CodeVerse Creatives
Esc
Numerical Aptitude Syllabus / Module 5 · Interest / Week 5

Week 5: Simple & Compound Interest

Interest is percentage change (Week 3) applied across time — and the gap between simple and compound interest, small in year one, widens every year after. This week builds both formulas cold, plus the CI-SI difference shortcut that turns a multi-step compound- interest question into one line.

Module 5 of 20 Week 5 of 20 ~2–3 Hours Practice Problems Included

By the end of this week, you'll be able to

  • Explain why compound interest overtakes simple interest, and by how much
  • Use the CI-SI difference shortcut for 2- and 3-year problems
  • Solve installment and effective-annual-rate problems
1 SI vs. CI — Why the Gap Compounds 2 The CI-SI Difference Shortcut 3 Installments 4 Effective Annual Rate 5 Practice Problems 6 Knowledge Check

1. SI vs. CI — Why the Gap Compounds

Both formulas describe money growing over time; they differ in what the interest is calculated on each period.

simple interest — interest always on the ORIGINAL principal
SI = (P x R x T) / 100

P = principal, R = rate % per year, T = time in years

Example: P=1000, R=10%, T=3 years
  SI = (1000 x 10 x 3)/100 = 300
  Amount = P + SI = 1,300

  Year 1 interest: 100. Year 2 interest: 100. Year 3: 100.
  -- interest is FLAT every year, always 10% of the original 1000
compound interest — interest on principal + all prior interest
Amount = P x (1 + R/100)^T
CI = Amount - P

Example: P=1000, R=10%, T=3 years
  Amount = 1000 x (1.10)^3 = 1000 x 1.331 = 1,331
  CI = 1,331 - 1,000 = 331

  Year 1 interest: 100 (10% of 1000)
  Year 2 interest: 110 (10% of 1,100 -- last year's amount)
  Year 3 interest: 121 (10% of 1,210)
  -- interest GROWS each year, because it's now earning
     interest on interest

At T=3, CI (331) exceeds SI (300) by 31 — small in absolute terms here, but the gap grows faster than linearly as either the rate or the time period increases, which is exactly why "compound interest is more expensive/valuable than it looks" is a genuine, non-obvious fact worth internalizing, not just a formula to apply.

2. The CI-SI Difference Shortcut

A very common question format gives the difference between CI and SI for 2 or 3 years and asks for the principal or rate — solving this from the full formulas is slow. Two shortcuts skip straight to the answer.

2-year shortcut
CI - SI (for 2 years) = P x (R/100)^2

Intuition: over 2 years, SI and CI agree on the interest for
year 1 AND on the "base" interest for year 2 -- they only
differ by the extra "interest on year 1's interest," which is
exactly R% of the year-1 interest (P x R/100), i.e. P(R/100)^2.

Worked: the difference between CI and SI on a sum for 2 years
at 10% is Rs. 50. Find the principal.
  50 = P x (10/100)^2 = P x 0.01
  P = 5,000
3-year shortcut
CI - SI (for 3 years) = P x (R/100)^2 x (3 + R/100)

Worked: find the difference between CI and SI on Rs. 8,000 for
3 years at 5% per annum.
  = 8000 x (0.05)^2 x (3 + 0.05)
  = 8000 x 0.0025 x 3.05
  = 61

Both shortcuts exist because CI and SI genuinely agree for the first year (there's no "interest on interest" yet to create a difference) — the entire gap accumulates starting in year 2, which is what makes a closed-form shortcut for exactly that gap possible.

3. Installments

"Pay off a loan in equal yearly installments" problems work backward from the compound-interest formula: each installment paid earlier has more time left to (hypothetically) earn interest before the debt is fully settled, so the equation accounts for that.

equal-installment setup (n years, rate R)
If a debt of amount A, at rate R%, is repaid in n equal
yearly installments of value x, the equation is:

A = x/(1+R/100) + x/(1+R/100)^2 + ... + x/(1+R/100)^n

Each term discounts one installment back to today's value --
this is exactly the "present value" idea: money paid n years
from now is worth less today than money paid now, by a factor
of (1+R/100)^n.

You won't usually need to solve this equation from scratch under exam time pressure — recognizing the shape (a sum of installments, each discounted by an increasing power of the rate) is usually enough to correctly set up a 2-installment problem, which is what most tests actually ask, rather than the general n-installment case.

4. Effective Annual Rate

When interest compounds more often than once a year (e.g. "10% per annum, compounded half-yearly"), the actual annual growth is higher than the stated rate — because interest is being compounded on itself more frequently.

half-yearly & quarterly compounding
Half-yearly: use HALF the rate, and DOUBLE the time period
  Amount = P x (1 + (R/2)/100)^(2T)

Quarterly: use a QUARTER the rate, and QUADRUPLE the time period
  Amount = P x (1 + (R/4)/100)^(4T)

Worked: P=10,000 at 10% p.a., compounded half-yearly, for 1 year
  Amount = 10,000 x (1 + 5/100)^2 = 10,000 x 1.1025 = 11,025

  Effective annual rate = (11,025-10,000)/10,000 x 100 = 10.25%
  -- higher than the stated 10%, because compounding happens
     twice within the year instead of once
More frequent compounding always means a higher effective rate

The stated (nominal) rate never changes — 10% p.a. is always 10% p.a. — but the effective rate you actually earn increases as compounding frequency increases, approaching (but never reaching) a theoretical maximum as compounding becomes continuous. This is why the compounding frequency, not just the headline rate, matters when comparing two interest offers.

5. Practice Problems

Practice

30 problems, no calculator — Easy, Medium & Tough

Use the CI-SI shortcuts wherever the question is exactly a 2- or 3-year difference. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.

Easy (1–10)

  1. Find the SI on Rs. 2,000 at 5% p.a. for 3 years.
  2. Find the SI on Rs. 5,000 at 8% p.a. for 2 years.
  3. Find the amount on Rs. 3,000 at 6% p.a. simple interest for 4 years.
  4. Find the CI on Rs. 1,000 at 10% p.a. for 1 year (compounded annually).
  5. Find the CI on Rs. 2,000 at 10% p.a. for 2 years.
  6. Find the rate of interest if Rs. 4,000 earns Rs. 800 SI in 4 years.
  7. Find the time in which Rs. 1,500 earns Rs. 450 SI at 10% p.a.
  8. Find the principal that earns Rs. 600 SI in 3 years at 8% p.a.
  9. Find the amount on Rs. 8,000 at 5% p.a. CI for 1 year.
  10. Find the SI on Rs. 6,000 at 12% p.a. for 6 months.
Easy — Answers

1) SI=(2000×5×3)/100=300. 2) SI=(5000×8×2)/100=800. 3) SI=720; amount=3,720. 4) CI=1000×0.1=100 (year 1 CI always equals SI). 5) CI=2000×(1.1²-1)=420. 6) Rate=(800×100)/(4000×4)=5%. 7) Time=(450×100)/(1500×10)=3 years. 8) Principal=(600×100)/(3×8)=2,500. 9) Amount=8000×1.05=8,400. 10) SI=(6000×12×0.5)/100=360.

Medium (11–20)

  1. Find the CI on Rs. 10,000 at 10% p.a. for 3 years, compounded annually.
  2. Find the difference between CI and SI on Rs. 5,000 at 10% p.a. for 2 years.
  3. A sum doubles itself in 5 years at simple interest. Find the rate percent per annum.
  4. Find the CI on Rs. 15,625 at 8% p.a. for 2 years.
  5. Find the amount on Rs. 10,000 at 8% p.a., compounded half-yearly, for 1 year.
  6. A sum of Rs. 12,000 becomes Rs. 13,440 in 2 years at simple interest. Find the rate.
  7. Find the principal that amounts to Rs. 6,000 in 4 years at 5% p.a. simple interest.
  8. Find the difference between CI and SI on Rs. 8,000 for 3 years at 5% p.a.
  9. What sum will amount to Rs. 1,331 in 3 years at 10% p.a. compound interest?
  10. A sum triples itself in 10 years at simple interest. Find the rate percent per annum.
Medium — Answers

11) CI=10000×(1.1³-1)=3,310. 12) CI-SI=5000×(0.1)²=50. 13) Amount doubles → SI=P over T years → R=100/5=20%. 14) CI=15625×(1.08²-1)=15625×0.1664=2,600. 15) Half-yearly: 10000×(1.04)²=10,816. 16) SI=1440 over 2 years → rate=(1440×100)/(12000×2)=6%. 17) 6000/1.2=5,000. 18) CI-SI=8000×(0.05)²×3.05=61. 19) P=1331/1.331=1,000. 20) Sum triples → SI=2P over T years → R=(2×100)/10=20%.

Tough (21–30)

  1. The compound interest on a sum for 2 years is Rs. 832 and the simple interest is Rs. 800. Find the rate of interest.
  2. A sum of money invested at compound interest amounts to Rs. 4,840 in 2 years and Rs. 5,324 in 3 years. Find the rate of interest.
  3. Find the effective annual rate of 10% p.a. compounded quarterly.
  4. The simple interest on a sum for 5 years at 8% p.a. is Rs. 1,200 more than the simple interest on the same sum for 3 years at the same rate. Find the sum.
  5. A sum of money at compound interest triples itself in 3 years. In how many years will it become 9 times itself?
  6. Find the compound interest on Rs. 20,000 for 1.5 years at 10% p.a., compounded half-yearly.
  7. A sum lent at compound interest for 2 years at 20% p.a. would fetch Rs. 482 more if the interest were compounded half-yearly instead of annually. Find the sum.
  8. A certain sum amounts to Rs. 7,350 in 2 years and Rs. 8,575 in 3 years at simple interest. Find the sum and the rate.
  9. Find the rate percent per annum if a sum becomes 8 times itself in 3 years at compound interest.
  10. The difference between compound interest and simple interest on a certain sum for 3 years at 10% p.a. is Rs. 31. Find the sum.
Tough — Answers

21) CI-SI=32=P(R/100)²; SI=800=PR×2/100 → PR=40,000. Dividing: R=32×100²/40000×... simplified: R=8%. 22) Interest for year 3 alone=5324-4840=484=4840×r/100 → r=10%. 23) (1.025)^4-1≈10.38%. 24) Difference=8%×2 extra years×P/100=1200 → 0.16P=1200 → P=7,500. 25) k³=3; want k^n=9=3² → n=3×2=6 years. 26) Half-yearly, 3 periods at 5%: 20000×1.05³=23,152.5; CI=3,152.5. 27) Annual: 1.44P; half-yearly (4 periods at 10%): 1.4641P; difference=0.0241P=482 → P=20,000. 28) SI/year=8575-7350=1225; Amount(2yr)=7350=P+2450 → P=4,900; rate=(1225×100)/4900=25%. 29) (1+r/100)³=8=2³ → r=100%. 30) P(R/100)²(3+R/100)=31, R=10%: 0.031P=31 → P=1,000.

6. Knowledge Check

Four quick questions. Expand each to check your answer.

Q1

Why does simple interest earn exactly the same amount every year, while compound interest earns more each successive year?

Simple interest is always calculated on the original, unchanging principal, so the same percentage of the same base produces the same amount every year. Compound interest is calculated on the principal PLUS all previously earned interest, so the base itself grows each year, making each year's interest larger than the last.

Q2

Why do CI and SI agree exactly in year 1, with the difference only appearing from year 2 onward?

In the first year, there's no prior interest yet for compound interest to compound on — both formulas are computing R% of the same original principal. The gap only opens starting in year 2, once compound interest begins earning interest on year 1's interest, which simple interest never accounts for.

Q3

Why does an investment compounded half-yearly at a stated 10% p.a. actually grow by more than 10% over one year?

Compounding twice within the year means the second half's 5% interest is calculated on a principal that already includes the first half's 5% gain, not the original amount — so a small amount of "interest on interest" accumulates within the same year, pushing the effective annual growth above the stated 10% nominal rate.

Q4

Why can the CI-SI difference shortcut for 2 years be written simply as P × (R/100)², without needing the full CI and SI formulas separately?

Since CI and SI agree entirely in year 1, the whole 2-year difference reduces to exactly the extra interest CI earns in year 2 on top of year 1's interest — which is R% of (P × R/100), i.e. P × (R/100)². Everything that's identical between the two formulas cancels out, leaving only this one small correction term.

← Back to Week 4: Profit, Loss & Discount Up next Week 6: Ratio, Proportion & Variation →