AFCAT Numerical Ability · Simple and Compound Interest

When a Sum Becomes n Times at Compound Interest

At compound interest equal stretches of time multiply the sum by equal factors, so a sum that becomes n times itself in T years becomes n² times in 2T years and n³ times in 3T years.

All Simple and Compound Interest notes

Why this matters

No AFCAT past-year question is on this page, but it is a big one and about a fifth of it is HARD. Every row is the same idea: the times add while the multiples multiply. The trap is the simple interest rule from earlier in the chapter, where the times add while the interest adds.

Concept 1 of 4: Whole powers: multiply the time

If a sum doubles in 5 years, the next 5 years double it again: 4 times itself in 10 years, 8 times in 15. Each equal period multiplies the sum by the same factor, so the periods add.

Definition

Becomes $n$ times in $T$ years, so becomes $n^k$ times in $kT$ years.
Write the target as a power of the given multiple: $125 = 5^3$ , $81 = 9^2$ , $1000 = 10^3$ .
The rate itself is never needed.

Multiples at compound interest

\left(1 + \tfrac{R}{100}\right)^T = n \;\Rightarrow\; \left(1 + \tfrac{R}{100}\right)^{kT} = n^k

Worked example

A sum becomes 4 times itself in 7 years at compound interest. In how many years will it become 64 times itself?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Simple and Compound Interest · Compound Interest: Sum Becoming n Times

A certain sum becomes

3

times itself in

6

years at compound interest. In how many years will it become

27

times?

Multiples multiply, they do not add

Doubling in 5 years makes 4 times itself in 10 years at compound interest. 'Twice the time, twice the interest' is the simple interest rule, and gives only 3 times.

Drill 27 more on whole powers: multiply the time

Concept 2 of 4: Through a common base

243 times and 9 times are both powers of 3. A sum that becomes

243 = 3^5

times itself in 10 years triples every 2 years, so 9 times (

3^2

) takes 4 years. Bring both multiples to one base.

Definition

Write both multiples as powers of one base $b$ . If $b^p$ takes $T$ years, then $b^q$ takes $\dfrac{q}{p}\,T$ years.
The time for one factor of $b$ is $\dfrac{T}{p}$ .
It works for a smaller target too: 81 times in 20 years means 9 times in 10 years and 3 times in 5.

Common base

b^p \text{ in } T \text{ years} \;\Rightarrow\; b^q \text{ in } \dfrac{q}{p}\,T \text{ years}

Worked example

A sum becomes 32 times itself in 15 years at compound interest. In how many years will it become 4 times itself?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Simple and Compound Interest · Compound Interest: Sum Becoming n Times

The time taken by a sum to become 8 times itself at compound interest is 9 years. Find the time required for the sum to become 16 times.

Scale the power, not the multiple

From 243 times in 10 years, 9 times takes

\tfrac25

of 10 years, which is 4 years. Taking

\tfrac{9}{243}

of 10 years treats the growth as a straight line.

Drill 14 more on through a common base

Concept 3 of 4: The rate from a multiple

If a sum becomes 9 times itself in 2 years, each year multiplies it by 3, since

3^2 = 9

. A yearly factor of 3 is a rate of 200%.

Definition

$1 + \tfrac{R}{100} = \sqrt[T]{n}$ , so $R = 100\left(\sqrt[T]{n} - 1\right)$ .
A factor of 2 is 100%, a factor of 3 is 200%, a factor of 5 is 400%.
When the root is not whole, simplify it first: $\sqrt[4]{9} = \sqrt3 \approx 1.732$ , a rate of about 73.2%.

Rate from a multiple

R = 100\left(\sqrt[T]{n} - 1\right)

Worked example

At what rate will a sum become 64 times itself in 3 years at compound interest?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Simple and Compound Interest · Compound Interest: Sum Becoming n Times

At compound interest a sum becomes

8

times in

6

years. What is the approximate annual rate of interest?

The rate is the factor minus 1

A yearly factor of 3 is a rate of 200%, not 300%. The rate is only the part added to the 1.

Drill 9 more on the rate from a multiple

Concept 4 of 4: The multiple after another time

Once the yearly factor is known, any time works: after

t

years the sum is (yearly factor)

^t

times itself. If ₹1,000 becomes ₹27,000 in 3 years, the factor is 3, and after 2 years the sum is 9 times itself.

Definition

A sum that becomes $n$ times itself in $T$ years is $n^{t/T}$ times itself after $t$ years.
The interest after $t$ years is that multiple minus 1, so the ratio of compound interest to principal is $\left(n^{t/T} - 1\right) : 1$ .
With money given: multiply the principal by the multiple, or divide a target amount by it to find the principal.

Multiple after t years

n^{t/T}

Worked example

₹2,000 becomes ₹18,000 in 2 years at compound interest. What will it be after 3 years?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Simple and Compound Interest · Compound Interest: Sum Becoming n Times

Rs.

10\,000

becomes Rs.

40\,000

4

years at compound interest. What will be the amount after

6

years at the same rate?

Subtract the principal for the interest

A sum that becomes 8 times itself has earned 7 times itself. The ratio of interest to principal is

7 : 1

, not

8 : 1

Drill 9 more on the multiple after another time

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)

Whole powers: multiply the time
Multiples at compound interest
$\left(1 + \tfrac{R}{100}\right)^T = n \;\Rightarrow\; \left(1 + \tfrac{R}{100}\right)^{kT} = n^k$
Through a common base
Common base
$b^p \text{ in } T \text{ years} \;\Rightarrow\; b^q \text{ in } \dfrac{q}{p}\,T \text{ years}$
The rate from a multiple
Rate from a multiple
$R = 100\left(\sqrt[T]{n} - 1\right)$
The multiple after another time
Multiple after t years
$n^{t/T}$

Watch out for (4)

Multiples multiply, they do not add
→ Whole powers: multiply the time
Scale the power, not the multiple
→ Through a common base
The rate is the factor minus 1
→ The rate from a multiple
Subtract the principal for the interest
→ The multiple after another time

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