CAT rarely tests mechanical formula substitution for interest; it tests the fundamental distinction between linear addition (SI) and exponential compounding (CI). This logic-first guide breaks down the Year-1 invariant, the interest-on-interest difference shortcuts, doubling races, and installment debt settlement.
Understand why SI is a fixed annual addition (AP) while CI is a constant multiplier (GP), and why SI₁ = CI₁ always holds for the first cycle.
In CAT arithmetic, Simple Interest (SI) is an Arithmetic Progression (fixed addition each year), while Compound Interest (CI) is a Geometric Progression (multiplication by a constant factor).
Principal P = ₹1,000 invested at R = 10% per annum.
Compare the balance growth over 3 years under SI vs. CI.
Under SI, interest is computed strictly on the original principal:
Year 1: ₹1,000 + ₹100 = ₹1,100
Year 2: ₹1,100 + ₹100 = ₹1,200
Year 3: ₹1,200 + ₹100 = ₹1,300
Linear addition: exactly ₹100 added every single year.
Under CI, previous interest is added to the principal to earn further interest:
Year 1: ₹1,000 × 1.10 = ₹1,100 (Same as SI!)
Year 2: ₹1,100 × 1.10 = ₹1,210 (+₹110)
Year 3: ₹1,210 × 1.10 = ₹1,331 (+₹121)
Exponential growth: the extra ₹10 in Year 2 is the interest earned on Year 1's ₹100!
The divergence begins strictly in Year 2, when CI begins earning interest on interest.
Derive why the 2-year difference is simply interest on the 1st year's interest, and apply the 3D₂ + Interest rule for 3-year problems.
Instead of expanding $(1 + R/100)^2$ algebraically, notice that the entire 2-year difference between CI and SI is simply the interest earned on the 1st year's interest.
Let Year 1 interest be $I_1 = \frac{P \times R}{100}$.
Subtracting total SI from total CI:
The difference between CI and SI on a sum for 2 years at 15% per annum is ₹144.
Find the principal sum P.
Convert 15% to fraction: $15/100 = 3/20$.
Solve for P directly:
⇒ Principal P = ₹6,400
Compare linear scaling (k - 1)T under SI against exponential power scaling under CI when determining how fast capital multiplies.
A sum doubles in T years. How long will it take to become 8 times? Notice how SI scales with linear factors while CI scales with exponents.
A sum of money doubles itself in 5 years.
How long will it take to become 8 times the original sum under:
Case (a) Simple Interest | Case (b) Compound Interest?
To double ($2P$), the interest earned is $2P - P = P$ in 5 years.
To become 8 times ($8P$), the interest needed is $8P - P = \mathbf{7P}$:
Time = (k − 1) × T
Time = (8 − 1) × 5 = 7 × 5 = 35 years
Under SI, each additional multiple requires another 5 years.
Under CI, the sum multiplies by 2 every 5 years:
P → 2P → 4P → 8P = 2³ × P
To reach $2^3$, exactly 3 compounding periods are needed:
35 years under SI vs. only 15 years under CI!
Master semi-annual and quarterly compounding using the successive percentage change shortcut without computing fractional powers.
When an annual rate of R% is compounded m times per year, the rate per cycle becomes R/m% and the number of periods becomes m × n. The effective annual yield rises because interest begins earning interest earlier.
Consider an investment of ₹10,000 at a nominal 20% per annum for 1 year:
1 cycle at 20%
Amount = ₹12,000
Yield = 20%
2 cycles of 6 months at 10%
Amount = ₹12,100
Yield = 21%
For semi-annual compounding at $R\%$, the rate per half-year is $r = R/2\%$.
Use the Successive Percentage Change formula ($a + b + \frac{ab}{100}$):
For $20\%$ annual ($r = 10\%$): $\text{EAR} = 10 + 10 + \frac{100}{100} = \mathbf{21\%}$.
Connect loan amortization to finite GP discounted cash flows, equating borrowed principal to present values of future payments.
When you borrow a principal P and repay it in equal installments of x, each installment paid in the future has a lower Present Value today because of compounding.
If rate is $R\%$, let the compounding multiplier be $m = 1 + \frac{R}{100}$.
The borrowed principal equals the sum of the discounted present values of all installments.
A sum of ₹10,500 is borrowed and to be paid back in 2 equal annual installments at 10% compound interest per annum.
Find the amount of each annual installment x.
Here $R = 10\% \implies m = 1.1 = 11/10$.
Combine over a common denominator (121):
Cross-multiply to isolate $x$:
Each annual installment is ₹6,050.
Deconstruct why earlier payments accrue interest in the lender's hands, deriving the debt settlement formula from first principles.
Under Simple Interest, an installment paid at the end of Year 1 sits in the hands of the lender until the final debt due date, earning interest for the remaining years.
To settle a debt of amount A due in n years at R% simple interest with equal annual installments of x:
What annual payment will discharge a debt of ₹6,450 due in 4 years at 5% simple interest per annum?
Here $A = 6,450$, $n = 4$ years, and $R = 5\%$:
Simplify the interest term: $\frac{x}{20} \times 6 = 0.3x$.
Each annual installment is ₹1,500.
A dedicated question set for Simple & Compound Interest is currently being authored. In the meantime, test your baseline score on the full Diagnostic:
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