AFCAT Numerical Ability · Simple and Compound Interest

Compound Interest Added Once a Year

Each year's interest is added to the sum and earns in its turn, so the amount is P(1 + R/100)ⁿ; the same factor read backwards gives the principal, the rate or the time.

All Simple and Compound Interest notes

Why this matters

This is the largest page of the chapter, and AFCAT asked it in 2016: the interest on a sum for three years. Most rows need only the factor (1 + R/100)ⁿ. Marks go when the rate or time is found by trial instead of by spotting a factor such as 1.21, 1.1025 or 1.331.

Concept 1 of 5: Amount and compound interest

₹5,000 at 10%: after one year it is ₹5,500. In the second year the interest is on ₹5,500, so ₹550, and the amount is ₹6,050. Simple interest would have given ₹6,000.

Definition

$A = P\left(1 + \tfrac{R}{100}\right)^n$ and the compound interest is $CI = A - P$ .
Factors worth knowing: $1.1^2 = 1.21$ , $1.1^3 = 1.331$ , $1.05^2 = 1.1025$ , $1.2^2 = 1.44$ , $1.08^2 = 1.1664$ , $1.12^2 = 1.2544$ .
Several investments: work out each one's interest and add. A part at simple interest still uses the simple interest formula.
A rise in the rate: work out the interest at each rate and subtract.

Compound amount

A = P\left(1 + \dfrac{R}{100}\right)^n, \quad CI = A - P

Worked example

Find the compound interest on ₹8,000 at 10% a year for 3 years.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2016 · AFCAT (I) 2016 — Numerical Ability · 04-208Easy

Example 1 · Simple and Compound Interest · Compound Interest: Annual

What would be the compound interest obtained on an amount of ₹4000 at the rate of 5% per annum after 3 yr?

Compound interest is not rate × years

At 10% for 3 years the compound interest is 33.1% of the sum, not 30%. The 30% figure is simple interest, and it is always among the options.

Drill 18 more on amount and compound interest

Concept 2 of 5: Principal, rate or time from the factor

Divide the amount by the principal. The result is

(1 + R/100)^n

; recognise it and the rate and time follow. 1.21 is 10% for 2 years, 1.1025 is 5% for 2 years, 1.331 is 10% for 3 years.

Definition

Principal, or the present worth of a sum due later: $P = \dfrac{A}{(1 + R/100)^n}$ .
From the interest: $P = \dfrac{CI}{(1 + R/100)^n - 1}$ .
Rate: take the $n$ th root of $\tfrac AP$ . More squares to know: $1.04^2 = 1.0816$ , $1.06^2 = 1.1236$ , $1.15^2 = 1.3225$ .
Time: keep multiplying by the yearly factor until you reach $\tfrac AP$ .

Present worth

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

Worked example

What sum amounts to ₹9,261 in 3 years at 5% compound interest?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Simple and Compound Interest · Compound Interest: Annual

The compound interest on ₹30000 at 7% per annum for a certain time is ₹4347. The time is

Divide by the factor

The present worth of ₹4,840 due in 2 years at 10% is

4840 \div 1.21 = 4000

. Taking 20% off ₹4,840 gives ₹3,872, which is wrong.

Drill 28 more on principal, rate or time from the factor

Concept 3 of 5: The same factor on another sum

At the same rate and for the same time every sum grows by the same factor. If ₹4,000 becomes ₹4,410 in 2 years, any sum becomes 1.1025 times itself in 2 years, and the rate need not be found.

Definition

Same rate, same time: new amount = new sum × $\dfrac{A_1}{P_1}$ . The interest scales the same way.
Same rate, a different time: find the yearly factor first (here $\sqrt{1.1025} = 1.05$ ), then raise it to the new time.
A 2-year factor squared is the 4-year factor.
A changed rate: find the rate, change it, and build the new factor.

Carrying the factor

A_2 = P_2 \times \dfrac{A_1}{P_1}

Worked example

₹8,000 becomes ₹9,331.20 in 2 years at compound interest. What will ₹12,500 become in 2 years, and in 3 years, at the same rate?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Simple and Compound Interest · Compound Interest: Annual

If Rs.

6,000

amounts to Rs.

7,260

2

years at compound interest, what will Rs.

9,000

amount to in

3

years at the same rate?

A different time needs the yearly factor

A sum that becomes 1.1664 times itself in 2 years becomes

1.08^3 = 1.259712

times itself in 3 years, not

1.1664 \times 1.5 = 1.7496

Drill 6 more on the same factor on another sum

Concept 4 of 5: Two amounts in a row, and one year's interest

Amounts after consecutive years differ by one year's interest on the earlier amount. So the later amount divided by the earlier one is

1 + R/100

, which gives the rate at once; dividing back gives the principal.

Definition

Rate: $R = \dfrac{A_{n+1} - A_n}{A_n} \times 100$ .
Principal: $P = \dfrac{A_n}{(1 + R/100)^n}$ .
The interest of year $k$ alone is $R\%$ of the amount at the start of that year, $P(1 + R/100)^{k-1}$ .
The compound interest for 3 years minus that for 2 years is the third year's interest.
Two amounts two years apart: their ratio is $(1 + R/100)^2$ .

Rate from consecutive amounts

1 + \dfrac{R}{100} = \dfrac{A_{n+1}}{A_n}

Worked example

A sum amounts to ₹6,050 in 2 years and to ₹6,655 in 3 years at compound interest. Find the rate and the sum.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Simple and Compound Interest · Compound Interest: Annual

A sum amounts to ₹2916 in 2 yr and ₹3149.28 in 3 yr at compound interest. The sum is

Divide by the earlier amount

From ₹6,050 to ₹6,655 is ₹605 on ₹6,050, which is 10%. Dividing by the later amount gives about 9.1%.

Drill 18 more on two amounts in a row, and one year's interest

Concept 5 of 5: Changing rates, part years, growth and depreciation

Each year multiplies the sum by its own factor, so a rate that changes is just a product of different factors. The same multiplying describes a town that grows or a machine that loses value every year.

Definition

Different rates in different years: $A = P\left(1 + \tfrac{R_1}{100}\right)\left(1 + \tfrac{R_2}{100}\right)\left(1 + \tfrac{R_3}{100}\right)$ .
A part year at annual compounding: the whole years compound, and the part year earns simple interest on the amount reached. $1\tfrac12$ years at 10% is $1.1 \times 1.05 = 1.155$ .
Growth of $r\%$ a year: $P\left(1 + \tfrac{r}{100}\right)^n$ . Depreciation of $r\%$ a year: $V\left(1 - \tfrac{r}{100}\right)^n$ .

Depreciation

V_n = V\left(1 - \dfrac{r}{100}\right)^n

Worked example

₹5,000 is invested for 3 years, at 10% in the first year, 12% in the second and 15% in the third, compounded annually. Find the amount.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 5 · Simple and Compound Interest · Compound Interest: Annual

What sum of money at compound interest will amount to ₹1260 in 3 yr, if the rate of interest is 5% for the first year, 20% for the second year and 25% for the third year?

A falling value multiplies by less than 1

A loss of 10% a year for 3 years leaves 72.9% of the value, not 70%.

The part year is simple

For

2\tfrac12

years at 10% compounded annually the factor is

1.21 \times 1.05

. Writing

1.1^{2.5}

treats the half year as compounded.

Drill 6 more on changing rates, part years, growth and depreciation

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 (5)

Amount and compound interest
Compound amount
$A = P\left(1 + \dfrac{R}{100}\right)^n, \quad CI = A - P$
Principal, rate or time from the factor
Present worth
$P = \dfrac{A}{\left(1 + \frac{R}{100}\right)^n}$
The same factor on another sum
Carrying the factor
$A_2 = P_2 \times \dfrac{A_1}{P_1}$
Two amounts in a row, and one year's interest
Rate from consecutive amounts
$1 + \dfrac{R}{100} = \dfrac{A_{n+1}}{A_n}$
Changing rates, part years, growth and depreciation
Depreciation
$V_n = V\left(1 - \dfrac{r}{100}\right)^n$

Watch out for (6)

Compound interest is not rate × years
→ Amount and compound interest
Divide by the factor
→ Principal, rate or time from the factor
A different time needs the yearly factor
→ The same factor on another sum
Divide by the earlier amount
→ Two amounts in a row, and one year's interest
A falling value multiplies by less than 1
→ Changing rates, part years, growth and depreciation
The part year is simple
→ Changing rates, part years, growth and depreciation

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