If on a certain amount, the difference between Compound interest and Simple interest at the rate 5% annual for three years is ₹ 183, then the Principal amount is -

Correct — option (1), ₹ 24,000.

Step 1 — write CI − SI for 3 years at rate r, as a decimal.
CI = P[(1 + r)³ − 1] = P(3r + 3r² + r³)
SI = P × 3r
CI − SI = P(3r² + r³) = P × r² × (3 + r)

Step 2 — put r = 5% = 0.05.
r² = 0.0025
3 + r = 3.05
r² × (3 + r) = 0.0025 × 3.05 = 0.007625

Step 3 — solve for P.
P × 0.007625 = 183
P = 183 ÷ 0.007625 = ₹ 24,000

In fractions: r = 1/20, so r²(3 + r) = (1/400) × (61/20) = 61/8,000, and 183 × 8,000 ÷ 61 = 3 × 8,000 = 24,000.

Check with P = ₹ 24,000:
Amount = 24,000 × 1.05³ = 24,000 × 1.157625 = ₹ 27,783
CI = 27,783 − 24,000 = ₹ 3,783
SI = 24,000 × 5 × 3 ÷ 100 = ₹ 3,600
CI − SI = ₹ 183

The idea to remember: for 3 years, CI − SI = P × r² × (3 + r); for 2 years, it is P × r².

Why the others are wrong

• (2)
  ₹ 8,000 — On ₹ 8,000 at 5% for 3 years, CI = ₹ 1,261 and SI = ₹ 1,200, so the difference is ₹ 61.

  The difference is proportional to the principal. ₹ 183 is three times ₹ 61, so it needs three times ₹ 8,000, which is ₹ 24,000.

• (3)
  ₹ 18,000 — On ₹ 18,000 at 5% for 3 years, CI = ₹ 2,837.25 and SI = ₹ 2,700, so the difference is ₹ 137.25, not ₹ 183.

  Each rupee of principal adds 0.007625 rupee to the difference here, so ₹ 183 needs 183 ÷ 0.007625 = ₹ 24,000.

• (4)
  ₹ 21,000 — On ₹ 21,000 at 5% for 3 years, CI ≈ ₹ 3,310.13 and SI = ₹ 3,150, so the difference is about ₹ 160.13, short of ₹ 183.

  The missing ₹ 22.875 corresponds to another ₹ 3,000 of principal: 3,000 × 0.007625 = ₹ 22.875.

Simple interest is paid on the principal alone: SI = P × R × T ÷ 100. Compound interest adds each year's interest to the principal, so later years earn interest on interest: Amount = P(1 + R/100)^T.

The gap between them comes only from interest on interest. For 2 years it is P × (R/100)²: one year's interest on the first year's interest.

For 3 years it is P × (R/100)² × (3 + R/100). The 3 counts three portions of interest on simple interest, one in year 2 and two in year 3; the extra R/100 is year 3's interest on year 2's interest on interest.

Both formulas assume compounding once a year.

Key facts

• Simple interest = P × R × T ÷ 100.
• Compound interest, compounded annually = P[(1 + R/100)^T − 1].
• For 2 years, CI − SI = P × (R/100)².
• For 3 years, CI − SI = P × (R/100)² × (3 + R/100).
• At 5% for 3 years, CI − SI = 0.007625 × P, so a difference of ₹ 183 needs P = ₹ 24,000.

Common traps

• Using the 2-year formula. P × r² = 183 gives P = ₹ 73,200, which leaves out the third year's interest on interest.
• Mixing percentage and decimal forms. With R = 5 as a whole number, use P × R² × (300 + R) ÷ 100³, not P × R² × (3 + R).
• Taking the amount as the interest. ₹ 27,783 includes the principal; the compound interest is ₹ 3,783.

UnlockIAS will link similar questions from RAS Pre 2018 and 2016 here once those papers are published on this site.

• practice — not a real PYQ

  The difference between compound interest (compounded annually) and simple interest on a sum at 10% per annum for 2 years is ₹ 50. The sum is –

  • (a)₹ 5,000
  • (b)₹ 500
  • (c)₹ 2,500
  • (d)₹ 50,000
  Answer(1) — For 2 years CI − SI = P × (0.1)² = 0.01P, so P = 50 ÷ 0.01 = ₹ 5,000.

  Option (2) divides by r = 0.1 instead of r²; option (3) divides by 2r² = 0.02; option (4) divides by r³ = 0.001.

• practice — not a real PYQ

  On ₹ 10,000 at 10% per annum, what is the difference between compound interest (compounded annually) and simple interest for 3 years?

  • (a)₹ 100
  • (b)₹ 300
  • (c)₹ 310
  • (d)₹ 3,310
  Answer(3) — CI = 10,000 × (1.331 − 1) = ₹ 3,310 and SI = ₹ 3,000, a difference of ₹ 310; the formula gives 10,000 × 0.01 × 3.1 = ₹ 310.

  Option (1) is the 2-year difference; option (2) leaves out the r³ term; option (4) is the compound interest itself.