Quantitative Aptitude PYQs

Shortcut Trick

Solve the bracketed part first: 39 + 24 = 63.

Group all positive integers: 92 + 58 + 63 + 41 = 254.

Group all negative integers: 66 + 73 = 139.

Calculate the final difference: 254 − 139 = 115.

∴ The correct answer is 115.

Alternate Method

Given Expression: 92 + 58 − 66 + (39 + 24) − 73 + 41

Follow the BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction):

⇒ 92 + 58 − 66 + 63 − 73 + 41

⇒ 150 − 66 + 63 − 73 + 41

⇒ 84 + 63 − 73 + 41

⇒ 147 − 73 + 41

⇒ 74 + 41

⇒ 115

∴ The correct answer is 115.

Additional Information

BODMAS Rule

The BODMAS acronym helps remember the order of operations: Brackets first, then Orders (powers/roots), followed by Division and Multiplication, and finally Addition and Subtraction.

Commutative Property

In expressions involving only addition and subtraction, terms can be rearranged or grouped as long as the sign preceding each number remains attached to it.

Simplification Tip

When dealing with multiple terms, adding all positive numbers together and all negative numbers together before taking the final difference reduces the risk of calculation errors.

Shortcut Trick

Apply BODMAS: 84 − (63 ÷ 7) + (36 × 2) − 19 + (45 ÷ 5)

⇒ 84 − 9 + 72 − 19 + 9

Note:

The terms −9 and +9 cancel each other out.

⇒ 84 + 72 − 19 = 156 − 19 = 137

∴ The correct answer is 137.

Alternate Method

Given: 84 − 63 ÷ 7 + (36 × 2) − 19 + 45 ÷ 5

Following the BODMAS rule (Bracket, Of, Division, Multiplication, Addition, Subtraction):

Step 1: Solve Bracket

⇒ 84 − 63 ÷ 7 + 72 − 19 + 45 ÷ 5

Step 2: Solve Division

⇒ 84 − 9 + 72 − 19 + 9

Step 3: Solve Addition

⇒ (84 + 72 + 9) − 9 − 19

⇒ 165 − 28

Step 4: Solve Subtraction

⇒ 137

∴ The correct answer is 137.

Additional Information

BODMAS Rule

It specifies the order of mathematical operations: Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Left to Right Rule

When operations of the same priority (like Division and Multiplication, or Addition and Subtraction) appear together, solve them from left to right.

Sign Convention

Always group positive numbers together and negative numbers together before the final subtraction to minimize calculation errors.

Shortcut Trick

Compare the base fraction 7/13 with the options using cross-multiplication.

For 7/13 and 4/7: Multiply 7 × 7 and 13 × 4.

7 × 7 = 49 and 13 × 4 = 52.

Since 52 > 49, the fraction 4/7 is greater than 7/13.

∴ The correct answer is 4/7.

Alternate Method

Convert the given fraction and the options into decimal values to compare their magnitudes.

Given: 7/13

⇒ Base fraction: 7/13 ≈ 0.5384

⇒ Option (1): 8/15 ≈ 0.5333 (Smaller)

⇒ Option (2): 21/40 = 0.5250 (Smaller)

⇒ Option (3): 14/27 ≈ 0.5185 (Smaller)

⇒ Option (4): 4/7 ≈ 0.5714 (Greater)

⇒ Comparing the values: 0.5714 > 0.5384.

∴ The correct answer is 4/7.

Additional Information

Cross-Multiplication Rule

For two fractions a/b and c/d, if a × d > b × c, then a c/d. If a × d < b × c, then a/b < c/d.

Comparison of Like Numerators

If two fractions have the same numerator, the fraction with the smaller denominator is the larger value.

LCD Method

Fractions can be compared by converting them to a Common Denominator using the Least Common Multiple (LCM) of the denominators.

Shortcut Trick

Count the decimal places: 0.625 (3 places) and 0.00625 (5 places).

Shift the decimal 5 places to the right for both numbers to remove the decimal from the divisor.

0.625 → 62500 and 0.00625 → 625.

Perform the division: 62500 ÷ 625 = 100.

∴ The correct answer is 100.

Alternate Method

Given:

Expression: 0.625 ÷ 0.00625

Method: Convert decimals into fractions for simplification.

⇒ 0.625 = 625/1000

⇒ 0.00625 = 625/100000

⇒ (625/1000) ÷ (625/100000)

⇒ (625/1000) × (100000/625)

⇒ (625 ÷ 625) × (100000 ÷ 1000)

⇒ 1 × 100

⇒ 100

∴ The correct answer is 100.

Additional Information

Scientific Notation

A number can be expressed as a × 10ⁿ. For 0.625, it is 6.25 × 10⁻¹, and for 0.00625, it is 6.25 × 10⁻³. Division results in subtraction of exponents.

Dividing by Powers of 10

To divide a decimal by 10, 100, or 1000, move the decimal point to the left by as many places as there are zeros in the divisor.

BODMAS Rule

In complex expressions, always follow the order: Brackets, Orders, Division/Multiplication, and Addition/Subtraction.

Shortcut Trick

Identify the expression as the Algebraic Identity: a³ − b³ − 3ab(a − b) = (a − b)³.

Here, a = 45 and b = 30, so (a − b) = 15.

The expression simplifies to 15³.

15³ = 15 × 15 × 15 = 3375.

∴ The correct answer is 3375.

Alternate Method

Given Expression: 45³ − 30³ − 3 × 45 × 30 × 15

Formula Used: (a − b)³ = a³ − b³ − 3ab(a − b)

⇒ Let a = 45 and b = 30.

⇒ Then (a − b) = 45 − 30 = 15.

⇒ Substituting these values into the identity:

⇒ 45³ − 30³ − 3 × 45 × 30 × (45 − 30) = (45 − 30)³

⇒ 45³ − 30³ − 3 × 45 × 30 × 15 = 15³

⇒ 15³ = 15 × 15 × 15

⇒ 225 × 15 = 3375

∴ The correct answer is 3375.

Additional Information

Cube of a Sum

The identity for the cube of a sum is (a + b)³ = a³ + b³ + 3ab(a + b).

Difference of Two Cubes

The factorized form is a³ − b³ = (a − b)(a² + ab + b²).

Sum of Two Cubes

The factorized form is a³ + b³ = (a + b)(a² − ab + b²).

Shortcut Trick

Second diagonal d₂ = (2 × Area) ÷ d₁ = (2 × 216) ÷ 18 = 24 cm.

In a rhombus, side a = √((d₁/2)² + (d₂/2)²).

Using Pythagorean triplet (9, 12, 15): Side = √(9² + 12²) = 15 cm.

∴ The correct answer is 15 cm.

Alternate Method

Given: Area of rhombus = 216 cm², Diagonal (d₁) = 18 cm.

Formula Used: Area = (1/2) × d₁ × d₂; Side (a) = √((d₁/2)² + (d₂/2)²)

Calculations:

⇒ 216 = (1/2) × 18 × d₂

⇒ 216 = 9 × d₂

⇒ d₂ = 24 cm

In a rhombus, diagonals bisect each other at right angles (90°).

⇒ Half of d₁ = 18/2 = 9 cm

⇒ Half of d₂ = 24/2 = 12 cm

Applying Pythagoras Theorem in the right-angled triangle:

⇒ Side² = 9² + 12²

⇒ Side² = 81 + 144 = 225

⇒ Side = √225 = 15 cm

∴ The correct answer is 15 cm.

Additional Information

Perimeter of a Rhombus

Since all sides of a rhombus are equal, the Perimeter = 4 × side.

Properties of Diagonals

The diagonals of a rhombus are perpendicular bisectors of each other and they also bisect the internal angles.

Area via Base and Height

The area can also be calculated as Base × Height, similar to a parallelogram.

Shortcut Trick

Product of diagonals (d₁ × d₂) = 2 × Area

d₂ = (2 × 120) ÷ 15 = 16 cm

Ratio of diagonals = 15 : 16

∴ The correct answer is 15 : 16.

Alternate Method

Given:

Area of the rhombus = 120 cm²

Length of one diagonal (d₁) = 15 cm

Formula Used:

Area of a Rhombus = ½ × d₁ × d₂

Calculations:

⇒ Area = ½ × d₁ × d₂

⇒ 120 = ½ × 15 × d₂

⇒ 120 × 2 = 15 × d₂

⇒ 240 = 15 × d₂

⇒ d₂ = 240 ÷ 15

⇒ d₂ = 16 cm

⇒ Ratio = d₁ : d₂ = 15 : 16

∴ The correct answer is 15 : 16.

Additional Information

Side and Diagonal Relationship

In a rhombus with side 'a' and diagonals d₁ and d₂, the relation is 4a² = d₁² + d₂².

Perimeter of Rhombus

The perimeter is calculated as 4 × side, as all four sides of a rhombus are equal in length.

Diagonal Properties

The diagonals of a rhombus always bisect each other at a 90-degree angle.

Shortcut Trick

The Least Common Multiple (LCM) of a set of numbers is the LCM of the LCMs of its subsets.

To find LCM(A, B, C, D), we calculate LCM(LCM(A, B), LCM(C, D)).

⇒ Required LCM = LCM(18, 24)

⇒ Multiples of 18: 18, 36, 54, 72... | Multiples of 24: 24, 48, 72...

∴ The correct answer is 72.

Alternate Method

Given:

⇒ LCM(A, B) = 18

⇒ LCM(C, D) = 24

Formula Used:

LCM by Prime Factorization: Product of the highest powers of all prime factors involved.

Calculations:

⇒ LCM(A, B, C, D) = LCM(18, 24)

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

⇒ Prime factors of 24 = 2 × 2 × 2 × 3 = 2³ × 3¹

⇒ LCM = 2^max(1,3) × 3^max(2,1)

⇒ LCM = 2³ × 3² = 8 × 9 = 72

∴ The correct answer is 72.

Additional Information

Associative Property of LCM

For any numbers a, b, c, and d, the LCM(a, b, c, d) is equal to LCM(LCM(a, b), LCM(c, d)).

Relation between HCF and LCM

For any two numbers x and y, HCF(x, y) × LCM(x, y) = x × y.

LCM of Fractions

LCM of fractions = LCM of Numerators ÷ HCF of Denominators.

Shortcut Trick

The Least Common Multiple (LCM) is always exactly divisible by the Highest Common Factor (HCF).

Check the options for divisibility by HCF (10):

135 ÷ 10 = 13.5 | 120 ÷ 10 = 12 | 166 ÷ 10 = 16.6 | 175 ÷ 10 = 17.5

Only 120 is a multiple of 10.

∴ The correct answer is 120.

Alternate Method

Given: HCF of two numbers = 10.

Property Used: For any two numbers, the LCM must be of the form (HCF × k), where k is a positive integer.

Let the two numbers be 10a and 10b, where a and b are co-prime numbers.

⇒ LCM = HCF × a × b

⇒ LCM = 10 × a × b

⇒ LCM ÷ 10 = a × b (must be an integer)

Checking the options:

⇒ 135 ÷ 10 = 13.5 (Not an integer)

⇒ 120 ÷ 10 = 12 (Integer)

⇒ 166 ÷ 10 = 16.6 (Not an integer)

⇒ 175 ÷ 10 = 17.5 (Not an integer)

∴ The correct answer is 120.

Additional Information

Fundamental Relation

For any two numbers, the product of the numbers is equal to the product of their HCF and LCM. (Product = HCF × LCM)

HCF of Co-prime Numbers

If two numbers are co-prime, their Highest Common Factor is always 1, and their LCM is simply their product.

Divisibility Rule

The HCF is the largest number that divides both given numbers exactly, while the LCM is the smallest number divisible by both.

Shortcut Trick

A loss of 30% means the Selling Price is (100 − 30) = 70% of the Cost Price.

SP = 70% of 5000

⇒ SP = 0.7 × 5000 = 3500

∴ The correct answer is ₹3500.

Alternate Method

Given:

Cost Price (CP) of the article = ₹5000

Loss Percentage = 30%

Formula Used:

Loss = (Loss% ÷ 100) × CP

Selling Price (SP) = CP − Loss

Calculations:

⇒ Loss = (30 ÷ 100) × 5000

⇒ Loss = 0.3 × 5000 = ₹1500

⇒ Selling Price (SP) = 5000 − 1500

⇒ Selling Price (SP) = ₹3500

∴ The correct answer is ₹3500.

Additional Information

Profit and Loss Basics

Profit occurs when SP > CP, and Loss occurs when CP > SP. Both are always calculated on the Cost Price unless stated otherwise.

Relationship Formula

SP = CP × [ (100 ± Profit/Loss %) ÷ 100 ]. Use plus for profit and minus for loss.

Marked Price and Discount

Selling Price is also calculated as Marked Price (MP) − Discount. Discount % is always calculated on the Marked Price.

Shortcut Trick

If 40% of the eggs are rotten, then the remaining 60% are NOT rotten.

Direct calculation: 60% of 500 = 0.60 × 500 = 300.

∴ The correct answer is 300 eggs.

Alternate Method

Given:

Total number of eggs = 500

Percentage of rotten eggs = 40%

Calculations:

⇒ Number of rotten eggs = 40% of 500

⇒ Number of rotten eggs = (40 ÷ 100) × 500

⇒ Number of rotten eggs = 40 × 5 = 200

⇒ Number of NOT rotten eggs = Total eggs − Rotten eggs

⇒ Number of NOT rotten eggs = 500 − 200 = 300

∴ The correct answer is 300 eggs.

Additional Information

Percentage Complement Rule

If a certain part of a whole is x%, the remaining part is always (100 − x)% of that whole.

Percentage to Fraction

To convert a percentage to a fraction, divide by 100. For example, 40% = 40/100 = 2/5 and 60% = 60/100 = 3/5.

Fraction of a Value

To find the value of a fraction of a total, multiply the fraction by the total: (3/5) × 500 = 300.

Shortcut Trick

Total maximum marks = 6 × 100 = 600

Total marks obtained = 50% of 600 = 300

Subject 1 marks = 30% of total marks obtained = 30% of 300 = 90

∴ The correct answer is 90.

Alternate Method

Given:

Number of subjects = 6

Marks per subject = 100

Total percentage scored = 50%

Subject 1 score percentage (relative to total marks obtained) = 30%

Formula used:

Marks Obtained = (Percentage ÷ 100) × Total Maximum Marks

Calculations:

⇒ Total Maximum Marks = 6 × 100 = 600

⇒ Total Marks Obtained = 50% of 600

⇒ Total Marks Obtained = (50 ÷ 100) × 600 = 300

⇒ Score in Subject 1 = 30% of 300

⇒ Score in Subject 1 = (30 ÷ 100) × 300

⇒ Score in Subject 1 = 30 × 3 = 90

∴ The correct answer is 90.

Additional Information

Percentage Calculation

To find the percentage of a quantity, use the formula: (Value ÷ Total Value) × 100. Conversely, to find a value from a percentage, use: (Percentage ÷ 100) × Total.

Successive Percentage Case

When taking a percentage of an already calculated percentage value (like 30% of the 50% obtained), the effective percentage of the original total is: (x × y) ÷ 100. Here: (30 × 50) ÷ 100 = 15% of 600.

Weighted Average

In problems involving multiple subjects with different maximum marks, the total percentage is calculated as (Sum of marks obtained ÷ Sum of total marks) × 100.

Shortcut Trick

Let Selling Price (SP) = 100.

Loss = 20% of SP = 20.

Cost Price (CP) = SP + Loss = 100 + 20 = 120.

Loss % on CP = (20 ÷ 120) × 100 = (1 ÷ 6) × 100 = 16.67%.

∴ The correct answer is 16.67 percent.

Alternate Method

Given: Loss calculated on Selling Price (SP) is 20%.

Formula Used: Loss % (on CP) = (Loss ÷ Cost Price) × 100; where Cost Price (CP) = Selling Price (SP) + Loss.

⇒ Let the Selling Price (SP) be ₹100.

⇒ According to the question, Loss = 20% of SP = (20 ÷ 100) × 100 = ₹20.

⇒ We know that Loss = Cost Price (CP) − Selling Price (SP).

⇒ CP = SP + Loss = 100 + 20 = ₹120.

⇒ Loss % on Cost Price = (Loss ÷ CP) × 100.

⇒ Loss % = (20 ÷ 120) × 100.

⇒ Loss % = (1 ÷ 6) × 100 = 16.666...

⇒ Loss % ≈ 16.67%.

∴ The correct answer is 16.67 percent.

Additional Information

Standard Loss Percentage

By default, Loss % is calculated as (CP − SP) ÷ CP × 100. It represents the loss relative to the initial investment.

Profit/Loss on Selling Price

When profit is calculated on SP, the CP is smaller (CP = SP − Profit). When loss is calculated on SP, the CP is larger (CP = SP + Loss).

Important Reciprocals

1/5 = 20% | 1/6 ≈ 16.67% | 1/7 ≈ 14.28% | 1/8 = 12.5%. These are useful for fast calculations in competitive exams.

Shortcut Trick

Assume Total Work (LCM of 8, 10) = 40 units.

Efficiency of (A+B+C) = 40 ÷ 8 = 5 units/day.

Efficiency of (A+B) = 40 ÷ 10 = 4 units/day.

Efficiency of C = 5 − 4 = 1 unit/day.

Required work = 3/8 × 40 = 15 units; Time = 15 ÷ 1 = 15 days.

∴ The correct answer is 15 days.

Alternate Method

Given: Time taken by A, B and C = 8 days; Time taken by A and B = 10 days.

⇒ Efficiency of C = Efficiency of (A + B + C) − Efficiency of (A + B)

⇒ Efficiency of C = 5 − 4 = 1 unit/day

⇒ Work to be completed = 3/8 × Total Work = 3/8 × 40 = 15 units

⇒ Time taken by C = 15 units ÷ 1 unit/day = 15 days

∴ The correct answer is 15 days.

Additional Information

Total Work and Efficiency

The core formula is Work = Efficiency × Time. Efficiency is the amount of work done in a single unit of time (e.g., one day).

Combined Efficiency

When people work together, their individual efficiencies are added to find the total group efficiency.

Partial Work

To find the time for a portion of work, multiply the total work by the required fraction and divide by the person's efficiency.

Shortcut Trick

Assume Total Work = LCM(16, 24, 12) = 48 units.

Sum of efficiencies of pairs = (M + R) + (R + P) + (P + M) = (48 ÷ 16) + (48 ÷ 24) + (48 ÷ 12) = 3 + 2 + 4 = 9 units/day.

Efficiency of (M + R + P) = 9 ÷ 2 = 4.5 units/day.

Efficiency of P = (M + R + P) − (M + R) = 4.5 − 3 = 1.5 units/day.

Time taken by P alone = 48 ÷ 1.5 = 32 days.

∴ The correct answer is 32 days.

Alternate Method

Given:

Time taken by (M + R) = 16 days

Time taken by (R + P) = 24 days

Time taken by (P + M) = 12 days

Formula Used:

Total Work = Time × Efficiency

Calculations:

⇒ Total efficiency of 2 × (M + R + P) = 3 + 2 + 4 = 9 units/day

⇒ Efficiency of (M + R + P) = 9 ÷ 2 = 4.5 units/day

⇒ Efficiency of P alone = Efficiency of (M + R + P) − Efficiency of (M + R)

⇒ Efficiency of P = 4.5 − 3 = 1.5 units/day

⇒ Time taken by P alone = 48 ÷ 1.5

⇒ Time = 32 days

∴ The correct answer is 32 days.

Additional Information

Combined Work Principle

When individuals work in pairs (A+B, B+C, C+A), their combined efficiency is 2 × (A+B+C). Individual efficiency is then found by subtracting the required pair from the total.

Work and Wages

Wages are always distributed in the ratio of the Work Done. If all persons work for the same duration, the wage ratio is the same as the Efficiency Ratio.

Pipes and Cisterns

The same logic applies where efficiency of an inlet pipe is positive and an outlet/leak pipe is negative. Total capacity is equivalent to Total Work.

Shortcut Trick

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

Speed in km/hr = 75 × (18 ÷ 5)

Speed = 15 × 18 = 270 km/hr.

∴ The correct answer is 270 km/hr.

Alternate Method

Given:

Speed = 75 m/s

Formula Used:

Speed (km/hr) = Speed (m/s) × (3600 ÷ 1000)

Calculations:

⇒ Speed in km/hr = 75 × (18 ÷ 5)

⇒ Speed = (75 ÷ 5) × 18

⇒ Speed = 15 × 18

⇒ Speed = 270 km/hr

∴ The correct answer is 270 km/hr.

Additional Information

km/hr to m/s Conversion

To convert speed from km/hr to m/s, multiply the value by 5/18 (e.g., 90 km/hr = 90 × 5/18 = 25 m/s).

Average Speed Formula

When a certain distance is covered at speed u and the same distance is covered at speed v, the Average Speed = (2 × u × v) ÷ (u + v).

Relative Speed

If two bodies move in the same direction, relative speed = u − v. In opposite directions, relative speed = u + v.

Shortcut Trick

Relative speed in opposite direction = Distance ÷ Time = 500 ÷ 5 = 100 m/s.

Ratio of speeds (Train : Man) = 9 :

1. Total ratio parts = 9 + 1 = 10 units.

10 units = 100 m/s ⇒ 1 unit = 10 m/s.

Speed of train = 9 units = 9 × 10 = 90 m/s.

Speed in km/hr = 90 × (18 ÷ 5) = 18 × 18 = 324 km/hr.

∴ The correct answer is 324 km/hr.

Alternate Method

Given:

Length of train (L) = 500 m

Time taken (t) = 5 sec

Let speed of man = x m/s, then speed of train = 9x m/s.

Formula Used:

Relative Speed (Opposite Direction) = (L₁ + L₂) ÷ Time

Calculations:

⇒ Relative Speed = Speed of train + Speed of man

⇒ 9x + x = 500 ÷ 5

⇒ 10x = 100

⇒ x = 10 m/s

⇒ Speed of train = 9 × 10 = 90 m/s

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

⇒ Speed of train = 90 × (18 ÷ 5) = 18 × 18

⇒ Speed of train = 324 km/hr

∴ The correct answer is 324 km/hr.

Additional Information

Relative Speed in Same Direction

When two objects move in the same direction with speeds S₁ and S₂, the relative speed is |S₁ − S₂|.

Relative Speed in Opposite Direction

When two objects move towards each other, the relative speed is the sum of their individual speeds (S₁ + S₂).

Unit Conversion Shortcut

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

Crossing a Point Object

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

Shortcut Trick

Assume Marked Price (X) = 100.

Purchase Price (Y) = 100 − 30% = 70.

Selling Price (Z) = 100 + 25% = 125.

Difference (Z − Y) = 125 − 70 = 55.

Required Percentage = (55 ÷ 70) × 100 = 550 ÷ 7 = 78.57%.

∴ The correct answer is 78.57 percent more.

Alternate Method

Given:

Marked Price of the article = Rs. X

Discount = 30% of X

Purchase Price (Y) = X − 30% of X

Selling Price (Z) = X + 25% of X

Let the Marked Price (X) be 100 units.

⇒ Purchase Price (Y) = 100 − (30% of 100)

⇒ Y = 100 − 30 = 70 units.

⇒ Selling Price (Z) = 100 + (25% of 100)

⇒ Z = 100 + 25 = 125 units.

To find how much percent more Z is than Y:

⇒ Difference = Z − Y = 125 − 70 = 55 units.

⇒ Percentage Increase = (Difference ÷ Base Value Y) × 100

⇒ Percentage = (55 ÷ 70) × 100

⇒ Percentage = (550 ÷ 7) % ≈ 78.57%.

∴ The correct answer is 78.57 percent more.

Additional Information

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). Selling Price (SP) = MP − Discount.

Percentage Change Formula

Percentage change = [(Final Value − Initial Value) ÷ Initial Value] × 100. Use positive for 'more' and negative for 'less'.

Profit Percentage

Profit % = (SP − CP) ÷ CP × 100, where CP (Cost Price) is the price at which the person bought the article.

Shortcut Trick

For two successive discounts a% and b%, the Net Discount % = a + b − (a × b)/100.

Net Discount % = 25 + 20 − (25 × 20)/100

Net Discount % = 45 − 500/100 = 45 − 5 = 40%

∴ The correct answer is 40%.

Alternate Method

Given:

First discount (d₁) = 25%

Second discount (d₂) = 20%

Formula Used:

Equivalent Discount = 100 − [100 × (100 − d₁)/100 × (100 − d₂)/100]

Calculations:

⇒ Let the Marked Price (MP) of the article be 100.

⇒ Price after 1^st discount = 100 − (25% of 100) = 100 − 25 = 75

⇒ Price after 2^nd discount = 75 − (20% of 75)

⇒ Price after 2^nd discount = 75 − (20/100 × 75) = 75 − 15 = 60

⇒ Total Net Discount = MP − Final Selling Price = 100 − 60 = 40

⇒ Net Discount % = (40 / 100) × 100 = 40%

∴ The correct answer is 40%.

Additional Information

Three Successive Discounts

For three discounts d₁, d₂, and d₃, find the equivalent of the first two using the formula, then apply the formula again with the third discount.

Relationship between MP and CP

MP/CP = (100 + Profit%)/(100 − Discount%). This is useful for finding the markup percentage.

Discount Calculation Rule

Discount is always calculated on the Marked Price (MP), while Profit or Loss is always calculated on the Cost Price (CP) unless stated otherwise.

Shortcut Trick

Formula for Net Discount for two successive discounts a and b = a + b − (a × b / 100)

Given: a = 50 and b = 30

Net Discount = 50 + 30 − (50 × 30 / 100)

Net Discount = 80 − 15 = 65%

∴ The correct answer is 65%.

Alternate Method

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

⇒ First discount = 50% of 100 = 50

⇒ Price after first discount = 100 − 50 = 50

⇒ Second discount = 30% of 50 = (30/100) × 50 = 15

⇒ Final Selling Price (SP) = 50 − 15 = 35

⇒ Total Discount = 100 − 35 = 65

⇒ Net Discount Percentage = (65 / 100) × 100 = 65%

∴ The correct answer is 65%.

Additional Information

Relationship between MP and SP

Selling Price (SP) = Marked Price (MP) × (100 − Discount%) / 100. Discount is always calculated on the Marked Price.

Three Successive Discounts

If three discounts a%, b%, and c% are given, the equivalent discount is calculated by finding the resultant of a and b first, and then using that result with c.

Successive Percentage Change

The formula x + y + (xy/100) is used for percentage increases (positive values) and decreases (negative values like discounts).