Section 7, arithmetic

Simple & Compound Interest

SI/CI turns up less often than profit-loss or speed questions, but it is quick marks: most questions are one formula plus careful multiplying factors. Learn the five patterns below and the chapter is done in two sittings.

How CAT asks it

Money split across two or three rates; the gap between CI and SI; two investors compared under different schemes; loans repaid in equal instalments; amounts after successive years.

Concepts you must know

  • SI = P × r × t ÷ 100, and Amount = P + SI. Simple interest is the same every year.
  • CI amount A = P(1 + r/100)ⁿ, and CI = A − P.
  • Half-yearly compounding: rate r/2 per period, 2n periods. Quarterly: r/4 per period, 4n periods.
  • CI − SI over 2 years = P(r/100)². Over 3 years = P(r/100)² × (3 + r/100).
  • Amounts in successive years form a GP: Aₙ₊₁ ÷ Aₙ = 1 + r/100.
  • Year-2 CI minus year-1 CI is the interest on year-1 interest, so r = (CI₂ − CI₁) ÷ CI₁ × 100.
  • Doubling: under SI, money doubles in 100/r years. Under CI, if it doubles in n years it becomes 4× in 2n and 8× in 3n; the rule of 72 gives the doubling time as about 72/r years.
  • Equal annual instalments X repaying a loan P at r% CI: P = X/(1 + r/100) + X/(1 + r/100)² + …, one term per instalment.
  • The two formulas to know by heart:
A=P(1+r100)nA = P\left(1 + \frac{r}{100}\right)^{n}
CI2 yr−SI2 yr=P(r100)2CI_{2\,\text{yr}} - SI_{2\,\text{yr}} = P\left(\frac{r}{100}\right)^{2}

Thought process

  1. Identify SI or CI, and the compounding period.
  2. Turn rates into factors: 10% → 11/10, 20% → 6/5, 25% → 5/4, 12.5% → 9/8.
  3. For split investments, alligate on the rates (Section 10).
  4. For comparisons, set the two amounts equal and cancel common factors before multiplying anything out.

Patterns CAT repeats

Numbered so your error log can point at them: “Pattern 2”.

  1. 1Split investment: total interest given; find how the money was split (alligation on rates).
  2. 2CI − SI gap for 2 or 3 years; find P or r.
  3. 3CI and SI for the same 2 years: r = 2 × (CI − SI) ÷ SI × 100, then P.
  4. 4Amounts after n and n + 1 years: their ratio is 1 + r/100.
  5. 5Two investors, two schemes: equal or related final amounts; find a principal.
  6. 6Equal instalments: a loan repaid in 2 or 3 equal annual payments.
  7. 7Doubling and tripling times under SI or CI.
  8. 8Different rates in different years: multiply the factors (1 + r₁/100)(1 + r₂/100)(1 + r₃/100).

Worked examples

Attempt each one before opening its solution, then compare methods.

  1. Example 1, split investment

    Priya invests ₹50,000, part at 6% and the rest at 10% simple interest. Her total interest after one year is ₹3,800. How much did she invest at 10%?

    Show solution

    Average rate = 3,800/50,000 = 7.6%. Alligation: 6% part : 10% part = (10 − 7.6) : (7.6 − 6) = 3 : 2. The 10% part is 2/5 of 50,000. Check: 1,800 + 2,000 = 3,800. Answer: ₹20,000.

  2. Example 2, CI − SI gap

    The difference between CI and SI on a sum for 2 years at 8% is ₹128. Find the sum.

    Show solution

    P × 0.08² = P × 0.0064 = 128. Answer: ₹20,000.

  3. Example 3, CI and SI given

    On a sum, the SI for 2 years is ₹2,400 and the CI for 2 years is ₹2,520. Find the rate and the sum.

    Show solution

    SI per year = 1,200. The extra 120 is one year’s interest on 1,200, so r = 10%. P = 1,200 ÷ 0.10. Answer: 10%; ₹12,000.

  4. Example 4, successive amounts

    A sum at CI grows to ₹13,310 in 3 years and ₹14,641 in 4 years. Find the rate and the sum.

    Show solution

    14,641/13,310 = 1.1, so r = 10%. P = 13,310 ÷ 1.331. Answer: 10%; ₹10,000.

  5. Example 5, half-yearly

    ₹40,000 is invested at 10% a year, compounded half-yearly, for 1 year. Find the amount.

    Show solution

    5% per half-year for 2 periods: 40,000 × 1.05² = 40,000 × 1.1025. Answer: ₹44,100 (interest ₹4,100, against ₹4,000 with yearly compounding).

  6. Example 6, equal instalments

    A loan of ₹21,000 is repaid in 2 equal annual instalments at 10% CI. Find each instalment.

    Show solution

    X/1.1 + X/1.21 = 21,000, so X × 2.1/1.21 = 21,000 and X = 12,100. Check: after year 1 the debt is 23,100; paying 12,100 leaves 11,000, which grows to 12,100. Answer: ₹12,100.

  7. Example 7, two investors

    Kabir invests ₹P at 8% CI for 2 years. Dev invests ₹10,000 more than Kabir at 8% SI for 2 years. Dev’s interest exceeds Kabir’s by ₹1,072. Find P.

    Show solution

    Kabir earns 0.1664P. Dev earns 0.16(P + 10,000) = 0.16P + 1,600. The gap 1,600 − 0.0064P = 1,072 gives P = 528/0.0064. Answer: ₹82,500.

  8. Example 8, doubling

    A sum doubles in 5 years at CI. When will it be 8 times? At what SI rate does a sum double in 8 years?

    Show solution

    8 = 2³, so three doubling periods. Under SI, doubling means total interest = P, so r × 8 = 100. Answer: 15 years; 12.5%.

Traps

  • Half-yearly compounding: halve the rate and double the periods. Doing only one is the most common error.
  • “Interest in the 3rd year” is not “interest for 3 years”.
  • In instalment questions, discount each payment separately; never split the loan equally.
  • Under SI, reaching 3× takes twice as long as reaching 2×. Under CI, the multiples compound.

Practice set

Timed, about 2.5 minutes per question. Log every miss in the error log.

  1. 1

    Find the SI on ₹8,000 at 7.5% for 4 years.

    Show answer

    ₹2,400: 8,000 × 0.075 × 4.

  2. 2

    Find the CI on ₹15,625 at 8% for 2 years.

    Show answer

    ₹2,600: 15,625 × 0.1664.

  3. 3

    Find CI − SI on ₹10,000 at 10% for 3 years.

    Show answer

    ₹310: 10,000 × 0.01 × 3.1 (CI 3,310 against SI 3,000).

  4. 4

    A sum at SI becomes ₹9,800 in 5 years and ₹12,005 in 8 years. Find the rate.

    Show answer

    12%: 3 years of SI = 2,205, so 735 a year; P = 9,800 − 5 × 735 = 6,125; 735/6,125 = 0.12.

  5. 5

    ₹1,00,000 is split between 5% and 8% SI and earns ₹6,800 a year. How much is at 8%?

    Show answer

    ₹60,000: average 6.8%, so 5% : 8% = 1.2 : 1.8 = 2 : 3.

  6. 6

    A city grows 5% a year and now has 4,41,000 people. What was its population two years ago?

    Show answer

    4,00,000: 4,41,000 ÷ 1.1025.

  7. 7

    On the same principal at annual CI, the interest for year 1 is ₹500 and for year 2 is ₹550. Find the rate and principal.

    Show answer

    10% and ₹5,000: 50/500 = 10%, and 500 ÷ 0.1.

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