Quantitative Aptitude PYQs

Shortcut Trick

Apply BODMAS: solve the bracket (29 + 19 = 48) and multiply (48 × 2 = 96).

The expression becomes: 76 + 85 − 49 + 96 − 91.

Combine all positive terms (76 + 85 + 96 = 257) and negative terms (49 + 91 = 140).

Final result: 257 − 140 = 117.

∴ The correct answer is 117.

Alternate Method

Given Expression: 76 + 85 − 49 + (29 + 19) × 2 − 91

Method Used: BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Step-by-step Calculation:

⇒ 76 + 85 − 49 + (48) × 2 − 91 (Solving Bracket)

⇒ 76 + 85 − 49 + 96 − 91 (Performing Multiplication)

⇒ 161 − 49 + 96 − 91 (Performing Addition from left)

⇒ 112 + 96 − 91 (Performing Subtraction)

⇒ 208 − 91 (Performing Addition)

⇒ 117 (Final Subtraction)

∴ The correct answer is 117.

Additional Information

BODMAS Rule

It is an acronym used to remember the order of operations in mathematics: Brackets (), Of (or Orders like roots/powers), Division, Multiplication, Addition, and Subtraction.

Grouping Positive and Negative Terms

To minimize calculation errors, sum all positive numbers and all negative numbers separately, then find their difference at the final step.

Shortcut Trick

Apply the BODMAS rule to solve the expression in order: Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Solve Of: (19 of 2) = 19 × 2 = 38.

Solve Division: 28 ÷ 4 = 7.

Simplify remaining terms: 92 − 36 + 7 + 38 − 14 + 17 = 104.

∴ The correct answer is 104.

Alternate Method

Given Expression: 92 − 36 + 28 ÷ 4 + (19 of 2) − 14 + 17

Formula Used: BODMAS Rule

⇒ 92 − 36 + 28 ÷ 4 + 38 − 14 + 17 (Solving Of first)

⇒ 92 − 36 + 7 + 38 − 14 + 17 (Solving Division)

⇒ (92 + 7 + 38 + 17) − (36 + 14) (Grouping Addition and Subtraction terms)

⇒ 154 − 50

⇒ 104

∴ The correct answer is 104.

Additional Information

BODMAS Rule

Defines the order of operations: Brackets (round, curly, square), Of (powers/multiplication), Division, Multiplication, Addition, and Subtraction.

The 'Of' Operator

The word 'of' functions as multiplication but must be performed before regular division and multiplication in the BODMAS hierarchy.

Grouping Strategy

For faster simplification, group all positive terms together and all negative terms together before the final subtraction.

Shortcut Trick

Count the decimal places in both numbers: 0.36 has 2 places and 0.00036 has 5 places.

Subtract the decimal places: 5 − 2 = 3 places difference.

Since the divisor has more decimal places, the result is 10³.

Result = 1000.

∴ The correct answer is 1000.

Alternate Method

Given: 0.36 ÷ 0.00036

Method: Convert decimal numbers into fractions for easier calculation.

⇒ 0.36 = 36/100

⇒ 0.00036 = 36/100000

⇒ Expression = (36/100) ÷ (36/100000)

⇒ Expression = (36/100) × (100000/36)

⇒ Expression = 100000 ÷ 100

⇒ 1000

∴ The correct answer is 1000.

Additional Information

Division of Decimals

When dividing by a decimal, move the decimal point in the divisor to the right until it becomes a whole number, and move it the same number of places in the dividend.

Power of 10 Rule

Multiplying a decimal by 10ⁿ shifts the decimal point n places to the right, while dividing by 10ⁿ shifts it n places to the left.

BODMAS Rule

In multi-step simplification questions, always follow the order of Brackets, Orders, Division, Multiplication, Addition, and Subtraction.

Shortcut Trick

Identify the standard algebraic identity: (a + b)² − (a − b)² = 4ab.

In the given expression, let a = 12.4 and b = 3.6.

The expression simplifies to: (4 × 12.4 × 3.6) ÷ (12.4 × 3.6).

The terms 12.4 × 3.6 in the numerator and denominator cancel each other out.

The remaining value is 4.

∴ The correct answer is 4.

Alternate Method

Given:

Expression = ((12.4 + 3.6)² − (12.4 − 3.6)²) / (12.4 × 3.6)

Formula Used:

Using the algebraic identity: (a + b)² − (a − b)² = 4ab

Calculations:

⇒ Let a = 12.4 and b = 3.6.

⇒ The numerator is in the form (a + b)² − (a − b)².

⇒ Numerator = 4 × a × b = 4 × 12.4 × 3.6.

⇒ The denominator is a × b = 12.4 × 3.6.

⇒ Expression = (4 × 12.4 × 3.6) ÷ (12.4 × 3.6).

⇒ Expression = 4.

∴ The correct answer is 4.

Additional Information

Sum of Squares Identity

(a + b)² + (a − b)² = 2(a² + b²). This is used when the middle sign is positive.

Difference of Squares

a² − b² = (a + b)(a − b). This identity is fundamental for factoring quadratic expressions.

BODMAS Rule

Always follow the order of Brackets, Orders, Division, Multiplication, Addition, and Subtraction for complex calculations.

Shortcut Trick

The given expression is in the form of the algebraic identity a³ + b³ + 3ab(a + b) = (a + b)³.

Here, let a = 10 and b = 6, which implies (a + b) = 16.

Substituting these values, the expression simplifies to (10 + 6)³.

⇒ 16³ = 4096.

∴ The correct answer is 4096.

Alternate Method

Given: (10)³ + (6)³ + 3 × 10 × 6 × 16

Method: Step-by-step simplification using arithmetic powers and BODMAS.

⇒ Calculate the cubes: 10³ = 1000 and 6³ = 216.

⇒ Calculate the product: 3 × 10 × 6 × 16 = 180 × 16.

⇒ 180 × 16 = 2880.

⇒ Total Sum = 1000 + 216 + 2880.

⇒ Total Sum = 4096.

∴ The correct answer is 4096.

Additional Information

Algebraic Cube Identities

The formula for the cube of a sum is (a + b)³ = a³ + b³ + 3ab(a + b). Similarly, the cube of a difference is (a − b)³ = a³ − b³ − 3ab(a − b).

Sum and Difference of Cubes

Do not confuse (a+b)³ with a³ + b³. The latter factors as (a + b)(a² − ab + b²), while a³ − b³ factors as (a − b)(a² + ab + b²).

Standard Cube Values

For faster calculations in competitive exams, it is recommended to memorize cubes up to 20, such as 11³ = 1331, 12³ = 1728, and 16³ = 4096.

Shortcut Trick

Area of Rhombus = (d₁ × d₂) ÷ 2

Product of Diagonals = 2 × Area = 2 × 54 = 108

Other Diagonal = 108 ÷ 9 = 12 cm

∴ The correct answer is 12 cm.

Alternate Method

Given:

Area of the rhombus = 54 cm²

Length of one diagonal (d₁) = 9 cm

Formula Used:

Area of a rhombus = (1/2) × d₁ × d₂

Calculations:

⇒ 54 = (1/2) × 9 × d₂

⇒ 54 × 2 = 9 × d₂

⇒ 108 = 9 × d₂

⇒ d₂ = 108 ÷ 9

⇒ d₂ = 12 cm

∴ The correct answer is 12 cm.

Additional Information

Properties of a Rhombus

All sides of a rhombus are equal, and the diagonals bisect each other at right angles (90°).

Side and Diagonals Relationship

The side 'a' of a rhombus can be found using the formula: 4a² = d₁² + d₂².

Perimeter of Rhombus

The perimeter is given by 4 × side. For this rhombus, side = √(4.5² + 6²) = 7.5 cm, so Perimeter = 30 cm.

Shortcut Trick

Sum of diagonals (d₁ + d₂) = 14 and Product (d₁d₂) = 2 × Area = 48.

By inspection, the numbers summing to 14 and multiplying to 48 are 6 and 8.

Half-diagonals are 3 and 4.

In a rhombus, side (s) forms a right-angled triangle with half-diagonals: s = √(3² + 4²).

Using the 3-4-5 Pythagorean triplet, side = 5 cm.

∴ The correct answer is 5 cm.

Alternate Method

Given:

Sum of diagonals (d₁ + d₂) = 14 cm

Area of rhombus = 24 cm²

Formula Used:

Area = (d₁ × d₂) / 2

4 × side² = d₁² + d₂²

Calculations:

⇒ Area = (d₁ × d₂) / 2 = 24

⇒ d₁ × d₂ = 48

⇒ (d₁ + d₂)² = d₁² + d₂² + 2d₁d₂

⇒ (14)² = d₁² + d₂² + 2(48)

⇒ 196 = d₁² + d₂² + 96

⇒ d₁² + d₂² = 100

⇒ 4 × side² = 100

⇒ side² = 25

⇒ side = √25 = 5 cm

∴ The correct answer is 5 cm.

Additional Information

Properties of a Rhombus

A rhombus is a quadrilateral with all sides equal. Its diagonals bisect each other at right angles (90°).

Side and Diagonal Relation

The side 's' and diagonals d₁ and d₂ are related by the formula: 4s² = d₁² + d₂².

Perimeter of Rhombus

Since all four sides are equal, the Perimeter = 4 × side.

Shortcut Trick

To find the LCM of (40, 18), check the multiples of the larger number 40.

Multiples of 40: 40, 80, 120, 160, 200, 240, 280, 320, 360.

Check divisibility by 18: 360 ÷ 18 = 20 (Perfectly divisible).

The first common multiple for both is 360.

∴ The correct answer is 360.

Alternate Method

Given:

LCM(A, B) = 40

C = 18

Formula Used:

LCM(A, B, C) = LCM(LCM(A, B), C)

Calculations:

⇒ LCM(A, B, C) = LCM(40, 18)

⇒ Prime factorization of 40 = 2 × 2 × 2 × 5 = 2³ × 5¹

⇒ Prime factorization of 18 = 2 × 3 × 3 = 2¹ × 3²

⇒ The LCM is the product of the highest power of every prime factor involved.

⇒ LCM = 2³ × 3² × 5¹

⇒ LCM = 8 × 9 × 5 = 360

∴ The correct answer is 360.

Additional Information

Associative Property of LCM

The Least Common Multiple is associative, meaning LCM(a, b, c) = LCM(LCM(a, b), c) = LCM(a, LCM(b, c)).

Relation between HCF and LCM

For any two positive integers x and y, the product of the numbers is equal to the product of their HCF and LCM: x × y = HCF(x, y) × LCM(x, y).

LCM of Fractions

The LCM of a set of fractions is calculated as (LCM of Numerators) ÷ (HCF of Denominators).

Shortcut Trick

The fundamental property of two numbers states: Product of two numbers = HCF × LCM.

Product = 15 × 360

Product = 5400

∴ The correct answer is 5400.

Alternate Method

Given:

Highest Common Factor (HCF) = 15

Least Common Multiple (LCM) = 360

Formula Used:

Product of numbers (n₁ × n₂) = HCF(n₁, n₂) × LCM(n₁, n₂)

Calculations:

⇒ Product = 15 × 360

⇒ Product = 15 × (300 + 60)

⇒ Product = (15 × 300) + (15 × 60)

⇒ Product = 4500 + 900

⇒ Product = 5400

∴ The correct answer is 5400.

Additional Information

HCF and LCM of Fractions

HCF = (HCF of Numerators) ÷ (LCM of Denominators). LCM = (LCM of Numerators) ÷ (HCF of Denominators).

Co-prime Numbers

If two numbers are co-prime, their HCF is always 1, and their LCM is equal to their product.

HCF/LCM Relationship

The LCM of any set of numbers is always exactly divisible by their HCF.

Shortcut Trick

If A is R% less than B, then B is more than A by:

[R ÷ (100 − R)] × 100

Given R = 50

⇒ [50 ÷ (100 − 50)] × 100 = [50 ÷ 50] × 100 = 100%

∴ The correct answer is 100%.

Alternate Method

Given:

A is 50% less than B.

Let the value of B be 100.

⇒ A = 100 − 50% of 100

⇒ A = 100 − 50 = 50

Now, we need to find how much percentage more B is compared to A.

⇒ Difference = B − A = 100 − 50 = 50

⇒ Percentage More = [Difference ÷ A] × 100

⇒ [50 ÷ 50] × 100

⇒ 1 × 100 = 100%

∴ The correct answer is 100%.

Additional Information

Percentage Increase Formula

Percentage Increase = [(New Value − Original Value) ÷ Original Value] × 100.

Reciprocal Relation

If A is 1/x less than B, then B is 1/(x−1) more than A. Here, 50% = 1/2, so B is 1/(2−1) = 1/1 = 100% more than A.

Base Change Property

Percentage values change based on the denominator (base). When comparing B to A, A becomes the base.

Shortcut Trick

Ratio of Cost Price (CP) to Selling Price (SP) = 500 : 550 = 10 : 11.

Profit = 11 − 10 = 1 unit.

Profit percentage = (Profit units ÷ CP units) × 100.

Profit percentage = (1 ÷ 10) × 100 = 10%.

∴ The correct answer is 10%.

Alternate Method

Given:

Cost Price (CP) of the article = ₹500

Selling Price (SP) of the article = ₹550

Formula Used:

Profit = Selling Price − Cost Price

Profit Percentage = (Profit ÷ Cost Price) × 100

Calculations:

⇒ Profit = 550 − 500 = ₹50

⇒ Profit Percentage = (50 ÷ 500) × 100

⇒ Profit Percentage = (1 ÷ 10) × 100

⇒ Profit Percentage = 10%

∴ The correct answer is 10%.

Additional Information

Loss Percentage

When Cost Price is greater than Selling Price, Loss % = [(CP − SP) ÷ CP] × 100.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). SP = MP × [ (100 − Discount %) ÷ 100 ].

Effective Profit with Discount

If an article is marked up by x% and then a discount of y% is given, net profit % = x − y − (xy ÷ 100).

Shortcut Trick

Value = 25% of 500 = 125.

Since 125 is 25% more than A, it represents 125% of A.

125% → 125

100% (Value of A) → (125 ÷ 125) × 100 = 100.

∴ The correct answer is 100.

Alternate Method

Given:

Percentage value = 25% of 500

Relation: This value is 25% greater than A.

Formula Used:

Number = Base × (1 + Percentage Increase/100)

Calculations:

⇒ 25% × 500 = (25/100) × 500 = 125

⇒ 125 = A + (25/100) × A

⇒ 125 = A + 0.25A

⇒ 125 = 1.25A

⇒ A = 125 ÷ 1.25

⇒ A = 12500 ÷ 125

⇒ A = 100

∴ The correct answer is 100.

Additional Information

Percentage Increase

To increase a quantity by x%, multiply it by (1 + x/100). For example, a 25% increase is equivalent to multiplying by 1.25 or 5/4.

Fractional Equivalents

Knowing common fractions like 25% = 1/4 or 20% = 1/5 helps in faster calculation. Here, A + A/4 = 5A/4 = 125.

Base Value Concept

Always identify the base (100%) on which the percentage is calculated. In this problem, A is the base value.

Shortcut Trick

Let the Selling Price (SP) = 100 units.

Loss = 30% of SP = 30 units.

Cost Price (CP) = SP + Loss = 100 + 30 = 130 units.

Actual Loss% (on CP) = (Loss ÷ CP) × 100 = (30 ÷ 130) × 100.

Loss% = 300 ÷ 13 ≈ 23.08%.

∴ The correct answer is 23.08 percent.

Alternate Method

Given:

Loss percentage on Selling Price = 30%

Formula Used:

1. Cost Price (CP) = Selling Price (SP) + Loss
2. Loss percentage on CP = (Loss ÷ CP) × 100

Calculations:

⇒ Let the Selling Price (SP) be ₹100.

⇒ Loss = 30% of ₹100 = ₹30.

⇒ Cost Price (CP) = SP + Loss = 100 + 30 = ₹130.

⇒ Loss Percentage on CP = (30 ÷ 130) × 100.

⇒ Loss Percentage = 300 ÷ 13 = 23.0769...

⇒ Loss Percentage ≈ 23.08%.

∴ The correct answer is 23.08 percent.

Additional Information

Profit/Loss on Selling Price

When profit or loss is calculated on Selling Price, it is called the apparent profit/loss. To find the actual profit/loss, it must always be calculated on the Cost Price.

Conversion Formulas

If Loss% on SP is L%, then Loss% on CP = [L ÷ (100 + L)] × 100. If Profit% on SP is P%, then Profit% on CP = [P ÷ (100 − P)] × 100.

Basic P&L Relations

Profit = SP − CP (when SP > CP). Loss = CP − SP (when CP > SP). In case of loss, CP is always greater than SP.

Shortcut Trick

Total Work = LCM(18, 27, 36) = 108 units

Individual Efficiencies: A = 6, B = 4, C = 3

Total Efficiency (A + B + C) = 6 + 4 + 3 = 13 units/day

Work to complete = 2/3 × 108 = 72 units

∴ The correct answer is 72/13 days.

Alternate Method

Given:

Time taken by A alone = 18 days

Time taken by B alone = 27 days

Time taken by C alone = 36 days

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Combined efficiency of A, B and C = 6 + 4 + 3 = 13 units/day

⇒ 2/3 part of the work = 2/3 × 108 = 72 units

⇒ Time taken to complete 72 units = Work ÷ Combined Efficiency

⇒ Required Time = 72 ÷ 13 days

∴ The correct answer is 72/13 days.

Additional Information

Work and Efficiency Relationship

The time taken to complete a work is inversely proportional to the efficiency of the person (Time ∝ 1/Efficiency).

Combined Work Formula

If A and B can do a work in x and y days respectively, together they can do it in (x × y) ÷ (x + y) days.

Work and Wages

When workers work for the same duration, their wages are distributed in the ratio of their efficiencies.

Shortcut Trick

Efficiency ratio of A and B is the inverse of their time ratio = 28 : 21 = 4 : 3.

The total wage of ₹4200 is divided into 4 + 3 = 7 parts based on efficiency.

Value of 1 part = 4200 ÷ 7 = ₹600.

The difference between their shares is (4 − 3) = 1 part.

Difference in amount = 1 × 600 = ₹600.

∴ The correct answer is ₹600.

Alternate Method

Given: A can complete work in 21 days, B in 28 days. Total contract amount = ₹4200.

Formula Used: Wages are distributed in the ratio of efficiency when the duration of work is the same for all individuals.

⇒ Total Efficiency parts = 4 + 3 = 7 units

⇒ Total amount for 7 units = ₹4200

⇒ Amount for 1 efficiency unit = 4200 ÷ 7 = ₹600

⇒ Share of A = 4 × 600 = ₹2400

⇒ Share of B = 3 × 600 = ₹1800

⇒ Difference in amounts = 2400 − 1800 = ₹600

∴ The correct answer is ₹600.

Additional Information

Wages and Efficiency

Wages are directly proportional to the amount of work done. If time is constant, Wage Ratio = Efficiency Ratio.

Work-Time Relation

Efficiency is inversely proportional to Time taken to complete a task (Efficiency = 1 ÷ Time).

Pipes and Cisterns

Similar to Time and Work, filling efficiency is treated as positive work and leakage/emptying efficiency is treated as negative work.

Shortcut Trick

When distance is constant, Speed is inversely proportional to Time (Speed ∝ 1/Time).

Ratio of Time (Walk : Cycle) = 4.5 hr : 4 hr = 9 : 8

Ratio of Speed (Walk : Cycle) = 8 : 9

Since Walking Speed 8 units = 8 km/hr, then 9 units = 9 km/hr.

∴ The correct answer is 9 km/hr.

Alternate Method

Given:

Walking Speed (S₁) = 8 km/hr

Time taken (T₁) = 4 hr 30 min = 4.5 hr

Time taken by Cycle (T₂) = 4 hr

Formula Used:

Distance = Speed × Time

Calculations:

⇒ Total Distance = S₁ × T₁

⇒ Distance = 8 × 4.5 = 36 km

⇒ Speed of Cycle (S₂) = Distance ÷ T₂

⇒ S₂ = 36 ÷ 4 = 9 km/hr

∴ The correct answer is 9 km/hr.

Additional Information

Constant Distance Property

When distance is fixed, the ratio of speeds is the inverse of the ratio of time taken: S₁ × T₁ = S₂ × T₂.

Unit Conversion

To convert minutes into hours, divide by 60 (e.g., 30 min = 30/60 = 0.5 hr). To convert km/hr to m/s, multiply by 5/18.

Average Speed

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

Shortcut Trick

Since the speed of the train remains constant, the time taken is directly proportional to the total distance covered.

Total Distance 1 = 150 + 250 = 400 m

Total Distance 2 = 150 + 500 = 650 m

Ratio of Distances = 400 : 650 = 8 : 13

Since 8 units = 8 seconds, then 13 units = 13 seconds.

∴ The correct answer is 13 sec.

Alternate Method

Given:

Length of train (L_t) = 150 m

Length of first bridge (L_b1) = 250 m

Time taken (t₁) = 8 sec

Length of second bridge (L_b2) = 500 m

Formula Used:

Speed = Total Distance ÷ Total Time

Total Distance = Length of Train + Length of Object (Bridge/Platform)

Calculations:

⇒ Total distance covered to cross the first bridge = 150 + 250 = 400 m

⇒ Speed of the train = 400 ÷ 8 = 50 m/s

⇒ Total distance covered to cross the second bridge = 150 + 500 = 650 m

⇒ Time taken to cross the second bridge = Total Distance ÷ Speed

⇒ Time = 650 ÷ 50

⇒ Time = 13 sec

∴ The correct answer is 13 sec.

Additional Information

Crossing a Stationary Point Object

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

Crossing a Stationary Long Object

When a train crosses a bridge, platform, or another stationary train, the distance covered is (Length of train + Length of object).

Unit Conversion

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

Shortcut Trick

Let Cost Price (CP) = 100. Loss = 30% ⇒ Selling Price (SP) = 70.

Discount = 7/10 of SP = 0.7 × 70 = 49.

Marked Price (MP) = SP + Discount = 70 + 49 = 119.

Required percentage = (CP / MP) × 100 = (100 / 119) × 100 ≈ 84.03%.

∴ The correct answer is 84.03%.

Alternate Method

Given:

Loss Percentage = 30%

Discount = (7/10) × Selling Price (SP)

Formula Used:

1. SP = CP − Loss
2. MP = SP + Discount

Calculations:

⇒ Let Cost Price (CP) = 100 units

⇒ Loss = 30% of 100 = 30 units

⇒ Selling Price (SP) = 100 − 30 = 70 units

⇒ Discount = (7/10) × 70 = 49 units

⇒ Marked Price (MP) = SP + Discount = 70 + 49 = 119 units

⇒ Required % = (CP / MP) × 100 = (100 / 119) × 100

⇒ Required % = 10000 / 119 ≈ 84.03%

∴ The correct answer is 84.03%.

Additional Information

Profit and Loss Components

Cost Price (CP) is the price at which an item is bought. Selling Price (SP) is the price at which it is sold. Profit = SP − CP and Loss = CP − SP.

Marked Price and Discount

Marked Price (MP) is the price printed on the article. Discount is always calculated on MP. SP = MP − Discount.

Relation between CP and MP

The ratio of CP to MP can be expressed as CP / MP = (100 − Discount%) / (100 + Profit%) or (100 − Loss%).

Shortcut Trick

Effective Discount % = x + y − (xy/100)

Effective Discount % = 30 + 20 − (30 × 20 / 100) = 50 − 6 = 44%

Net Discount = 44% of Marked Price

Net Discount = 44% of ₹500 = (44/100) × 500 = ₹220

∴ The correct answer is ₹220.

Alternate Method

Given: Marked Price (MP) = ₹500, Successive discounts = 30% and 20%.

⇒ First discount = 30% of ₹500 = (30/100) × 500 = ₹150

⇒ Value after first discount = 500 − 150 = ₹350

⇒ Second discount = 20% of ₹350 = (20/100) × 350 = ₹70

⇒ Total Net Discount = First discount + Second discount = 150 + 70 = ₹220

⇒ Alternatively, Final Selling Price (SP) = 350 − 70 = ₹280

⇒ Net Discount = Marked Price (MP) − Selling Price (SP) = 500 − 280 = ₹220

∴ The correct answer is ₹220.

Additional Information

Successive Discount Formula

For three successive discounts of a%, b%, and c%, calculate the equivalent for the first two, then use that result with the third.

Selling Price and Discount

Selling Price = Marked Price × (1 − d₁/100) × (1 − d₂/100) ... × (1 − d_n/100).

Net Discount Value

Net Discount = Marked Price − Final Selling Price.

Shortcut Trick

Net Discount = [a + b − (a × b / 100)]%

Here, a = 10% and b = 30%

Net Discount = [10 + 30 − (10 × 30 / 100)]%

Net Discount = [40 − 3]% = 37%

∴ The correct answer is 37%.

Alternate Method

Given values: First discount = 10%, Second discount = 30%

Assume the Marked Price (MP) of the article is 100 units.

⇒ First discount = 10% of 100 = 10 units

⇒ Price after first discount = 100 − 10 = 90 units

⇒ Second discount = 30% of 90 = 27 units

⇒ Selling Price (SP) = 90 − 27 = 63 units

⇒ Total Discount = 100 − 63 = 37 units

⇒ Net Discount % = (37 / 100) × 100 = 37%

∴ The correct answer is 37%.

Additional Information

Successive Discounts Formula

For two successive discounts of x% and y%, the single equivalent discount = [x + y − (xy / 100)]%.

Fraction Method

Final Price = Marked Price × (100−d₁)/100 × (100−d₂)/100. This is useful for complex percentages.

Successive Percentage Change

For successive increases of x% and y%, the net change = [x + y + (xy / 100)]%. Note the change in sign compared to discounts.