Quantitative Aptitude PYQs

Shortcut Trick

Use the successive percentage formula: Net Profit% = a + b + (a × b) ÷ 100.

Here, a = 10% (P to R) and b = 5% (R to Q).

Net Profit% = 10 + 5 + (10 × 5) ÷ 100 = 15 + 50 ÷ 100 = 15 + 0.5 = 15.5%.

∴ The correct answer is 15.5%.

Alternate Method

Given: Profit P to R = 10%, Profit R to Q = 5%.
Formulas: SP = CP × (100 + P%) ÷ 100 | Profit% = (Profit ÷ CP) × 100.

⇒ Let the Cost Price (CP) for P = 100.

⇒ Price at which P sells to R = 100 × (110 ÷ 100) = 110.

⇒ Price at which R sells to Q = 110 × (105 ÷ 100) = 115.5.

⇒ If P sells directly to Q at R's selling price, then P's SP = 115.5.

⇒ Profit for P = 115.5 − 100 = 15.5.

⇒ Profit Percentage = (15.5 ÷ 100) × 100 = 15.5%.

∴ The correct answer is 15.5%.

Additional Information

Successive Percentage Changes

When a value is increased by x% and then again by y%, the overall increase is given by (x + y + xy/100)%.

Effective Profit

In a chain of sales, the effective cost to the final buyer relative to the original seller's cost is the product of all multiplying factors (1 + P%).

Shortcut Trick

Difference in distance = 15 km − 10 km = 5 km

Difference in fare = ₹330 − ₹230 = ₹100

Per-km charge = Difference in fare ÷ Difference in distance = 100 ÷ 5 = ₹20

Base fare = Total fare for 10 km − (10 × Per-km charge) = 230 − (10 × 20) = 230 − 200 = ₹30

∴ The correct answer is Base fare = ₹30; Per-kilometre Charge = ₹20.

Alternate Method

Given:

Fare for 10 km Raju = ₹230

Fare for 15 km Meena = ₹330

Formula Used:

Total Fare = Base Fare + (Distance × Rate per km)

Let Base Fare = B and Charge per km = R.

For Raju (10 km):

⇒ B + 10R = 230 → (Equation 1)

For Meena (15 km):

⇒ B + 15R = 330 → (Equation 2)

Subtract Equation 1 from Equation 2:

⇒ (B + 15R) − (B + 10R) = 330 − 230

⇒ 5R = 100

⇒ R = 100 ÷ 5 = 20

Substitute R = 20 in Equation 1:

⇒ B + 10(20) = 230

⇒ B + 200 = 230

⇒ B = 230 − 200 = 30

∴ The correct answer is Base fare = ₹30; Per-kilometre Charge = ₹20.

Additional Information

Linear Equations in Two Variables

A system of equations of the form ax + by = c can be solved using Elimination or Substitution methods to find specific constants.

Fixed vs Variable Cost

The base fare represents the fixed cost (independent of usage), while the per-km charge represents the variable cost (dependent on distance).

Rate of Change

The per-kilometre charge is essentially the slope of the linear fare equation, calculated as Change in Fare ÷ Change in Distance.

Shortcut Trick

The Surface Area of a Sphere is calculated using the formula: 4 × π × r².

Given radius (r) = 22 cm.

Surface Area = 4 × π × (22)² = 4 × 484 × π = 1936π cm².

∴ The correct answer is 1936π cm².

Alternate Method

Given:

Radius of the sphere (r) = 22 cm

Formula Used:

Surface Area (A) = 4πr²

Calculations:

⇒ Surface Area = 4 × π × (22 cm)²

⇒ Surface Area = 4 × π × 484 cm²

⇒ Surface Area = 1936π cm²

∴ The correct answer is 1936π cm².

Additional Information

Volume of a Sphere

The total volume or capacity of a sphere is given by the formula: V = (4/3) × π × r³.

Hemisphere Properties

For a hemisphere of radius r, the Curved Surface Area (CSA) is 2πr² and the Total Surface Area (TSA) is 3πr².

Ratio of Areas

If the radii of two spheres are in the ratio a:b, then their surface areas are in the ratio a²:b².

Shortcut Trick

Total votes = 100%

Third candidate % = 100% − (40% + 30%) = 30%

Number of votes = 30% of 36,000 = 0.30 × 36,000

∴ The correct answer is 10,800 votes.

Alternate Method

Given:

Total votes polled = 36,000

Votes for First candidate = 40%

Votes for Second candidate = 30%

Calculations:

⇒ Total percentage of votes = 100%

⇒ Percentage for Third candidate = 100% − (40% + 30%)

⇒ Percentage for Third candidate = 30%

⇒ Number of votes = (30/100) × 36,000

⇒ Number of votes = 30 × 360

⇒ Number of votes = 10,800

∴ The correct answer is 10,800 votes.

Additional Information

Percentage Calculation

To find the percentage of a number, use the formula: (Percentage/100) × Total Value.

Vote Distribution

In an election with no invalid votes, the sum of percentages obtained by all candidates always equals 100%.

Base Value Concept

The total number of votes polled acts as the base (100%) for calculating individual shares.

Shortcut Trick

For mean proportional (M) of two numbers A and B, the formula is M² = A × B.

Given M = 18 and B = 27.

⇒ 18² = A × 27

⇒ 324 = A × 27

⇒ A = 324 ÷ 27 = 12.

∴ The correct answer is 12.

Alternate Method

Given:

First number = A

Second number = 27

Mean Proportional = 18

Formula Used:

If b is the mean proportional between a and c, then a : b :: b : c or b = √(a × c).

Calculations:

⇒ A / 18 = 18 / 27

⇒ A = (18 × 18) ÷ 27

⇒ A = 324 ÷ 27

⇒ A = 12

∴ The correct answer is 12.

Additional Information

Third Proportional

The third proportional of two numbers a and b is defined as c such that a : b = b : c, which gives c = b² ÷ a.

Fourth Proportional

The fourth proportional of a, b, and c is d such that a : b = c : d, which gives d = (b × c) ÷ a.

Continued Proportion

Numbers a, b, c, d... are in continued proportion if a/b = b/c = c/d...

Shortcut Trick

Relative Speed (Opposite Direction) = 18 + 20 = 38 m/s.

Total Distance (Sum of Lengths) = 457 + 113 = 570 m.

Crossing Time = Total Distance ÷ Relative Speed = 570 ÷ 38.

∴ The correct answer is 15 sec.

Alternate Method

Given:

Length of Train A (L₁) = 457 m

Length of Train B (L₂) = 113 m

Speed of Train A (S₁) = 18 m/s

Speed of Train B (S₂) = 20 m/s

Formula Used:

Time = (L₁ + L₂) ÷ (S₁ + S₂) [When moving in opposite directions]

Calculations:

⇒ Total distance to be covered = 457 + 113 = 570 m

⇒ Relative speed = 18 + 20 = 38 m/s

⇒ Time taken = 570 ÷ 38

⇒ Time taken = 15 sec

∴ The correct answer is 15 sec.

Additional Information

Relative Speed in Same Direction

When two objects move in the same direction, their relative speed is the difference of their speeds: Relative Speed = |S₁ − S₂|.

Conversion of Speed Units

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Train Crossing a Point Object

When a train crosses a pole or a stationary man, the distance covered is equal to the length of the train itself.

Shortcut Trick

The smallest number divisible by a set of numbers is their LCM.

Number + 20 = LCM(12, 20, 15, 18, 24)

LCM of (12, 20, 15, 18, 24) = 360

Required number = 360 − 20 = 340

∴ The correct answer is 340.

Alternate Method

Given: Divisors are 12, 20, 15, 18, and 24. The number is increased by 20.

Formula Used: Required Number = LCM(a, b, c, d, e) − Increased Value

⇒ Prime factorization of divisors:

⇒ 12 = 2² × 3

⇒ 20 = 2² × 5

⇒ 15 = 3 × 5

⇒ 18 = 2 × 3²

⇒ 24 = 2³ × 3

⇒ LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

⇒ Let the required number be x.

⇒ x + 20 = 360

⇒ x = 360 − 20

⇒ x = 340

∴ The correct answer is 340.

Additional Information

LCM (Least Common Multiple)

The smallest positive integer that is divisible by each of the given numbers. Found by taking the highest power of all prime factors involved.

LCM Adjustment Rules

If a number is increased by K, the required number is LCM − K. If a number is decreased by K, the required number is LCM + K.

HCF vs LCM

Use HCF when finding the largest number that divides given numbers; use LCM when finding the smallest number that is divided by given numbers.

Shortcut Trick

Divide the numerical coefficients and subtract the exponents of matching variables directly.

Numbers: 10 ÷ 2 = 5

Exponents of a: 5 − 2 = 3 ⇒ a³

Exponents of b: 6 − 3 = 3 ⇒ b³

Combining all parts gives 5 a³ b³.

∴ The correct answer is 5 a³ b³.

Alternate Method

Given:

Expression: (10 × a⁵ × b⁶) ÷ (2 × a² × b³)

Formula Used:

Law of Exponents for division: x^m ÷ xⁿ = x^m−n

Calculation:

⇒ Separate the constants and variables:

⇒ (10 ÷ 2) × (a⁵ ÷ a²) × (b⁶ ÷ b³)

⇒ Apply the division law of exponents by subtracting the denominator powers from the numerator powers:

⇒ 5 × a⁵⁻² × b⁶⁻³

⇒ 5 × a³ × b³

⇒ 5 a³ b³

∴ The correct answer is 5 a³ b³.

Additional Information

Laws of Exponents (Multiplication)

When multiplying terms with the same base, add their exponents: a^m × aⁿ = a^m+n

Power of a Power Rule

When raising a power to another power, multiply the exponents: (a^m)ⁿ = a^m×n

Negative Exponents

A negative exponent indicates the reciprocal of the base raised to a positive power: a⁻ⁿ = 1 / aⁿ

Zero Exponent Rule

Any non-zero base raised to the power of zero is always equal to one: a⁰ = 1

Shortcut Trick

Convert capacity: 25.872 litres = 25872 cm³.

Given height (h) = 3r, so Volume (V) = 3πr³.

r³ = 25872 ÷ (3 × 22/7) = 25872 × 7 ÷ 66 = 2744 ⇒ r = 14 cm.

Shortcut for Curved Surface Area (CSA): CSA = 2V ÷ r.

CSA = (2 × 25872) ÷ 14 = 25872 ÷ 7 = 3696 cm².

∴ The correct answer is 3696 cm².

Alternate Method

Given:

Capacity of the cylindrical vessel (V) = 25.872 litres = 25872 cm³

Height (h) = 3 × Radius (r)

Value of π = 22/7

Formula Used:

Volume of a cylinder = πr²h

Curved Surface Area (CSA) = 2πrh

Calculations:

⇒ V = πr²h

⇒ 25872 = (22/7) × r² × (3r)

⇒ 25872 = (66/7) × r³

⇒ r³ = (25872 × 7) ÷ 66

⇒ r³ = 392 × 7 = 2744

⇒ r = √³(2744) = 14 cm

⇒ h = 3 × 14 = 42 cm

⇒ CSA = 2 × (22/7) × 14 × 42

⇒ CSA = 2 × 22 × 2 × 42

⇒ CSA = 44 × 84 = 3696 cm²

∴ The correct answer is 3696 cm².

Additional Information

Volume of Cylinder

The total space enclosed within a cylinder is given by V = πr²h.

Total Surface Area (TSA)

TSA of a cylinder includes the curved surface and two circular bases: TSA = 2πr(h + r).

Unit Conversion Rule

1 Litre is equal to 1000 cm³; always convert to consistent units before calculation.

Shortcut Trick

Multiply Original Ratio by (100 + Increment)% for each person.

New ratio → 18 × 145 : 5 × 186 : 10 × 189

Simplifying → 2610 : 930 : 1890

Divide by 30 → 87 : 31 : 63

∴ The correct answer is 87:31:63.

Alternate Method

Given: Original ratio of salaries X : Y : Z = 18 : 5 : 10

Formula: New Value = Old Value × (100 + Increase %)/100

⇒ New Salary of X = 18 × (145/100) = 26.1

⇒ New Salary of Y = 5 × (186/100) = 9.3

⇒ New Salary of Z = 10 × (189/100) = 18.9

⇒ New Ratio = 26.1 : 9.3 : 18.9

⇒ New Ratio = 261 : 93 : 189 (Multiplying by 10)

⇒ New Ratio = 87 : 31 : 63 (Dividing by 3)

∴ The correct answer is 87:31:63.

Additional Information

Successive Percentage Change

If a value is changed by a% and then by b%, the net change is [a + b + (a × b)/100]%.

Ratio Scaling

Multiplying or dividing all terms of a ratio by the same non-zero number does not change the ratio.

Compound Ratio

The compound ratio of (a : b) and (c : d) is (ac : bd).

Shortcut Trick

Discount = Marked Price (MP) − Selling Price (SP) = 8,730 − 5,940 = 2,790.

Discount% = (Discount ÷ MP) × 100.

Discount% = (2,790 ÷ 8,730) × 100 ≈ 31.958%.

Rounding to two decimal places, we get 31.96%.

∴ The correct answer is 31.96%.

Alternate Method

Given → Marked Price (MP) = ₹8,730 and Selling Price (SP) = ₹5,940.
Formulas: Discount = MP − SP | Discount% = (Discount ÷ MP) × 100.

⇒ Calculate Discount: 8,730 − 5,940 = 2,790.
⇒ Calculate Discount Percentage: (2,790 ÷ 8,730) × 100.
⇒ 279,000 ÷ 8,730 ≈ 31.9587%.
⇒ Final rounded value = 31.96%.

∴ The correct answer is 31.96%.

Additional Information

Marked Price & Discount

The Marked Price (MP) is the price printed on the product. Discount is always calculated on the MP unless specified otherwise.

Relation between SP and MP

SP = MP × (100 − Discount%) ÷ 100. This formula helps find any missing value when the other two are known.

Shortcut Trick

Let total capacity of cistern = 16 units (LCM of 8 and 16).

Combined efficiency of P and Q = 16 ÷ 8 = 2 units/hour.

Efficiency of P = 16 ÷ 16 = 1 unit/hour.

Efficiency of Q = 2 − 1 = 1 unit/hour.

Time required for Q to fill one-fourth of cistern = (16 ÷ 4) ÷ 1 = 4 hours.

∴ The correct answer is 4 hours.

Alternate Method

⇒ Efficiency of Q = (Efficiency of P + Q) − (Efficiency of P)

⇒ Efficiency of Q = 2 − 1 = 1 unit/hour

⇒ One-fourth of the total work = 16 ÷ 4 = 4 units

⇒ Time taken by Q to complete 4 units of work = 4 ÷ 1 = 4 hours

∴ The correct answer is 4 hours.

Additional Information

Pipes & Cisterns Efficiency Rule

The rate of work done by an individual pipe is its efficiency. For an outlet pipe or leak, efficiency is taken as negative, while for an inlet pipe, it is taken as positive.

Work-Time Reciprocal Relation

If a pipe can fill a tank in x hours, the part of the tank filled in 1 hour is 1/x.

Shortcut Trick

When equal distances are covered at different speeds, the Average Speed is the Harmonic Mean of the speeds.

Formula for 3 equal segments: Average Speed = 3 × v₁ × v₂ × v₃ ÷ (v₁v₂ + v₂v₃ + v₃v₁)

Average Speed = (3 × 20 × 30 × 60) ÷ (20×30 + 30×60 + 60×20)

Average Speed = 108000 ÷ (600 + 1800 + 1200) = 108000 ÷ 3600 = 30 km/hr.

∴ The correct answer is 30 km/hr.

Alternate Method

Given: Total distance = 180 km | Segment 1 = 60 km at 20 km/hr | Segment 2 = 60 km at 30 km/hr | Segment 3 = 60 km at 60 km/hr.

Formula Used: Average Speed = Total Distance ÷ Total Time

⇒ Time for 1^st segment (t₁) = 60 ÷ 20 = 3 hours

⇒ Time for 2^nd segment (t₂) = 60 ÷ 30 = 2 hours

⇒ Time for 3^rd segment (t₃) = 60 ÷ 60 = 1 hour

⇒ Total Time = 3 + 2 + 1 = 6 hours

⇒ Average Speed = 180 ÷ 6 = 30 km/hr

∴ The correct answer is 30 km/hr.

Additional Information

Average Speed for Two Equal Distances

When two equal distances are covered at speeds 'u' and 'v', Average Speed = (2uv) ÷ (u + v).

Weighted Average Speed

If distances d₁, d₂ are covered at speeds v₁, v₂, then Average Speed = (d₁ + d₂) ÷ (d₁/v₁ + d₂/v₂).

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Shortcut Trick

New member age = New average + (Change in average × Old number of members)

Here, New average = 30 years, Change = 30 − 29 = 1 year, Old count = 34.

Age of teacher = 30 + (1 × 34) = 30 + 34 = 64 years.

∴ The correct answer is 64 years.

Alternate Method

Given:

Initial students = 34 | Initial average age = 29 years

New average after adding teacher = 30 years

Formula Used:

Total Age = Average Age × Number of persons

Teacher age = (Total age of 35 persons) − (Total age of 34 students)

Calculations:

⇒ Total age of 34 students = 34 × 29 = 986 years

⇒ Total people including teacher = 34 + 1 = 35

⇒ Total age of 35 persons = 35 × 30 = 1050 years

⇒ Teacher age = 1050 − 986 = 64 years

∴ The correct answer is 64 years.

Additional Information

Average Displacement Method

When a new person joins and the average increases, the new person brings the "new average" plus the "increment" distributed over all previous members.

Weighted Average Formula

A_combined = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is the count and x is the individual average/value.

Average Basics

Average = (Sum of all observations) ÷ (Total number of observations). To find the sum, multiply the average by the count.

Shortcut Trick

Check the HCF of each pair; relatively prime numbers must have HCF = 1.

(68, 85) → Both divisible by 17.

(102, 153) → Both divisible by 51.

(65, 91) → Both divisible by 13.

(52, 105) → 52 = 2² × 13 and 105 = 3 × 5 × 7. No common factor except 1.

∴ The correct answer is (52, 105).

Alternate Method

Relatively Prime (or Coprime) numbers are pairs whose Highest Common Factor (HCF) is 1.

⇒ For (68, 85): 68 = 17 × 4, 85 = 17 × 5. HCF = 17.

⇒ For (102, 153): 102 = 51 × 2, 153 = 51 × 3. HCF = 51.

⇒ For (65, 91): 65 = 13 × 5, 91 = 13 × 7. HCF = 13.

⇒ For (52, 105): 52 = 2 × 2 × 13, 105 = 3 × 5 × 7. HCF = 1.

∴ The correct answer is (52, 105).

Additional Information

Definition of Coprime Numbers

Two numbers are relatively prime if they have no common positive factors other than 1.

Consecutive Numbers

Any two consecutive integers are always relatively prime to each other (e.g., 20 and 21).

Prime Numbers Pair

Any two distinct prime numbers are always relatively prime (e.g., 7 and 11).

Shortcut Trick

Let total work (capacity) be LCM of 32, 36, and 20 = 1440 units.

Efficiency of A = 45, B = 40, and A+B+C = 72.

Efficiency of C = 72 − (45 + 40) = −13 units/min.

Since 13 units/min corresponds to 13 cubic units/min, 1 unit = 1 cubic unit.

∴ The correct answer is 1440 cubic units.

Alternate Method

Given: Time for Pipe A = 32 min, Pipe B = 36 min, Pipe A+B+C = 20 min. Rate of Pipe C = 13 cubic units/min.

⇒ Total efficiency of A and B = 45 + 40 = 85 units/min

⇒ Efficiency of Pipe C = (A+B+C) − (A+B) = 72 − 85 = −13 units/min

⇒ Negative sign indicates emptying.

⇒ Actual emptying rate = 13 cubic units/min

⇒ 13 units = 13 cubic units → 1 unit = 1 cubic unit

⇒ Total Capacity = 1440 × 1 = 1440 cubic units

∴ The correct answer is 1440 cubic units.

Additional Information

Efficiency and Time

Efficiency is inversely proportional to the time taken to complete the work. Efficiency = Total Work ÷ Time.

Inlet and Outlet Pipes

Inlet pipes have positive efficiency as they fill the tank, while outlet (emptying) pipes have negative efficiency.

Tank Capacity Formula

Capacity = Time taken by pipe × Flow rate (Rate at which water enters or leaves per minute).

Shortcut Trick

Apply the difference of squares identity: a² − b² = (a + b)(a − b).

Let a = (x + 2) and b = (x − 3).

Sum (a + b) = (x + 2) + (x − 3) = 2x − 1.

Difference (a − b) = (x + 2) − (x − 3) = 5.

Required result = 5 × (2x − 1) = 10x − 5.

∴ The correct answer is 10x − 5.

Alternate Method

Given:

The algebraic expression to simplify is (x + 2)² − (x − 3)².

Formula Used:

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

Calculations:

⇒ Expand the first term: (x + 2)² = x² + 2 × x × 2 + 2² = x² + 4x + 4

⇒ Expand the second term: (x − 3)² = x² − 2 × x × 3 + 3² = x² − 6x + 9

⇒ Subtract the two expanded expressions: (x² + 4x + 4) − (x² − 6x + 9)

⇒ Remove brackets and change signs: x² + 4x + 4 − x² + 6x − 9

⇒ Combine like terms: (x² − x²) + (4x + 6x) + (4 − 9)

⇒ Result = 10x − 5

∴ The correct answer is 10x − 5.

Additional Information

Standard Algebraic Identities

These identities help simplify complex polynomial expressions: (a + b)(a − b) = a² − b² and (a + b)² = a² + 2ab + b².

Combining Like Terms

Only terms with the same variables and exponents can be added or subtracted during simplification.

Distributive Property

When a negative sign is outside a bracket, every term inside the bracket must change its sign: −(a − b) = −a + b.

Shortcut Trick

When an item price increases by x% and consumption decreases by x%, the net change is always a decrease.

Net percentage change = −(x² / 100)%

Net change = −(20² / 100)% = −(400 / 100)% = −4%

Negative sign indicates a decrease.

∴ The correct answer is 4%.

Alternate Method

Given:

Increase in Price = 20%

Decrease in Consumption = 20%

Formula Used:

Total Expenditure = Price × Consumption

Net Change % = a + b + (ab / 100)

Calculations:

⇒ Let initial price be 100 and consumption be 100.

⇒ Initial Expenditure = 100 × 100 = 10,000

⇒ New Price = 120 (after 20% increase)

⇒ New Consumption = 80 (after 20% decrease)

⇒ New Expenditure = 120 × 80 = 9,600

⇒ Decrease in Expenditure = 10,000 − 9,600 = 400

⇒ Percentage Decrease = (400 / 10,000) × 100 = 4%

∴ The correct answer is 4%.

Additional Information

Successive Percentage Change

If a value is changed by x% and then by y%, the net change is given by [x + y + (xy / 100)]%.

Expenditure and Budget

If price increases by r%, to keep expenditure constant, consumption must decrease by [r / (100 + r)] × 100%.

Product Stability Rule

When the product of two variables is constant, an increase in one leads to a proportional decrease in the other.

Shortcut Trick

Age ratio (Father : Son): Present = 2 : 1 | 15 years ago = 3 : 1

Make ratio differences equal: 2:1 (diff 1) → 4:2 (diff 2) | 3:1 (diff 2)

Ratio change: 4 → 3 (1 unit) = 15 years

Father's present age = 4 units = 4 × 15 = 60 years

∴ The correct answer is 60 years.

Alternate Method

Given:

Present age of Son = x

Present age of Father = 2x

Formula Used:

Age 15 years ago = (Present Age − 15)

Calculations:

⇒ (2x − 15) = 3 × (x − 15)

⇒ 2x − 15 = 3x − 45

⇒ 3x − 2x = 45 − 15

⇒ x = 30

⇒ Father's present age = 2x = 2 × 30

⇒ Father's present age = 60 years

∴ The correct answer is 60 years.

Additional Information

Ratio of Ages

If the age of A and B are in the ratio a:b, then their ages can be assumed as ax and bx.

Age Difference Property

The difference between the ages of two persons remains constant at any point in time.

Future and Past Ages

If present age is x, then age 'n' years ago was (x−n) and age 'n' years hence will be (x+n).

Shortcut Trick

Net percentage change for 2 successive 20% increases = 20 + 20 + (20 × 20)/100 = 44%.

Total Amount = Principal × (100 + Net %)/100.

⇒ Amount = 3,500 × 1.44 = ₹5,040.

∴ The correct answer is ₹5,040.

Alternate Method

Given:

Principal (P) = ₹3,500

Rate (R) = 20% per annum

Time (T) = 2 years (compounded annually)

Formula Used:

Amount (A) = P(1 + R/100)^T

Calculations:

⇒ A = 3,500 × (1 + 20/100)²

⇒ A = 3,500 × (1 + 1/5)²

⇒ A = 3,500 × (6/5) × (6/5)

⇒ A = 3,500 × 36/25

⇒ A = 140 × 36

⇒ A = 5,040

∴ The correct answer is ₹5,040.

Additional Information

Compound Interest (CI) Formula

CI = Amount − Principal = P[(1 + R/100)^T − 1].

Effective Rate of Interest

For 2 years, the effective rate is R + R + (R × R)/100.

Difference between CI and SI for 2 years

Difference (D) = P(R/100)².