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Simple and Compound Interest for CUET General Aptitude

About 10 min to learn · 60 questions

Interest questions in CUET's General Aptitude Test are short, number-friendly and very learnable, because almost every one follows a handful of patterns. This lesson covers the simple interest formula and its rearrangements, time in months and days, doubling, compound interest and growth factors, half-yearly and quarterly compounding, changing rates, working back from amounts, and the shortcuts for the difference between compound and simple interest, with worked examples and traps.

Syllabus area
General Mental Ability, Numerical Ability
On this page · 8 sections
  1. 01The core idea
  2. 02Simple interest
  3. 03Compound interest
  4. 04CI minus SI
  5. 05Exam shortcuts
  6. 06Common mistakes
  7. 07Quick revision
  8. 08Check your understanding

The core idea

Simple interest is charged only on the original principal, so it is the same every year and grows in a straight line. Compound interest adds each year's interest to the principal, so every year's interest is a little bigger than the last, and the amount grows by a multiplier.

Four words run through every question: the principal PP, the money lent; the rate RR, per cent per year; the time TT, in years; and the amount AA, principal plus interest. Many wrong options are the amount when the interest is asked for, or the other way round.

Simple interest

What is simple interest and how do you calculate it?

Simple interest is P×R×T100\tfrac{P \times R \times T}{100}, and the amount is the principal plus the interest. Because it is the same every year, three years of interest is exactly three times one year's.

Simple interest

SI=P×R×T100A=P+SI\text{SI} = \frac{P \times R \times T}{100} \qquad A = P + \text{SI}

Example 1 · Interest and amount

Nandini deposits ₹6,500 at 8% a year simple interest for 4 years. Find the interest and the amount.

  1. One year's interest: 8% of 6,500 is 520.

  2. Four years: 4×520=2,0804 \times 520 = 2{,}080.

  3. Amount: 6,500+2,080=8,5806{,}500 + 2{,}080 = 8{,}580.

Answer

Interest ₹2,080, amount ₹8,580

How do you find the principal, rate or time in simple interest?

Rearrange the same formula: P=100×SIRTP = \tfrac{100 \times \text{SI}}{RT}, R=100×SIPTR = \tfrac{100 \times \text{SI}}{PT} and T=100×SIPRT = \tfrac{100 \times \text{SI}}{PR}. If you are given the amount, subtract the principal first, or work out one year's interest.

Example 2 · Finding the rate

A loan of ₹14,000 grows to ₹17,780 at simple interest in 3 years. What is the rate?

  1. Interest: 17,780−14,000=3,78017{,}780 - 14{,}000 = 3{,}780.

  2. One year's interest: 3,780÷3=1,2603{,}780 \div 3 = 1{,}260.

  3. Rate: 1,26014,000×100=9%\tfrac{1{,}260}{14{,}000} \times 100 = 9\%.

Answer

9% a year

Example 3 · Two amounts, principal unknown

A sum amounts to ₹7,700 in 2 years and ₹8,750 in 5 years at simple interest. Find the sum.

  1. Three extra years give 8,750−7,700=1,0508{,}750 - 7{,}700 = 1{,}050, so one year's interest is 350.

  2. Step back two years from year 2: 7,700−2×350=7,0007{,}700 - 2 \times 350 = 7{,}000.

Answer

₹7,000

The rate is 3507,000=5%\tfrac{350}{7{,}000} = 5\%.

How do you handle time given in months or days?

Convert the time into years first, because the rate is per year: divide months by 12 and days by 365. So 9 months is 34\tfrac{3}{4} year, 1 year 4 months is 43\tfrac{4}{3} years, and 146 days is 146365=25\tfrac{146}{365} = \tfrac{2}{5} year.

Watch out

"2 years 6 months" is 2.5 years, not 2.6, because 6 months is half a year. Likewise 1 year 3 months is 1.25 years, not 1.3.

What is the quickest way to solve "a sum doubles in n years" questions?

At simple interest a sum doubles when the interest equals the principal, so R×T=100R \times T = 100. It becomes kk times itself when R×T=100(k−1)R \times T = 100(k - 1).

  • Doubles in 16 years: R=100÷16=6.25%R = 100 \div 16 = 6.25\%.
  • At 5%, it triples in 200÷5=40200 \div 5 = 40 years.
  • If it doubles in 4 years, the interest is PP in 4 years, so becoming 5 times itself needs interest 4P4P, which takes 16 years, not 20.

Key idea

At compound interest the pattern is different: a sum that doubles in nn years becomes 4 times in 2n2n years and 8 times in 3n3n, because each doubling starts from the bigger amount.

Compound interest

What is compound interest and how is it different from simple interest?

Compound interest adds each year's interest to the principal, so the next year's interest is earned on a bigger balance. The amount after nn years is P(1+R100)nP(1 + \tfrac{R}{100})^n, and the interest is that amount minus PP. In the first year CI and SI are equal; after that CI is larger.

Compound interest

A=P(1+R100)nCI=A−PA = P\left(1 + \frac{R}{100}\right)^n \qquad \text{CI} = A - P

Example 4 · Two years compounded

Find the compound interest on ₹12,500 at 8% a year for 2 years, compounded annually.

  1. After year 1: 12,500×1.08=13,50012{,}500 \times 1.08 = 13{,}500.

  2. After year 2: 13,500×1.08=14,58013{,}500 \times 1.08 = 14{,}580.

  3. CI: 14,580−12,500=2,08014{,}580 - 12{,}500 = 2{,}080.

Answer

₹2,080

Simple interest would be ₹2,000.

RateFactor for 2 yearsFactor for 3 years
4%1.08161.124864
5%1.10251.157625
8%1.16641.259712
10%1.211.331
20%1.441.728

Exam shortcut

A fraction often makes the arithmetic cleaner: 5% is 2120\tfrac{21}{20}, 4% is 2625\tfrac{26}{25}, 10% is 1110\tfrac{11}{10} and 20% is 65\tfrac{6}{5}. ₹24,000 at 5% for 3 years is 24,000×9,2618,000=27,78324{,}000 \times \tfrac{9{,}261}{8{,}000} = 27{,}783, because 203=8,00020^3 = 8{,}000.

How do you calculate compound interest compounded half-yearly or quarterly?

Divide the yearly rate by the number of periods in a year, and multiply the years by the same number. Half-yearly: half the rate, twice the periods. Quarterly: a quarter of the rate, four times the periods.

  • 12% a year compounded half-yearly for 1121\tfrac{1}{2} years: 6% a period for 3 periods.
  • 20% a year compounded quarterly for 6 months: 5% a period for 2 periods.

Example 5 · Half-yearly

Find the amount on ₹10,000 at 8% a year compounded half-yearly for 1 year.

  1. 4% a half-year, for 2 half-years.

  2. Amount: 10,000×1.04×1.04=10,81610{,}000 \times 1.04 \times 1.04 = 10{,}816.

Answer

₹10,816

Compounded yearly it would be ₹10,800.

Watch out

Half-yearly means half the rate and twice the periods. Doing only one of the two, such as using the full yearly rate every half-year, gives a planted option.

What if the rate of interest changes every year?

Multiply by each year's growth factor in turn. Do not add the rates, and do not average them.

Example 6 · Two different rates

A deposit of ₹5,000 earns 4% in the first year and 10% in the second, compounded annually. Find the interest.

  1. Amount: 5,000×1.04×1.10=5,200×1.10=5,7205{,}000 \times 1.04 \times 1.10 = 5{,}200 \times 1.10 = 5{,}720.

  2. CI: 5,720−5,000=7205{,}720 - 5{,}000 = 720.

Answer

₹720

Adding the rates as simple interest gives ₹700, and averaging to 7% twice gives ₹724.50. Both are wrong.

How do you find the rate or principal from amounts in successive years?

At compound interest, the growth from one year to the next is the rate itself: R=A3−A2A2×100R = \tfrac{A_3 - A_2}{A_2} \times 100. Once you know the rate, go backwards by dividing by the growth factor, never by subtracting.

Example 7 · Working back

A sum at compound interest amounts to ₹4,840 after 2 years and ₹5,324 after 3 years. Find the rate and the sum.

  1. Growth in year 3: 5,324−4,840=4845{,}324 - 4{,}840 = 484, on 4,840, so R=10%R = 10\%.

  2. Sum: 4,840÷1.21=4,0004{,}840 \div 1.21 = 4{,}000.

Answer

10% and ₹4,000

CI minus SI

What is the shortcut for the difference between CI and SI for 2 years?

For 2 years at R%R\%, compounded annually, CI − SI is P(R100)2P(\tfrac{R}{100})^2. The difference is just the second year's interest on the first year's interest.

Difference for 2 years

CI−SI=P(R100)2\text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^2

Example 8 · From the two interests

The CI on a sum for 2 years is ₹2,244 and the SI at the same rate is ₹2,200. Find the rate and the sum.

  1. One year's SI is 1,100. The extra 44 is interest on that 1,100 for a year.

  2. Rate: 441,100×100=4%\tfrac{44}{1{,}100} \times 100 = 4\%.

  3. Sum: 2,200×1004×2=27,500\tfrac{2{,}200 \times 100}{4 \times 2} = 27{,}500.

Answer

4% and ₹27,500

Exam shortcut

  • On ₹16,000 at 5%: 16,000×0.0025=4016{,}000 \times 0.0025 = 40.
  • From the 2-year SI: the difference is half the SI times R100\tfrac{R}{100}. SI of ₹3,000 at 6% gives 1,500×0.06=901{,}500 \times 0.06 = 90.
  • Backwards: a difference of ₹90 at 6% means P=90÷0.0036=25,000P = 90 \div 0.0036 = 25{,}000.

What is the formula for the difference between CI and SI for 3 years?

For 3 years, CI − SI is P(R100)2(3+R100)P(\tfrac{R}{100})^2(3 + \tfrac{R}{100}). It is not three times the 2-year difference; the extra R100\tfrac{R}{100} is interest earned on interest-on-interest.

Difference for 3 years

CI−SI=P(R100)2(3+R100)\text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right)

On ₹6,000 at 10%: 6,000×0.01×3.1=1866{,}000 \times 0.01 \times 3.1 = 186. Check: CI is 1,986 and SI is 1,800.

Key idea

These shortcuts are for annual compounding. For half-yearly compounding over one year, use the 2-year shortcut with periods and the half-yearly rate: ₹60,000 at 10% a year compounded half-yearly for 1 year gives 60,000×0.052=15060{,}000 \times 0.05^2 = 150.

Exam shortcuts

  • SI is linear. Find one year's interest and multiply.
  • Doubling at SI. R×T=100R \times T = 100. For kk times, 100(k−1)100(k - 1).
  • Growth factors. Learn the 2- and 3-year factors for 4%, 5%, 8%, 10% and 20%.
  • Successive amounts. The growth from one year to the next is the rate.
  • CI − SI. P(R100)2P(\tfrac{R}{100})^2 for 2 years, times (3+R100)(3 + \tfrac{R}{100}) for 3.

Common mistakes

  • Interest or amount? Both are usually among the options.
  • Months as decimals. 2 years 6 months is 2.5 years.
  • "Doubles" means interest PP, not 2P2P.
  • A fraction of the amount, not the principal. Interest of 27\tfrac{2}{7} of the amount is 25\tfrac{2}{5} of the principal.
  • Half the rate but not twice the periods, or the other way round.
  • Adding or averaging changing rates.
  • Subtracting to go backwards. Divide by the growth factor.
  • Forgetting the root. R2=49R^2 = 49 means 7%, and 21% growth over 2 years is 10% a year.
  • Three times the 2-year difference for 3 years.
  • Part payments. After a repayment, next year's interest is on what is left.

Quick revision

Simple interest

  • SI=PRT100\text{SI} = \tfrac{PRT}{100}, A=P+SIA = P + \text{SI}
  • Time in years: months ÷ 12, days ÷ 365
  • Doubles: RT=100RT = 100. kk times: RT=100(k−1)RT = 100(k - 1)

Compound interest

  • A=P(1+R100)nA = P(1 + \tfrac{R}{100})^n
  • Half-yearly: R2\tfrac{R}{2}, 2n2n periods. Quarterly: R4\tfrac{R}{4}, 4n4n
  • Changing rates: multiply the factors
  • 1.12=1.211.1^2 = 1.21, 1.13=1.3311.1^3 = 1.331, 1.052=1.10251.05^2 = 1.1025

CI minus SI

  • 2 years: P(R100)2P(\tfrac{R}{100})^2
  • 3 years: P(R100)2(3+R100)P(\tfrac{R}{100})^2(3 + \tfrac{R}{100})
  • First year: CI = SI

Interest is built on percentages, and growth and depreciation use the same multipliers there.

Next · 07Quick revision

Check your understanding

60 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Simple interestQ1 of 60Q1

Simple interest

24 questions

Question 1

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Meenakshi deposits ₹8,000 in a savings scheme that pays simple interest at 6% per annum. How much interest does she earn in 3 years?

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AnswerB · ₹1,440

  1. SI=P×R×T100=8000×6×3100\text{SI} = \frac{P \times R \times T}{100} = \frac{8000 \times 6 \times 3}{100}.

  2. SI = ₹1,440.

Worth noting

₹480 is only one year's interest, and ₹9,440 is the amount (principal + interest), not the interest.

Question 2

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A bank charged Rafiq ₹1,350 as simple interest on a loan taken for 3 years at 5% per annum. How much did he borrow?

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AnswerA · ₹9,000

  1. P=SI×100R×T=1350×1005×3P = \frac{\text{SI} \times 100}{R \times T} = \frac{1350 \times 100}{5 \times 3}.

  2. P=13500015=9000P = \frac{135000}{15} = 9000, so he borrowed ₹9,000.

Worth noting

Leaving out the 3 years gives ₹27,000; adding the interest to the principal gives ₹10,350, which is the amount repaid, not the loan.

Question 3

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Harpreet invested ₹12,000 and received ₹2,880 as simple interest after 4 years. What was the rate of interest per annum?

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AnswerB · 6%

  1. R=SI×100P×T=2880×10012000×4R = \frac{\text{SI} \times 100}{P \times T} = \frac{2880 \times 100}{12000 \times 4}.

  2. R=28800048000=6%R = \frac{288000}{48000} = 6\% per annum.

Worth noting

₹2,880 is 24% of ₹12,000, but that is the interest for all 4 years together. Per year it is 24÷4=6%24 \div 4 = 6\%.

Question 4

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Joseph's deposit of ₹5,000 grows to ₹6,600 at 8% per annum simple interest. For how many years was the money deposited?

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AnswerB · 4 years

  1. Interest = ₹6,600 − ₹5,000 = ₹1,600.

  2. Interest for one year = 8% of ₹5,000 = ₹400.

  3. Time = 1600400=4\frac{1600}{400} = 4 years.

Worth noting

Using the amount ₹6,600 as if it were the interest gives 16.5 years. Subtract the principal first.

Question 5

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A sum of money doubles itself in 8 years at simple interest. What is the rate of interest per annum?

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AnswerC · 12.5%

  1. Doubling means the interest earned equals the principal P.

  2. So P=P×R×8100P = \frac{P \times R \times 8}{100}, which gives R=1008=12.5%R = \frac{100}{8} = 12.5\%.

Worth noting

The interest is P, not 2P. Treating the interest as 2P gives 25%.

Question 6

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Kavya borrows ₹15,000 from a cooperative society at 7% per annum simple interest. What amount must she repay after 2 years?

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AnswerC · ₹17,100

  1. SI=15000×7×2100=2100\text{SI} = \frac{15000 \times 7 \times 2}{100} = 2100.

  2. Amount = ₹15,000 + ₹2,100 = ₹17,100.

Worth noting

₹2,100 is only the interest. ₹17,173.50 is what compound interest would give; simple interest is charged on the original ₹15,000 every year.

Question 7

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A sum lent at simple interest becomes 95\frac{9}{5} of itself in 10 years. What is the rate of interest per annum?

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AnswerA · 8%

  1. Interest = 95P−P=45P\frac{9}{5}P - P = \frac{4}{5}P, which is 80% of P.

  2. Over 10 years that is 80÷10=8%80 \div 10 = 8\% per annum.

Worth noting

Taking the whole 95\frac{9}{5} (180%) as interest gives 18%, and 80% is the interest for all 10 years together. The sum itself is part of the 95\frac{9}{5}.

Question 8

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Rohit earns ₹960 as simple interest on ₹6,000 in 2 years. At the same rate, how much interest will the same sum earn in 5 years?

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AnswerB · ₹2,400

  1. Simple interest is the same every year: ₹960 ÷ 2 = ₹480 per year.

  2. In 5 years: 5 × ₹480 = ₹2,400.

Worth noting

₹4,800 multiplies the 2-year interest by 5. ₹8,400 is the amount, not the interest.

Question 9

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At what rate per annum simple interest will a sum become three times itself in 20 years?

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AnswerB · 10%

  1. Three times itself means the interest is 2P, that is, 200% of P.

  2. Per year: 200÷20=10%200 \div 20 = 10\%.

Worth noting

5% is the rate for doubling in 20 years; tripling needs interest of 2P, not P.

Question 10

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At simple interest, a sum becomes twice itself in 6 years. In how many years will it become four times itself?

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AnswerB · 18 years

  1. Twice itself means the interest in 6 years is P.

  2. Four times itself means the interest needed is 3P.

  3. Simple interest grows evenly, so 3P needs 3×6=183 \times 6 = 18 years.

Worth noting

12 years is compound-interest thinking (doubling twice). 24 years multiplies 6 by 4, forgetting that the principal itself is already one of the four parts.

Question 11

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Ishaan takes a short loan of ₹24,000 at 9% per annum simple interest and repays it after 10 months. How much interest does he pay?

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AnswerC · ₹1,800

  1. Time must be in years: 10 months = 1012=56\frac{10}{12} = \frac{5}{6} year.

  2. Interest for one full year = 9% of ₹24,000 = ₹2,160.

  3. Interest for 56\frac{5}{6} year = 2160×562160 \times \frac{5}{6} = ₹1,800.

Worth noting

Writing 10 months as 0.10 year gives ₹216, and ₹2,160 is a whole year's interest. A year has 12 months, so 10 months is 10/1210/12 of a year.

Question 12

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Fatima keeps ₹36,500 in a deposit that pays 10% per annum simple interest. She withdraws it after 73 days. What interest does she earn? (Take 1 year = 365 days.)

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AnswerA · ₹730

  1. 73 days = 73365=15\frac{73}{365} = \frac{1}{5} year.

  2. Interest for one year = 10% of ₹36,500 = ₹3,650.

  3. Interest for 15\frac{1}{5} year = 36505\frac{3650}{5} = ₹730.

Worth noting

₹3,650 is a full year's interest, and ₹2,66,450 treats 73 days as 73 years. Days must be turned into a fraction of 365.

Question 13

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Tenzin borrows ₹20,000 at 12% per annum simple interest and clears the loan with a single payment at the end of 2 years 6 months. How much does he pay?

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AnswerC · ₹26,000

  1. 2 years 6 months = 2.5 years.

  2. SI=20000×12×2.5100=6000\text{SI} = \frac{20000 \times 12 \times 2.5}{100} = 6000.

  3. Payment = ₹20,000 + ₹6,000 = ₹26,000.

Worth noting

Writing 2 years 6 months as 2.6 years gives ₹26,240. Six months is half a year, so the time is 2.5 years.

Question 14

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Aditi invests ₹10,000, part of it at 6% and the rest at 8% per annum simple interest. Her total interest for one year is ₹720. How much did she invest at 6%?

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AnswerA · ₹4,000

  1. If all ₹10,000 were at 8%, the interest would be ₹800.

  2. The actual interest is ₹720, which is ₹80 less.

  3. Each ₹100 moved to 6% loses ₹2, so the amount at 6%=(80÷2)×1006\% = (80 \div 2) \times 100 = ₹4,000.

Worth noting

₹6,000 is the part invested at 8%, and ₹9,000 treats all the interest as earned at 8%. Check: 6% of 4,000+8% of 6,000=240+4806\% \text{ of } 4{,}000 + 8\% \text{ of } 6{,}000 = 240 + 480 = ₹720.

Question 15

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A bank pays simple interest at 5% per annum for the first 2 years, 6% per annum for the next 3 years and 8% per annum after that. How much interest does a deposit of ₹12,000 earn in 7 years?

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AnswerC · ₹5,280

  1. Total rate-years = 5×2+6×3+8×2=10+18+16=445 \times 2 + 6 \times 3 + 8 \times 2 = 10 + 18 + 16 = 44.

  2. Interest = 44% of ₹12,000 = ₹5,280.

Worth noting

₹3,360 stops after 5 years and misses the last 2 years at 8%. ₹6,720 wrongly uses 8% for all 7 years.

Question 16

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At simple interest, a sum amounts to ₹9,440 in 3 years and to ₹10,400 in 5 years. What is the sum?

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AnswerB · ₹8,000

  1. The extra 2 years add ₹10,400 − ₹9,440 = ₹960 of interest.

  2. So one year's interest = ₹480.

  3. Sum = ₹9,440 − 3 × ₹480 = ₹9,440 − ₹1,440 = ₹8,000.

Worth noting

₹960 is the interest for 2 years, not 1. Taking it as one year's interest gives ₹6,560; subtracting it only once gives ₹8,480.

Question 17

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A sum lent at simple interest amounts to ₹17,600 after 2 years and to ₹20,800 after 6 years. What is the rate of interest per annum?

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AnswerB · 5%

  1. Interest for the 4 extra years = ₹20,800 − ₹17,600 = ₹3,200, so one year's interest = ₹800.

  2. Principal = ₹17,600 − 2 × ₹800 = ₹16,000.

  3. Rate = 80016000×100=5%\frac{800}{16000} \times 100 = 5\%.

Worth noting

₹3,200 is 20% of ₹16,000, but that covers 4 years. The rate is always per year and on the principal, not on an amount.

Question 18

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Had the rate of simple interest been 2% higher, Nikhil would have earned ₹540 more on his deposit over 3 years. How much did he deposit?

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AnswerA · ₹9,000

  1. The extra ₹540 comes only from the extra 2% for 3 years.

  2. P×2×3100=540\frac{P \times 2 \times 3}{100} = 540, so P=540×1006=9000P = \frac{540 \times 100}{6} = 9000.

  3. He deposited ₹9,000.

Worth noting

Dividing by 2 but forgetting the 3 years gives ₹27,000. The extra interest is 2% for EACH of the 3 years, so 6% of P in all.

Question 19

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Sneha borrows a sum at 7% per annum simple interest and immediately lends it at 10% per annum simple interest. She gains ₹1,800 in 3 years. How much did she borrow?

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AnswerA · ₹20,000

  1. Her gain comes from the difference in rates: 10%−7%=3%10\% - 7\% = 3\% per year.

  2. Over 3 years she gains 3×3%=9%3 \times 3\% = 9\% of the sum.

  3. 9% of the sum = ₹1,800, so the sum = 1800×100÷91800 \times 100 \div 9 = ₹20,000.

Worth noting

Forgetting the 3 years gives ₹60,000, and using 2% as the gap gives ₹30,000. The gain each year is only the 3% gap, and it builds for 3 years.

Question 20

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What principal will earn ₹1,500 as simple interest in 1 year 3 months at 8% per annum?

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AnswerB · ₹15,000

  1. 1 year 3 months = 1312=541\frac{3}{12} = \frac{5}{4} years.

  2. P=1500×1008×54=15000010=15000P = \frac{1500 \times 100}{8 \times \frac{5}{4}} = \frac{150000}{10} = 15000.

  3. The principal is ₹15,000.

Worth noting

Ignoring the 3 months gives ₹18,750. Reading 1 year 3 months as 1.5 years gives ₹12,500; 3 months is a quarter of a year.

Question 21

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Divide ₹11,000 into two parts so that the simple interest on the first part at 6% per annum for 5 years equals the simple interest on the second part at 5% per annum for 4 years. What is the first part?

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AnswerA · ₹4,400

  1. Interest on first part = a×6×5÷100=30a÷100a \times 6 \times 5 \div 100 = 30a \div 100; on second part = b×5×4÷100=20b÷100b \times 5 \times 4 \div 100 = 20b \div 100.

  2. 30a=20b30a = 20b, so a:b=2:3a : b = 2 : 3.

  3. First part = 25×11000\frac{2}{5} \times 11000 = ₹4,400 (second part ₹6,600).

Worth noting

Comparing only the rates (6a=5b)(6a = 5b) gives ₹5,000; the time must be multiplied in too. ₹6,600 is the second part.

Question 22

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The simple interest on a loan is 1625\frac{16}{25} of the principal, and the number of years equals the rate per cent per annum. What is the rate?

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AnswerB · 8%

  1. Let the rate be R% and the time R years. Then SI=P×R×R100\text{SI} = \frac{P \times R \times R}{100}.

  2. R2100=1625\frac{R^2}{100} = \frac{16}{25}, so R2=64R^2 = 64.

  3. R=8%R = 8\%, and the time is 8 years.

Worth noting

Stopping at R2=64R^2 = 64 gives the tempting 64%. Take the square root.

Question 23

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The simple interest on a sum for 6 years is 38\frac{3}{8} of the amount at the end of those 6 years. What is the rate of interest per annum?

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AnswerC · 10%

  1. Amount = P+IP + I and I=38I = \frac{3}{8} of the amount, so P=58P = \frac{5}{8} of the amount.

  2. Hence IP=35\frac{I}{P} = \frac{3}{5}, which is 60% of P.

  3. Over 6 years this is 60÷6=10%60 \div 6 = 10\% per annum.

Worth noting

The fraction is of the AMOUNT, not the principal. Treating 38\frac{3}{8} as a fraction of P gives 37.5÷6=6.25%37.5 \div 6 = 6.25\%.

Question 24

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Zoya invests ₹10,000 at 8% per annum simple interest for 3 years. She then invests the whole amount she receives at 10% per annum simple interest for 2 more years. How much does she have at the end?

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AnswerD · ₹14,880

  1. After 3 years: interest = 24% of ₹10,000 = ₹2,400, so she has ₹12,400.

  2. Next 2 years: interest = 20% of ₹12,400 = ₹2,480.

  3. Final amount = ₹12,400 + ₹2,480 = ₹14,880.

Worth noting

₹14,400 applies the 10% to the original ₹10,000 instead of the ₹12,400 she reinvested. ₹4,880 is the total interest, not what she has.

Compound interest

24 questions

Question 25

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Arjun deposits ₹10,000 at 10% per annum compound interest, compounded annually. What will the amount be after 2 years?

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AnswerD · ₹12,100

  1. After year 1: ₹10,000 + 10% = ₹11,000.

  2. After year 2: ₹11,000 + 10% of ₹11,000 = ₹11,000 + ₹1,100 = ₹12,100.

Worth noting

₹12,000 is the simple-interest amount; compound interest charges the second year's 10% on ₹11,000, not ₹10,000. ₹2,100 is the interest, not the amount.

Question 26

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What is the compound interest on ₹5,000 at 8% per annum for 2 years, compounded annually?

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AnswerC · ₹832

  1. Amount = 5000×(1.08)2=5000×1.1664=58325000 \times (1.08)^2 = 5000 \times 1.1664 = 5832.

  2. CI = ₹5,832 − ₹5,000 = ₹832.

Worth noting

₹800 is the simple interest. ₹5,832 is the amount; the question asks only for the interest.

Question 27

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Lalremsiami invests ₹8,000 at 5% per annum, compounded annually. What amount will she receive after 3 years?

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AnswerD · ₹9,261

  1. Amount = 8000×(2120)3=8000×926180008000 \times \left(\frac{21}{20}\right)^3 = 8000 \times \frac{9261}{8000}.

  2. Amount = ₹9,261.

Worth noting

₹9,200 is simple interest for 3 years; ₹8,820 stops after 2 years; ₹1,261 is the interest, not the amount. Writing 1.05 as 21/2021/20 makes the cube easy because 203=8,00020^3 = 8{,}000.

Question 28

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The compound interest (compounded annually) on a deposit is ₹1,320 for the second year and ₹1,452 for the third year. What is the rate of interest per annum?

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AnswerB · 10%

  1. The third year's interest is the second year's interest plus interest on it: ₹1,452 − ₹1,320 = ₹132.

  2. Rate = 1321320×100=10%\frac{132}{1320} \times 100 = 10\%.

Worth noting

The ₹132 is interest on the earlier year's ₹1,320, so divide by 1,320. Dividing by 1,452 gives about 9.1%.

Question 29

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At 10% per annum compounded annually, the interest on a deposit for the first year is ₹500. What is the interest for the third year?

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AnswerD · ₹605

  1. Each year's interest is 10% more than the previous year's, because it is earned on a bigger balance.

  2. Second year: ₹500 × 1.1 = ₹550.

  3. Third year: ₹550 × 1.1 = ₹605.

Worth noting

₹600 adds a fixed ₹50 every year, which is not how compounding works: the increase itself grows. ₹500 would be right only for simple interest, and ₹550 is the second year's interest.

Question 30

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A deposit at 10% per annum, compounded annually, amounts to ₹24,200 after 2 years. What will it amount to after 3 years?

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AnswerC · ₹26,620

  1. Each year the balance is multiplied by 1.1.

  2. Amount after 3 years = ₹24,200 × 1.1 = ₹26,620.

Worth noting

Adding ₹2,000 (the first year's interest) or ₹2,200 (the second year's) gives the wrong totals. The third year's interest is 10% of the current ₹24,200.

Question 31

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What is the compound interest on ₹25,000 at 20% per annum for 2 years, compounded annually?

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AnswerC · ₹11,000

  1. Amount = 25000×(1.2)2=25000×1.44=3600025000 \times (1.2)^2 = 25000 \times 1.44 = 36000.

  2. CI = ₹36,000 − ₹25,000 = ₹11,000.

Worth noting

₹10,000 is the simple interest. The extra ₹1,000 is the second year's 20% on the first year's ₹5,000 interest.

Question 32

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Gurpreet deposits ₹18,000 at 5% per annum, compounded annually. What will the amount be after 2 years?

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AnswerD · ₹19,845

  1. After year 1: ₹18,000 + ₹900 = ₹18,900.

  2. After year 2: ₹18,900 + 5% of ₹18,900 = ₹18,900 + ₹945 = ₹19,845.

Worth noting

₹19,800 is the simple-interest amount. ₹1,845 is the interest, not the amount.

Question 33

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Kiran invests ₹16,000 at 10% per annum, compounded half-yearly. What is the amount after 1 year?

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AnswerC · ₹17,640

  1. Half-yearly: rate per half-year = 10%÷2=5%10\% \div 2 = 5\%, and 1 year = 2 half-years.

  2. Amount = 16000×(1.05)2=16000×1.1025=1764016000 \times (1.05)^2 = 16000 \times 1.1025 = 17640.

  3. The amount is ₹17,640.

Worth noting

₹17,600 compounds only once (annually). ₹19,360 wrongly uses 10% for each half-year; the yearly rate must be halved.

Question 34

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Find the compound interest on ₹20,000 for 1121\frac{1}{2} years at 20% per annum, compounded half-yearly.

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AnswerC · ₹6,620

  1. Rate per half-year = 10%; number of half-years in 1121\frac{1}{2} years = 3.

  2. Amount = 20000×(1.1)3=20000×1.331=2662020000 \times (1.1)^3 = 20000 \times 1.331 = 26620.

  3. CI = ₹26,620 − ₹20,000 = ₹6,620.

Worth noting

₹6,000 is simple interest; ₹6,400 compounds yearly (20% for a year, then 10% for the half-year); ₹14,560 uses 20% for each half-year.

Question 35

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At what rate per annum, compounded annually, will ₹6,250 amount to ₹7,290 in 2 years?

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AnswerB · 8%

  1. (1+R100)2=72906250=729625(1 + \frac{R}{100})^2 = \frac{7290}{6250} = \frac{729}{625}.

  2. Taking square roots: 1+R100=2725=1.081 + \frac{R}{100} = \frac{27}{25} = 1.08.

  3. R=8%R = 8\% per annum.

Worth noting

8.32% treats the growth as simple interest (16.64%÷2)(16.64\% \div 2). With compounding, take the square root of the growth factor instead of halving.

Question 36

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A sum invested at 10% per annum, compounded annually, amounts to ₹14,520 after 2 years. What was the sum?

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AnswerB · ₹12,000

  1. Amount = P×(1.1)2=1.21PP \times (1.1)^2 = 1.21P.

  2. P=145201.21=12000P = \frac{14520}{1.21} = 12000, so the sum was ₹12,000.

Worth noting

₹12,100 divides by 1.2 (simple interest for 2 years). ₹13,200 goes back only one year. ₹11,616 subtracts 20% of ₹14,520, but percentages must be undone by dividing.

Question 37

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A bank pays compound interest at 10% for the first year and 15% for the second year. What does a deposit of ₹20,000 amount to after 2 years?

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AnswerC · ₹25,300

  1. After year 1: ₹20,000 × 1.10 = ₹22,000.

  2. After year 2: ₹22,000 × 1.15 = ₹25,300.

Worth noting

₹25,000 adds 10%+15%10\% + 15\% as simple interest. ₹25,312.50 uses the average rate 12.5% twice; with changing rates, multiply the year factors one by one.

Question 38

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Deepa's deposit earns compound interest at 5% in the first year, 10% in the second and 20% in the third. What is the compound interest on ₹10,000 over the 3 years?

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AnswerC · ₹3,860

  1. Amount = 10000×1.05×1.10×1.2010000 \times 1.05 \times 1.10 \times 1.20.

  2. 10,000×1.05=10,50010{,}000 \times 1.05 = 10{,}500; × 1.10=11,5501.10 = 11{,}550; × 1.20=13,8601.20 = 13{,}860.

  3. CI = ₹13,860 − ₹10,000 = ₹3,860.

Worth noting

₹3,500 adds the rates as simple interest; ₹3,310 wrongly averages the rates to 10% each year; ₹13,860 is the amount.

Question 39

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A sum doubles itself in 5 years at compound interest, compounded annually. In how many years will it become 8 times itself?

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AnswerA · 15 years

  1. At compound interest, the sum doubles every 5 years.

  2. 8=2×2×28 = 2 \times 2 \times 2, so it must double three times.

  3. Time = 3×5=153 \times 5 = 15 years.

Worth noting

35 years is the simple-interest answer (interest 7P at P per 5 years). 40 years just multiplies 5 by 8.

Question 40

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Meenal invests ₹50,000 in a scheme paying 12% per annum, compounded annually. How much interest does she earn in the second year alone?

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AnswerB · ₹6,720

  1. After year 1 the balance is ₹50,000 + ₹6,000 = ₹56,000.

  2. Second-year interest = 12% of ₹56,000 = ₹6,720.

Worth noting

₹6,000 is the first year's interest; ₹12,720 is the total for both years; ₹7,526.40 is the third year's interest.

Question 41

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Find the compound interest on ₹16,000 for 9 months at 20% per annum, compounded quarterly.

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AnswerB · ₹2,522

  1. Rate per quarter = 20%÷4=5%20\% \div 4 = 5\%; 9 months = 3 quarters.

  2. Amount = 16000×(1.05)3=16000×1.157625=1852216000 \times (1.05)^3 = 16000 \times 1.157625 = 18522.

  3. CI = ₹18,522 − ₹16,000 = ₹2,522.

Worth noting

₹2,400 is simple interest for 9 months. ₹11,648 uses 20% for each quarter. ₹18,522 is the amount.

Question 42

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A deposit earns ₹1,025 as compound interest in 2 years at 5% per annum, compounded annually. How much was deposited?

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AnswerA · ₹10,000

  1. CI = P[(1.05)2−1]=P(1.1025−1)=0.1025PP[(1.05)^2 - 1] = P(1.1025 - 1) = 0.1025P.

  2. P=10250.1025=10000P = \frac{1025}{0.1025} = 10000, so ₹10,000 was deposited.

Worth noting

₹10,250 divides by 0.10 (simple interest for 2 years). ₹11,025 is the amount at the end, not the deposit.

Question 43

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Hari has ₹10,000 to invest for 2 years. Scheme P pays 12% per annum simple interest; Scheme Q pays 11% per annum compounded annually. Which scheme earns more interest, and by how much?

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AnswerA · Scheme P, by ₹79

  1. Scheme P: SI = 12%×2=24%12\% \times 2 = 24\% of ₹10,000 = ₹2,400.

  2. Scheme Q: amount = 10000×(1.11)2=10000×1.2321=1232110000 \times (1.11)^2 = 10000 \times 1.2321 = 12321, so CI = ₹2,321.

  3. Scheme P earns ₹2,400 − ₹2,321 = ₹79 more.

Worth noting

₹121 is only the compounding bonus at 11% (10,000×0.112)(10{,}000 \times 0.11^2). It does not make up for the 1% lower rate over 2 years, which costs ₹200.

Question 44

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A sum at compound interest, compounded annually, amounts to ₹13,230 after 2 years and to ₹13,891.50 after 3 years. What is the sum?

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AnswerB · ₹12,000

  1. The third year's interest = ₹13,891.50 − ₹13,230 = ₹661.50, earned on ₹13,230.

  2. Rate = 661.513230×100=5%\frac{661.5}{13230} \times 100 = 5\%.

  3. Sum = 13230(1.05)2=132301.1025=12000\frac{13230}{(1.05)^2} = \frac{13230}{1.1025} = 12000, that is, ₹12,000.

Worth noting

Subtracting ₹661.50 twice (₹11,907) treats the interest as simple. Each earlier year's interest is smaller, so divide by 1.05 twice instead.

Question 45

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At what rate per annum, compounded annually, will ₹15,625 amount to ₹17,576 in 3 years?

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AnswerA · 4%

  1. (1+R100)3=1757615625(1 + \frac{R}{100})^3 = \frac{17576}{15625}.

  2. Notice 15625=25315625 = 25^3 and 17576=26317576 = 26^3, so 1+R100=2625=1.041 + \frac{R}{100} = \frac{26}{25} = 1.04.

  3. R=4%R = 4\% per annum.

Worth noting

Treating the growth of ₹1,951 as simple interest over 3 years gives about 4.16%. With compounding the cube root is needed, and spotting the cubes 25³ and 26³ makes it quick.

Question 46

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The compound interest on a sum for 3 years at 10% per annum, compounded annually, is ₹4,965. What is the sum?

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AnswerA · ₹15,000

  1. CI = P[(1.1)3−1]=P(1.331−1)=0.331PP[(1.1)^3 - 1] = P(1.331 - 1) = 0.331P.

  2. 0.331P=49650.331P = 4965, so P=49650.331=15000P = \frac{4965}{0.331} = 15000.

  3. The sum is ₹15,000.

Worth noting

₹16,550 divides by 0.3 (simple interest for 3 years). ₹19,965 is the amount at the end, not the sum lent.

Question 47

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Shabnam borrows ₹30,000 at 10% per annum, compounded annually. She repays ₹13,000 at the end of the first year. How much must she pay at the end of the second year to clear the loan?

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AnswerB · ₹22,000

  1. End of year 1: ₹30,000 × 1.1 = ₹33,000 is owed.

  2. After paying ₹13,000, ₹20,000 remains.

  3. End of year 2: ₹20,000 × 1.1 = ₹22,000.

Worth noting

₹23,300 charges two years' interest on the full ₹30,000, ignoring that the ₹13,000 payment stopped earning interest. ₹20,000 forgets the second year's interest.

Question 48

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A sum earns compound interest at 10% in the first year and 20% in the second year. The total compound interest for the 2 years is ₹4,800. What is the sum?

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AnswerA · ₹15,000

  1. Amount = P×1.1×1.2=1.32PP \times 1.1 \times 1.2 = 1.32P, so CI = 0.32P0.32P.

  2. 0.32P=48000.32P = 4800, so P=48000.32=15000P = \frac{4800}{0.32} = 15000.

  3. The sum is ₹15,000.

Worth noting

₹16,000 adds the rates to 30% as if they were simple interest. The second year's 20% is also earned on the first year's interest, which makes 32% in all.

Difference between SI and CI

12 questions

Question 49

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Find the difference between the compound interest (compounded annually) and the simple interest on ₹10,000 at 5% per annum for 2 years.

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AnswerA · ₹25

  1. For 2 years, CI − SI = P(R100)2P\left(\frac{R}{100}\right)^2.

  2. = 10000×(5100)2=10000×1400=2510000 \times \left(\frac{5}{100}\right)^2 = 10000 \times \frac{1}{400} = 25, so the difference is ₹25.

Worth noting

The difference is the interest on the first year's interest (5% of ₹500). ₹1,025 is the whole CI, not the difference.

Question 50

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Anjali puts ₹30,000 in a deposit paying 10% per annum compound interest (compounded annually), and Rahul puts ₹30,000 in one paying 10% per annum simple interest. After 2 years, how much more interest has Anjali earned?

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AnswerA · ₹300

  1. Extra interest = P(R100)2=30000×1100P\left(\frac{R}{100}\right)^2 = 30000 \times \frac{1}{100}.

  2. Anjali earns ₹300 more.

Worth noting

The extra ₹300 is 10% of the first year's ₹3,000 interest, earned in the second year. ₹6,300 is Anjali's total interest, not the difference.

Question 51

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The simple interest on a sum for 2 years at 8% per annum is ₹1,600. What is the difference between the compound interest (compounded annually) and the simple interest on it for the same time and rate?

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AnswerB · ₹64

  1. One year's SI = ₹1,600 ÷ 2 = ₹800.

  2. The difference is the second year's interest on that ₹800: 8% of ₹800 = ₹64.

Worth noting

₹128 takes 8% of the whole ₹1,600. Only the first year's ₹800 of interest earns interest in the second year.

Question 52

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Imran deposits ₹7,500 at 8% per annum for 1 year. What is the difference between the compound interest (compounded annually) and the simple interest?

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AnswerA · ₹0

  1. With annual compounding, the first year's interest is 8% of ₹7,500 = ₹600 under both methods.

  2. Interest on interest starts only from the second year, so the difference is ₹0.

Worth noting

₹48 comes from applying P(R/100)2P(R/100)^2, which is the formula for 2 years, not 1.

Question 53

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What is the difference between the compound interest (compounded annually) and the simple interest on ₹8,000 at 10% per annum for 3 years?

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AnswerD · ₹248

  1. For 3 years, CI − SI = P(R100)2(3+R100)P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right).

  2. = 8000×1100×3.1=80×3.1=2488000 \times \frac{1}{100} \times 3.1 = 80 \times 3.1 = 248.

  3. Check: CI = 8,000×1.331−8,0008{,}000 \times 1.331 - 8{,}000 = ₹2,648; SI = ₹2,400; difference = ₹248.

Worth noting

₹80 is the 2-year difference and ₹240 simply triples it. The 3-year formula has the extra R/100R/100 because interest is also earned on interest-on-interest.

Question 54

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The compound interest (compounded annually) on a sum for 2 years at 12% per annum exceeds the simple interest by ₹72. What is the sum?

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AnswerC · ₹5,000

  1. P(12100)2=72P\left(\frac{12}{100}\right)^2 = 72, so P×0.0144=72P \times 0.0144 = 72.

  2. P=720.0144=5000P = \frac{72}{0.0144} = 5000, so the sum is ₹5,000.

Worth noting

The rate must be squared. Dividing by 0.12 only once gives ₹600; squaring 12 but forgetting to square the 100 gives ₹60,000.

Question 55

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The difference between the compound interest (compounded annually) and the simple interest on ₹40,000 for 2 years is ₹100. What is the rate of interest per annum?

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AnswerC · 5%

  1. 40000×(R100)2=10040000 \times \left(\frac{R}{100}\right)^2 = 100, so (R100)2=1400\left(\frac{R}{100}\right)^2 = \frac{1}{400}.

  2. R100=120\frac{R}{100} = \frac{1}{20}, so R=5%R = 5\%.

Worth noting

R2=25R^2 = 25 gives R=5R = 5, not 25; do not forget the square root. 0.25% is just 100 as a percentage of 40,000.

Question 56

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The compound interest on a sum for 2 years at 10% per annum, compounded annually, is ₹2,520. What is the simple interest on the same sum for the same time and rate?

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AnswerD · ₹2,400

  1. CI for 2 years at 10%=P[(1.1)2−1]=0.21P10\% = P[(1.1)^2 - 1] = 0.21P, so P=25200.21=12000P = \frac{2520}{0.21} = 12000.

  2. SI = 12,000×10×2÷10012{,}000 \times 10 \times 2 \div 100 = ₹2,400.

Worth noting

₹1,200 is only one year's SI. ₹2,268 takes 10% off the CI, but the gap is ₹120 (10% of the first year's ₹1,200), not ₹252.

Question 57

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Find the difference between the compound interest (compounded half-yearly) and the simple interest on ₹20,000 at 10% per annum for 1 year.

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AnswerA · ₹50

  1. Half-yearly: 5% per half-year for 2 half-years. CI = 20000×(1.05)2−20000=205020000 \times (1.05)^2 - 20000 = 2050.

  2. SI for 1 year = 10% of ₹20,000 = ₹2,000.

  3. Difference = ₹2,050 − ₹2,000 = ₹50.

Worth noting

₹200 uses P(R/100)2P(R/100)^2 with R=10R = 10. With half-yearly compounding the periods are half-years, so the shortcut uses R=5R = 5: 20,000×0.05220{,}000 \times 0.05^2 = ₹50.

Question 58

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The difference between the compound interest (compounded annually) and the simple interest on a sum for 3 years at 20% per annum is ₹1,536. What is the sum?

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AnswerA · ₹12,000

  1. CI − SI = P(20100)2(3+20100)=P×0.04×3.2=0.128PP\left(\frac{20}{100}\right)^2\left(3 + \frac{20}{100}\right) = P \times 0.04 \times 3.2 = 0.128P.

  2. 0.128P=15360.128P = 1536, so P=15360.128=12000P = \frac{1536}{0.128} = 12000.

  3. The sum is ₹12,000.

Worth noting

₹38,400 uses the 2-year formula. ₹12,800 uses 3 in place of (3+0.2)(3 + 0.2), forgetting the extra R/100R/100 term.

Question 59

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On a certain sum, the simple interest for 2 years is ₹1,800 and the compound interest (compounded annually) for 2 years at the same rate is ₹1,881. What is the sum?

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AnswerB · ₹10,000

  1. One year's SI = ₹900. The extra ₹81 is interest on that ₹900 for one year.

  2. Rate = 81900×100=9%\frac{81}{900} \times 100 = 9\%.

  3. Sum = 1800×1009×2=10000\frac{1800 \times 100}{9 \times 2} = 10000, that is, ₹10,000.

Worth noting

₹20,000 forgets that ₹1,800 is for 2 years. ₹10,450 uses the CI ₹1,881 in the simple-interest formula.

Question 60

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At 10% per annum, the difference between the compound interest (compounded annually) and the simple interest on a sum for 2 years is ₹50. What is the difference for 3 years?

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AnswerC · ₹155

  1. 3-year difference ÷ 2-year difference = P(R/100)2(3+R/100)P(R/100)2=3+R100=3.1\frac{P(R/100)^2(3 + R/100)}{P(R/100)^2} = 3 + \frac{R}{100} = 3.1.

  2. 3-year difference = ₹50 × 3.1 = ₹155.

  3. Check: the sum is 50÷0.0150 \div 0.01 = ₹5,000; CI = ₹1,655, SI = ₹1,500, difference = ₹155.

Worth noting

₹150 multiplies by 3 and misses the R/100R/100. ₹650 uses 3+R=133 + R = 13 instead of 3+R/100=3.13 + R/100 = 3.1.

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