3. Shortcuts & Speed Tricks

The tricks below are not alternative theory — every one of them is a direct consequence of the formulas already derived in Section 2. Their purpose is purely speed: in the exam hall, you will not have time to write out a full derivation for every question, so the goal is to recognise the question type within a few seconds and jump straight to the one-line shortcut. Practice each trick with at least five to ten of your own numbers until the pattern becomes automatic, rather than something you have to consciously recall mid-exam.

Trick 1 — Use the CI–SI difference formula directly; never compute CI and SI separately for 2/3-year “difference” questions. If asked for the difference between CI and SI over 2 years, just apply Diff = PR²/10000 — do not calculate the full CI and SI amounts and subtract. Example: P = 12,000, R = 5%, 2 years → Diff = 12000 × 25/10000 = Rs. 30 (computed in one line).

Trick 2 — Assume P = 100 whenever the question gives only rate/time and asks for a ratio, percentage, or “how much more” type answer. Since CI/SI questions are usually scale-invariant (the answer doesn’t depend on the actual principal when only rates/ratios are asked), let P = 100 to avoid big multiplications. Example: “CI exceeds SI by what percent of SI for a 2-year, 10% deposit?” Assume P=100. SI = 20, CI = 21, so CI exceeds SI by 1, which is 1/20 = 5% of SI.

Trick 3 — For successive-year CI, use the multiplying-factor chain instead of expanding powers. Amount after each year is obtained by multiplying by (1 + R/100) repeatedly — track the running amount rather than raising to a power from scratch each time, especially when a question asks for CI in “the 3rd year alone” (this is just the interest earned only in year 3, i.e., A₃ − A₂). Example: P = 10,000, R = 10%. A₁=11,000, A₂=12,100, A₃=13,310. CI for the 3rd year alone = 13,310 − 12,100 = Rs. 1,210 (this is simply SI at 10% on A₂, a very fast mental shortcut).

Trick 4 — Ratio method for two different SI deposits/time periods. If a sum under SI becomes k times itself in T years, then R = (k−1) × 100/T directly, skipping the full formula. Example: A sum becomes 1.5 times itself (SI = 0.5P) in 6 years. R = 0.5 × 100/6 = 8.33% p.a.

Trick 5 — Doubling/tripling time shortcut for SI. A sum doubles under SI in T years ⟺ R = 100/T. Tripling ⟺ SI = 2P ⟺ R = 200/T. Memorise these two relations to answer “in how many years will a sum double at R% SI” instantly: T = 100/R. Example: At what rate will a sum double itself under SI in 8 years? R = 100/8 = 12.5%.

Trick 6 — CI doubling/multiplying-time shortcut using exponents (no logs needed for clean multiples). If a sum becomes k times itself in CI in T years at a fixed rate, then to become k^m times itself takes m×T years (because each “doubling period” simply repeats the same multiplying factor). Example: A sum becomes 2 times itself in 3 years under CI. In how many years will it become 8 times itself? Since 8 = 2³, time = 3 × 3 = 9 years.

Trick 7 — Successive percentage / net rate shortcut for 2-year CI or growth-then-depreciation problems. Use the net-effect formula from the Percentage chapter: Net% = a + b + ab/100 (with signs) instead of computing two separate steps. Example: CI at 10% for 2 years — net = 10 + 10 + (10×10)/100 = 20 + 1 = 21%, i.e., CI on Rs. 100 for 2 years at 10% = Rs. 21 (matches Trick 2’s example — same shortcut, applied from a different angle).

Trick 8 — When compounding is half-yearly/quarterly, immediately rewrite R and T before doing anything else. Make this the very first line you write on your rough sheet: “R → R/2, T → 2T” (half-yearly) or “R → R/4, T → 4T” (quarterly). This single habit eliminates the single most common CI error in the exam.

Trick 9 — For CI over a fractional year, split mentally into “whole-year CI” + “SI on the last accumulated amount for the fraction,” and do NOT raise the base to a fractional power. Example: CI at 10% for 2½ years on Rs. 8,000. First find A after 2 years = 8000 × 1.21 = 9,680. Then add SI on 9,680 for ½ year at 10% = 484. Final amount = Rs. 10,164; CI = Rs. 2,164 — all mental, no fractional exponent needed.

Trick 10 — For installment (CI) problems with only 2 installments, use the direct sum-of-two-fractions shortcut rather than the general summation formula. For n = 2: P = x/(1+r) + x/(1+r)². Factor out 1/(1+r): P = [x/(1+r)] × [1 + 1/(1+r)] = [x(2+r)]/(1+r)². This two-step factoring is faster than plugging into a calculator-style series for n = 2, the most commonly tested case.

Trick 11 — Recognise “amount in year n” vs “amount in year n+1” ratio questions instantly. If a question gives the CI amount after 2 years and after 3 years (as in Example 2.7.1), never re-derive P and R from scratch with two equations — simply divide the two amounts to get (1 + R/100) in one step, since the extra year is just one more multiplication by the same factor. This same division trick works for any two consecutive years, e.g., amount after 5 years divided by amount after 4 years also gives (1 + R/100) directly.

Trick 12 — For “sum triples/quadruples under SI” questions, generalise the doubling-time shortcut instead of setting up SI = PRT/100 from scratch. If a sum becomes k times itself under SI in T years, then SI earned = (k−1)P, so R = 100(k−1)/T. This single generalised line replaces the doubling formula (Trick 5, where k=2) and the tripling formula (Example 2.12.2, where k=3) with one memorised relation. Example: A sum becomes 4 times itself under SI in 12 years. R = 100 × 3/12 = 25% p.a.

Trick 13 — For multi-year, changing-rate CI problems, treat each year as an independent percentage-change step and chain the multiplying factors mentally — never expand algebraically. Convert each year’s rate into a multiplying factor (e.g., 10% → 1.1, 20% → 1.2, −10% depreciation → 0.9) and multiply the factors together first before applying to the principal; this avoids re-writing the base principal at every intermediate step and reduces silly arithmetic slips. Example: Rs. 8,000 grows 25% in year 1 and depreciates 20% in year 2. Combined factor = 1.25 × 0.8 = 1.00, so the final amount is unchanged at Rs. 8,000 — a calculation many aspirants get wrong by adding percentages instead of multiplying factors.