2.7 Finding Principal, Rate, or Time Given CI Data

These are “reverse engineering” problems — apply the standard CI/amount formula and solve for the unknown algebraically.

Solved Example 2.7.1: A sum amounts to Rs. 9,680 in 2 years and Rs. 10,648 in 3 years at CI, compounded annually. Find the rate of interest and the sum.

Solution: The amount in the 3rd year is obtained by multiplying the 2-year amount by (1 + R/100) exactly once (one more compounding step): 1 + (R)/(100) = (10648)/(9680) = 1.1 ⇒ R = 10% Now, Sum = 9680/(1.1)² = 9680/1.21 = Rs. 8,000.

Solved Example 2.7.2: At what rate of CI per annum will a sum of Rs. 5,000 amount to Rs. 6,050 in 2 years?

Solution: CI = 6050 − 5000 = 1050. Using (1+r)² − 1 = 1050/5000 = 0.21, so (1+r)² = 1.21, 1+r = 1.1, r = 10%. Rate = 10% p.a.

Solved Example 2.7.3: A sum amounts to Rs. 4,400 in 1 year and Rs. 5,324 in 3 years at compound interest, compounded annually. Find the rate of interest and the sum (edge case: the two given years are not consecutive — they are 2 years apart).

Solution: Going from year 1 to year 3 is exactly 2 more compounding steps, so the ratio of the amounts gives (1+R/100)² directly, not (1+R/100)¹: (1+(R)/(100))^2 = (5324)/(4400) = 1.21 ⇒ 1+(R)/(100) = 1.1 ⇒ R = 10% Now, Sum = A₁/(1+R/100) = 4400/1.1 = Rs. 4,000. Always check how many compounding steps separate the two given years before taking the ratio — using the wrong power here is a common slip.

Solved Example 2.7.4: In what time will Rs. 15,625 amount to Rs. 17,576 at 4% p.a. compound interest, compounded annually? (edge case: finding T instead of P or R)

Solution: (1+(4)/(100))^T = (17576)/(15625) = 1.124864 Since (1.04)³ = 1.124864 exactly, T = 3 years. Recognising perfect powers like this (by testing T = 2, 3, 4 mentally) is much faster than using logarithms, which SSC/RRB exams never require.