Quantitative Aptitude PYQs

Shortcut Trick

Solve the bracketed terms first and then group positive and negative values.

Expression → 82 + 66 − 70 + 73 − 54 + 36

Sum of positive values → 82 + 66 + 73 + 36 = 257

Sum of negative values → 70 + 54 = 124

Final Result → 257 − 124 = 133

∴ The correct answer is 133.

Alternate Method

Given Expression: 82 + 66 − (41 + 29) + 73 − 54 + 36

Method Used: Apply the BODMAS rule (Bracket, Of, Division, Multiplication, Addition, Subtraction).

⇒ 82 + 66 − (70) + 73 − 54 + 36

⇒ 148 − 70 + 73 − 54 + 36

⇒ 78 + 73 − 54 + 36

⇒ 151 − 54 + 36

⇒ 97 + 36

⇒ 133

∴ The correct answer is 133.

Additional Information

BODMAS Rule

It defines the order of operations: Brackets first, then Of (powers/roots), then Division and Multiplication, and finally Addition and Subtraction.

Grouping Method

In multi-step addition and subtraction, it is often faster and more accurate to sum all positive terms and negative terms separately before the final subtraction.

Sign Convention

When a negative sign precedes a bracket, the signs of all terms inside the bracket are reversed when the bracket is opened: −(a + b) = −a − b.

Shortcut Trick

Solve using BODMAS: 93 − 48 + 44 + 37 − 19 + 16

Sum of positive numbers: 93 + 44 + 37 + 16 = 190

Sum of negative numbers: 48 + 19 = 67

Final difference: 190 − 67 = 123

∴ The correct answer is 123.

Alternate Method

Given Expression: 93 − 48 + 22 × 2 + (37 ÷ 1) − 19 + 16

Method: Apply BODMAS rule sequentially.

⇒ 93 − 48 + 22 × 2 + 37 − 19 + 16 (Solving Bracket and Division: 37 ÷ 1 = 37)

⇒ 93 − 48 + 44 + 37 − 19 + 16 (Solving Multiplication: 22 × 2 = 44)

⇒ (93 + 44 + 37 + 16) − (48 + 19) (Grouping Addition and Subtraction terms)

⇒ 190 − 67 = 123

∴ The correct answer is 123.

Additional Information

VBODMAS Rule

It represents the order: Vinculum, Bracket, Of, Division, Multiplication, Addition, and Subtraction. It is essential for solving complex arithmetic expressions.

Grouping Method

To minimize errors in long Addition/Subtraction strings, group all positive terms and negative terms separately before the final subtraction.

Shortcut Trick

For proper fractions where the difference between numerator and denominator is constant (Gap = 1 in all cases), the fraction with smaller values is smaller.

Given reference: 5/6 (Gap = 1)

Options: 10/11 (Gap = 1), 4/5 (Gap = 1), 9/10 (Gap = 1), 7/8 (Gap = 1)

Ordering: 4/5 < 5/6 < 7/8 < 9/10 < 10/11

∴ The correct answer is 4/5.

Alternate Method

Given: Reference fraction = 5/6. Options to check for smaller value: 10/11, 4/5, 9/10, 7/8.

Formula Used: Cross-multiplication method: For two fractions a/b and c/d, if (a × d) > (b × c), then a/b > c/d.

Step 1: Compare 5/6 with 10/11

⇒ 5 × 11 = 55

⇒ 6 × 10 = 60

⇒ 55 < 60 ∴ 5/6 < 10/11 (10/11 is larger)

Step 2: Compare 5/6 with 4/5

⇒ 5 × 5 = 25

⇒ 6 × 4 = 24

⇒ 25 > 24 ∴ 5/6 > 4/5 (4/5 is smaller)

Step 3: Compare 5/6 with 9/10

⇒ 5 × 10 = 50

⇒ 6 × 9 = 54

⇒ 50 < 54 ∴ 5/6 < 9/10 (9/10 is larger)

Step 4: Compare 5/6 with 7/8

⇒ 5 × 8 = 40

⇒ 6 × 7 = 42

⇒ 40 < 42 ∴ 5/6 < 7/8 (7/8 is larger)

∴ The correct answer is 4/5.

Additional Information

Cross Multiplication Rule

To compare a/b and c/d, calculate the products ad and bc. If ad > bc, then a/b > c/d. If ad < bc, then a/b < c/d.

Proper Fraction Gap Method

In proper fractions (Numerator < Denominator), if the difference (D − N) is the same for all, the fraction with the largest values is the largest and the one with the smallest values is the smallest.

Common Denominator Method

Convert all fractions to a common denominator by taking the LCM of all denominators, then compare the resulting numerators.

Shortcut Trick

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

Identify the values: a = 15.3 and b = 5.3.

Required result = (15.3 − 5.3)²

Required result = (10)² = 100

∴ The correct answer is 100.

Alternate Method

Given:

Expression: (15.3 + 5.3)² − 4 × 15.3 × 5.3

Formula Used:

Standard expansion: (a + b)² = a² + 2ab + b² or direct calculation.

Calculations:

⇒ (20.6)² − 4 × 15.3 × 5.3

⇒ 424.36 − 4 × 81.09

⇒ 424.36 − 324.36

⇒ 100

∴ The correct answer is 100.

Additional Information

Binomial Squared Identities

(a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². These are the foundations for complex simplification.

Relationship between Sum and Difference

(a + b)² − (a − b)² = 4ab. This identity is extremely useful in Quantitative Aptitude to find the product when sum and difference are given.

Sum of Squares Formula

(a + b)² + (a − b)² = 2(a² + b²). This relates the squares of the sum and difference to the sum of the individual squares.

Shortcut Trick

The perimeter of a square is the total boundary length, which is 4 times its side.

Given side = 8 cm.

Perimeter = 4 × 8 = 32 cm.

∴ The correct answer is 32 cm.

Alternate Method

Given:

Side of the square (s) = 8 cm

Formula Used:

Perimeter of Square = Sum of all sides = 4 × side

Calculations:

⇒ Perimeter = 4 × s

⇒ Perimeter = 4 × 8 cm

⇒ Perimeter = 32 cm

∴ The correct answer is 32 cm.

Additional Information

Area of a Square

The total space occupied by the square is given by the formula: Area = side².

Diagonal of a Square

The length of the line connecting opposite vertices is: Diagonal = side × √2.

Properties of a Square

A square is a regular quadrilateral where all four sides are equal and all four angles are 90°.

Shortcut Trick

Area of rhombus = (d₁ × d₂) / 2

84 = (14 × d₂) / 2

84 = 7 × d₂ ⇒ d₂ = 12 cm

Required ratio (d₁ : d₂) = 14 : 12 = 7 : 6

∴ The correct answer is 7 : 6.

Alternate Method

Given:

Area of the rhombus = 84 cm²

One diagonal (d₁) = 14 cm

Formula Used:

Area of a rhombus = (1/2) × (Product of its diagonals)

Calculations:

⇒ 84 = (1/2) × 14 × d₂

⇒ 84 = 7 × d₂

⇒ d₂ = 84 / 7

⇒ d₂ = 12 cm

⇒ Ratio of diagonals = d₁ : d₂

⇒ Ratio = 14 : 12

⇒ Ratio = 7 : 6

∴ The correct answer is 7 : 6.

Additional Information

Rhombus Properties

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

Side-Diagonal Relationship

The relationship between side 'a' and diagonals d₁, d₂ is given by 4a² = d₁² + d₂².

Perimeter of Rhombus

The perimeter of a rhombus is 4 × side length (a).

Shortcut Trick

Side of Rhombus (a) = 60 ÷ 4 = 15 cm.

Semi-diagonals (d₁/2, d₂/2) and side (a) form a Pythagorean Triplet.

Semi-diagonal 1 = 18 ÷ 2 = 9 cm.

Using triplet (9, 12, 15): Semi-diagonal 2 = 12 cm.

Second Diagonal (d₂) = 12 × 2 = 24 cm.

Sum = 18 + 24 = 42 cm.

∴ The correct answer is 42 cm.

Alternate Method

Given:

Perimeter (P) = 60 cm

First Diagonal (d₁) = 18 cm

Formula Used:

Side (a) = Perimeter ÷ 4

Relation: 4a² = d₁² + d₂²

Calculations:

⇒ Side of rhombus (a) = 60 ÷ 4 = 15 cm

⇒ Let the second diagonal be d₂.

⇒ 4 × (15)² = (18)² + d₂²

⇒ 4 × 225 = 324 + d₂²

⇒ 900 = 324 + d₂²

⇒ d₂² = 900 − 324

⇒ d₂² = 576

⇒ d₂ = √576 = 24 cm

⇒ Sum of diagonals = d₁ + d₂ = 18 + 24 = 42 cm

∴ The correct answer is 42 cm.

Additional Information

Area of Rhombus

The area can be calculated using the diagonals: Area = 1/2 × d₁ × d₂.

Diagonals of a Rhombus

The diagonals of a rhombus bisect each other at Right Angles (90°).

Side-Diagonal Relationship

The relationship between side 'a' and diagonals 'd₁, d₂' is given by 4a² = d₁² + d₂².

Shortcut Trick

Net change in the sum of ages = Age of joining boy − Sum of ages of leaving boys

Net change = 32 − (18 + 19) = 32 − 37 = −5 years

New group size = 16 − 2 + 1 = 15 boys

New Average = (Initial Total Sum + Net Change) ÷ New Count = (16 × 20 − 5) ÷ 15 = 21 years

∴ The correct answer is 21 years.

Alternate Method

Given:

Initial number of boys = 16

Initial average age = 20 years

Ages of boys who left = 18 years and 19 years

Age of boy who joined = 32 years

Formula Used:

Average = Total Sum of Observations ÷ Total Number of Observations

Calculations:

⇒ Initial Total Age of 16 boys = 16 × 20 = 320 years

⇒ Total age after two boys left = 320 − (18 + 19) = 320 − 37 = 283 years

⇒ Total age after one boy joined = 283 + 32 = 315 years

⇒ New total number of boys = 16 − 2 + 1 = 15 boys

⇒ New average age = 315 ÷ 15 = 21 years

∴ The correct answer is 21 years.

Additional Information

Combined Average

If two groups have n₁ and n₂ members with averages A₁ and A₂, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Effect of Change on Average

When a new value is added, the new average increases if the value is greater than the old average and decreases if it is smaller.

Net Deviation Rule

The sum of deviations of all observations from their mean (average) is always zero.

Shortcut Trick

Property: Product of HCF and LCM of two numbers is equal to the Product of the numbers (A × B).

Required Product = (2⁵ × 3²) × (2⁸ × 3⁴)

Using law of exponents a^m × aⁿ = a^m+n, we add the powers of the same bases.

Product = 2⁽⁵⁺⁸⁾ × 3⁽²⁺⁴⁾ = 2¹³ × 3⁶

∴ The correct answer is 2¹³ × 3⁶.

Alternate Method

Given:

Number A = 2⁵ × 3²

Number B = 2⁸ × 3⁴

Formula Used:

HCF × LCM = Product of the numbers

Calculations:

⇒ HCF of (A, B) = 2^min(5,8) × 3^min(2,4) = 2⁵ × 3²

⇒ LCM of (A, B) = 2^max(5,8) × 3^max(2,4) = 2⁸ × 3⁴

⇒ Product = (2⁵ × 3²) × (2⁸ × 3⁴)

⇒ Product = 2⁽⁵⁺⁸⁾ × 3⁽²⁺⁴⁾

⇒ Product = 2¹³ × 3⁶

∴ The correct answer is 2¹³ × 3⁶.

Additional Information

HCF of Prime Factors

To find the Highest Common Factor (HCF), take the smallest power of each common prime factor present in the numbers.

LCM of Prime Factors

To find the Least Common Multiple (LCM), take the highest power of every prime factor present in any of the numbers.

Relationship for Two Numbers

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. This property is only applicable to two numbers.

Shortcut Trick

The prime numbers strictly between 5 and 12 are 7 and 11.

For any set of prime numbers, the Least Common Multiple (LCM) is always their product because they have no common factors other than 1.

LCM = 7 × 11 = 77.

∴ The correct answer is 77.

Alternate Method

Given: The range of numbers is between 5 and 12.

Formula Used: LCM (a, b) = Product of numbers if they are coprime.

⇒ List the integers between 5 and 12: 6, 7, 8, 9, 10, 11.

⇒ Determine which of these are prime numbers: 7 and 11.

⇒ 6 is composite (2 × 3), 8 is composite (2³), 9 is composite (3²), and 10 is composite (2 × 5).

⇒ 7 and 11 are prime as they have no divisors other than 1 and themselves.

⇒ To find the LCM of 7 and 11, list their multiples:

⇒ Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, ...

⇒ Multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, ...

⇒ The Least Common Multiple is 77.

∴ The correct answer is 77.

Additional Information

Prime Numbers and LCM

For any set of distinct prime numbers, their HCF is always 1 and their LCM is always the product of those numbers.

The Product Rule

For any two numbers a and b: LCM(a, b) × HCF(a, b) = a × b. Since HCF of two different primes is 1, LCM = a × b.

Prime Numbers (1 to 20)

The prime numbers up to 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Note that 1 is neither prime nor composite.

Shortcut Trick

Determine the ratio of B to A directly by cancelling common terms.

B : A = 6X : 8X = 6 : 8 = 3 : 4

Required percentage = (3 ÷ 4) × 100

Since 3/4 is equivalent to 75% in standard fraction-to-percentage conversion.

∴ The correct answer is 75%.

Alternate Method

Given:

Value of A = 8X

Value of B = 6X

Formula Used:

Required Percentage = (Value to be compared ÷ Base Value) × 100

⇒ Required Percentage = (B ÷ A) × 100

⇒ Required Percentage = (6X ÷ 8X) × 100

⇒ Required Percentage = (6 ÷ 8) × 100

⇒ Required Percentage = (3 ÷ 4) × 100

⇒ Required Percentage = 0.75 × 100

⇒ Required Percentage = 75%

∴ The correct answer is 75%.

Additional Information

Fraction to Percentage Conversion

Common fractions like 1/4 = 25%, 1/2 = 50%, 3/4 = 75%, and 1/5 = 20% are essential for quick calculations in competitive exams.

Base of Comparison

In the phrase 'X is what percent of Y', the value following 'of' (which is Y) is always taken as the denominator (base).

Percentage Change Formula

Percentage Increase or Decrease = (Difference ÷ Original Value) × 100. In this case, B is 25% less than A.

Shortcut Trick

Ratio of Cost Price (CP) to Selling Price (SP) = 8000 : 6000 = 4 : 3.

Loss in units = 4 − 3 = 1 unit.

Loss % = (Loss ÷ CP) × 100 = (1 ÷ 4) × 100 = 25%.

∴ The correct answer is 25%.

Alternate Method

Given:

Cost Price (CP) of the article = ₹8000

Selling Price (SP) of the article = ₹6000

Formula Used:

Loss = CP − SP

Loss % = (Loss ÷ CP) × 100

Calculations:

⇒ Loss = 8000 − 6000

⇒ Loss = ₹2000

⇒ Loss % = (2000 ÷ 8000) × 100

⇒ Loss % = (1 ÷ 4) × 100

⇒ Loss % = 25%

∴ The correct answer is 25%.

Additional Information

Profit Percentage

When SP is greater than CP, Profit % = [(SP − CP) ÷ CP] × 100.

Selling Price Calculation

If loss percentage is given, SP = CP × (100 − Loss%) ÷ 100.

Cost Price Calculation

If profit percentage is given, CP = (SP × 100) ÷ (100 + Profit%).

Shortcut Trick

The ratio of Cost Price (CP) to Selling Price (SP) is the inverse of the given percentage values.

CP : SP = 10 : 25 = 2 : 5

Profit = 5 − 2 = 3 units

Profit Percentage = (Profit ÷ CP) × 100 = (3 ÷ 2) × 100 = 150%

∴ The correct answer is 150%.

Alternate Method

Given:

25% of Cost Price (CP) = 10% of Selling Price (SP)

Formula Used:

Profit % = [(SP − CP) ÷ CP] × 100

Calculations:

⇒ (25 ÷ 100) × CP = (10 ÷ 100) × SP

⇒ 25 × CP = 10 × SP

⇒ CP ÷ SP = 10 ÷ 25

⇒ CP ÷ SP = 2 ÷ 5

⇒ Let Cost Price = 2 units and Selling Price = 5 units

⇒ Profit = SP − CP = 5 − 2 = 3 units

⇒ Profit % = (Profit ÷ CP) × 100

⇒ Profit % = (3 ÷ 2) × 100

⇒ Profit % = 1.5 × 100 = 150%

∴ The correct answer is 150%.

Additional Information

Basic Profit and Loss

Profit occurs when SP > CP. Profit = SP − CP. Loss occurs when CP > SP. Loss = CP − SP.

Selling Price Formula

When profit percentage is known: SP = CP × (100 + Profit %) ÷ 100.

Marked Price and Discount

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

Shortcut Trick

Find the target value: 180% of 120 = 1.8 × 120 = 216

The number to be subtracted from 420 to get 216 is the difference.

Required Number = 420 − 216 = 204

∴ The correct answer is 204.

Alternate Method

Given:

Initial number = 420

Target value = 180% of 120

Calculations:

⇒ Let the number to be subtracted be x.

⇒ According to the question:

⇒ 420 − x = 180% × 120

⇒ 420 − x = (180 ÷ 100) × 120

⇒ 420 − x = 1.8 × 120

⇒ 420 − x = 216

⇒ x = 420 − 216

⇒ x = 204

∴ The correct answer is 204.

Additional Information

Percentage Calculation

To find x% of y, use the formula: (x × y) ÷ 100. This represents a portion of the whole value.

Converting Percentage to Decimal

Divide the percentage by 100 to get the decimal equivalent. For example, 180% = 1.80.

Word Problem Translation

Terms like of usually imply multiplication, while subtracted from indicates the order: Base − Value.

Shortcut Trick

Remaining fraction after first day = 1 − 4/7 = 3/7

Remaining fraction after second day = 3/7 × (1 − 2/5) = 3/7 × 3/5 = 9/35

According to the question, 9/35 units = 18

Total work (1 unit) = 18 × 35 ÷ 9 = 70 units

∴ The correct answer is 70 units.

Alternate Method

Given:

Work done on the first day = 4/7 of total work

Work done on the second day = 2/5 of remaining work

Final remaining work = 18 units

⇒ Let the total work be W units.

⇒ Work completed on the first day = 4/7 × W

⇒ Remaining work after the first day = W − 4W/7 = 3W/7

⇒ Work completed on the second day = 2/5 × (3W/7) = 6W/35

⇒ Remaining work after the second day = 3W/7 − 6W/35

⇒ Taking LCM of 7 and 35 (which is 35): (15W − 6W) ÷ 35 = 9W/35

⇒ Since the remaining work is given as 18 units:

⇒ 9W/35 = 18

⇒ W = (18 × 35) ÷ 9

⇒ W = 2 × 35 = 70 units

∴ The correct answer is 70 units.

Additional Information

Fractional Remainder Concept

When a part f₁ is removed from a whole, the remainder is (1 − f₁). If another part f₂ is removed from this remainder, the new remainder is (1 − f₁) × (1 − f₂).

Successive Reduction Formula

This approach is equivalent to the successive percentage decrease formula often used in population depreciation: Final Value = Initial Value × (1 − r₁/100) × (1 − r₂/100).

LCM Method for Total Work

To avoid fractions, assume the total work as the LCM of the denominators (7 and 5), which is 35 units, and solve for parts.

Shortcut Trick

Ratio of work done (A : B) = (2 × 5) : (6 × 2) = 10 : 12 = 5 : 6

Total wage ratio parts = 5 + 6 = 11 parts

Value of 11 parts = ₹2750

Difference between shares (A and B) = 6 − 5 = 1 part

Value of 1 part = 2750 ÷ 11 = 250

∴ The correct answer is ₹250.

Alternate Method

Given:

Work duration of A = 2 days | Efficiency of A = 5 units/day

Work duration of B = 6 days | Efficiency of B = 2 units/day

Total wages received = ₹2750

Formula Used:

Total Work = Time × Efficiency

Share of Wage ∝ Total Work Done

Calculations:

⇒ Total work done by A = 2 × 5 = 10 units

⇒ Total work done by B = 6 × 2 = 12 units

⇒ Total work done by both = 10 + 12 = 22 units

⇒ Share of A = (10 ÷ 22) × 2750 = 1250

⇒ Share of B = (12 ÷ 22) × 2750 = 1500

⇒ Difference in shares = 1500 − 1250 = 250

∴ The correct answer is ₹250.

Additional Information

Work and Wages Principle

Wages are always distributed in proportion to the actual work done by each person, not just their individual efficiencies.

Efficiency Formula

Efficiency = Total Work ÷ Time Taken. If work is constant, Efficiency is inversely proportional to Time.

Ratio Distribution

To find an individual share: Individual share = (Individual Ratio Part ÷ Sum of Ratio Parts) × Total Amount.

Shortcut Trick

When two objects move in opposite directions, their relative speed is the sum of their individual speeds.

Since speeds are equal (v), Relative Speed = v + v = 2v.

Total distance to cross each other = Sum of lengths = 238 + 238 = 476 m.

Relative Speed = Total Distance ÷ Time ⇒ 2v = 476 ÷ 14 = 34 m/s.

Speed of each train (v) = 34 ÷ 2 = 17 m/s.

∴ The correct answer is 17 m/s.

Alternate Method

Given:

Length of first train (L₁) = 238 m

Length of second train (L₂) = 238 m

Time taken (t) = 14 seconds

Let the speed of each train be v m/s.

Formula Used:

Relative Speed (Opposite Direction) = Total Distance ÷ Total Time

v₁ + v₂ = (L₁ + L₂) ÷ t

Calculations:

⇒ Total Distance = 238 + 238 = 476 m

⇒ Relative Speed = v + v = 2v

⇒ 2v = 476 ÷ 14

⇒ 2v = 34

⇒ v = 34 ÷ 2

⇒ v = 17 m/s

∴ The correct answer is 17 m/s.

Additional Information

Relative Speed Concepts

When moving in the same direction, relative speed = |v₁ − v₂|. In opposite directions, relative speed = v₁ + 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.

Crossing Objects

When a train crosses a stationary object of length L_p (like a platform), the total distance covered is Length of Train + L_p.

Shortcut Trick

Assume total distance as LCM of speeds (40, 50) × denominator (3) = 200 × 3 = 600 km.

1/3^rd distance (200 km) at 40 km/hr → Time₁ = 200 ÷ 40 = 5 hrs.

Remaining distance (400 km) at 50 km/hr → Time₂ = 400 ÷ 50 = 8 hrs.

Average Speed = Total Distance ÷ Total Time = 600 ÷ (5 + 8) = 600/13 km/hr.

∴ The correct answer is 600/13 km/hr.

Alternate Method

Given: Part 1 speed (v₁) = 40 km/hr, Part 2 speed (v₂) = 50 km/hr, Part 1 distance = 1/3 of total, Part 2 distance = 2/3 of total.

Formula: Average Speed = Total Distance ÷ Total Time

⇒ Let the total distance be d km.

⇒ Time₁ = (d/3) ÷ 40 = d ÷ 120 hrs

⇒ Time₂ = (2d/3) ÷ 50 = 2d ÷ 150 = d ÷ 75 hrs

⇒ Total Time = (d ÷ 120) + (d ÷ 75)

⇒ LCM of 120 and 75 is 600. Total Time = (5d + 8d) ÷ 600 = 13d ÷ 600 hrs

⇒ Average Speed = d ÷ (13d/600) = (d × 600) ÷ 13d = 600/13 km/hr

∴ The correct answer is 600/13 km/hr.

Additional Information

Average Speed for Equal Distances

When two equal distances are covered at speeds x and y, the Average Speed is given by 2xy ÷ (x + y).

Harmonic Mean of Speeds

Average Speed is always the harmonic mean of the individual speeds when the distances covered are equal.

Weighted Average Speed Formula

For different distances d₁, d₂, ... and speeds v₁, v₂, ..., the Average Speed = (d₁ + d₂ + ...) ÷ (d₁/v₁ + d₂/v₂ + ...).

Shortcut Trick

Ratio of Cost Price (CP) to Marked Price (MP) = 3 : 8

Ratio of Marked Price (MP) to Selling Price (SP) = 8 : 5

Combined ratio CP : MP : SP = 3 : 8 : 5

Profit = SP − CP = 5 − 3 = 2 units

Profit % = (Profit ÷ CP) × 100 = (2 ÷ 3) × 100 = 200/3 %

∴ The correct answer is (200/3)%.

Alternate Method

Given:

Marked Price (MP) = (8/3) × Cost Price (CP)

Selling Price (SP) = (5/8) × Marked Price (MP)

Formula Used:

Profit % = [(Selling Price − Cost Price) ÷ Cost Price] × 100

Calculations:

⇒ Let Cost Price (CP) = x

⇒ Marked Price (MP) = (8/3) × x = 8x/3

⇒ Selling Price (SP) = (5/8) × (8x/3)

⇒ SP = 5x/3

⇒ Profit = SP − CP = (5x/3) − x = 2x/3

⇒ Profit % = [(2x/3) ÷ x] × 100

⇒ Profit % = (2/3) × 100 = 200/3 %

∴ The correct answer is (200/3)%.

Additional Information

Marked Price (MP)

The price printed on an article or the label price is called the Marked Price. Discount is always calculated on the MP.

Relation between CP and MP

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

Selling Price (SP)

Selling Price is the final price after applying a discount to the Marked Price (SP = MP − Discount).

Shortcut Trick

Use the CP-MP Ratio formula: CP / MP = (100 − Discount%) / (100 + Profit%)

CP / 6000 = (100 − 30) / (100 + 50) = 70 / 150 = 7 / 15

CP = (7 / 15) × 6000 = 7 × 400 = ₹2800

∴ The correct answer is ₹2800.

Alternate Method

Given: Marked Price (MP) = ₹6000, Discount = 30%, Profit = 50%

Formula: Selling Price (SP) = MP × (100 − D%)/100 | Cost Price (CP) = SP × 100/(100 + P%)

⇒ Selling Price (SP) = 6000 − 30% of 6000

⇒ SP = 6000 − (30/100) × 6000 = 6000 − 1800 = ₹4200

⇒ Profit is 50%, so SP = 150% of CP

⇒ 4200 = CP × (150/100)

⇒ CP = 4200 × (2/3)

⇒ CP = 1400 × 2 = ₹2800

∴ The correct answer is ₹2800.

Additional Information

Relationship between CP and MP

The ratio CP/MP = (100 − D%)/(100 + P%) is the most efficient way to solve problems involving both discount and profit.

Successive Discounts

Equivalent discount for two successive discounts a% and b% = (a + b − ab/100)%.

Mark-up Percentage

Mark-up % = [(Marked Price − Cost Price) / Cost Price] × 100.