Alternative check:
To verify, pick any convenient value for R, for example R = 10. Then P should be ₹50. Compute SI using the formula: SI = (50 * 10 * 2) / 100 = (1000) / 100 = 10 The interest is ₹10, which matches SI = R = 10. Try another value, say R = 8. Then: SI = (50 * 8 * 2) / 100 = 800 / 100 = 8 Again, interest equals ₹8, which is consistent with SI = R. This confirms that P = ₹50 is correct for all valid R values.

Why Other Options Are Wrong:
₹100: Substituting P = 100 leads to SI = (100 * R * 2) / 100 = 2R, which is double the required interest.
₹200: With P = 200, SI becomes 4R, far larger than the given interest R.
₹50/R: This wrongly introduces division by R and does not satisfy the equation for general R values.
₹100/R: This also fails for general R and leads to inconsistent interest values when tested with specific R examples.

Common Pitfalls:
Learners sometimes confuse the symbol R being used in two different roles and mistakenly try to substitute a random numeric value before solving algebraically. Others forget to cancel R from both sides and end up with an expression that still contains R in the principal. A safe strategy is to work symbolically until the very end and cancel common nonzero factors carefully. Always double check the result by substituting one or two trial values for R to ensure consistency.

Final Answer:
The principal amount for which the simple interest at R% per annum for 2 years equals ₹R is ₹50.