Compound Interest (CI) is what happens when your interest starts earning interest of its own. Think of it like a snowball rolling downhill — it picks up more snow (interest) as it grows, so the growth accelerates each period.
Here is a concrete picture. Say you deposit ₹10,000 at 10% per year.
The extra ₹310 over SI is the "interest on interest" — the compounding effect.
In daily life, your fixed deposits, PPF returns, and loan EMIs all use compound interest. For UP Police Constable exam, CI problems are usually straightforward — they test whether you know the core formula and can spot a pattern quickly. The exam rarely asks you to compute (1.1)^5 from scratch; instead it gives numbers that work out cleanly (like ₹53,240 at 10% for 3 years, which is just 40,000 × 1.331).
The key insight you need before solving any CI problem: always identify what you are given and what you need to find. The four variables are Principal (P), Rate (R), Time (n), Amount (A). Know three, find the fourth. Every UP Police CI question reduces to this.
One more thing to internalize early: CI − SI relationship. For 2 years, CI − SI = P(R/100)². For 3 years, CI − SI = P(R/100)²(3 + R/100). These are faster than applying the full formula.
Where:
A = Final Amount (मूलधन + ब्याज)P = Principal (मूलधन)R = Rate of interest per annum (%)n = Number of yearsCompound Interest = A − P
When interest is compounded half-yearly, you split the rate by 2 and double the time:
Quarterly compounding — divide rate by 4, multiply time by 4:
UP Police exam rarely goes beyond half-yearly, but know the pattern: divide the rate by the compounding frequency, multiply the time by the same.
These are the (1 + R/100)^n values that actually appear in UP Police questions:
| Rate | n=1 | n=2 | n=3 | |------|-----|-----|-----| | 5% | 1.05 | 1.1025 | 1.157625 | | 10% | 1.1 | 1.21 | 1.331 | | 20% | 1.2 | 1.44 | 1.728 |
You do not need to know these for arbitrary rates. The exam uses 5%, 10%, and 20% almost exclusively. Memorize these 9 values and you can solve most problems in 20–30 seconds.
Flip the formula:
Example: A = ₹53,240, R = 10%, n = 3.
P = 53240 / 1.331 = 40,000. Done. (Divide, don't multiply.)
Look — if a sum doubles in t years under CI, then in 2t years it becomes 4 times (not twice as much as double). This is because:
P × k = 2P, so k = 2P × k² = P × 4 = 4PThis is the pattern: in each additional t-year period, the amount multiplies by the same factor k. So if ₹3,000 becomes ₹6,000 in 2 years, in 4 years it becomes ₹6,000 × 2 = ₹12,000. Interest = ₹12,000 − ₹3,000 = ₹9,000.
This technique appears in UP Police papers more than you'd expect.
If SI for 2 years is known, find P first:
Then apply the CI formula normally. You will not need to memorize a separate formula for this — just convert SI to P, then compute CI.
If you are given A₁ (amount after t years) and A₂ (amount after 2t years):
Then: P = A₁ / k = A₁²/ A₂
Example: A₁ = ₹4,460 (after 3 years), A₂ = ₹6,690 (after 6 years).
k = 6690/4460 = 1.5. So P = 4460/1.5 = 4460 × (4460/6690) = 4460²/6690 ≈ ₹2,973.33.
When a loan is repaid in equal annual instalments of value X at rate R%:
where r = R/100.
For 2 instalments of ₹1,764 at 5%:
1764/1.05 = 16801764/1.1025 = 1600(Use the answer key as provided — always match your working to the given options.)
Given P, A, and R — find n:
Then match both sides to standard powers. Example: P = ₹1,000, A = ₹1,331, R = 10%.
(1.1)^n = 1.331 = (1.1)³, so n = 3 years.
Before touching any CI problem, immediately write down the multiplier for the given rate and time from your memorized table. For R=10%, n=2: write "×1.21". For R=5%, n=3: write "×1.157625". Now every problem becomes a single multiplication or division. No formula derivation needed.
Example: "What P gives A = ₹4,840 at 10% for 2 years?" — You know multiplier = 1.21. So P = 4840/1.21 = 4000. Done in 8 seconds vs. 40 seconds writing out the formula fresh each time.
If a principal doubles in t years at CI, it becomes 4× in 2t years, 8× in 3t years. This is because each t-year block multiplies by the same factor.
When a problem says "₹3,000 becomes ₹6,000 in 2 years" and asks for the 4-year amount — do not calculate the rate. Just write: 4 years = 2 doubling periods, so amount = 3000 × 4 = ₹12,000. This avoids R calculation entirely. Standard method: 90 seconds. This pattern: 10 seconds.
When a problem gives you SI and asks for CI, derive P from SI first: P = (SI × 100) / (R × T). Use the result directly in the CI multiplier table.
Example: SI for 2 years at 5% = ₹1,600. P = (1600 × 100)/(5 × 2) = ₹16,000. Then CI for 3 years at 5% = 16000 × 0.157625 = ₹2,522. If you had tried to guess P or work backwards from CI directly, you'd waste 60+ seconds. This two-step route takes 20 seconds.
When given amounts at two different time periods that are multiples of each other (e.g., 3 years and 6 years), use: P = A₁² / A₂.
Example: A after 3 years = ₹4,460; A after 6 years = ₹6,690. P = (4460 × 4460) / 6690 = 19,891,600 / 6690 ≈ ₹2,973.33. This skips finding R entirely. Standard approach (find R, then P): 3+ steps. This shortcut: 1 step.
When A/P simplifies to a recognizable power of (1 + R/100), read off n directly.
Example: P = ₹1,000, A = ₹1,331, R = 10%. Compute A/P = 1.331. Recognize 1.331 = (1.1)³. So n = 3. No logarithms, no trial and error. This works for all standard exam values since UP Police uses only clean numbers. Matching takes 5 seconds; iterative trial takes 45 seconds.
Read the problem. Label what you know: P, A, R, n, CI (= A − P).
Step 1 — What are you finding?
A = P(1 + R/100)^n directly. Look up multiplier from memory.(1 + R/100).Step 2 — Check for a pattern before computing:
P = A₁²/A₂.Step 3 — Verify with options: CI problems in UP Police always have clean answers. If your number does not match an option exactly, check whether you multiplied instead of divided (or vice versa). The most common exam error is confusing A with CI — always confirm whether the question asks for the amount or the interest.
Total target time per CI question: under 60 seconds.
Why this question: This is the simplest type — it tests whether you know that CI = Amount − Principal. Many candidates overcomplicate it.
Solving path: Amount = ₹16,854. Principal = ₹15,000. CI = 16,854 − 15,000 = ₹1,854. One subtraction. Option (d) is correct. Do not get distracted by the 6% rate — it is irrelevant once you have both A and P.
Why this question: Tests reverse application — given the final amount and rate, find the original principal. Standard exam format.
Solving path: Recognize R = 10%, n = 3. Multiplier = 1.331. So P = 53,240 / 1.331. Compute: 53,240 / 1.331 = 40,000 (verify: 40,000 × 1.331 = 53,240). Option (b).
Why this question: Another reverse-formula question — simpler because n = 2 and the multiplier 1.21 is very clean.
Solving path: R = 10%, n = 2. Multiplier = 1.21. P = 4840/1.21 = 4,000. Option (a). Cross-check: 4000 × 1.21 = 4,840. Confirmed.
Why this question: Tests the doubling-period pattern. If you try to find R and apply the formula, you waste 2+ minutes. The shortcut is immediate.
Solving path: ₹3,000 → ₹6,000 in 2 years. The amount multiplies by 2 every 2 years. In 4 years: ₹6,000 × 2 = ₹12,000. CI = 12,000 − 3,000 = ₹9,000. Option (b).
Why this question: Combines SI and CI — a two-step problem that trips up candidates who do not isolate P first.
Solving path: SI = ₹1,600, R = 5%, T = 2 years. P = (1600 × 100)/(5 × 2) = ₹16,000. Now CI for 3 years at 5%: multiplier = 1.157625. CI = 16,000 × 0.157625 = ₹2,522. Option (a).
Why this question: Tests the two-amount shortcut. Trying to find R explicitly is time-consuming and error-prone.
Solving path: A₁ = ₹4,460 (3 years), A₂ = ₹6,690 (6 years). Ratio = 6690/4460 = 1.5 = growth factor for 3 years. P = A₁ / 1.5 = 4460/1.5 ≈ ₹2,973.33. Alternatively: P = A₁²/A₂ = (4460 × 4460)/6690 ≈ ₹2,973.33. Option (c).
Why this question: Tests finding time — requires recognizing that A/P is a perfect power of the rate multiplier.
Solving path: CI = ₹331, P = ₹1,000, so A = ₹1,331. R = 10%. A/P = 1331/1000 = 1.331 = (1.1)³. So n = 3 years. Option (a).
Confusing Amount with CI. The formula gives you A (total amount). If the question asks for CI, you must subtract P. Half the wrong answers in CI come from submitting A instead of A − P.
Applying SI formula to a CI problem. The two formulas look similar at a glance — SI uses P × R × T / 100, CI uses P(1 + R/100)^n. When the question says "compound interest", reach for the multiplier table, not the linear SI formula.
Dividing rate by 100 twice. When computing 1 + R/100, the R is already in percentage form. If R = 10%, you write 1 + 10/100 = 1.1. Do not write 1 + 0.10/100.
Forgetting to halve the rate for half-yearly compounding. Half-yearly means rate ÷ 2 AND time × 2. Candidates often only double the time but forget to halve the rate, inflating the answer.
Using the doubling rule incorrectly. The doubling rule says the multiplier repeats every t years — not that the interest doubles. If ₹3,000 doubles in 2 years, the amount in 4 years is ₹12,000 (multiplied by 4), not ₹9,000 (doubled interest added again).
Not checking units of time. If the rate is annual but the time given is in months, convert months to years first. A 6-month period with an annual rate of 10% means R = 10%, n = 0.5 — or equivalently, treat as half-yearly with rate 5% for 1 period.