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AFCAT Numerical Ability · Formula sheet

Simple and Compound Interest formulas

32 formulas and 37 common traps for AFCAT Numerical Ability Simple and Compound Interest, grouped by subtopic.

Full notes with worked examples

Simple Interest: the Formula, Both Ways

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Simple interest and amount

Simple interest

SI=P×R×T100,A=P+SISI = \dfrac{P \times R \times T}{100}, \quad A = P + SI

Finding the principal, rate or time

Finding the rate or time

R=100×SIP×T,T=100×SIP×RR = \dfrac{100 \times SI}{P \times T}, \quad T = \dfrac{100 \times SI}{P \times R}

Rates per month, quarter or half-year

Matching the units

SI=P×(rate per period)×(number of periods)100SI = \dfrac{P \times (\text{rate per period}) \times (\text{number of periods})}{100}

Common traps

Turn months into years

At 12% a year, 8 months is 812\tfrac{8}{12} of a year, so the interest on ₹5,000 is ₹400. Putting T=8T = 8 gives ₹4,800.

Interest or amount

₹6,000 at 5% for 2 years earns ₹600 and amounts to ₹6,600. Read which one is asked: the options usually carry both.

Take the interest out of the amount first

₹6,000 amounting to ₹7,440 has earned ₹1,440. Dividing the whole ₹7,440 by 6000×46000 \times 4 as if it were the interest gives a rate of 31%.

A rate per period is not a yearly rate

At 3% a quarter the yearly rate is 12%. Using 3% as the yearly rate makes the interest a quarter of what it should be.

Compounding words at simple interest

'Calculated half-yearly' changes nothing at simple interest. Only compound interest grows faster when it is added more often; that comes later in the chapter.

Finding the Principal or Rate from Two Facts

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A sum given at two different times

One year's interest

I1=A2−A1t2−t1,P=A1−t1I1I_1 = \dfrac{A_2 - A_1}{t_2 - t_1}, \quad P = A_1 - t_1 I_1

Find the rate in one case, use it in another

Rate from a complete case

R=100 (A−P)P×TR = \dfrac{100\,(A - P)}{P \times T}

Common traps

Take off the right number of years

With ₹1,610 after 3 years and ₹70 a year, the principal is 1610−3×70=14001610 - 3 \times 70 = 1400. Taking off only one year's interest, or the years of the later date, gives a wrong principal that is still among the options.

'A further' period

'₹2,200 after 2 years and ₹2,600 after a further 4 years' puts the second date at year 6. The ₹400 is four years' interest, not six.

Change the rate, not the amount

If the rate is doubled, a rate of 4% becomes 8% and the interest doubles. Doubling the old amount, principal included, is a different and wrong number.

Changing the Rate, Time or Sum, and Splitting a Sum

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Interest scales with the sum, the rate and the time

Scaling the interest

SI2=SI1×P2R2T2P1R1T1SI_2 = SI_1 \times \dfrac{P_2 R_2 T_2}{P_1 R_1 T_1}

A difference of two interests gives the sum

Sum from a difference

P=100×difference(R2−R1) TP = \dfrac{100 \times \text{difference}}{(R_2 - R_1)\,T}

One sum split between two rates

Splitting a sum

x⋅R1T1100+(S−x)⋅R2T2100=Ix \cdot \dfrac{R_1 T_1}{100} + (S - x) \cdot \dfrac{R_2 T_2}{100} = I

Common traps

Scale, do not add

Raising the rate from 5% to 7% multiplies the interest by 75\tfrac75. Adding ₹2 to the interest, or 2% of the interest, is not the same thing.

Multiply the rate difference by the time

Rates of 5% and 8% for 2 years differ by 6% of the sum, not 3%. Forgetting the time doubles the sum you find.

Answer the part that is asked

If xx is the part at the lower rate, the question may ask for the part at the higher rate, S−xS - x. Both are among the options.

Equal interest means the inverse ratio

Equal interest at 12% and at 4% puts more money at the lower rate: the parts at 12% and at 4% are in the ratio 4:12=1:34 : 12 = 1 : 3, not 3:13 : 1.

When a Sum Doubles, Triples or Becomes n Times at Simple Interest

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The multiple and the rate

Multiple and rate

R×T=100 (n−1)R \times T = 100\,(n - 1)

From one multiple to another

Time to a new multiple

T2=T1×n2−1n1−1T_2 = T_1 \times \dfrac{n_2 - 1}{n_1 - 1}

The multiple, amount or interest after another time

Multiple after t years

1+t×n−1T1 + t \times \dfrac{n - 1}{T}

Common traps

n times means n minus 1 as interest

A sum that becomes 4 times itself in 12 years has earned 300%, so the rate is 25%. Using 400% gives 3313%33\tfrac13\%.

Count interest, not the multiple

A sum doubling in 3 years at simple interest becomes 4 times itself in 9 years, because 4 times needs three principals of interest. 6 years is the compound interest answer.

Add the 1 back

After 10 years at 15% a year a sum has earned 1.5 times itself, so it is 2.5 times itself. Stopping at 1.5 answers 'how much interest', not 'how many times'.

Compound Interest Added Once a Year

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Amount and compound interest

Compound amount

A=P(1+R100)n,CI=A−PA = P\left(1 + \dfrac{R}{100}\right)^n, \quad CI = A - P

Principal, rate or time from the factor

Present worth

P=A(1+R100)nP = \dfrac{A}{\left(1 + \frac{R}{100}\right)^n}

The same factor on another sum

Carrying the factor

A2=P2×A1P1A_2 = P_2 \times \dfrac{A_1}{P_1}

Two amounts in a row, and one year's interest

Rate from consecutive amounts

1+R100=An+1An1 + \dfrac{R}{100} = \dfrac{A_{n+1}}{A_n}

Changing rates, part years, growth and depreciation

Depreciation

Vn=V(1−r100)nV_n = V\left(1 - \dfrac{r}{100}\right)^n

Common traps

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.

Divide by the factor

The present worth of ₹4,840 due in 2 years at 10% is 4840÷1.21=40004840 \div 1.21 = 4000. Taking 20% off ₹4,840 gives ₹3,872, which is wrong.

A different time needs the yearly factor

A sum that becomes 1.1664 times itself in 2 years becomes 1.083=1.2597121.08^3 = 1.259712 times itself in 3 years, not 1.1664×1.5=1.74961.1664 \times 1.5 = 1.7496.

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%.

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 2122\tfrac12 years at 10% compounded annually the factor is 1.21×1.051.21 \times 1.05. Writing 1.12.51.1^{2.5} treats the half year as compounded.

When a Sum Becomes n Times at Compound Interest

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Whole powers: multiply the time

Multiples at compound interest

(1+R100)T=n  ⇒  (1+R100)kT=nk\left(1 + \tfrac{R}{100}\right)^T = n \;\Rightarrow\; \left(1 + \tfrac{R}{100}\right)^{kT} = n^k

Through a common base

Common base

bp in T years  ⇒  bq in qp T yearsb^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(nT−1)R = 100\left(\sqrt[T]{n} - 1\right)

The multiple after another time

Multiple after t years

nt/Tn^{t/T}

Common traps

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.

Scale the power, not the multiple

From 243 times in 10 years, 9 times takes 25\tfrac25 of 10 years, which is 4 years. Taking 9243\tfrac{9}{243} of 10 years treats the growth as a straight line.

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.

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 7:17 : 1, not 8:18 : 1.

The Difference Between Compound and Simple Interest

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Two years: P(R/100)²

Two-year gap

CI−SI=P(R100)2CI - SI = P\left(\dfrac{R}{100}\right)^2

The rate from the two-year gap

Rate from the gap

R=100CI−SIPR = 100\sqrt{\dfrac{CI - SI}{P}}

Three years

Three-year gap

CI−SI=P(R100)2(3+R100)CI - SI = P\left(\dfrac{R}{100}\right)^2\left(3 + \dfrac{R}{100}\right)

From one interest to the other

Sum from compound interest

P=CI(1+R100)n−1P = \dfrac{CI}{\left(1 + \frac{R}{100}\right)^n - 1}

Both interests given: interest on interest

Rate from both interests

R100=CI−SISI/2\dfrac{R}{100} = \dfrac{CI - SI}{SI / 2}

Common traps

Square the rate as a fraction

At 5% the factor is 0.052=0.00250.05^2 = 0.0025, not 0.025. Using 0.025 makes the gap ten times too big.

R squared is not R

If R2=49R^2 = 49, the rate is 7%, not 49% and not 4.9%.

Three years is not three times two

At 20% on ₹1,000 the two-year gap is ₹40, but the three-year gap is ₹128. Tripling ₹40 misses the interest earned on interest on interest.

Divide by the compound share

At 5% for 2 years compound interest is 10.25% of the sum, so ₹1,230 of it means a sum of ₹12,000. Dividing by 10% instead treats it as simple interest and overstates the sum.

Halve the simple interest first

With ₹1,000 and ₹1,050 for 2 years, the rate is 50500=10%\tfrac{50}{500} = 10\%. Dividing by the whole ₹1,000 gives 5%.

Compounding Half-Yearly, Quarterly or Monthly

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Rate per period and number of periods

Compounded k times a year

A=P(1+R100k)knA = P\left(1 + \dfrac{R}{100k}\right)^{kn}

Annual against half-yearly, and CI against SI

Gap between two compoundings

gap=P×(factor1−factor2)\text{gap} = P \times (\text{factor}_1 - \text{factor}_2)

Effective annual rate

E=(1+R100k)k−1E = \left(1 + \dfrac{R}{100k}\right)^k - 1

From an effective rate back to a period rate

Period rate from the effective rate

rate per period=1+E100k−1\text{rate per period} = \sqrt[k]{1 + \dfrac{E}{100}} - 1

Common traps

Change both the rate and the count

For 2 years at 12% compounded half-yearly the factor is 1.0641.06^4. Halving the rate but keeping 2 periods, or keeping 12% for 4 periods, are both offered as options.

The two-year shortcut is annual only

P(R/100)2P(R/100)^2 is the gap between compound and simple interest for 2 years compounded annually. With half-yearly compounding, work out both interests in full.

Never below the quoted rate

12% compounded monthly is about 12.68% a year. An option below 12% cannot be its effective rate.

Take the root, do not divide

21% a year is 10% a half-year compounded, not 10.5%. Halving a compound rate is the simple interest habit.

Paying Back a Loan in Instalments

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Equal instalments

Two equal instalments

x=loan×(1+r)22+rx = \text{loan} \times \dfrac{(1 + r)^2}{2 + r}

The loan from its payments

Loan from its payments

loan=∑kxk(1+r)k\text{loan} = \sum_k \dfrac{x_k}{(1 + r)^k}

Carrying the balance forward

Balance after a year

Bk=Bk−1(1+r)−xkB_k = B_{k-1}(1 + r) - x_k

Common traps

An instalment is not loan ÷ number

₹4,200 in two equal instalments at 10% is ₹2,420 each, not ₹2,100 plus some simple interest. Each payment pays off a different amount of today's money.

Each payment has its own factor

₹3,630 due in 2 years at 10% is worth 3630÷1.21=30003630 \div 1.21 = 3000 today. Dividing every payment by 1.1 overstates the loan.

Grow before you subtract

A payment at the end of a year comes after that year's interest. Taking it off first and then adding interest charges less interest than the loan really owes.

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