Aptitude · Compound Interest · Model 1

Finding whichever of the five is missing

Q: How do you handle a time of 2½ years?

Compound for the two whole years, then apply simple interest for the half year. At 10% that is P × (1.1)² × (1 + 5/100). Raising 1.1 to the power 2.5 is not the intended convention and gives a slightly different answer.

The direct model: principal, rate, time, amount and interest, with any one hidden. Forwards it is a power; backwards it is a root or a logarithm — except that exam numbers are always chosen so you never actually need either.

Work any of the five quantities, forwards or backwards →

Going backwards, divide out the power rather than reaching for a root. Exam rates make (1 + R/100)^T a recognisable fraction: 1.21, 1.331, 1.44.

01 The idea

The five quantities, and which are easy to recover

Model 1 is every question that hands you three or four of P, R, T, A, CI and asks for the rest. Forwards — principal, rate and time given — it is one application of the formula and nothing else.

Backwards splits into two cases. Recovering the principal from an amount is easy: divide by the multiplier. If ₹1,331 is the amount after 3 years at 10%, the principal is 1331 ÷ 1.331 = ₹1,000.

Recovering the rate or the time is harder in principle, because you need a root or a logarithm. In practice you never do, because setters choose numbers that make the multiplier a recognisable value. If A/P comes to 1.21 you are looking at 10% for two years; 1.331 is 10% for three; 1.44 is 20% for two.

So the working skill in Model 1 is recognising these multipliers on sight. A short list covers nearly every question you will meet, and it converts what looks like a logarithm problem into pattern matching.

Compute A/P first. If it is 1.21, 1.331, 1.44 or 1.728, you already know the rate and the time without solving anything.

Growth multiplierA/P = (1 + R/100)^T. The single number that carries all the information about rate and time together.

Recognisable powers1.1² = 1.21, 1.1³ = 1.331, 1.2² = 1.44, 1.2³ = 1.728, 1.05² = 1.1025. Exam questions are built from these.

Fractional yearsFor a time like 2½ years, compound for the whole years and apply simple interest for the fraction — the standard convention, and a common source of error.

02 Worked example

₹1,331 after 3 years at 10% — find the principal

The reverse direction, which is where Model 1 questions actually live. A sum amounts to ₹1,331 in 3 years at 10% per annum compound interest. Find the principal and the compound interest.

Write the multiplierThree years at 10% means multiplying by 1.1 three times.(1 + 10/100)³ = (11/10)³ = 1331/1000 = 1.331

Divide the amount by itThe amount is the principal times the multiplier, so the principal is the amount divided by it.P = 1331 ÷ 1.331 = ₹1,000

Prefer the fraction to the decimalKeeping 11/10 as a fraction makes the division exact and obvious.P = 1331 × 1000/1331 = ₹1,000

Get the interest by subtractionAs always in this chapter, the interest is the amount less the principal.CI = 1331 − 1000 = ₹331

Check it forwardsPush the principal back through and confirm you land on the amount.1000 × 1.1 × 1.1 × 1.1 = 1331 ✓

Notice that recognising 1331 as 11³ is what makes this a ten-second question. Exam setters use 1,331, 1,210, 1,728 and 1,225 precisely because they are clean powers, so a number that looks arbitrary is usually a signal. If a division comes out untidy, check whether you have the right number of years before assuming the arithmetic is just ugly.

03 The method

The five cases, and the fractional-year convention

Four of the five are a single line. The fifth — a time that is not a whole number of periods — has a convention you have to know.

A = P(1 + R/100)^T, so P = A ÷ (1 + R/100)^T, and the multiplier A/P determines R and T together.

For a fractional time, compound the whole years and add simple interest for the fraction. For 2½ years at 10%: A = P(1.1)² × (1 + 5/100) — two years compounded, then half a year’s simple interest at 5%. This is the standard exam convention; raising 1.1 to the power 2.5 is not what is wanted.

A/P	Means	Recognise it as
1.1025	`5% for 2 years`	`(21/20)²`
1.21	`10% for 2 years`	`(11/10)²`
1.331	`10% for 3 years`	`(11/10)³`
1.44	`20% for 2 years`	`(6/5)²`
1.728	`20% for 3 years`	`(6/5)³`
1.2544	`12% for 2 years`	`(28/25)²`
Untidy result	`re-check T`	usually a wrong period count

05 Cheat sheet

Model 1 on one page

The five cases and the two conventions. The last row is the one most often got wrong.

Missing	Route	Worked
Amount	`P(1+R/100)^T`	1000 at 10%, 3 yr → 1,331
Compound interest	`A − P`	1,331 − 1,000 = 331
Principal	`A ÷ (1+R/100)^T`	1,331 / 1.331 = 1,000
Rate or time	`recognise A/P`	1.331 → 10%, 3 years
Half-yearly	`R/2 over 2T periods`	10% 2 yr → 1,215.51
Fractional year	`compound whole, simple for the part`	2½ yr → (1.1)² × 1.05
Power for a fractional year	`wrong`	not 1.1^2.5

Learn the standard multipliers1.21, 1.331, 1.44, 1.728 and 1.1025 cover most questions. Recognising A/P as one of these removes the need for any root or logarithm.

Keep rates as fractions10% is 11/10 and 20% is 6/5. Fractions cancel exactly against exam principals; decimals invite rounding for no benefit.

Fractional years use the mixed conventionCompound the whole years, then apply simple interest for the leftover fraction. Raising the multiplier to a fractional power is not the intended method.

06 Where & why

Where Model 1 shows up

The bread-and-butter compound interest question, set as a speed item rather than a thinking one.

Every placement paper

Two- or three-year direct calculation

Almost always at 10% or 20% so the power stays clean. Answerable in seconds if you know the multiplier.

Bank PO · SSC

Principal from a given amount

The reverse direction, and the reason the standard multipliers are worth memorising — recognising 1,331 as 11³ is the whole question.

Half-yearly variants

The same question with the rate halved

Set to check whether you adjust both the rate and the period count rather than just one of them.

Fractional-year questions

2½ or 1¾ years

The mixed convention is the entire content. Students who use a fractional power get a close but wrong answer.

Model 1 rewards recognition over calculation. Ten minutes spent learning the standard multipliers will save more exam time than any amount of practice at long multiplication.

07 Interview questions

What gets asked

Ten, including the two conventions that separate a right answer from a nearly-right one.

A sum amounts to ₹1,331 in 3 years at 10% compound interest. Find the principal.

₹1,000. The multiplier is (11/10)³ = 1.331, so the principal is 1331 ÷ 1.331 = ₹1,000. Recognising 1,331 as 11 cubed makes this immediate.

How would you find the rate if you were told a sum grew from ₹10,000 to ₹12,100 in 2 years?

10%. The multiplier is 12100/10000 = 1.21, and 1.21 is 1.1 squared, so the rate is 10% per annum. In principle this needs a square root; in practice the multiplier is always recognisable.

What if the multiplier is not one you recognise?

Then re-check the number of periods before assuming you need a root. An untidy multiplier is usually a sign that the time was misread, or that the compounding is half-yearly rather than annual. Genuine logarithms are essentially never required.

How do you handle a time of 2½ years?

Compound for the two whole years, then apply simple interest for the half year. At 10% that is P × (1.1)² × (1 + 5/100). Raising 1.1 to the power 2.5 is not the intended convention and gives a slightly different answer.

Find the CI on ₹10,000 at 12% for 2 years.

₹2,544. The multiplier is (28/25)² = 1.2544, so the amount is ₹12,544 and the interest is ₹2,544. The 1.2544 is worth adding to your list of recognisable multipliers, since 12% appears often.

Why keep rates as fractions rather than decimals?

Because they cancel exactly. 10% is 11/10, so three years is 1331/1000 and dividing ₹1,331 by it is instant. As a decimal, 1.331 invites rounding and hides the structure that makes the question easy.

A sum doubles in 4 years at compound interest. What is the multiplier per year?

The fourth root of 2, about 1.1892, so roughly 18.92% a year. This is one of the rare cases where the answer is genuinely irrational, which is why exams phrase doubling questions as “when does it become 4 times” instead — that needs no root at all.

₹4,000 at 25% for 2 years, compounded annually. Find the amount.

₹6,250. Twenty-five per cent is 5/4, so two years is 25/16, and 4000 × 25/16 = ₹6,250. Recognising 25% as 5/4 turns an awkward-looking rate into the easiest kind of arithmetic.

What is the amount on ₹1,000 at 10% for 2 years compounded half-yearly?

Hard

(a) Compute and tabulate (1 + R/100)² and (1 + R/100)³ as exact fractions for R = 5, 10, 12, 12.5, 20 and 25. (b) A sum grows from ₹12,800 to ₹16,200 under compound interest. Using your table, identify the rate and the time. (c) Explain why exam setters choose these particular rates, and what that tells you to do when a multiplier does not appear in your table.

Follow-up

Building the table yourself is worth far more than being given it, and part (b) then falls out by recognition rather than by algebra. Part (c) is the strategic point: an unrecognised multiplier is evidence about the question, not a signal to reach for logarithms.

Show the hint

For (b), work out 16200/12800 as a fraction in lowest terms and compare it against your table.