SSC CGL Maths · Simple and Compound Interest

Compound Interest

Under compound interest each year's interest joins the principal, so the amount is multiplied by the same growth factor every year.

All Simple and Compound Interest notes

Why this matters

Eighteen questions in the bank, seventeen of them from the 2025 shifts: compound interest was almost absent from the earlier papers. Twelve are EASY. Know the growth factors of 10%, 20% and 12% by heart and most of these become one multiplication or one division.

Concept 1 of 4: Compound interest amount and principal

In year 2 the lender charges interest on last year's interest too. So the sum grows by a fixed percent of itself each year, like a price that rises by 10% and then by 10% of the new price.

Definition

Amount: $A = P\left(1 + \frac{r}{100}\right)^n$ for $n$ years compounded yearly. Compound interest: $CI = A - P$ .
Principal from the amount: $P = A \div \left(1 + \frac{r}{100}\right)^n$ .
Fast route, 2 years: the total percent is $2r + \frac{r^2}{100}$ . At 10% that is 21%; at 20% it is 44%.
Fast route, 3 years: go year by year. At 10%, 1000 becomes 1100, 1210, 1331.
Write the factor as a fraction when it is friendly: 10% is $\frac{11}{10}$ , 20% is $\frac{6}{5}$ , 12% is $\frac{28}{25}$ , 4% is $\frac{26}{25}$ .
Profit or growth that is reinvested every year is compound interest by another name.

Compound interest, yearly

A = P\left(1 + \frac{r}{100}\right)^n \qquad CI = A - P \qquad \text{2 years: } \left(2r + \frac{r^2}{100}\right)\%

rrate percent per year
nnumber of years, compounded yearly

Worked example

Find the compound interest on ₹6,000 at 5% per annum for 3 years, compounded annually.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 1 · Q53Easy

Example 1 · Simple and Compound Interest · Compound Interest

What is the compound interest on Rs. 8000 for 2 year at

10\%

per annum?

Doubling the first year's interest

Compound interest for 2 years is more than twice one year's interest, because year 2 also charges interest on year 1's interest. At 10% on ₹5,000 it is ₹1,050, not ₹1,000.

Giving the amount when compound interest is asked

P(1 + r)^n

is the amount. The compound interest is the amount minus the principal, and both figures usually sit among the options.

Drill 5 more on compound interest amount and principal

Concept 2 of 4: Rate or time from the growth factor

Divide the amount by the principal and you get the growth factor raised to the number of years. Recognise the power and you have both the rate and the time.

Definition

$\frac{A}{P} = \left(1 + \frac{r}{100}\right)^n$ .
Time: match the ratio to a known power. $1.21 = 1.1^2$ , $1.331 = 1.1^3$ , $1.44 = 1.2^2$ , $1.728 = 1.2^3$ , $1.2544 = 1.12^2$ , $1.1025 = 1.05^2$ .
Rate: take the $n$ th root of the ratio. For 2 years, $1 + \frac{r}{100} = \sqrt{A/P}$ .
Steady growth of production or population is the same rule: each year's figure is the last one times the same factor. Find the factor from two neighbouring years, then apply it again.

Growth factor

\left(1 + \frac{r}{100}\right)^n = \frac{A}{P}

Worked example

In how many years will ₹8,000 become ₹10,648 at 10% per annum compound interest?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 17 Sep 2025, Shift 2 · Q57Easy

Example 2 · Simple and Compound Interest · Compound Interest

If ₹5000 amounts to ₹6050 over 2 years with compound interest, what is the annual interest rate?

Dividing the total growth by the years

A 21% rise over 2 years at compound interest is 10% a year, not 10.5%:

1.1^2 = 1.21

. Splitting the total growth evenly is the simple interest method.

Reading constant growth as a constant addition

A constant growth rate means the same percent each year, so the increase itself grows. From 2,500 to 3,000 is 20%, and the next year rises by 600, not 500.

Drill 5 more on rate or time from the growth factor

Concept 3 of 4: Interest year by year and consecutive amounts

Each year's interest is the rate on the amount at the start of that year. So the jump from one year's amount to the next is that rate on the earlier amount.

Definition

Consecutive amounts $A_n$ and $A_{n+1}$ : the rate is $\frac{A_{n+1} - A_n}{A_n} \times 100$ . Then the principal is $A_n$ divided by the factor $n$ times.
Interest for the $k$ th year alone $= \frac{r}{100} \times$ the amount after $k - 1$ years $= P\left(1 + \frac{r}{100}\right)^{k-1} \times \frac{r}{100}$ .
The interest for the 3rd year is not the compound interest for 3 years; it is one year's slice.
Under simple interest, by contrast, the jump is the same every year and is a percent of the principal.

Year-by-year compound interest

r = \frac{A_{n+1} - A_n}{A_n} \times 100 \qquad I_k = P\left(1 + \frac{r}{100}\right)^{k-1} \times \frac{r}{100}

$A_n$ the amount after $n$ years
$I_k$ the interest earned in year $k$ alone

Worked example

A sum at compound interest amounts to ₹7,200 after 2 years and ₹8,640 after 3 years. Find the rate and the sum.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 3 · Q60Easy

Example 3 · Simple and Compound Interest · Compound Interest

If the amount at the end of the 2nd year and 3rd year on a certain principal at compound interest is ₹12,000 and ₹13,200 respectively, find the rate of interest per annum.

Using the simple interest method on compound amounts

Under compound interest the gap between two consecutive amounts is the rate on the earlier amount, not on the principal. Stepping back by equal gaps, as for simple interest, gives the wrong principal.

Drill 3 more on interest year by year and consecutive amounts

Concept 4 of 4: Compound and simple interest on the same sum

For one year the two are equal. After that compound interest pulls ahead, by exactly the interest earned on earlier interest.

Definition

For the same $P$ , $r$ and $n$ years compounded yearly: $CI = SI$ when $n = 1$ , and $CI > SI$ after.
Given CI, find SI: get the principal from the CI factor, $P = \frac{CI}{(1 + r/100)^n - 1}$ , then $SI = \frac{P r n}{100}$ .
The gap for 2 years: $CI - SI = P\left(\frac{r}{100}\right)^2$ , the interest on the first year's interest.
The gap for 3 years: $CI - SI = P\left(\frac{r}{100}\right)^2\left(3 + \frac{r}{100}\right)$ .

Gap between compound and simple interest

CI - SI = P\left(\frac{r}{100}\right)^2 \ (2 \text{ years}) \qquad CI - SI = P\left(\frac{r}{100}\right)^2\left(3 + \frac{r}{100}\right) \ (3 \text{ years})

Worked example

The compound interest on a sum at 10% per annum for 2 years is ₹1,050. Find the simple interest on it at the same rate for the same time.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 13 Sep 2025, Shift 2 · Q61Moderate

Example 4 · Simple and Compound Interest · Compound Interest

If the compound interest on a certain sum at

16\frac{2}{3}\%

per annum for 3 years is ₹1270, find the simple interest on the same sum at the same rate and for the same period.

Expecting a gap in the first year

Compounded yearly, compound and simple interest are equal for the first year. The gap starts in year 2, which is why the 2-year gap is just the interest on the first year's interest.

Drill 1 more on compound and simple interest on the same sum

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Compound interest amount and principal
Compound interest, yearly
$A = P\left(1 + \frac{r}{100}\right)^n \qquad CI = A - P \qquad \text{2 years: } \left(2r + \frac{r^2}{100}\right)\%$
Rate or time from the growth factor
Growth factor
$\left(1 + \frac{r}{100}\right)^n = \frac{A}{P}$
Interest year by year and consecutive amounts
Year-by-year compound interest
$r = \frac{A_{n+1} - A_n}{A_n} \times 100 \qquad I_k = P\left(1 + \frac{r}{100}\right)^{k-1} \times \frac{r}{100}$
Compound and simple interest on the same sum
Gap between compound and simple interest
$CI - SI = P\left(\frac{r}{100}\right)^2 \ (2 \text{ years}) \qquad CI - SI = P\left(\frac{r}{100}\right)^2\left(3 + \frac{r}{100}\right) \ (3 \text{ years})$

Watch out for (6)

Doubling the first year's interest
→ Compound interest amount and principal
Giving the amount when compound interest is asked
→ Compound interest amount and principal
Dividing the total growth by the years
→ Rate or time from the growth factor
Reading constant growth as a constant addition
→ Rate or time from the growth factor
Using the simple interest method on compound amounts
→ Interest year by year and consecutive amounts
Expecting a gap in the first year
→ Compound and simple interest on the same sum

Test yourself on Simple and Compound Interest

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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