Aptitude · Simple Interest · Model 3

Two photographs of the same money, and the difference between them

Q: How would you sanity-check a Model 3 answer under exam pressure?

Push your principal and rate through the condition you did not use to derive them. For the worked example, 7800 + 9 × 1300 = ₹19,500, matching the second amount. If it does not match, the usual cause is stripping the wrong number of years when reversing to the principal.

Here the question stops giving you one situation and starts giving you two: the same sum at two different times, or two rates, or two sums swapped between rates. You are not meant to solve either condition. You are meant to subtract them.

Give two timelines and watch the principal cancel itself out →

When the same principal appears in both conditions, subtracting them cancels it. What is left is pure interest — and pure interest is one division away from everything else.

01 The idea

Do not solve both conditions. Subtract them.

A sum amounts to ₹13,000 in 4 years and ₹19,500 in 9 years. The instinct is to write two equations in P and R and grind. That works, and it is slow, and it is where the arithmetic errors live.

Look at what the two amounts have in common instead. Both are the same principal plus some interest. So when you subtract one from the other, the principal cancels exactly, and the ₹6500 left over is interest and nothing else — specifically, the interest of the five years that separate the two photographs.

From there it unwinds. Five years of interest is ₹6500, so one year is ₹1300. The ₹13,000 was the principal plus four such years, so the principal is 13,000 − 5200 = ₹7800. And ₹1300 a year on ₹7800 is 16.67%. No formula was written down.

The same move handles every variant. If two conditions differ only in the rate, the difference in interest is caused only by the rate difference. If two equal sums sit at different rates, the gap between their interests depends on the rate gap alone. Find what is common, subtract it away, and read what the difference is telling you.

Difference of amounts = difference of interests, whenever the principal is the same. That one line is the whole model.

Two-condition questionAny question giving two situations for comparison rather than one to solve — two timelines, two rates, two sums, or amounts swapped between rates.

The extra yearsThe gap between the two times. The difference between the two amounts is exactly the interest earned in those years, which is why dividing by them gives one year's interest.

Difference of ratesWhen only the rate changes, the extra interest is P × ΔR × T / 100 — so a rate rise of d% over T years adds (d × T)% of the principal, whatever the original rate was.

02 Worked example

₹13,000 in 4 years, ₹19,500 in 9 — find both unknowns

Two unknowns, and neither condition alone can give you either. A certain sum amounts to ₹13,000 in 4 years and to ₹19,500 in 9 years at simple interest. Find the principal and the rate per annum.

Write the two conditions side by sideSame sum, same rate, two different times. That is the licence to compare them directly.after 4 years → 13,000 after 9 years → 19,500

Subtract to kill the principalEach amount is the principal plus its interest. The principal is the same in both, so it vanishes and only interest survives.19,500 − 13,000 = 6,500 ← interest for the 5 extra years

Get one year’s interestFive years of interest is ₹6500, and simple interest earns the same figure every year.6,500 / 5 = 1,300 per year

Reverse out of the first amountThe ₹13,000 contained four years of interest on top of the principal. Take them off.P = 13,000 − 4 × 1,300 = 13,000 − 5,200 = 7,800

Read the rate off the principalOne year earned ₹1300 on a principal of ₹7800. Express it as a percentage.R = 1,300 / 7,800 × 100 = 16.67% per annum

Check it from the other end before moving on: 7800 + 9 × 1300 = 7800 + 11,700 = ₹19,500, which is the second amount exactly. Every Model 3 answer can be verified against the condition you did not use to find it, and that check costs about five seconds. Use it — it catches the off-by-one-year error that this model invites.

03 The method

The six shapes, and the one instinct behind all of them

Model 3 questions look varied because the examiner changes what differs between the two conditions. The response is always the same: find what is shared, subtract it, interpret the remainder.

Difference of amounts = difference of interests when the principal is shared. When only the rate differs by d over T years: extra interest = P × d × T / 100, which is (d × T)% of the principal.

Never solve both conditions. If you find yourself with two simultaneous equations in P and R, you have missed the subtraction. The only thing the two conditions are for is their difference — and the difference always has one fewer unknown than either condition on its own.

What differs between the conditions	What the difference tells you	Then
The time only	Interest of the extra years	Divide by the extra years for one year
The rate only	(ΔR × T)% of the principal	One division gives the principal
Two equal sums, different rates	(ΔR × T)% of each sum	Same division, the rate gap is all that matters
Two different sums, same rate	Interest ratio follows the sums	Find one year's interest on each
Amounts swapped between two rates	ΔR% of the difference of the sums	Gives the gap between the two sums
A stated interest difference in rupees	A known percentage of the principal	Unitary step to 100%

05 Cheat sheet

Model 3 on one page

Every row is a subtraction. The only thing that changes is what you are left holding afterwards.

Case	Route	Worked
Two timelines	`ΔA / ΔT = one year`	6500/5 = 1300 a year
Principal, from two timelines	`P = A₁ − T₁ × (one year)`	13,000 − 5,200 = 7,800
Rate, from two timelines	`R = (one year)/P × 100`	1300/7800 = 16.67%
Rate rises by d%	`extra = (d × T)% of P`	4% for 9 yr = 36% of 30,000 = 10,800
Two equal sums, rates differ by d	`ΔSI = (d × T)% of P`	1% for 6 yr → 6% = 4200, P = 70,000
Interest difference given in rupees	`P = ΔSI × 100/(d × T)`	280 × 100/14 = 2,000
Amounts swapped	`Δyearly = d% of (x − y)`	5% of the gap = 40, so gap = 800

Only the difference of the rates mattersIn an equal-sums comparison the actual rates are decoration. 8.5% against 9.5% behaves exactly like 2% against 3% — the gap of 1% is all that enters the arithmetic.

Check against the unused conditionFind the principal from the first timeline, then verify it against the second. It costs five seconds and catches the off-by-one-year slip this model invites.

A shrinking sum is impossibleIf the later amount is smaller than the earlier one, or the implied principal comes out negative, the two conditions cannot describe one sum at one rate. Re-read before calculating.

06 Where & why

Where Model 3 shows up

This is the model that carries the marks in simple interest, because it rewards seeing the structure rather than remembering the formula.

Bank PO · SBI · RRB Mains

The two-timeline question

“Amounts to X in a years and Y in b years.” The single most reliably set hard simple interest question in banking papers, and a ten-second question once the subtraction is instinct.

SSC CGL · CPO

Rate-change and equal-sums comparisons

“If the rate were 4% higher…” and “two equal sums at 8.5% and 9.5%…”. Both collapse to (ΔR × T)% of the principal.

TCS Digital · Accenture

Rates written with a variable

“At (a + 2.5)% instead of (a + 6)%.” Alarming to look at, but the a cancels in the subtraction and only the 3.5% gap survives.

Instalments and partnership

The same cancelling trick

Both chapters compare a quantity under two arrangements. The habit of subtracting what is shared before calculating anything transfers directly.

The tell for this model is the word and joining two situations, or the word if introducing a second one. When you see either, do not reach for the formula — reach for the difference.

07 Interview questions

The questions this model attracts

Ten, from the core cancellation to the two places where the technique needs care.

A sum amounts to ₹13,000 in 4 years and ₹19,500 in 9 years. How do you start?

Subtract the amounts. Both contain the same principal, so the difference of ₹6500 is pure interest — the interest of the 5 years between the two readings. Dividing gives ₹1300 a year, and from there the principal is 13,000 − 4 × 1300 = ₹7800 and the rate is 16.67%.

Why does subtracting work at all?

Because each amount is principal plus interest, and the principal is identical in both conditions. Subtracting removes it exactly, leaving a quantity with one fewer unknown. It is the same reason elimination works on simultaneous equations — you are just doing it in one line instead of four.

₹30,000 amounts to ₹40,500 in 9 years. If the rate rose by 4%, what is the new amount?

You do not need the original rate. A rise of 4% for 9 years adds 36% of the principal, and 36% of ₹30,000 is ₹10,800. So the new amount is 40,500 + 10,800 = ₹51,300. Computing the old rate first is legal but wasted work.

Two equal sums earn interest at 8.5% and 9.5% for 6 years, and the second earns ₹4200 more. Find each sum.

Only the 1% gap matters. Over 6 years a 1% difference produces 6% of the principal, and that 6% is ₹4200. So the principal is 4200 × 100/6 = ₹70,000. The actual rates never enter the calculation, which is what makes the question quick.

A sum is invested at (a + 2.5)% and would have earned ₹280 more at (a + 6)% over 4 years. Find the sum.

The a cancels: the rate gap is 6 − 2.5 = 3.5%. Over 4 years that is 14% of the principal, and 14% is ₹280, so the principal is ₹2000. Questions written with a variable rate are testing whether you notice that only the difference survives.

What is the swapped-amounts question and how do you handle it?

Two parts of a sum sit at two rates; swapping them changes the total interest by a stated figure. The change comes only from the rate gap acting on the difference between the two parts. If the gap is 5% and swapping adds ₹40 a year, then 5% of the difference is ₹40, so the parts differ by ₹800. Combine that with the original total interest to finish.

Work that swapped example through.

Parts x at 20% and y at 25% give a yearly interest of ₹250; swapped they give ₹290. Subtracting, 0.05x − 0.05y = 40, so x − y = ₹800. Substituting into 20x + 25y = 25,000 gives y = ₹200 and x = ₹1000. Check: 20% of 1000 plus 25% of 200 is 200 + 50 = ₹250, and swapped it is 250 + 40 = ₹290.

A sum of ₹4500 becomes ₹5400 in 5 years. How long for ₹7500 to become ₹9300 at the same rate?

First case: interest ₹900 on ₹4500 over 5 years is 20% total, so 4% a year. Second case needs interest of ₹1800, and 4% of ₹7500 is ₹300 a year, so 1800/300 = 6 years. The two sums differ, so here you do need the rate — the shortcut is finding it as a percentage first.

When does the subtraction trick not apply?

Hard

Sandeep invests one part of his money at 20% and the rest at 25% simple interest, receiving ₹250 of interest in a year. Had the two amounts been swapped, he would have received ₹40 more. (a) Find how much was invested at 20%. (b) Verify your answer by computing both the original and the swapped interest. (c) Explain why the ₹40 depends only on the difference between the two invested amounts, and state what the answer would be if the ₹40 were ₹0 instead.

Follow-up

Part (c) is the insight. Swapping moves the 5% rate gap from one pile to the other, so the change is 5% of the difference between the piles — which means a ₹40 change and a ₹0 change tell you something immediate about how the money was split.

Show the hint

Write the two totals as equations and subtract them; see which quantity survives and which cancels.