Quantitative Aptitude PYQs

Shortcut Trick

Solve brackets and multiplication first, then group positive and negative terms.

132 − 44 + 42 − 18

⇒ (132 + 42) − (44 + 18)

⇒ 174 − 62 = 112

∴ The correct answer is 112.

Alternate Method

Given expression: 66 × 2 − 44 + (23 + 19) − 18

Formula Used: BODMAS Rule (Bracket, Of, Division, Multiplication, Addition, Subtraction)

⇒ 66 × 2 − 44 + (42) − 18 (Solving the Bracket)

⇒ 132 − 44 + 42 − 18 (Performing Multiplication)

⇒ 132 + 42 − 44 − 18 (Rearranging Addition terms)

⇒ 174 − 62 (Performing Addition for both groups)

⇒ 112 (Final Subtraction)

∴ The correct answer is 112.

Additional Information

BODMAS Rule

This rule dictates the hierarchy of operations: Brackets first, then Orders (powers/roots), followed by Division and Multiplication, and finally Addition and Subtraction.

Digital Sum Verification

To verify, find the digital sum of the expression: (6+6)×2 − (4+4) + (2+3+1+9) − (1+8) → 3×2 − 8 + 6 − 9 = 4. The digital sum of 112 is 1+1+2 = 4, confirming the result.

Shortcut Trick

Solve the expression using the VBODMAS priority order.

Perform Of calculation first: (28 of 2) = 56.

Perform Division and Multiplication: 36 ÷ 6 = 6 and 17 × 2 = 34.

Final calculation: 64 + 6 − 56 + 34 = 48.

∴ The correct answer is 48.

Alternate Method

Given: 64 + 36 ÷ 6 − (28 of 2) + 17 × 2

Using the BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction):

⇒ 64 + 36 ÷ 6 − (28 × 2) + 17 × 2

⇒ 64 + 36 ÷ 6 − 56 + 17 × 2 (Solved Of inside bracket)

⇒ 64 + 6 − 56 + 17 × 2 (Solved Division: 36 ÷ 6 = 6)

⇒ 64 + 6 − 56 + 34 (Solved Multiplication: 17 × 2 = 34)

⇒ 70 − 56 + 34 (Solved Addition: 64 + 6 = 70)

⇒ 14 + 34 = 48 (Solved Subtraction then final Addition)

∴ The correct answer is 48.

Additional Information

VBODMAS Rule

The acronym stands for Vinculum, Brackets, Of, Division, Multiplication, Addition, and Subtraction, representing the hierarchy of mathematical operations.

The 'Of' Operator

The Of operator is mathematically equivalent to multiplication but must be executed before division and standard multiplication.

Left-to-Right Precedence

When multiple operators of the same level (like addition and subtraction) appear, they are generally solved from left to right for consistency.

Shortcut Trick

Identify the factors of the denominators: 1×2, 2×3, 3×4, ..., 9×10.

For a telescoping series where the difference between factors is 1:

Sum = (1 ÷ Difference) × [(1 ÷ First Factor) − (1 ÷ Last Factor)]

Sum = (1 ÷ 1) × [1/1 − 1/10] = 1 − 0.1 = 9/10.

∴ The correct answer is 9/10.

Alternate Method

Given Expression: (1/2) + (1/6) + (1/12) + (1/20) + ... + (1/90)

Method: Convert each term into partial fractions.

⇒ 1/(1×2) + 1/(2×3) + 1/(3×4) + 1/(4×5) + ... + 1/(9×10)

⇒ (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + (1/4 − 1/5) + ... + (1/9 − 1/10)

⇒ 1 − 1/2 + 1/2 − 1/3 + 1/3 − 1/4 + ... + 1/9 − 1/10

⇒ On cancelling intermediate terms: 1 − 1/10

⇒ (10 − 1) / 10

⇒ 9/10

∴ The correct answer is 9/10.

Additional Information

Telescoping Series Formula

For a series of the form Σ 1/[n(n+k)], the sum is (1/k) × [Σ(1/n − 1/(n+k))]. All intermediate terms cancel out except the first 'k' and last 'k' terms.

Sum of First 'n' Terms

The sum of the first 'n' terms of the series 1/(1×2) + 1/(2×3) + ... + 1/[n(n+1)] is always given by the direct formula: n / (n+1).

Partial Fraction Decomposition

It is a method used to break down complex rational expressions into sums of simpler fractions, which is highly useful in calculus and infinite series.

Shortcut Trick

Convert the recurring decimals into their expanded form for direct comparison.

Target value: 0.63̅ = 0.6333...

Option 1: 0.64 = 0.6400... (Greater)

Option 2: 0.67̅ = 0.6777... (Greater)

Option 3: 0.61̅ = 0.6111... (Smaller)

Option 4: 0.69 = 0.6900... (Greater)

∴ The correct answer is 0.61̅.

Alternate Method

Given: The decimal value is 0.63̅.

Formula Used: To convert a mixed recurring decimal to a fraction: (Full number − Non-recurring part) ÷ (9 for every repeating digit followed by 0 for every non-repeating digit after decimal).

⇒ Convert 0.63̅ to a fraction: (63 − 6) ÷ 90 = 57/90

⇒ Check Option 1: 0.64 = 64/100 = 57.6/90 (Greater than 57/90)

⇒ Check Option 2: 0.67̅ = (67 − 6) ÷ 90 = 61/90 (Greater than 57/90)

⇒ Check Option 3: 0.61̅ = (61 − 6) ÷ 90 = 55/90 (Smaller than 57/90)

⇒ Check Option 4: 0.69 = 69/100 = 62.1/90 (Greater than 57/90)

⇒ Comparing the numerators with a common denominator of 90: 55 < 57.

∴ The correct answer is 0.61̅.

Additional Information

Pure Recurring Decimal

A decimal in which all the figures after the decimal point are repeated. Example: 0.3̅ = 3/9 = 1/3.

Mixed Recurring Decimal

A decimal in which at least one figure after the decimal point is not repeated and then some figures are repeated. Example: 0.12̅ = (12 − 1)/90 = 11/90.

Comparison of Decimals

Compare the digits from left to right. The first place where digits differ determines which decimal is larger (e.g., 0.63 vs 0.61, since 3 > 1, 0.63 is larger).

Shortcut Trick

Convert the mixed fractions and options into decimal values for quick comparison.

Range: 2 1/3 ≈ 2.33 and 2 5/6 ≈ 2.83.

Option 1: 18/7 ≈ 2.57 (Lies within the range 2.33 to 2.83).

Option 2 & 3: 17/9 ≈ 1.89 and 20/11 ≈ 1.82 (Both are less than 2.33).

Option 4: 38/13 ≈ 2.92 (This is greater than 2.83).

∴ The correct answer is 18/7.

Alternate Method

Given:

Lower Limit: 2 1/3 = 7/3

Upper Limit: 2 5/6 = 17/6

Method: Convert all fractions to common denominators or use cross-multiplication to verify the inequality.

⇒ Convert 7/3 and 18/7 to compare: (7 × 7) = 49 and (18 × 3) = 54. Since 54 > 49, then 18/7 > 7/3.

⇒ Convert 18/7 and 17/6 to compare: (18 × 6) = 108 and (17 × 7) = 119. Since 108 < 119, then 18/7 < 17/6.

⇒ Since 7/3 < 18/7 < 17/6, the value lies in the required interval.

⇒ Checking other options: 17/9 = 1.88, 20/11 = 1.81, and 38/13 = 2.92. None fall between 2.33 and 2.83.

∴ The correct answer is 18/7.

Additional Information

Comparison of Fractions

To compare two fractions a/b and c/d, check the products (a × d) and (b × c). If ad > bc, then a c/d.

Mixed Fraction Conversion

A mixed fraction a b/c is converted to an improper fraction using the formula: (a × c + b) / c.

Density Property of Rational Numbers

Between any two rational numbers, there are infinitely many other rational numbers that can be found by taking the average or using common denominators.

Shortcut Trick

In a Rhombus, the Area is given by half the product of its Diagonals.

Second Diagonal (d₂) = (2 × Area) ÷ d₁

d₂ = (2 × 112) ÷ 14 = 224 ÷ 14 = 16 cm

∴ The correct answer is 16 cm.

Alternate Method

Given:

Area of the Rhombus = 112 cm²

Length of one Diagonal (d₁) = 14 cm

Formula Used:

Area of Rhombus = (1 ÷ 2) × d₁ × d₂

Calculations:

⇒ 112 = (1 ÷ 2) × 14 × d₂

⇒ 112 = 7 × d₂

⇒ d₂ = 112 ÷ 7

⇒ d₂ = 16 cm

∴ The correct answer is 16 cm.

Additional Information

Side Length of a Rhombus

The side 'a' can be calculated using the diagonals: a = (1 ÷ 2) × √(d₁² + d₂²).

Relationship between Side and Diagonals

For any Rhombus, the sum of the squares of the Diagonals is equal to four times the square of its side: d₁² + d₂² = 4a².

Perimeter of a Rhombus

Since all sides are equal in a Rhombus, the perimeter is simply 4 × side.

Shortcut Trick

Side of the Rhombus = 100 ÷ 4 = 25 cm.

Half of the given diagonal (d₁/2) = 40 ÷ 2 = 20 cm.

Recognize the Pythagorean triplet (15, 20, 25) ⇒ Half of other diagonal (d₂/2) = 15 cm.

Total second diagonal (d₂) = 15 × 2 = 30 cm.

Required ratio = 40 : 30 = 4 : 3.

∴ The correct answer is 4:3.

Alternate Method

Given: Perimeter = 100 cm, Diagonal₁ (d₁) = 40 cm.

Formula: Side (a) = Perimeter ÷ 4; (Side)² = (d₁/2)² + (d₂/2)².

⇒ Side (a) = 100 ÷ 4 = 25 cm.

⇒ (25)² = (40 ÷ 2)² + (d₂ ÷ 2)²

⇒ 625 = 20² + (d₂ ÷ 2)²

⇒ 625 = 400 + (d₂ ÷ 2)²

⇒ (d₂ ÷ 2)² = 625 − 400 = 225

⇒ d₂ ÷ 2 = √225 = 15 cm

⇒ Diagonal₂ (d₂) = 15 × 2 = 30 cm.

⇒ Ratio = d₁ : d₂ = 40 : 30 = 4 : 3.

∴ The correct answer is 4:3.

Additional Information

Area of Rhombus

The area of a rhombus is calculated as half the product of its diagonals: Area = 1/2 × d₁ × d₂.

Diagonals Property

The diagonals of a rhombus bisect each other at right angles (90°) and also bisect the interior angles.

Relation between Side and Diagonals

The sum of the squares of the diagonals is equal to four times the square of the side: d₁² + d₂² = 4 × a².

Shortcut Trick

For any two numbers, the product of the numbers is equal to the product of their HCF and LCM.

Other number = (Product of HCF × LCM) ÷ Given number

Other number = 6930 ÷ 154 = 45

∴ The correct answer is 45.

Alternate Method

Given:

Product of HCF and LCM = 6930

One number (n₁) = 154

Formula Used:

n₁ × n₂ = HCF × LCM

Calculations:

⇒ 154 × n₂ = 6930

⇒ n₂ = 6930 ÷ 154

⇒ n₂ = 3465 ÷ 77 (Dividing by 2)

⇒ n₂ = 495 ÷ 11 (Dividing by 7)

⇒ n₂ = 45

∴ The correct answer is 45.

Additional Information

HCF and LCM Relation

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

Co-prime Numbers

Two numbers are said to be co-prime if their Highest Common Factor (HCF) is 1. For such numbers, their LCM is simply the product of the numbers.

Divisibility by HCF

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

Shortcut Trick

For two numbers in the ratio a : b, if a and b are co-prime, then HCF = LCM ÷ (a × b).

Here, Ratio = 2 : 3 and LCM = 60.

HCF = 60 ÷ (2 × 3) = 60 ÷ 6 = 10.

Numbers = (2 × 10) and (3 × 10) = 20 and 30.

∴ The correct answer is 20, 30.

Alternate Method

Given:

Ratio of two numbers = 2 : 3

Least Common Multiple (LCM) = 60

Formula Used:

If two numbers are ax and bx, their LCM is x × LCM(a, b).

Calculations:

⇒ Let the two numbers be 2x and 3x, where x is the Highest Common Factor (HCF).

⇒ LCM of 2x and 3x = 6x.

⇒ According to the question, 6x = 60.

⇒ x = 60 ÷ 6 = 10.

⇒ First number = 2x = 2 × 10 = 20.

⇒ Second number = 3x = 3 × 10 = 30.

∴ The correct answer is 20, 30.

Additional Information

Product Rule of Two Numbers

The product of two numbers is always equal to the product of their HCF and LCM: n₁ × n₂ = HCF × LCM.

LCM of Co-prime Numbers

If two numbers are co-prime (their HCF is 1), then their LCM is simply the product of the two numbers.

HCF as a Component

Any two numbers can be written as Hx and Hy, where H is the HCF and x, y are co-prime integers.

Shortcut Trick

Let B = 1 unit.

A is 5 times more than B ⇒ A = 1 + 5 = 6 units.

Difference = 6 − 1 = 5 units.

Required percentage = (5 / 6) × 100 = 83.33%.

∴ The correct answer is 83.33%.

Alternate Method

Given:

Let the value of B be 100.

According to the question, A is 5 times more than B.

⇒ Value of A = B + 5 × B

⇒ Value of A = 100 + 500 = 600

⇒ Difference = A − B = 600 − 100 = 500

⇒ Percentage B is less than A = (Difference / A) × 100

⇒ Required % = (500 / 600) × 100

⇒ Required % = 5/6 × 100 = 83.33%

∴ The correct answer is 83.33%.

Additional Information

Difference between "Times" and "Times More"

If A is n times B, then A = n × B. If A is n times more than B, then A = B + (n × B) = (n + 1) × B.

Percentage Less Formula

Percentage Less = [(Higher Value − Lower Value) / Higher Value] × 100.

Standard Fraction Conversion

1/6 ≈ 16.66% | 5/6 ≈ 83.33% | 1/3 ≈ 33.33% | 2/3 ≈ 66.66%.

Shortcut Trick

Ratio of Cost Price (CP) to Selling Price (SP) = 800 : 1000 = 4 : 5.

Profit = 5 − 4 = 1 unit on base of 4 units.

Profit percentage = (1 ÷ 4) × 100 = 25%.

∴ The correct answer is 25%.

Alternate Method

Given:

Cost Price (CP) = ₹800

Selling Price (SP) = ₹1000

Formula Used:

Profit = Selling Price − Cost Price

Profit Percentage = (Profit ÷ Cost Price) × 100

Calculations:

⇒ Profit = 1000 − 800 = ₹200

⇒ Profit Percentage = (200 ÷ 800) × 100

⇒ Profit Percentage = (1 ÷ 4) × 100

⇒ Profit Percentage = 25%

∴ The correct answer is 25%.

Additional Information

Loss Percentage

When the Cost Price is greater than the Selling Price, a loss occurs. Loss % = (Loss ÷ CP) × 100, where Loss = CP − SP.

Marked Price and Discount

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

Calculating SP from Profit %

The Selling Price can be directly found using the formula: SP = CP × (100 + Profit%) ÷ 100.

Shortcut Trick

Ratio of numbers (x : y) is found by: (1/4) × 70% × x = (1/5) × 50% × y

⇒ 7/40 × x = 1/10 × y ⇒ 7x = 4y ⇒ x : y = 4 : 7

Difference (7 − 4) = 3 units = 240 ⇒ 1 unit = 80

Sum (4 + 7) = 11 units = 11 × 80 = 880

Required value = 150% of 880 = 1.5 × 880 = 1320

∴ The correct answer is 1320.

Alternate Method

Given:

First number (A) condition: 1/4 × 70% × A

Second number (B) condition: 1/5 × 50% × B

Difference between numbers (B − A) = 240

Calculation:

⇒ (1/4) × (7/10) × A = (1/5) × (1/2) × B

⇒ 7/40 × A = 1/10 × B

⇒ 7A = 4B ⇒ A/B = 4/7

⇒ Let A = 4k and B = 7k

⇒ Difference = 7k − 4k = 3k = 240

⇒ k = 240 ÷ 3 = 80

⇒ Sum = A + B = 4k + 7k = 11k = 11 × 80 = 880

⇒ Required value = 150% of 880 = (150/100) × 880

⇒ Required value = 1.5 × 880 = 1320

∴ The correct answer is 1320.

Additional Information

Percentage to Fraction Conversion

Converting percentages to simplified fractions (e.g., 70% = 7/10, 50% = 1/2) speeds up calculations in ratio-based problems.

Ratio Property

If aA = bB, then the ratio A : B is b : a. This inverse relationship is fundamental in solving comparison equations.

150 Percent Calculation

Calculating 150% of a value is equivalent to adding half of the value to itself (Value + 0.5 × Value).

Shortcut Trick

Total difference in Selling Prices = 2000 − 1200 = 800.

Ratio of Profit to Loss = 19 : 1 (Total units = 20).

Value of 1 unit (Loss) = 800 ÷ 20 = 40.

Cost Price (CP) = 1200 + 40 = 1240.

Profit % at 2000 = (760 ÷ 1240) × 100 = 61.29%.

∴ The correct answer is 61.29%.

Alternate Method

Given:

Selling Price 1 (SP₁) = ₹2000

Selling Price 2 (SP₂) = ₹1200

Profit = 19 × Loss

Formula Used:

Profit = SP − CP | Loss = CP − SP

Profit % = (Profit ÷ CP) × 100

⇒ Let the Cost Price be x.

⇒ Profit at ₹2000 = 2000 − x

⇒ Loss at ₹1200 = x − 1200

⇒ According to the question: 2000 − x = 19 × (x − 1200)

⇒ 2000 − x = 19x − 22800

⇒ 20x = 24800

⇒ x = 1240

⇒ Cost Price (CP) = ₹1240

⇒ Profit at ₹2000 = 2000 − 1240 = 760

⇒ Profit % = (760 ÷ 1240) × 100 = 61.29%

∴ The correct answer is 61.29%.

Additional Information

Profit and Loss Basics

Profit occurs when Selling Price is greater than Cost Price, while Loss occurs when Cost Price is greater than Selling Price.

Percentage Calculation

Both Profit Percentage and Loss Percentage are always calculated on the Cost Price unless otherwise specified in the question.

Relation between Profit and SP

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

Shortcut Trick

Assume Total Work as LCM of (9, 12, 18) = 36 units.

Efficiency of A = 36 ÷ 9 = 4 units/day.

Efficiency of B = 36 ÷ 12 = 3 units/day.

Efficiency of C = 36 ÷ 18 = 2 units/day.

Combined efficiency = 4 + 3 + 2 = 9 units/day.

Time taken = Total Work ÷ Total Efficiency = 36 ÷ 9 = 4 days.

∴ The correct answer is 4 days.

Alternate Method

Given:

Time taken by A = 9 days

Time taken by B = 12 days

Time taken by C = 18 days

Formula Used:

Efficiency = Total Work ÷ Time Taken

Total Work = LCM of days taken by individuals.

Calculations:

⇒ Total efficiency of A, B, and C = 4 + 3 + 2 = 9

⇒ Time taken by A, B, and C together = Total Work ÷ Combined Efficiency

⇒ Time = 36 ÷ 9

⇒ Time = 4 days

∴ The correct answer is 4 days.

Additional Information

Combined Work Rate

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

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete the work (Efficiency ∝ 1/Time).

Work and Wages

When multiple people work together, wages are distributed in the ratio of their efficiencies or the amount of work done by each.

Shortcut Trick

When workers work for the same duration, their wage ratio is the inverse of their time ratio.

Time ratio of A : B = 25 : 30 = 5 : 6.

Efficiency (Wage) ratio A : B = 6 : 5.

Total units = 6 + 5 = 11 units → ₹6600.

Difference in shares = 6 − 5 = 1 unit → ₹6600 ÷ 11.

∴ The correct answer is ₹600.

Alternate Method

Given:

Time taken by A = 25 days

Time taken by B = 30 days

Total wage = ₹6600

Formula Used:

Efficiency = Total Work ÷ Time taken; Wage is distributed in the ratio of efficiency when time is constant.

Calculations:

⇒ Total Efficiency of (A + B) = 6 + 5 = 11 units.

⇒ Share of A = (6 ÷ 11) × 6600 = ₹3600.

⇒ Share of B = (5 ÷ 11) × 6600 = ₹3000.

⇒ Difference in their amounts = 3600 − 3000 = ₹600.

∴ The correct answer is ₹600.

Additional Information

Wages and Efficiency

Wages are directly proportional to the amount of work done. If workers work for the same number of days, the ratio of their wages is equal to the ratio of their efficiencies.

Inverse Variation

Efficiency is inversely proportional to the time taken to complete the work (E = W/T). For two people, E₁:E₂ = T₂:T₁.

Work and Wages Case

If a person finishes a work in D days, their one-day work (efficiency) is 1/D. The total wage is divided based on the total work contribution of each person.

Shortcut Trick

Speed of boat in still water = 24 km/hr

Upstream speed (Against the flow) = 16 km/hr

Speed of stream = Still water speed − Upstream speed

Speed of stream = 24 − 16 = 8 km/hr

∴ The correct answer is 8 km/hr.

Alternate Method

Given:

Speed of the boat in still water (u) = 24 km/hr

Upstream speed of the boat (U) = 16 km/hr

Formula Used:

Upstream Speed (U) = Speed of boat in still water (u) − Speed of stream (v)

Calculations:

⇒ 16 = 24 − v

⇒ v = 24 − 16

⇒ v = 8 km/hr

∴ The correct answer is 8 km/hr.

Additional Information

Downstream Speed

It is the speed of the boat in the direction of the stream. Formula: Downstream Speed (D) = u + v.

Still Water Speed Formula

If downstream (D) and upstream (U) speeds are known, speed in still water (u) = (D + U) ÷ 2.

Stream Speed Formula

If downstream (D) and upstream (U) speeds are known, speed of stream (v) = (D − U) ÷ 2.

Shortcut Trick

The difference in time can be calculated by dividing the difference in distance by the constant speed.

Difference in distance = 540 km − 225 km = 315 km

Time difference = 315 km ÷ 15 km/hr = 21 hr

∴ The correct answer is 21 hr.

Alternate Method

Given:

First distance (D₁) = 540 km

Second distance (D₂) = 225 km

Speed (S) = 15 km/hr

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Time taken for first distance (T₁) = 540 ÷ 15

⇒ T₁ = 36 hr

⇒ Time taken for second distance (T₂) = 225 ÷ 15

⇒ T₂ = 15 hr

⇒ Difference between time = T₁ − T₂

⇒ Difference = 36 hr − 15 hr = 21 hr

∴ The correct answer is 21 hr.

Additional Information

Speed-Distance-Time Relation

The fundamental relation is Distance = Speed × Time. When speed is constant, distance is directly proportional to time.

Average Speed

Total distance covered divided by the total time taken. For two equal distances at speeds x and y, Average Speed = 2xy ÷ (x + y).

Relative Speed

When two objects move in the same direction, relative speed = S₁ − S₂. When they move in opposite directions, relative speed = S₁ + S₂.

Shortcut Trick

Apply the Successive Percentage formula for net profit:

Net Profit % = M − D − (M × D ÷ 100)

Where Markup (M) = 80% and Discount (D) = 40%

Profit % = 80 − 40 − (80 × 40 ÷ 100) = 40 − 32 = 8%

∴ The correct answer is 8%.

Alternate Method

Given:

⇒ Let the Cost Price (CP) of the article be 100.

⇒ Marked Price (MP) is 80% more than CP.

⇒ Discount offered is 40% on MP.

Calculations:

⇒ Marked Price (MP) = 100 + (80% of 100) = 180

⇒ Discount Amount = 40% of 180 = (40 ÷ 100) × 180 = 72

⇒ Selling Price (SP) = MP − Discount = 180 − 72 = 108

⇒ Profit = SP − CP = 108 − 100 = 8

⇒ Profit Percentage = (Profit ÷ CP) × 100 = (8 ÷ 100) × 100 = 8%

∴ The correct answer is 8%.

Additional Information

CP, MP, Profit and Discount Relation

The relationship between cost, marked price, profit, and discount is given by: CP × (100 + P%) = MP × (100 − D%).

Markup Percentage

Markup percentage is calculated on the cost price: Markup % = [(MP − CP) ÷ CP] × 100.

Successive Discounts

If two successive discounts of d₁% and d₂% are given, the equivalent single discount is [d₁ + d₂ − (d₁ × d₂ ÷ 100)]%.

Shortcut Trick

Use the ratio-unitary method where Marked Price (MP) is 100%.

If Discount is 20%, then Selling Price (SP) = 80% of MP.

Checking Option (3): 80% = ₹3600 ⇒ 1% = ₹45 ⇒ 100% = ₹4500.

Since the result matches the required MP, Option (3) is correct.

∴ The correct answer is Selling price = Rs. 3600 and Discount = 20 percent.

Alternate Method

Given:

For the correct case, Selling Price (SP) = ₹3600 and Discount = 20%.

Formula Used:

Marked Price (MP) = (SP × 100) ÷ (100 − Discount%)

Calculations:

⇒ Marked Price = (3600 × 100) ÷ (100 − 20)

⇒ MP = 360000 ÷ 80

⇒ MP = ₹4500

Similarly, checking other options:

⇒ Option 1: MP = 4200 ÷ 0.88 ≈ ₹4772.7

⇒ Option 2: MP = 3800 ÷ 0.86 ≈ ₹4418.6

⇒ Option 4: MP = 4100 ÷ 0.81 ≈ ₹5061.7

∴ The correct answer is Selling price = Rs. 3600 and Discount = 20 percent.

Additional Information

Marked Price (MP)

Also known as the list price or printed price, it is the value before any discount is applied.

Discount Calculation

Discount is always calculated on the Marked Price. Formula: Discount = MP − SP.

Successive Discounts

If two discounts a% and b% are given, the equivalent discount is (a + b − (ab/100))%.

Shortcut Trick

Let the Cost Price (CP) of the article be 100.

Based on the 25% mark-up, the Marked Price (MP) becomes 125.

With an 18% Profit, the Selling Price (SP) becomes 100 + 18 = 118.

The Discount offered is MP − SP = 125 − 118 = 7.

Discount % = (Discount ÷ MP) × 100 = (7 ÷ 125) × 100 = 5.6%.

∴ The correct answer is 5.6%.

Alternate Method

Given:

Mark-up percentage = 25%

Profit percentage = 18%

Calculations:

⇒ Let CP = 100.

⇒ MP = CP + 25% of CP = 100 + 25 = 125.

⇒ SP = CP + 18% of CP = 100 + 18 = 118.

⇒ Discount Amount = MP − SP = 125 − 118 = 7.

⇒ Discount % = (Discount Amount ÷ MP) × 100

⇒ Discount % = (7 ÷ 125) × 100 = 5.6%.

∴ The correct answer is 5.6%.

Additional Information

Relationship between CP and MP

The ratio of Marked Price to Cost Price is given by: MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%).

Successive Percentage Formula

The net Profit % can also be calculated using the formula: P = M − D − (M × D) ÷ 100, where M is Mark-up % and D is Discount %.

Effective Discount

For two successive discounts of d₁% and d₂%, the single equivalent discount is: d₁ + d₂ − (d₁ × d₂) ÷ 100.