Simple Interest for RRB Group D — Formula, Tricks, and PYQ Solutions

beginner 18 min read

Concept

Simple Interest (SI) is the most straightforward form of interest calculation — you earn (or pay) interest only on the original amount you put in (or borrowed), never on the interest that has already accumulated. The original amount is called the Principal (P), the percentage charged per year is the Rate (R), and the duration is Time (T).

Here is the cleanest analogy: imagine you lend a friend Rs. 1,000 for 3 years at 10% per year. Each year, they owe you exactly Rs. 100 — not Rs. 100 in year one, then Rs. 110 in year two (that would be compound interest). In SI, the interest "meter" runs at a flat, fixed pace. Year 1 adds Rs. 100. Year 2 adds another Rs. 100. Year 3 adds another Rs. 100. Total interest: Rs. 300. Final amount returned: Rs. 1,300.

That flat-rate behaviour is what makes SI problems tractable in an exam — the interest grows in a straight line. When you plot Amount against Time in SI, you get a straight line starting at the Principal. In compound interest, that same plot is a curve. If you ever need to distinguish the two in a word problem, look for that linearity clue.

For RRB Group D, SI questions are typically one or two steps. The examiners test four things:

  1. Finding one missing variable (P, R, T, or SI itself) when three are known.
  2. Finding the original principal when two amounts at different time points are given.
  3. Profit/loss when someone borrows at one rate and lends at another.
  4. "Doubling" or "becomes a fraction of itself" questions.

None of these require anything beyond the core formula and a bit of algebraic rearrangement. You do not need a calculator — every value is chosen to divide cleanly if you set it up right.


Deep Dive

The Core Formula

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

Amount=P+SI=P(1+RT100)Amount = P + SI = P\left(1 + \frac{RT}{100}\right)

Where:

Everything in SI problems comes from this one equation. The four variables are P, R, T, and SI. Any question gives you three and asks for the fourth.

Rearranged Forms (Keep These Ready)

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

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

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

You don't need to memorise all four separately. Memorise one triangle: SI at the top, P × R × T at the bottom. Cover the variable you want — the rest tells you what to do (multiply or divide).

Finding Principal from Two Amounts

Look — this is a question type that trips up beginners. You're told: "A sum amounts to Rs. X in a years and Rs. Y in b years." Both amounts are at SI. Here's the insight:

The difference in Amount between any two time points equals the SI earned in that gap:

SIper year=YXbaSI_{per\ year} = \frac{Y - X}{b - a}

Once you have SI per year, multiply by a years to get SI for the first period, then subtract from X to get Principal.

Example: Amounts to Rs. 2,420 in 2 years and Rs. 2,600 in 3 years.

Doubling / Multiplying Problems

"At what rate will a sum double in n years?"

If the sum doubles, Amount = 2P, so SI = P. Plug into the formula:

P=P×R×n100    R=100nP = \frac{P \times R \times n}{100} \implies R = \frac{100}{n}

For the sum to double in 12 years: R = 100/12 = 8.33%.

General pattern: If the sum becomes k times itself in n years: SI=(k1)P    R=(k1)×100nSI = (k-1)P \implies R = \frac{(k-1) \times 100}{n}

If it becomes 5/4 of itself: k = 5/4, so k − 1 = 1/4. R = (1/4 × 100)/n.

Difference Between CI and SI (2 Years)

One question in the PYQ set uses this. For 2 years:

CISI=P×(R100)2CI - SI = P \times \left(\frac{R}{100}\right)^2

This is worth remembering as a standalone shortcut for 2-year problems. You do not need to compute CI and SI separately and subtract — that takes 5 steps. This formula does it in 2.

Borrowing-and-Lending Profit

When someone borrows at rate r₁ and lends at rate r₂ (where r₂ > r₁) on the same principal for the same time:

Gain=P×(r2r1)×T100Gain = \frac{P \times (r_2 - r_1) \times T}{100}

This is just the SI formula applied to the rate difference. You can compute it in one line instead of computing two separate interest values.


Memory Tricks & Shortcuts

patternThe 100/n Doubling Rule

When a sum doubles at simple interest, SI equals the Principal itself. Instead of setting up the full equation, directly use R = 100/n where n is the number of years. For n = 12, R = 100/12 = 8.33%. For n = 10, R = 10%. For n = 8, R = 12.5%. This instantly solves any "double in X years" question without algebra. Standard setup: 4 steps. This rule: 1 step. Speed gain: reduces ~40 seconds of algebra to a 5-second recall.

patternTwo-Amount Subtraction for Principal

When you are given amounts at two different time points (e.g., Rs. A after t₁ years and Rs. B after t₂ years), SI per year = (B − A)/(t₂ − t₁). Then Principal = A − (t₁ × SI per year). This avoids setting up two simultaneous equations with unknowns P and R. Standard method using two equations: 6-8 lines. This subtraction method: 2 lines. Speed gain: saves approximately 45-60 seconds.

eliminationRate-Difference Shortcut for Profit

When borrowing at r₁% and lending at r₂%, the gain per year is simply P × (r₂ − r₁)/100. Do not compute total interest paid and total interest received as separate steps. Example: Borrow Rs. 15,000 at 12%, lend at 15%, for 3 years. Rate difference = 3%. Gain = 15000 × 3 × 3/100 = 15000 × 9/100 = Rs. 1,350. Standard method: compute 5400 and 6750 separately, then subtract. This shortcut: one multiplication. Saves approximately 30 seconds and removes one subtraction error point.

patternCI minus SI for 2 Years: One-Shot Formula

For the difference between CI and SI over exactly 2 years, use: Difference = P × (R/100)². For P = 8,000 and R = 10%: 8000 × (0.1)² = 8000 × 0.01 = Rs. 80. Without this formula, you'd compute CI = 8000 × 1.1² − 8000 = 1,680 and SI = 8000 × 10 × 2/100 = 1,600, then subtract: 4 separate calculations vs. one multiplication. Saves approximately 40 seconds.

substitutionSI as Fraction of Principal

When the question says "a sum becomes k/m of itself in n years," the SI earned = (k/m − 1) × P = ((k−m)/m) × P. Then R = ((k−m)/m × 100)/n. Example: sum becomes 5/4 of itself in 5 years. SI = (5/4 − 1)P = P/4. R = (1/4 × 100)/5 = 100/20 = 5%. You can substitute P = 100 as a dummy value and the percentage works out directly. Reduces risk of fraction arithmetic errors in the exam hall.


Fast-Solving Framework

When you see an SI question in the exam hall, run through this decision path:

Step 1 — Identify what's given and what's asked. Count the knowns: P, R, T, SI. Three are given, one is missing. If two Amounts at different times are given instead of P directly, use the subtraction trick first to find SI-per-year, then Principal.

Step 2 — Check for a pattern type:

Step 3 — Plug and solve. Always confirm units: if T is in months, convert to years (T_months / 12). If R is given as per quarter, convert to per annum. RRB Group D almost never uses non-annual rates, but check.

Step 4 — Verify with options. If your answer does not match any option, re-check whether you used Amount or SI in the formula. The most common error is substituting Amount where SI is required.


Solved PYQs

Why this question: Direct application of the base formula, solving for R. This is the most common SI question type in RRB Group D.

Previous Year Questionपिछले वर्ष का प्रश्न
If the simple interest on Rs. 3,600 for 4 years is Rs. 1,440, what is the rate of interest per annum?
यदि Rs. 3,600 पर 4 साल का साधारण ब्याज Rs. 1,440 है, तो प्रति वर्ष ब्याज दर क्या है?
  1. 8%
  2. 10%
  3. 12%
  4. 15%
  1. 8%
  2. 10%
  3. 12%
  4. 15%
Solutionसमाधान
Using SI = PRT/100, we have 1440 = (3600 × R × 4)/100. Solving: 1440 = 144R, therefore R = 1440/144 = 10% per annum.
साधारण ब्याज = मूलधन×दर×समय/100 का उपयोग करते हुए, 1440 = (3600 × दर × 4)/100। हल करने पर: 1440 = 144×दर, अतः दर = 1440/144 = 10% प्रति वर्ष।

Solving path: Plug known values into SI = PRT/100. You get 1440 = (3600 × R × 4)/100 = 144R. So R = 1440/144 = 10. Pick option B. Time taken with this approach: under 20 seconds.


Why this question: Tests whether you can extract Principal from two Amount values without getting confused by the missing Rate. The subtraction trick is the only efficient method here.

Previous Year Questionपिछले वर्ष का प्रश्न
A certain sum at simple interest amounts to Rs. 2,420 in 2 years and Rs. 2,600 in 3 years. The sum is:
एक निश्चित राशि साधारण ब्याज पर 2 साल में Rs. 2,420 और 3 साल में Rs. 2,600 हो जाती है। वह राशि है:
  1. Rs. 2,060
  2. Rs. 2,080
  3. Rs. 2,100
  4. Rs. 2,240
  1. Rs. 2,060
  2. Rs. 2,080
  3. Rs. 2,100
  4. Rs. 2,240
Solutionसमाधान
SI for 1 year = 2600 - 2420 = Rs. 180. SI for 2 years = 180 × 2 = Rs. 360. Principal = Amount after 2 years - SI for 2 years = 2420 - 360 = Rs. 2060.
1 वर्ष का साधारण ब्याज = 2600 - 2420 = 180 रुपये। 2 वर्ष का साधारण ब्याज = 180 × 2 = 360 रुपये। मूलधन = 2 वर्ष बाद राशि - 2 वर्ष का साधारण ब्याज = 2420 - 360 = 2060 रुपये।

Solving path: SI per year = 2600 − 2420 = Rs. 180. SI for 2 years = Rs. 360. Principal = 2420 − 360 = Rs. 2,060. Pick option A. This is a 3-step mental calculation — no formula rearrangement needed.


Why this question: Borrowing-and-lending profit is a classic exam pattern. The rate-difference shortcut cuts the work in half.

Previous Year Questionपिछले वर्ष का प्रश्न
A person borrows Rs. 15,000 at 12% per annum simple interest and lends it at 15% per annum simple interest. His gain in 3 years is:
एक व्यक्ति Rs. 15,000 प्रति वर्ष 12% साधारण ब्याज पर उधार लेता है और उसे प्रति वर्ष 15% साधारण ब्याज पर उधार दे देता है। 3 साल में उसका फायदा है:
  1. Rs. 1,200
  2. Rs. 1,350
  3. Rs. 1,500
  4. Rs. 1,800
  1. Rs. 1,200
  2. Rs. 1,350
  3. Rs. 1,500
  4. Rs. 1,800
Solutionसमाधान
Interest paid = (15000 × 12 × 3)/100 = Rs. 5,400. Interest received = (15000 × 15 × 3)/100 = Rs. 6,750. Gain = 6,750 - 5,400 = Rs. 1,350.
दिया गया ब्याज = (15000 × 12 × 3)/100 = 5,400 रुपये। प्राप्त ब्याज = (15000 × 15 × 3)/100 = 6,750 रुपये। लाभ = 6,750 - 5,400 = 1,350 रुपये।

Solving path: Rate difference = 15% − 12% = 3%. Gain = (15000 × 3 × 3)/100 = 135000/100 = Rs. 1,350. Pick option B. The standard method of computing two separate interest values and subtracting takes twice as long and introduces an extra subtraction step where errors occur.


Why this question: Doubling problems appear regularly and are completely solvable in one step if you know the rule.

Previous Year Questionपिछले वर्ष का प्रश्न
At what rate percent per annum will a sum of money double itself in 12 years at simple interest?
साधारण ब्याज पर कोई राशि 12 साल में दोगुनी हो जाए, इसके लिए प्रति वर्ष ब्याज दर कितने प्रतिशत होनी चाहिए?
  1. 8.33%
  2. 10%
  3. 12%
  4. 12.5%
  1. 8.33%
  2. 10%
  3. 12%
  4. 12.5%
Solutionसमाधान
For money to double, Amount = 2P where P is principal. SI = 2P - P = P. Using SI = PRT/100: P = (P × R × 12)/100. Solving: 1 = 12R/100, therefore R = 100/12 = 8.33%.
पैसे के दोगुना होने के लिए, राशि = 2P जहाँ P मूलधन है। साधारण ब्याज = 2P - P = P। साधारण ब्याज = मूलधन×दर×समय/100 का उपयोग करते हुए: P = (P × दर × 12)/100। हल करने पर: दर = 100/12 = 8.33%।

Solving path: Sum doubles → SI = P. Apply R = 100/n = 100/12 = 8.33%. Pick option A. Do not set up the full equation on paper; this is a one-second mental calculation once you have the rule.


Why this question: CI vs SI for 2 years is a cross-topic question. Knowing the shortcut formula P(R/100)² makes this a 10-second question rather than a 2-minute one.

Previous Year Questionपिछले वर्ष का प्रश्न
The difference between compound interest and simple interest on Rs. 8,000 for 2 years at 10% per annum is:
Rs. 8,000 पर 2 साल के लिए 10% प्रति वर्ष की दर से चक्रवृद्धि ब्याज और साधारण ब्याज का अंतर है:
  1. Rs. 80
  2. Rs. 100
  3. Rs. 120
  4. Rs. 160
  1. Rs. 80
  2. Rs. 100
  3. Rs. 120
  4. Rs. 160
Solutionसमाधान
For 2 years, the difference between CI and SI = P(R/100)². Here, P = 8000, R = 10%. Difference = 8000 × (10/100)² = 8000 × 0.01 = Rs. 80.
2 वर्षों के लिए, चक्रवृद्धि ब्याज और साधारण ब्याज के बीच अंतर = मूलधन×(दर/100)²। यहाँ, मूलधन = 8000, दर = 10%। अंतर = 8000 × (10/100)² = 8000 × 0.01 = 80 रुपये।

Solving path: Difference = 8000 × (10/100)² = 8000 × 0.01 = Rs. 80. Pick option A. If you had computed CI step-by-step (8000 × 1.21 = 9680, CI = 1680, SI = 1600, difference = 80), it would take about 90 seconds. The formula takes 10.


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