◆ QA · Arithmetic

Simple & Compound Interest, formulas + CAT PYQs

SI, CI, the CI−SI gap, half-yearly & quarterly compounding, installments, population growth and depreciation, a perennial CAT favourite that rewards clean formula discipline.

18formulas

19CAT PYQs

★★★priority

Formula Sheet CAT PYQs

Formula & Concept Sheet

A-to-Z. Everything you need for this chapter, distilled. Notation: P = Principal, A = Amount, I = Interest, r% = rate per year, n = number of years.

1Simple Interest (SI)

Interest is paid only on the original principal every year, it never compounds.
Amount = Principal + Interest.

SI = P×r×n / 100 A = P + SI = P(1 + rn/100)

2Compound Interest (CI)

Each year's interest is added to the principal, so the base grows, interest on interest.
Amount after n years at r% p.a. compounded annually:

A = P(1 + r/100)ⁿ CI = A − P = P[(1 + r/100)ⁿ − 1]

3CI − SI difference

For the same P, r and n, with a = r/100:
2 years: (CI)₂ − (SI)₂ = Pa²
3 years: (CI)₃ − (SI)₃ = Pa²(a + 3)

a = r/100; (CI−SI)₂ = P(r/100)² (CI−SI)₃ = P(r/100)²·(r/100 + 3)

4Half-yearly compounding

Principal changes every six months (semi-annually).
Halve the rate, double the number of periods.

A = P(1 + (r/2)/100)²ⁿ

5Quarterly compounding

Principal changes every three months.
Quarter the rate, quadruple the periods.

A = P(1 + (r/4)/100)⁴ⁿ

6Fractional-year compounding

When time is in fractions, e.g. 2¾ years: use the whole-year power, then SI-style growth for the leftover part.

A = P(1 + r/100)² × (1 + (¾r)/100)

7Varying yearly rates

If rate is R₁ in year 1, R₂ in year 2, R₃ in year 3, multiply each year's factor.

A = P(1+R₁/100)(1+R₂/100)(1+R₃/100)

8Population growth

P = original population, r% = annual growth rate, P′ = population after n years.
Same engine as compound interest.

P′ = P(1 + r/100)ⁿ

9Depreciation / decline

When a value falls at r% per year (machines, populations declining), use a minus sign.

Value after n yrs = P(1 − r/100)ⁿ

10Effective annual rate

Compounding more often than yearly makes the real (effective) rate higher than the nominal rate.
For k compoundings per year:

Effective rate = [(1 + r/(100k))ᵏ − 1]×100%

11Multiplying time, SI

A principal that becomes X times in T years at simple interest will become Y times in:

Years = T × (Y − 1)/(X − 1)

12Multiplying time, CI

A principal that becomes X times in T years at compound interest becomes Y times in T×n years, where:

Years = T×n, where Xⁿ = Y

13Two CI multiples

If a sum at CI becomes x times in n₁ years and y times in n₂ years, then:

x^(1/n₁) = y^(1/n₂)

14Installments (equal repayments)

An installment paid earlier earns more "discount", so it grows by more compounding before the closure date.
For a loan P repaid in equal installments x at r% over the years: amount the loan grows to = sum of each installment grown to the closure year.

P(1+r/100)ⁿ = x(1+r/100)ⁿ⁻¹ + … + x

15Present value of an installment

The value of an installment x due after t years, discounted to today at CI:

PV = x / (1 + r/100)ᵗ

16Net / spread interest

Borrow at one rate, lend at a higher rate: the net interest retained is the difference.

Net interest = (lend rate − borrow rate)×Principal

17Successive CI per year (ratio of interests)

Under CI, each year's interest equals the previous year's interest multiplied by (1 + r/100).
Handy for finding r from two consecutive years' interests:

r% = (Iₖ₊₁ − Iₖ)/Iₖ × 100

18SI vs CI quick facts

Year 1: SI = CI (no compounding yet).
From year 2 onward, CI > SI and the gap widens.
Doubling at r% SI takes 100/r years; at CI use rule-of-72 estimate (≈72/r).

Practice questions generated · up to 100

Original easy-hard warm-up drills (not CAT PYQs). Pick the levels, generate a set, reveal answers.

CAT Previous-Year Questions

Real CAT questions with worked solutions from the book. Difficulty: Easy Moderate Hard. Click any to reveal the solution.

CAT 2018

Hard CAT 2018 TITA

John borrowed ₹2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal installments, the first after one year and the second after another year. The first installment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each installment, in ₹, is

Show solution

₹1,21,000. By the end of 2 years, 2,10,000 amounts to 2,10,000 × 1.1 × 1.1 = ₹2,54,100. Let each installment be x. John pays the first installment at the end of year 1; the unpaid balance attracts another 10%, so by year 2 it has grown to 1.1x. Thus 1.1x + x = 2,54,100 ⇒ 2.1x = 2,54,100 ⇒ x = ₹1,21,000.

Hard CAT 2018 TITA

Gopal borrows ₹X from Ankit at 8% annual interest. He then adds ₹Y of his own money and lends ₹(X + Y) to Ishan at 10% annual interest. At the end of the year, after returning Ankit's dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent ₹(X + 2Y) to Ishan at 10%, then the net interest retained by him would have increased by ₹150. If all interests are compounded annually, find the value of X + Y.

Show solution

₹4,000. Gopal receives 10% of (X + Y) from Ishan and pays 8% of X to Ankit, so Gopal gains 2% of X + 10% of Y, and Ankit gains 8% of X. These are equal: 2%X + 10%Y = 8%X ⇒ 6%X = 10%Y ⇒ X : Y = 5 : 3. Lending X+2Y instead of X+Y raises the retained interest by the extra 10% of Y = ₹150 ⇒ Y = ₹1500. Then X = (5/3)×1500 = 2500. So X + Y = 2500 + 1500 = ₹4,000.

CAT 2019

Moderate CAT 2019

Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in ₹) after a year, if Bina's interest income exceeds Amala's by ₹250?

(1) 6350
(2) 6000
(3) 7000
(4) 7250

Show solution

(4) 7250. Let investments be 300x, 400x, 500x. Interest incomes = 300x×6/100 = 18x, 400x×5/100 = 20x, 500x×4/100 = 20x. Bina exceeds Amala: 20x − 18x = 2x = 250 ⇒ x = 125. Total = 18x + 20x + 20x = 58x = 58×125 = ₹7250.

Moderate CAT 2019 TITA

A person invested a total amount of ₹15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3% respectively. If the total annual interest income is ₹76000, then the amount (in ₹ lakh) invested in the fixed deposit was

Show solution

9 lakh. Let the two ratio-deposits be 200x and 100x. Their interest = 200x×4/100 + 100x×3/100 = 8x + 3x = 11x. Fixed-deposit amount = 15,00,000 − 300x, its interest = (15,00,000 − 300x)×6/100 = 90,000 − 18x. Total: 90,000 − 18x + 11x = 90,000 − 7x = 76,000 ⇒ 7x = 14,000 ⇒ x = 2000. Fixed deposit = 15,00,000 − 300×2000 = 9,00,000 = 9 lakh.

Moderate CAT 2019 TITA

Amal invests ₹12000 at 8% interest, compounded annually, and ₹10000 at 6% interest, compounded semi-annually, both investments being for one year. Bimal invests his money at 7.5% simple interest for one year. If Amal and Bimal get the same amount of interest, then the amount, in rupees, invested by Bimal is

Show solution

₹20920. Amal's amount after 1 year = 12000×1.08 + 10000×(1.03)² = 12960 + 10609 = 23569. Amal's interest = 23569 − 22000 = 1569. Let Bimal invest 100b: his interest = 100b×7.5×1/100 = 7.5b. Set 7.5b = 1569 ⇒ b = 209.2 ⇒ Bimal invested 100b = ₹20920.

CAT 2020

Moderate CAT 2020 TITA

Veeru invested ₹10000 at 5% simple annual interest, and exactly after two years, Joy invested ₹8000 at 10% simple annual interest. How many years after Veeru's investment, will their balances, i.e., principal plus accumulated interest, be equal?

Show solution

12 years. When Joy invests (2 years in), Veeru's balance = 10000 + 10000×5×2/100 = 11000, Joy's = 8000. Gap = 3000. Each year Veeru earns 10000×5/100 = 500, Joy earns 8000×10/100 = 800, so Joy closes 800 − 500 = 300 of the gap per year. Time to close 3000 = 3000/300 = 10 years after Joy starts = 12 years after Veeru's investment.

Moderate CAT 2020 TITA

For the same principal amount, the compound interest for two years at 5% per annum exceeds the simple interest for three years at 3% per annum by ₹1125. Then the principal amount in rupees is:

Show solution

₹90000. Let principal = x. CI for 2 years at 5% = x(1.05)² − x = x(1.1025 − 1) = 0.1025x. SI for 3 years at 3% = x×3×3/100 = 0.09x. Difference: 0.1025x − 0.09x = 0.0125x = 1125 ⇒ x = 1125/0.0125 = ₹90000.

Moderate CAT 2020 TITA

A person invested a certain amount of money at 10% annual interest, compounded half-yearly. After one and a half years, the interest and principal together became ₹18522. The amount, in rupees, that the person had invested is

Show solution

₹16000. Half-yearly rate = 10/2 = 5%; 1½ years = 3 half-year periods. Amount = P(105/100)³, so 18522 = P×(105/100)³ ⇒ P = 18522×100³/105³ = 18522/1.157625 = ₹16000.

CAT 2021

Hard CAT 2021 · Slot 1

Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹806.25 and ₹866.72 respectively, the interest accrued, in INR, during the fourth year is nearest to

(1) 931.72
(2) 926.84
(3) 929.48
(4) 934.65

Show solution

(1) 931.72. Under CI, each year's interest is the previous year's times (1 + r/100). So r% = (866.72 − 806.25)/806.25 × 100 = 60.47/806.25 × 100 ≈ 7.5%. Fourth-year interest = 866.72 × 1.075 ≈ ₹931.72.

Hard CAT 2021 · Slot 3

Bank A offers 6% interest rate per annum compounded half yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in ₹, to Rupa is

(1) 1436
(2) 2346
(3) 2436
(4) 3436

Show solution

(3) 2436. Let Bank B rate = R%, so Bank C = 2R%. Raju invests P in Bank B for n years; SI in B = C.I. in A for one year: P·R·n/100 = P[(1.03)² − 1] ⇒ R·n/100 = (1.03)² − 1 = 0.0609 ⇒ R×n = 6.09. Rupa invests 10000 in Bank C at 2R% for 2n years: interest = 10000×(2R)×(2n)/100 = 10000×4×R×n/100 = 100×4×6.09 = ₹2436.

CAT 2022

Easy CAT 2022 · Slot 1

Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is:

(1) 60%
(2) 62.5%
(3) 37.5%
(4) 40%

Show solution

(3) 37.5%. Let parts be x and y. x×15×4/100 = y×12×3/100 ⇒ 60x = 36y ⇒ x/y = 36/60 = 3/5. First part as % of savings = x/(x+y) = 3/(3+5) = 3/8 = 37.5%.

Moderate CAT 2022 · Slot 2 TITA

Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is:

Show solution

20 years. Let total capital = 15x (LCM-friendly). Parts: 3x at 6%, 5x at 10%, 7x at 1%. Annual interest = (3x×6 + 5x×10 + 7x×1)/100 = (18x + 50x + 7x)/100 = 75x/100. For n years: (75x/100)·n ≥ 15x ⇒ 0.75n ≥ 15 ⇒ n ≥ 20. Minimum = 20 years.

Moderate CAT 2022 · Slot 3

Nitu has an initial capital of ₹20,000. Out of this, she invests ₹8,000 at 5.5% in bank A, ₹5,000 at 5.6% in bank B and the remaining amount at x% in bank C; each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been:

(1) 900
(2) 800
(3) 1000
(4) 700

Show solution

(2) 800. Bank C amount = 20000 − (8000 + 5000) = 7000. Combined interest = 5% of 20000 = 1000. So 8000×5.5/100 + 5000×5.6/100 + 7000×x/100 = 1000 ⇒ 440 + 280 + 70x = 1000 ⇒ 70x = 280 ⇒ x = 4%. Entire 20000 at 4% = 20000×4×1/100 = ₹800.

CAT 2023

Hard CAT 2023 · Slot 1 TITA

Anil invests ₹22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is

Show solution

₹20808. Half-yearly rate = 2%. Anil's amount = 22000(1 + 2/100)¹² = 22000(51/50)¹². Sunil invests x for 5 years (10 half-years): x(51/50)¹⁰, then 1 year at 10% SI multiplies by (1 + 10/100) = 11/10. Equate: 22000(51/50)¹² = x(51/50)¹⁰×(11/10) ⇒ 22000(51/50)² = x×(11/10) ⇒ 22000×(51/50)²×(10/11) = x ⇒ 20000×(2601/2500) = x ⇒ x = ₹20808.

Hard CAT 2023 · Slot 2

Anil borrows ₹2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays ₹10,320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year. Then, the total interest, in rupees, paid over the three years is nearest to

(1) 40,991
(2) 45,311
(3) 33,130
(4) 51,311

Show solution

(4) 51,311. Half-yearly rate = 4%; each year = 2 periods, factor (1.04)² = 1.0816 per year. Let x be paid at the end of year 3. Value equation grown to year 3: 2,00,000×(1.04)⁶ = 10,320×(1.04)⁴ + x. Solving gives the closing payment x ≈ ₹2,40,991. Total interest = (10,320 + 2,40,991) − 2,00,000 = ₹51,311.

CAT 2024 & 2025, recent

Fresh questions distributed from the real CAT 2024 & CAT 2025 papers into this chapter.

ModerateCAT 2024 · Slot 1

An amount of Rs 10000 is deposited in bank A for a certain number of years at a simple interest of 5% per annum. On maturity, the total amount received is deposited in bank B for another 5 years at a simple interest of 6% per annum. If the interests received from bank A and bank B are in the ratio 10 : 13, then the investment period, in years, in bank A is

(1) 4
(2) 3
(3) 6
(4) 5

Show solution

(3) 6. Let the period in bank A be n years. Interest from A = 10000 × 5 × n / 100 = 500n. Amount on maturity = 10000 + 500n. Interest from B = (10000 + 500n) × 6 × 5 / 100 = 3000 + 150n. Ratio 500n : (3000 + 150n) = 10 : 13 ⇒ 6500n = 30000 + 1500n ⇒ 5000n = 30000 ⇒ n = 6.

ModerateCAT 2024 · Slot 2

Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is

(1) 20860
(2) 20640
(3) 20480
(4) 20808

Show solution

(4) 20808. Half-yearly rate = 2%. Anil's amount = 22000(1.02)¹². Sunil invests X for 5 years (10 periods): X(1.02)¹⁰, then 1 year at 10% SI → X(1.02)¹⁰(1.1). Equate: X(1.02)¹⁰(1.1) = 22000(1.02)¹² ⇒ X = 22000(1.02)²/1.1 = 22000 × 1.0404/1.1 = 20808.

HardCAT 2025 · Slot 1

At a certain simple rate of interest, a given sum amounts to Rs 13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to

(1) 3096
(2) 3221
(3) 3180
(4) 3150

Show solution

(2) 3221. SI for 3.5 yr = 18960 − 13920 = 5040 ⇒ SI/yr = 1440. Principal P = 13920 − 3×1440 = 9600; rate = 1440/9600 = 15% p.a. → 7.5% per half-year, 4 periods over 2 years. CI = 9600[(1.075)⁴ − 1] ≈ 9600 × 0.335469 ≈ 3221.

HardCAT 2025 · Slot 2

A loan of Rs 1000 is fully repaid by two installments of Rs 530 and Rs 594, paid at the end of first and second year, respectively. If the interest is compounded annually, then the rate of interest, in percentage, is

(1) 8
(2) 10
(3) 11
(4) 9

Show solution

(1) 8. Present value of installments = loan: 1000 = 530/(1+r) + 594/(1+r)². Let R = 1+r: 1000R² = 530R + 594 ⇒ 1000R² − 530R − 594 = 0 ⇒ R = 1.08 ⇒ r = 8%.