Simple Interest for UP Police Constable — Formula, Tricks & PYQs

beginner 18 min read

Concept

Simple Interest (साधारण ब्याज) is the most direct way to calculate the cost of borrowing money — or the return on lending it. The core idea is this: interest is charged only on the original principal, every single year, at a fixed rate. It never compounds, never snowballs.

Think of it like a daily-wage worker getting paid a flat ₹500 every day regardless of how many days he has already worked. Day 1 = ₹500. Day 10 = ₹500. Day 100 = ₹500. The wage never changes because of accumulated days. Similarly, in Simple Interest, the interest for Year 3 is exactly the same as Year 1 — because it is always calculated on the same principal.

This makes Simple Interest linear. Plot it on a graph and you get a straight line. That linearity is exactly why UP Police Constable paper loves it — the calculations are clean, they fit in a 60-second window, and the traps are predictable (more on those in Common Mistakes).

Three variables drive everything:

Everything in this chapter is a rearrangement of one formula. Once that formula is in your muscle memory, you stop "solving" these questions — you just read and write.

A real-world analogy: the government's Kisan Credit Card or a moneylender in a village often charges simple interest. If you borrow ₹10,000 at 10% per annum for 2 years, you owe ₹2,000 extra at the end — ₹1,000 per year, flat. No interest on interest. That is Simple Interest.


Deep Dive

The Core Formula

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

And the total amount received or repaid:

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

These two lines contain every Simple Interest question on any government exam. The rest is just algebra — isolating whichever variable is unknown.

Rearranged for Each Unknown

| Unknown | Formula | |---|---| | SI | P×R×T100\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} |

You do not need to memorise all four. Memorise the triangle method: write P, R, T in a triangle, multiply across the bottom (P × R × T), divide by 100 to get SI. Then cover whatever you want to find, and the triangle tells you what to divide into what.

"Becomes N Times" — The Key Pattern for UP Police

This is the most exam-relevant variant. When a question says "a sum becomes N times," you must immediately translate:

This is because Amount = P + SI, so SI = Amount − P = NP − P = (N − 1)P.

Now plug into the SI formula:

(N1)P=P×R×T100(N-1)P = \frac{P \times R \times T}{100}

Cancel P from both sides:

(N1)=R×T100(N-1) = \frac{R \times T}{100}

So: R×T=100(N1)R \times T = 100(N-1)

Look at how clean this is. If a sum triples (N = 3) in 25 years: R=100×(31)25=20025=8%R = \frac{100 \times (3-1)}{25} = \frac{200}{25} = 8\%

If a sum becomes 4 times (N = 4) in 20 years: R=100×(41)20=30020=15%R = \frac{100 \times (4-1)}{20} = \frac{300}{20} = 15\%

Both of these are actual PYQs from 2024. The pattern is identical.

"Same Time, Different Multiple" Variant

When a question says "it becomes 3 times at 5%, at what rate will it become 6 times in the same time?" — here is the clean approach:

Step 1: Find T from the first condition. 2P=P×5×T1002P = \frac{P \times 5 \times T}{100}T=40T = 40 years

Step 2: Use T in the second condition. 5P=P×R×401005P = \frac{P \times R \times 40}{100}R=50040=12.5%R = \frac{500}{40} = 12.5\%

The key insight: T stays fixed. You find T from condition 1, then plug into condition 2.

Watch/Object Settlement Problems

These appear regularly. The logic is simple:

  1. Calculate total amount due (P + SI)
  2. Subtract what was actually paid in cash
  3. The remainder is the value of the object

If A lends ₹30,000 at 10% for 5 years, total due = 30,000 + 15,000 = ₹45,000. B pays ₹40,000 cash + a watch. Watch value = ₹45,000 − ₹40,000 = ₹5,000.

Units Consistency

One trap the paper sets: mixing months and years, or giving rate as "per month" when you expect "per annum." Always check:


Memory Tricks & Shortcuts

patternPRT Triangle

Draw an inverted triangle. Write SI at the top. Below it, on the left write P×R×T, and on the right write 100. To find SI: multiply P, R, T and divide by 100. To find any of P, R, or T: SI × 100, then divide by the product of the other two. This visual takes 2 seconds to recall versus hunting through rearranged formulas. Standard rearrangement time: 20s. Triangle recall: 3s.

patternN-Times Shortcut

When a sum becomes N times: immediately write R × T = 100 × (N − 1). Don't set up the full SI equation. For "triples in 25 years": R × 25 = 100 × 2 = 200, so R = 8%. You skip two algebraic steps. Standard method: set up SI = 2P, substitute, cancel P, isolate R — 4 steps. Shortcut: one multiplication, one division — 2 steps. Time saved: roughly 25 seconds per question.

estimationPercentage-of-Principal Mental Calculation

Compute R × T first. This gives you "what percentage of P is the SI." For ₹37,500 at 2% for 4 years: R × T = 8. So SI = 8% of 37,500 = 8 × 375 = ₹3,000. You never write the fraction 37500 × 2 × 4 / 100. Instead, 2 × 4 = 8, then 1% of 37500 = 375, then 8 × 375 = 3,000. Standard full-product method: 6 digit-multiplication steps. This shortcut: 3 steps. Saves 15-20 seconds.

patternAmount = P × (1 + RT/100) — One-Shot Calculation

Instead of finding SI first and then adding P, compute the multiplier directly. P = ₹6,000, R = 12%, T = 2 years. Multiplier = 1 + (12 × 2)/100 = 1 + 0.24 = 1.24. Amount = 6000 × 1.24 = ₹7,440. No intermediate SI calculation needed. Standard two-step method: calculate SI = ₹1,440, then add ₹6,000. One-shot: single multiplication. Eliminates one addition step, reduces carry errors.

patternSame-Time Rate Scaling

When a sum becomes X times at rate R, and the question asks what rate makes it Y times in the same time: the SI needed changes from (X−1)P to (Y−1)P, while T stays fixed. So the new rate = R × (Y−1)/(X−1). For "3 times at 5%, same T, becomes 6 times": new rate = 5 × (6−1)/(3−1) = 5 × 5/2 = 12.5%. No need to find T at all. Standard method: find T (2 steps), substitute into new equation (2 steps). This shortcut: 1 ratio multiplication. Saves 30 seconds.


Fast-Solving Framework

When you see a Simple Interest question in the exam hall, run this decision tree in 5 seconds:

Step 1 — What type is this?

Step 2 — Units check. Is time in months? Is rate monthly? Convert before touching the formula.

Step 3 — Which variable is missing? Isolate it using the triangle. SI × 100 on top, product of the other two below.

Step 4 — Arithmetic shortcut. Compute R × T first, then take that percentage of P using 1% anchor.

If any answer option looks wildly off — say, an amount less than the principal — eliminate immediately without calculation. This alone saves 10-15 seconds on badly-set traps.


Solved PYQs

Why this question: Tests the basic Amount = P + SI formula and the one-shot multiplier approach. Most common question type in UP Police Constable.

Previous Year Questionपिछले वर्ष का प्रश्न2026
A shopkeeper lends ₹6,000 at 12% per annum simple interest for 2 years. What is the total amount he will receive at the end of the period?
एक दुकानदार ₹6,000 को 12% वार्षिक साधारण ब्याज की दर से 2 वर्षों के लिए उधार देता है। अवधि के अंत में उसे कुल कितनी राशि प्राप्त होगी?
  1. ₹7,400
  2. ₹1,567
  3. ₹6,000
  4. ₹7,440
  1. ₹7,440
  2. ₹7,400
  3. ₹1,567
  4. ₹6,000
Solutionसमाधान
SI = P × R × T / 100 = 6000 × 12 × 2 / 100 = ₹1,440. Total amount = 6000 + 1440 = ₹7,440.

Solving path: R × T = 12 × 2 = 24. So SI = 24% of 6,000 = ₹1,440. Amount = 6,000 + 1,440 = ₹7,440. Or one-shot: 6,000 × 1.24 = ₹7,440. Eliminate option C (₹6,000) immediately — that would mean zero interest.


Why this question: Hindi-language SI question — tests whether you read मूल राशि (principal), दर (rate), and समय (time) correctly. Pure formula application.

Previous Year Questionपिछले वर्ष का प्रश्न2026
साधारण ब्याज ज्ञात कीजिए जब मूल राशि ₹ 12,800 है, ब्याज दर 7% प्रति वर्ष है और समय अवधि 3 वर्ष है।
साधारण ब्याज ज्ञात कीजिए जब मूल राशि ₹ 12,800 है, ब्याज दर 7% प्रति वर्ष है और समय अवधि 3 वर्ष है।
  1. ₹ 2,898
  2. ₹ 2,688
  3. ₹ 2,900
  4. ₹ 2,709
  1. ₹ 2,898
  2. ₹ 2,709
  3. ₹ 2,900
  4. ₹ 2,688
Solutionसमाधान
SI = P×R×T/100 = 12800×7×3/100 = 268800/100 = ₹2,688.

Solving path: R × T = 7 × 3 = 21. SI = 21% of 12,800. 1% of 12,800 = 128. 21 × 128 = 2,688. Answer: ₹2,688. Cross-check: none of the wrong options (₹2,898, ₹2,900, ₹2,709) are anywhere near a simple arithmetic slip — they are designed to catch you if you misread rate or time.


Why this question: "Becomes N times" pattern — the highest-frequency SI variant in UP Police 2024 paper.

Previous Year Questionपिछले वर्ष का प्रश्न2024
The amount will triple in 25 years on simple interest. Find the rate of interest.
साधारण ब्याज पर यह राशि 25 साल में तीन गुना हो जाएगी, तो ब्याज दर ज्ञात कीजिए।
  1. 15%
  2. 12%
  3. 10%
  4. 8%
  1. 10%
  2. 15%
  3. 12%
  4. 8%
Solutionसमाधान
If principal triples, interest = 2P. Using SI formula: 2P = P×R×25/100, so R = 200/25 = 8%.

Solving path: "Triples" → SI = 2P. Use R × T = 100 × (3−1) = 200. T = 25 years, so R = 200/25 = 8%. Direct. Do not set up the full equation with an actual value of P.


Why this question: Tests "becomes 4 times" — same pattern, different multiplier. Confirms you can generalise, not just memorise one case.

Previous Year Questionपिछले वर्ष का प्रश्न2024
A sum of money becomes four times in 20 years at simple interest. Find the rate of interest.
साधारण ब्याज पर 20 वर्षों में एक धनराशि चार गुना हो जाती है। ब्याज दर ज्ञात कीजिए।
  1. 5%
  2. 20%
  3. 8%
  4. 15%
  1. 20%
  2. 8%
  3. 15%
  4. 5%
Solutionसमाधान
If principal P becomes 4P, interest = 3P. SI = P×R×T/100 → 3P = P×R×20/100 → R = 300/20 = 15%.

Solving path: "4 times" → SI = 3P. R × T = 100 × 3 = 300. T = 20 years. R = 300/20 = 15%. Answer is 15%. Note: option A (5%) is a trap — 5% × 20 years = 100%, which only doubles the principal, not quadruples it.


Why this question: Watch/object settlement — a recurring story-format question that students waste time on by overcomplicating.

Previous Year Questionपिछले वर्ष का प्रश्न2024
A lends ₹30,000 to B at the rate of 10% per annum on simple interest. After 5 years, B returned ₹40,000 and a watch to settle the dues. Find the price of the watch.
A, B को 10% वार्षिक की दर से साधारण ब्याज पर ₹30,000 उधार देता है। 5 वर्षों के बाद B बकाया राशि का निपटान करने के लिए ₹40,000 और एक घड़ी वापस करता है। घड़ी का मूल्य ज्ञात कीजिए।
  1. ₹3,200
  2. ₹3,000
  3. ₹4,800
  4. ₹5,000
  1. ₹4,800
  2. ₹3,200
  3. ₹5,000
  4. ₹3,000
Solutionसमाधान
SI = 30000×10×5/100 = 15000. Total due = 30000+15000 = 45000. B paid 40000 cash + watch. Watch value = 45000-40000 = ₹5,000.

Solving path: SI = 30,000 × 10 × 5 / 100 = ₹15,000. Total due = 30,000 + 15,000 = ₹45,000. B paid ₹40,000 cash. Remaining = ₹45,000 − ₹40,000 = ₹5,000 = price of watch. The question is linear once you compute total due first.


Why this question: Same-time rate scaling — the advanced pattern that combines two conditions. Tests whether you can link two "becomes X times" scenarios.

Previous Year Questionपिछले वर्ष का प्रश्न2024
A sum of money becomes three times at 5% per annum simple interest. At what rate % will it become six times in the same time?
एक धनराशि साधारण ब्याज पर 5% वार्षिक दर से 3 गुना हो जाती है। उसी समय में यह धनराशि किस दर % पर 6 गुना हो जाएगी?
  1. 10.5%
  2. 14%
  3. 12.5%
  4. 11%
  1. 10.5%
  2. 14%
  3. 11%
  4. 12.5%
Solutionसमाधान
P becomes 3P means SI = 2P. Time T = 2P/(P×5/100) = 40 years. For 6P: SI = 5P. Rate = 5P×100/(P×40) = 500/40 = 12.5%.

Solving path using scaling trick: New rate = 5 × (6−1)/(3−1) = 5 × 5/2 = 12.5%. Alternatively: 3 times at 5% means T = 40 years. For 6 times: SI needed = 5P. R = 5P × 100 / (P × 40) = 12.5%. Both routes confirm 12.5%.


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