Interest (SI & CI)
๐ Log in to trackSimple and compound interest โ a near-certain slot in every shift. Tier 1 favourites: direct SI, amount at CI with clean chips, CIโSI differences and doubling chains. Tier 2 adds instalments and multi-year rate cases. Everything runs on the multiplier-chip habit.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
5 exam-level questions worked step by step.
68 questions โ untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (68 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
Direct SI โ find P, R or T
very commonThree of principal, rate, time and SI (or the amount) are given; the fourth is asked.
How to solve: SI = PRT/100; if an amount is given take SI = A โ P first. 'n times in T years' gives R = 100(nโ1)/T โ the interest is only (nโ1) times the principal.
Example: A sum of money doubles itself in 10 years at simple interest. In how many years will it become triple itself?
R = 100/10 = 10%; tripling needs 200% interest โ 20 years.
Amount at compound interest (chips)
very commonP, R and a small whole number of years are given, compounded annually โ amount or CI asked.
How to solve: A = P(1 + R/100)^T using fraction chips (10% โ 11/10). Net per cents: 2 years at 10% โ 21%, at 20% โ 44%. For P from A, divide by the chip power.
Example: What sum will amount to โน6,655 in 3 years at 10% per annum compound interest?
6655 รท 1.1ยณ = 6655 รท 1.331 = โน5,000.
CI โ SI difference (2 or 3 years)
very common'The difference between CI and SI for 2 (or 3) years is โนd' โ find the sum or the rate.
How to solve: 2 years: P(R/100)ยฒ. 3 years: P(R/100)ยฒ(3 + R/100). Reverse the factor to get P. With both CI and SI given, R = 200 ร difference/SI for 2 years.
Example: The difference between the compound interest and simple interest on a sum for 2 years at 10% per annum is โน50. The sum is:
P = 50 ร 10000/100 = โน5,000.
Half-yearly / quarterly compounding
very common'Compounded half-yearly or quarterly', usually for 1โ2 years including halves like 1ยฝ years.
How to solve: Halve the rate and double the periods (quarter them for quarterly): A = P(1 + R/200)^(2T). Count periods carefully โ 1ยฝ years half-yearly is 3 periods.
Example: Find the compound interest on โน10,000 at 20% per annum for 1ยฝ years, compounded half-yearly.
3 half-years at 10%: 10000 ร 1.1ยณ = 13310 โ CI โน3,310.
Doubling / tripling chain
common'Doubles in T years โ when is it 4, 8, 16 times?' at CI, or the same chain at SI.
How to solve: CI: powers of 2 โ 8 times in 3T. SI: linear โ (nโ1) rule, R = 100(nโ1)/T. Mixing the two growth laws is the trap.
Example: A sum doubles itself in 8 years at compound interest. In how many years will it become 8 times itself?
8 = 2ยณ โ 3 ร 8 = 24 years.
Equal annual instalments
common'A loan is repaid in 2 or 3 equal annual instalments at R% interest' โ find the instalment or the loan.
How to solve: Discount each instalment: P = ฮฃ x/(1+r)^k. 10%, 2 years: P = 210x/121. Verify with a zero-balance table. SI version: P = nx โ (r/100)xยทn(nโ1)/2.
Example: A loan of โน3,310 at 10% per annum compound interest is repaid in 3 equal annual instalments. Each instalment is:
3310 = x(10/11 + 100/121 + 1000/1331) โ x = โน1,331.
Different rates in different years
common'8% in the first year, 10% in the second', compounded annually.
How to solve: Multiply year-wise chips: A = P(1+rโ)(1+rโ). A negative year (loss) uses its own chip. Never average the rates.
Example: โน25,000 is invested at 10% for the first year and 20% for the second, compounded annually. The amount is:
25000 ร 1.1 ร 1.2 = โน33,000.
Rate from consecutive amounts
commonAmounts after two consecutive years are given โ at CI they fix the ratio, at SI the difference.
How to solve: CI: Aโ/Aโ = 1 + R/100. SI: Aโ โ Aโ = one year's interest = PR/100; subtract from the amount for P. Non-consecutive gaps: divide by the year gap.
Example: A sum amounts to โน4,840 in 2 years and โน5,324 in 3 years at compound interest. The rate per annum is:
5324/4840 = 1.1 โ 10%.