Simple Interest for Agniveer Army CEE — Formula, Tricks & Solved PYQs

beginner 18 min read

Concept

Simple Interest (SI) is the interest calculated only on the original principal — never on the accumulated interest. Every year, the interest earned is the same fixed amount. Think of it like a daily wage job: you get the same pay every day regardless of how long you have been working. Contrast this with compound interest, where interest keeps adding to the base, like a snowball rolling downhill.

Here is the clearest way to see it: you deposit ₹1000 at 10% per year for 3 years. Under simple interest, you earn ₹100 every single year — year 1, year 2, year 3. Total interest = ₹300. The principal stays at ₹1000 throughout. That is the defining property.

The four variables you will always deal with:

The reason SI is a favourite in competitive exams is that it tests whether you can rearrange a single formula under time pressure. The formula itself is trivial. The challenge is reading the question carefully enough to plug in the right values — and manipulating it quickly when the unknown is R or T rather than SI.

In the Agniveer Army CEE, SI questions are direct. They rarely go beyond two-step problems. If you can do the basic SI = (P × R × T) / 100 confidently in both directions (finding SI, finding P, finding R, finding T), you will not drop a single mark here.


Deep Dive

The Core Formula and Its Four Rearrangements

The master equation is:

SI=P×R×T100SI = \frac{P \times R \times T}{100}

And the amount formula:

A=P+SI=P(1+R×T100)A = P + SI = P\left(1 + \frac{R \times T}{100}\right)

From the SI formula, you get four working rearrangements — memorise all four, not just the first:

| Find | Formula | |------|---------| | SI | P×R×T{100}\frac{P \times R \times T}\{100\} | | P | SI×100R×T\frac{SI \times 100}{R \times T} | | R | SI×100P×T\frac{SI \times 100}{P \times T} | | T | SI×100P×R\frac{SI \times 100}{P \times R} |

Look — when the question gives you Amount (A) and asks for SI, always subtract first: SI = A - P. Do not try to work backward from A directly.

The "Doubles / Triples" Pattern

A common Agniveer question type: "A sum doubles in N years. In how many years does it become M times?"

Here is the logic. If a sum doubles in N years, the SI in N years equals P (the principal itself). So:

SI=PP×R×N100=PR=100NSI = P \Rightarrow \frac{P \times R \times N}{100} = P \Rightarrow R = \frac{100}{N}

Now, for the sum to become M times, the SI needed = (M - 1) × P. Use the same rate:

T=(M1)P×100P×R=(M1)×100R=(M1)×NT = \frac{(M-1)P \times 100}{P \times R} = \frac{(M-1) \times 100}{R} = (M-1) \times N

Shortcut result: If a sum doubles in N years at SI, it becomes M times in (M − 1) × N years.

Doubles in 8 years → becomes 5 times in (5 − 1) × 8 = 32 years. Done in 5 seconds.

Two-Investment Problems

When two amounts are invested at different rates for the same time period, calculate SI separately on each and add. There is no trick smarter than this — just do two clean multiplications and add.

SItotal=P1×R1×T100+P2×R2×T100SI_{total} = \frac{P_1 \times R_1 \times T}{100} + \frac{P_2 \times R_2 \times T}{100}

Since T is common, you can factor it out: SItotal=T{100}(P1R1+P2R2)SI_{total} = \frac{T}\{100\}(P_1 R_1 + P_2 R_2).

Profit from Re-lending

When someone borrows at rate R1R_1 and lends at rate R2R_2 (where R2>R1R_2 > R_1), their profit over time T is simply:

Profit=P×(R2R1)×T100Profit = \frac{P \times (R_2 - R_1) \times T}{100}

This is because both are applied to the same principal for the same time. You do not need to calculate two separate interests and subtract — one combined multiplication is enough.

CI vs SI Difference for 2 Years

This is a hybrid question that shows up in the PYQ set. The formula you need to know:

CISI (for 2 years)=P×R21002=PR210000CI - SI \text{ (for 2 years)} = P \times \frac{R^2}{100^2} = \frac{P R^2}{10000}

This is derived from the fact that CI for 2 years = P[(1+R{100})21]P\left[\left(1+\frac{R}\{100\}\right)^2 - 1\right] and SI for 2 years = 2PR{100}\frac{2PR}\{100\}. Expand the CI expression and subtract SI — the extra term is exactly PR2{10000}\frac{PR^2}\{10000\}.

If you are given the difference and asked to find P:

P=Difference×10000R2P = \frac{Difference \times 10000}{R^2}

Units and Time Conversion

Always check the time unit. If T is given in months, convert: T (years) = months / 12. If the rate is per month, either keep everything in months or convert to annual. Mixing units is the single biggest source of wrong answers in this topic.


Memory Tricks & Shortcuts

patternPRT Anchor — One Formula, Four Unknowns

Write the formula as a triangle: P × R × T at the top, 100 × SI at the bottom. Cover whatever you want to find — the remaining three form the expression. This is the same memory device used for speed-distance-time. Once you fix this visual in your head, you never need to re-derive. Cover SI → you see PRT/100. Cover P → you see (SI × 100)/(R × T). Standard working time without the triangle: 15-20 seconds of formula recall. With it: 3 seconds.

patternDoubles Pattern — (M−1)×N Rule

When a sum doubles in N years, it becomes M times in exactly (M − 1) × N years. This works because interest per year is constant (P/N), so every additional multiple of P takes exactly N more years. Example: doubles in 8 years → triples in 16, becomes 5× in 32, becomes 10× in 72. Standard method: derive rate first (100/N = 12.5%), then use T = SI×100/(P×R) — about 40 seconds. Pattern method: (5−1)×8 = 32 — under 5 seconds.

estimationRate-Finding Shortcut via Percentage Growth

When a principal grows from P to A over T years, the total percentage growth = [(A−P)/P] × 100. Annual rate = total growth% / T years. Example: ₹6000 → ₹7200 in 2 years. Growth = 1200/6000 × 100 = 20%. Annual rate = 20/2 = 10%. This avoids writing the formula entirely. Standard method: set up SI = PRT/100 and solve algebraically — 30 seconds. This method: 10 seconds.

patternRe-lending Profit — Use Rate Difference Directly

Instead of computing two interest amounts and subtracting, use: Profit = P × (R₂ − R₁) × T / 100. Example: borrow ₹8000 at 15%, lend at 18%, 3 years. Profit = 8000 × (18−15) × 3 / 100 = 8000 × 3 × 3 / 100 = 72000/100 = ₹720. Standard method (two separate calculations): 45 seconds. Rate-difference method: 15 seconds.

patternCI−SI for 2 Years — P×R²/10000

The difference between CI and SI for exactly 2 years is always P×R²/10000. If you know the difference and rate, isolate P instantly: P = Difference × 10000 / R². Example: difference = ₹225, R = 15%. P = 225 × 10000 / 225 = ₹10000. Standard method: write full CI expansion, expand, subtract SI term by term — 60 seconds minimum. Formula method: 10 seconds.


Fast-Solving Framework

Read the question and immediately identify: what is the unknown (SI, P, R, or T)?

Step 1. Is Amount (A) given instead of SI? If yes, compute SI = A − P before anything else.

Step 2. Is the unknown SI? → Direct plug-in: (P × R × T) / 100. Done.

Step 3. Is the unknown T or R? → Use T = (SI × 100) / (P × R) or R = (SI × 100) / (P × T).

Step 4. Is the unknown P? → Use P = (SI × 100) / (R × T).

Step 5. Does the question say "doubles in N years, find time to become M times"? → Answer is (M − 1) × N. No formula needed.

Step 6. Two investments at different rates, same time? → Calculate SI separately, add.

Step 7. Borrow at one rate, lend at another? → Profit = P × (R₂ − R₁) × T / 100.

Step 8. CI vs SI difference for 2 years? → P × R² / 10000 = Difference.

Time units: if months are given, divide by 12 before applying the formula.


Solved PYQs

Why this question: This is the most direct application of the SI formula — the kind you should solve in under 20 seconds.

Previous Year Questionपिछले वर्ष का प्रश्न2024
What is the simple interest on Rs 2500 for 3 years at an annual interest rate of 5%?
  1. 125
  2. 750
  3. 375
  4. 255
Solutionसमाधान
SI = (P × R × T) / 100 = (2500 × 5 × 3) / 100 = 37500 / 100 = Rs 375.

Solving path: P = 2500, R = 5, T = 3. SI = (2500 × 5 × 3) / 100 = 37500 / 100 = 375. No traps here — standard plug-in.


Why this question: Tests your ability to extract T when Amount is given — a very common question structure in Agniveer CEE.

Previous Year Questionपिछले वर्ष का प्रश्न
A person borrowed ₹15000 at 12% simple interest per annum. After how many years will the amount become ₹25800?
एक व्यक्ति ने ₹15000 प्रति वर्ष 12% साधारण ब्याज पर उधार लिए। कितने वर्षों बाद कुल राशि ₹25800 हो जाएगी?
  1. 5 years
  2. 6 years
  3. 7 years
  4. 8 years
  1. 5 वर्ष
  2. 6 वर्ष
  3. 7 वर्ष
  4. 8 वर्ष
Solutionसमाधान
Simple Interest = Amount - Principal = 25800 - 15000 = ₹10800. Using SI = PRT/100: 10800 = 15000 × 12 × T/100. Solving: T = 10800 × 100/(15000 × 12) = 1080000/180000 = 6 years.
साधारण ब्याज = मिश्रधन - मूलधन = 25800 - 15000 = ₹10800। साधारण ब्याज सूत्र का उपयोग करते हुए: 10800 = 15000 × 12 × T/100। हल करने पर: T = 6 वर्ष।

Solving path: SI = 25800 − 15000 = 10800. Now use T = (SI × 100) / (P × R) = (10800 × 100) / (15000 × 12) = 1080000 / 180000 = 6 years. The trap is forgetting to subtract P from A first.


Why this question: Two-step problem — find rate from first scenario, apply to second. Tests whether you stay organised under pressure.

Previous Year Questionपिछले वर्ष का प्रश्न
If ₹6000 amounts to ₹7200 in 2 years at simple interest, what will be the simple interest on ₹9000 for 3 years at the same rate?
यदि ₹6000 साधारण ब्याज पर 2 वर्षों में ₹7200 हो जाते हैं, तो उसी दर से ₹9000 पर 3 वर्षों का साधारण ब्याज क्या होगा?
  1. ₹2700
  2. ₹3000
  3. ₹3300
  4. ₹3600
  1. ₹2700
  2. ₹3000
  3. ₹3300
  4. ₹3600
Solutionसमाधान
SI on ₹6000 for 2 years = 7200 - 6000 = ₹1200. Rate = (SI × 100)/(P × T) = (1200 × 100)/(6000 × 2) = 10% per annum. SI on ₹9000 for 3 years = (9000 × 10 × 3)/100 = ₹2700.
₹6000 पर 2 वर्षों का साधारण ब्याज = 7200 - 6000 = ₹1200। दर = (साधारण ब्याज × 100)/(मूलधन × समय) = (1200 × 100)/(6000 × 2) = 10% प्रति वर्ष। ₹9000 पर 3 वर्षों का साधारण ब्याज = ₹2700।

Solving path: SI on ₹6000 for 2 years = 7200 − 6000 = 1200. Rate = (1200 × 100) / (6000 × 2) = 10%. Now SI on ₹9000 for 3 years = (9000 × 10 × 3) / 100 = 2700. Answer: ₹2700.


Why this question: Tests rate-finding. Many candidates stumble because the SI is not a round number — use exact arithmetic, not approximation.

Previous Year Questionपिछले वर्ष का प्रश्न
At what rate percent per annum will ₹2000 amount to ₹2662 in 3 years at simple interest?
किस वार्षिक ब्याज दर पर ₹2000 साधारण ब्याज से 3 वर्षों में ₹2662 हो जाएंगे?
  1. 10%
  2. 11%
  3. 12%
  4. 13%
  1. 10%
  2. 11%
  3. 12%
  4. 13%
Solutionसमाधान
Simple Interest = 2662 - 2000 = ₹662. Using SI = PRT/100: 662 = 2000 × R × 3/100. Solving: R = (662 × 100)/(2000 × 3) = 66200/6000 = 11.033%. Approximately 11%.
साधारण ब्याज = 2662 - 2000 = ₹662। साधारण ब्याज सूत्र का उपयोग करते हुए: 662 = 2000 × R × 3/100। हल करने पर: R = 11%।

Solving path: SI = 2662 − 2000 = 662. R = (662 × 100) / (2000 × 3) = 66200 / 6000 ≈ 11.03%. The options say 11% — match to the nearest. The question is testing recognition that R is "approximately 11%".


Why this question: The "doubles in N years" pattern — if you know the (M−1)×N rule, this is a 5-second question.

Previous Year Questionपिछले वर्ष का प्रश्न
A sum of money at simple interest doubles itself in 8 years. In how many years will it become 5 times itself at the same rate?
साधारण ब्याज पर कोई राशि 8 वर्षों में दोगुनी हो जाती है। उसी दर पर वह राशि कितने वर्षों में 5 गुनी हो जाएगी?
  1. 24 years
  2. 28 years
  3. 32 years
  4. 36 years
  1. 24 वर्ष
  2. 28 वर्ष
  3. 32 वर्ष
  4. 36 वर्ष
Solutionसमाधान
If principal doubles in 8 years, SI = P in 8 years. Rate = (P × 100)/(P × 8) = 12.5% per annum. For amount to become 5P, SI needed = 4P. Time = (4P × 100)/(P × 12.5) = 400/12.5 = 32 years.
यदि मूलधन 8 वर्षों में दोगुना हो जाता है, तो साधारण ब्याज = P। दर = 12.5% प्रति वर्ष। मिश्रधन 5P बनने के लिए, साधारण ब्याज = 4P चाहिए। समय = 32 वर्ष।

Solving path: Doubles in 8 years → (M−1)×N = (5−1)×8 = 32 years. The longer route: R = 100/8 = 12.5%. For 5P, SI needed = 4P. T = (4P × 100) / (P × 12.5) = 400/12.5 = 32. Both paths reach 32 years.


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