Basic Concepts: Simple Interest & Compound Interest | CSAT Preparation - UPSC PDF Download

What are Interest Rates?

Interest and interest rates are fundamental concepts in finance and everyday commerce. A small change in interest rates can produce large effects on loans, savings, investment returns and the wider economy.

Interest is the amount charged by a lender from a borrower on the principal loan sum. It is the cost of borrowing money.

Interest rate is the percentage at which interest is charged on the principal sum per unit time (commonly per annum).

The rate at which interest is charged depends on two factors:

1. The value of money doesn’t remain same over time. It changes with time. The net worth of ₹ 100 today will not be same tomorrow i.e. If 5 pens could be bought presently with a INR 100 note then in future, maybe only 4 pens can be bought with the same ₹ 100 note. The reason behind this the inflation or price rise. So, the interest rate includes this factor of inflation
2. The credibility of the borrower, if there is more risk and chance of default on the borrower’s part then more interest will be charged. And, if there is less chance of payment failure on the part of borrower then the rate of interest would be lower.

Because interest rates affect lending, saving, investment and prices across the economy, they are central to financial decision-making and form a standard topic in quantitative sections of competitive examinations and standard school curricula.

There are two basic methods of charging interest:

• Simple Interest
• Compound Interest

Simple interest is interest calculated only on the original principal for the entire period. Interest does not itself earn further interest.

• Simple interest does not get added to the principal; it remains separate and fixed for a given principal, rate and time.
• The formula for simple interest is: SI = (P × R × T) / 100.

Here, P is the principal (original amount), R is the annual rate of interest (in percent), and T is the time in years.

• The total amount to be repaid after T years is A = P + SI.

Example: Find the simple interest on ₹68,000 at 20% p.a. for 2 years.

Solution: Compute SI using the formula SI = (P × R × T) / 100.

Substitute values: P = 68,000; R = 20; T = 2.

SI = (68,000 × 20 × 2) / 100.

SI = (68,000 × 40) / 100.

SI = 2,720,000 / 100 = ₹27,200.

The simple interest for 2 years is ₹27,200 and the amount repaid would be ₹68,000 + ₹27,200 = ₹95,200.

The following results and shortcuts are commonly used with simple interest (illustrated in the images that follow): present worth relations, proportional interest calculations, time and rate manipulations, and methods to combine several investments at different rates. Consult the diagrams for common formulae and examples.

Q1: Nitu has an initial capital of ₹20,000. Out of this, she invests ₹8,000 at 5.5% in bank A, ₹5,000 at 5.6% in bank B and the remaining amount at x% in bank C, each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been:

a. 700
b. 800
c. 900
d. 1000

Sol: Option 'b' is correct.
Explanation:

Let the rate in bank C be x% per annum.

Interest from A = 8,000 × 5.5 / 100 = ₹440.

Interest from B = 5,000 × 5.6 / 100 = ₹280.

Amount invested in C = 20,000 - 8,000 - 5,000 = ₹7,000.

Interest from C = 7,000 × x / 100 = 70x.

Total interest = 440 + 280 + 70x = 720 + 70x.

Total interest equals 5% of initial capital = 20,000 × 5 / 100 = ₹1,000.

So, 720 + 70x = 1,000.

70x = 280.

x = 4.

If entire capital 20,000 were invested at 4%, interest = 20,000 × 4 / 100 = ₹800.

Q2: Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is ____________.

Sol: Let the total capital be 15x (LCM of denominators 5 and 3 makes calculations easier).

One-fifth = 3x invested at 6% 
→ annual interest = 3x × 6 / 100 = 0.18x.

One-third = 5x invested at 10% 
→ annual interest = 5x × 10 / 100 = 0.5x.

Remaining = 15x - 3x - 5x = 7x invested at 1%
 → annual interest = 7x × 1 / 100 = 0.07x.

Total annual interest = 0.18x + 0.5x + 0.07x = 0.75x.

We need n such that n × annual interest ≥ initial capital (15x).

So, n × 0.75x ≥ 15x.

n ≥ 15 / 0.75 = 20.

Minimum number of years required = 20 years.

Q3: Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is:

a. 37.5%
b. 62.5%
c. 60%
d. 40%

Sol: Option 'b' is correct.

Explanation:

Let the two parts be x and y.

Interest on first part = x × 15% × 4 = x × 0.15 × 4 = 0.6x.

Interest on second part = y × 12% × 3 = y × 0.12 × 3 = 0.36y.

Given 0.6x = 0.36y ⇒ x / y = 0.36 / 0.6 = 36 / 60 = 3 / 5.

So x : y = 3 : 5; total savings = 8 parts; percentage invested in first part = (3/8) × 100% = 37.5%.

• Compound interest (CI) is interest calculated on the principal and on previously earned interest. It is the standard method used by banks and financial institutions.
• If an amount P is invested at R% per annum compounded once per year, then after n years the amount is P(1 + R/100)ⁿ.
• Example: If ₹100 is deposited for 3 years at 6% p.a. compounded annually:
• The amount after 1 year = ₹100 × 1.06 = ₹106.
• The amount after 3 years = ₹100 × (1.06)³ = ₹119.1016.

Compound Interest - Installments and Normal Present Value Reasoning

When a loan is repaid in equal instalments at regular intervals and interest is charged at a compound rate, the instalment value can be found by equating the present value of all instalments to the loan amount. The present value of each instalment depends on the compounding period and rate. The general formula for an instalment (annuity) uses discount factors and standard annuity formulae shown in the image below.

Example: One can purchase a flat for ₹55,000 or pay a cash down payment of ₹4,275 and the rest in three equal instalments. The society charges interest at 16% p.a. compounded half-yearly. If the flat is purchased under the instalment plan, find the value of each instalment.

Sol:

Cost of flat = ₹55,000.

Down payment = ₹4,275.

Loan amount = 55,000 - 4,275 = ₹50,725.

Annual nominal rate = 16% compounded half-yearly.

Half-yearly rate = 16 / 2 = 8% per half-year.

Number of half-yearly instalments = 3.

Let each instalment be x.

Present value of first instalment (paid after one half-year) = x / (1.08).

Present value of second instalment (after two half-years) = x / (1.08)².

Present value of third instalment (after three half-years) = x / (1.08)³.

Equate sum of present values to loan: x[1/1.08 + 1/1.08² + 1/1.08³] = 50,725.

Discount factors computed:

1/1.08 = 0.9259259.

1/1.08² = 0.8573388.

1/1.08³ = 0.7938320.

Sum of discount factors ≈ 2.577097.

x × 2.577097 = 50,725.

x ≈ 50,725 / 2.577097 ≈ ₹19,689. (rounded to nearest rupee)

• Compounded annually: 
  Amount = P (1 + R/100)ⁿ
• Compounded half-yearly:
  Amount = P (1 + (R/2)/100)²ⁿ
• Compounded quarterly: 
  Amount = P (1 + (R/4)/100)⁴ⁿ
• If rates differ year to year (R1% for year 1, R2% for year 2, etc.): 
  Amount = P ∏(1 + R_i/100)
• Present worth of ₹x due after n years at rate R%: 
  Present worth = x / (1 + R/100)ⁿ
• If a sum becomes x times in n years under CI, the annual rate R = 100 (x^1/n - 1)
• If P amounts to A₁ after T years and to A₂ after T + 1 years at CI, then annual rate R = (A₂ - A₁) / A₁ × 100

Example 1: A man invests ₹5,000 for 3 years at 5% p.a. compounded yearly. Income tax at 20% on the interest earned is deducted at the end of each year. Find the amount at the end of the third year.

Sol: Principal at start = ₹5,000.

Annual compound rate = 5%.

At end of first year, gross interest = 5,000 × 0.05 = ₹250.

Income tax on this interest = 20% of 250 = ₹50.

Net interest added = 250 - 50 = ₹200.

Amount at end of year 1 = 5,000 + 200 = ₹5,200.

At end of second year, gross interest = 5,200 × 0.05 = ₹260.

Income tax = 20% of 260 = ₹52.

Net interest added = 260 - 52 = ₹208.

Amount at end of year 2 = 5,200 + 208 = ₹5,408.

At end of third year, gross interest = 5,408 × 0.05 = ₹270.40.

Income tax = 20% of 270.40 = ₹54.08.

Net interest added = 270.40 - 54.08 = ₹216.32.

Amount at end of year 3 = 5,408 + 216.32 = ₹5,624.32.

Example 2: A sum of ₹12,000 deposited at compound interest becomes double after 5 years. After 20 years, how much will it become?

Sol: Let annual growth factor = (1 + r/100).

Given (1 + r/100)⁵ = 2.

After 20 years the factor = (1 + r/100)²⁰ = ((1 + r/100)⁵)⁴ = 2⁴ = 16.

Therefore amount after 20 years = 12,000 × 16 = ₹192,000.

Shortcut reasoning: Doubling every 5 years means four doublings in 20 years, so 12,000 × 2 × 2 × 2 × 2 = 192,000.

Example 3: John borrowed Rs. 2,10,000 from a bank at 10% per annum compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment is Rs _________?

Sol: Principal borrowed = ₹2,10,000.

Amount due after 2 years = 2,10,000 × 1.1 × 1.1 = 2,10,000 × (1.1)² = ₹2,54,100.

Let each instalment be x. The first instalment is paid at end of year 1 and will accumulate with interest for one more year; so its value at end of year 2 equals x × 1.1.

The second instalment is paid at end of year 2 and its value at end of year 2 is x.

Equate total to amount due: 1.1x + x = 2,54,100.

2.1x = 2,54,100.

x = 2,54,100 / 2.1 = ₹1,21,000.

Example 4: Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit's dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y.

Sol: Interest paid to Ankit on X at 8% = 0.08X.

Interest received from Ishan when lending (X + Y) at 10% = 0.10(X + Y) = 0.10X + 0.10Y.

Net interest retained in first case = 0.10X + 0.10Y - 0.08X = 0.02X + 0.10Y.

It is given that this equals Ankit's interest income 0.08X, so:

0.02X + 0.10Y = 0.08X.

Rearrange: 0.10Y = 0.06X ⇒ X = (0.10/0.06) Y = (10/6) Y = (5/3) Y.

In the second scenario, Gopal lends (X + 2Y) at 10% so interest received = 0.10X + 0.20Y.

Net interest retained in second case = 0.10X + 0.20Y - 0.08X = 0.02X + 0.20Y.

Increase in retained interest = (0.02X + 0.20Y) - (0.02X + 0.10Y) = 0.10Y = 150.

So Y = 1,500.

From X = (5/3)Y ⇒ X = (5/3) × 1,500 = 2,500.

Therefore X + Y = 2,500 + 1,500 = ₹4,000.

Depreciation is the decrease in the value of an asset (for example machinery, electronics or buildings) over time due to wear, obsolescence or other causes. If the rate of depreciation is r% per annum and the current value is A₀, then after t years the value A₁ is given by:

A₁ = A₀ (1 - r/100)^t

Example 1: The value of the laptop depreciates at the rate of 30 % per annum. The present value of the laptop if Rs. 40000. What will be the worth of laptop two years from now?

a. Rs. 28000

b. Rs. 16000

c. Rs. 24400

d. Rs. 19600

Sol: Option 'd' is correct.

Explanation (stepwise):

Initial value A₀ = ₹40,000.

Depreciation rate r = 30% per annum.

After 2 years: A₁ = 40,000 × (1 - 0.30)² = 40,000 × (0.70)².

A₁ = 40,000 × 0.49 = ₹19,600.

Example 3: The value of a residential flat constructed at a cost of Rs100000 is depreciating at the rate of 10% per annum. What will be its value 3 years after construction?

Sol: Initial value V0 = ₹100,000.

Depreciation rate = 10% ⇒ multiplier each year = 0.9.

Value after 3 years = 100,000 × (0.9)³ = 100,000 × 0.729 = ₹72,900.

Population growth or decline often follows multiplicative (successive) change similar to compound interest. The same formulae apply when modelling population change.

Let P0 be the initial population, r% the annual growth (or decline) rate, n the number of years, and P1 the population after n years.

When population increases:

P₁ = P₀ (1 + r/100)ⁿ

When population decreases:

P₁ = P₀ (1 - r/100)ⁿ

Example 1 : The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years ?

Sol: P₀ = 125,000.

r = 2% ⇒ multiplier = 1.02.

P after 3 years = 125,000 × (1.02)³ = 125,000 × 1.061208 = 132,651 (rounded to nearest whole number).

Example 2: The rate of annual growth of population is 5%. If the present population of a city is 140000, then after three years the population of that city will be approximately?

a. 153000
b. 164000
c. 162000
d. 134000

Sol: Option 'c' is correct

Explanation: The rate of annual growth of population = 5%
The present population of a city = 140000
Formula used: Population of town = P(1+r/100)
According to the question Population of town after three years = P(1+r/100)
=140000(1+5/100)³
= 140000(105/100)³
= (140000 × 21/20 × 21/20 × 21/20) 
= (1296540/8) 
= 162067.5 ∼ 162000
The required population after three years is 162000

Example 3: The population of the city is increasing at the rate of 8%. If the current population of the city is 100000 then what will be the population after 2 years?
a. 115540
b. 126440
c. 116640
d. 126640

Sol: Option 'c' is correct

Explanation: Principal, P = 10000
Rate, R = 8%
Time, n = 2 years
A = 100000 * (1 + 8/100)²
A= 100000 x (108/100) × (108/100)
A = 116640
The population after 2 years will be 116640.