Quantitative Aptitude PYQs

Shortcut Trick

Discount = Marked Price − Selling Price

Discount = 657 − 526 = 131

Discount% = (Discount ÷ MP) × 100

Discount% = (131 ÷ 657) × 100 = 19.9391...

Rounding off to two decimal places, we get 19.94%.

∴ The correct answer is 19.94%.

Alternate Method

Given → Marked Price (MP) = ₹657, Selling Price (SP) = ₹526.
Formulas: Discount = MP − SP | Discount% = (Discount ÷ MP) × 100

→ Discount = 657 − 526 = 131
→ Discount% = (131 ÷ 657) × 100
→ Discount% = 13100 ÷ 657
→ Discount% ≈ 19.939%
→ Rounded Discount% = 19.94%

∴ The correct answer is 19.94%.

Additional Information

Marked Price and Discount

The Marked Price is the price printed on the label. Discount is always calculated on the Marked Price unless stated otherwise. Discount% = (MP − SP)/MP × 100.

Selling Price Calculation

Selling Price can also be calculated if discount percentage is known: SP = MP × (100 − Discount%)/100.

Shortcut Trick

If each number in a series is multiplied by a constant (k), the new average is simply the old average multiplied by that same constant.

Old Average = 7

Multiplier (k) = 3

New Average = 7 × 3 = 21

∴ The correct answer is 21.

Alternate Method

Given:

Total numbers (n) = 13

Original average = 7

⇒ Sum of 13 numbers = Average × Count

⇒ Original Sum = 7 × 13 = 91

If each number is multiplied by 3, the total sum also increases 3 times:

⇒ New Sum = 91 × 3 = 273

⇒ New Average = New Sum ÷ Count

⇒ New Average = 273 ÷ 13

⇒ New Average = 21

∴ The correct answer is 21.

Additional Information

Effect of Basic Operations on Average

If every term is increased or decreased by x, the average increases or decreases by x respectively. If divided by x, the average is divided by x.

Weighted Average

When combining two groups, the weighted average = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂), where n is the count and a is the average.

Average of Consecutive Numbers

For any Arithmetic Progression (AP), the average is always the middle term or (First Term + Last Term) ÷ 2.

Shortcut Trick

Ratio of SP₁ to SP₂ is 1000 : 1400 = 5 : 7.

Since SP₁ represents a 5% loss, 5 units = (100 − 5)% = 95%.

Then 1 unit = 95 ÷ 5 = 19%.

SP₂ represents 7 units = 7 × 19% = 133%.

The profit percentage x = 133% − 100% = 33%.

∴ The correct answer is 33.

Alternate Method

Given: Selling Price 1 (SP₁) = ₹1,000 | Loss% = 5% | Selling Price 2 (SP₂) = ₹1,400.
Formulas: CP = SP × 100 ÷ (100 − Loss%) | Profit% = (SP − CP) ÷ CP × 100.

⇒ Calculate CP: CP = 1000 × 100 ÷ (100 − 5) = 1000 × 100 ÷ 95.
⇒ CP = 100000 ÷ 95 = ₹1052.63 (approx) or 20000 ÷ 19.
⇒ Profit = SP₂ − CP = 1400 − (20000 ÷ 19).
⇒ Profit = (26600 − 20000) ÷ 19 = 6600 ÷ 19.
⇒ Profit% (x) = (6600 ÷ 19) ÷ (20000 ÷ 19) × 100.
⇒ x = (6600 ÷ 20000) × 100 = 33.

∴ The correct answer is 33.

Additional Information

Cost Price (CP)

The price at which an article is purchased. It is the base for calculating Profit or Loss percentages.

Selling Price (SP)

The price at which an article is sold. If SP > CP, there is a profit. If SP < CP, there is a loss.

Shortcut Trick

Income ratio = 5 : 8 | Expenditure ratio = 1 : 4

Difference in income parts = 8 − 5 = 3 | Difference in expenditure parts = 4 − 1 = 3

Since differences are same, (5 − 1) units = (8 − 4) units = 4 units = savings

⇒ 4 units = ₹75,000 → 1 unit = ₹18,750

Mohan income = 5 units = 5 × ₹18,750 = ₹93,750

∴ The correct answer is ₹93,750.

Alternate Method

Given: Ratio of Incomes = 5 : 8, Ratio of Expenditures = 1 : 4, Savings = ₹75,000 each.

Formula: Income − Expenditure = Savings

⇒ Let Mohan income be 5x and David income be 8x.

⇒ Expenditure of Mohan = 5x − 75,000

⇒ Expenditure of David = 8x − 75,000

⇒ (5x − 75,000) / (8x − 75,000) = 1 / 4

⇒ 4 × (5x − 75,000) = 1 × (8x − 75,000)

⇒ 20x − 3,00,000 = 8x − 75,000

⇒ 12x = 2,25,000

⇒ x = 18,750

⇒ Mohan income = 5x = 5 × 18,750 = 93,750

∴ The correct answer is ₹93,750.

Additional Information

Income, Expenditure and Savings Relationship

The fundamental relation is Income = Expenditure + Savings. If savings are constant, the change in income directly affects the expenditure.

Ratio Balancing Method

When a constant value is added or subtracted from both terms of a ratio, we balance the difference between ratio parts to solve quickly without variables.

Cross Multiplication Trick

For two ratios (a:b and c:d) and a constant difference (k), the variable can be found by: x = [k × (d − c)] / [ad − bc].

Shortcut Trick

Ratio of dimensions = 6 : 5 : 4 ⇒ Volume units = 6 × 5 × 4 = 120 units³

120 units³ = 810000 √2 ⇒ 1 unit³ = 6750 √2 = 3375 × (2 √2) = 15³ × (√2)³

Side ratio constant (1 unit) = 15 √2 cm

Total Surface Area units = 2 × (30 + 20 + 24) = 148 units²

Surface Area = 148 × (15 √2)² = 148 × (225 × 2) = 148 × 450 = 66600 cm²

∴ The correct answer is 66600 cm².

Alternate Method

Given:

Ratio of Length, Breadth and Height = 6 : 5 : 4

Volume of the Cuboid = 810000 √2 cm³

Formula Used:

Volume of Cuboid = Length × Breadth × Height

Total Surface Area = 2 × (LB + BH + HL)

Calculations:

⇒ Let Length = 6x, Breadth = 5x, and Height = 4x

⇒ Volume = 6x × 5x × 4x = 120x³

⇒ 120x³ = 810000 √2

⇒ x³ = 6750 √2 = 3375 × 2 × √2 = 3375 × (√2)³

⇒ x³ = 15³ × (√2)³ ⇒ x = 15 √2

⇒ Total Surface Area = 2 × (6x × 5x + 5x × 4x + 4x × 6x)

⇒ Surface Area = 2 × (30x² + 20x² + 24x²) = 2 × 74x² = 148x²

⇒ Surface Area = 148 × (15 √2)² = 148 × (225 × 2) = 148 × 450

⇒ Surface Area = 66600 cm²

∴ The correct answer is 66600 cm².

Additional Information

Diagonal of a Cuboid

The longest rod that can be placed inside a cuboid is its diagonal, calculated as d = √(L² + B² + H²).

Lateral Surface Area (LSA)

The area of the four vertical walls of a cuboid is given by LSA = 2 × Height × (Length + Breadth).

Volume and Surface Area of a Cube

For a cube with side 'a', the Volume is a³ and the Total Surface Area is 6a².

Shortcut Trick

Ratio of articles P : Q : R = 3 : 4 : 5

Total Value ratio = (280 × 3) : (180 × 4) : (100 × 5) = 840 : 720 : 500 = 2060 units

2060 units = ₹6,180 ⇒ 1 unit = 3

Articles of type Q = 4 × 3 = 12

∴ The correct answer is 12 articles.

Alternate Method

Given: Prices of P, Q, R are ₹280, ₹180, ₹100. Ratio of quantity = 3 : 4 : 5. Total expenditure = ₹6,180.

Calculations:

⇒ Let the number of articles be 3x, 4x, and 5x respectively.

⇒ (280 × 3x) + (180 × 4x) + (100 × 5x) = 6180

⇒ 840x + 720x + 500x = 6180

⇒ 2060x = 6180

⇒ x = 6180 ÷ 2060 = 3

⇒ Number of articles of type Q = 4x = 4 × 3 = 12

∴ The correct answer is 12 articles.

Additional Information

Ratio and Value Relationship

When quantities are given in a ratio (a:b:c) and their individual values are (v₁, v₂, v₃), the total value is calculated as (a × v₁ + b × v₂ + c × v₃) × k.

Unitary Method in Ratios

Finding the value of one 'part' or 'unit' by dividing the total actual value by the sum of ratio-product parts allows for quick calculation of any individual component.

Shortcut Trick

Net discount for 10% and 15% = 10 + 15 − (10 × 15) ÷ 100 = 23.5%.

Net discount for 23.5% and 7% = 23.5 + 7 − (23.5 × 7) ÷ 100.

= 30.5 − 1.645 = 28.855%.

The approximate single discount is 28.85%.

∴ The correct answer is 28.85%.

Alternate Method

Given → Discounts are d₁ = 7%, d₂ = 10%, and d₃ = 15%.
Formulas: Single Equivalent Discount = 100 − [100 × (100 − d₁)/100 × (100 − d₂)/100 × (100 − d₃)/100]

⇒ Assume Marked Price (MP) = 100.
⇒ Price after 7% discount = 100 × 0.93 = 93.
⇒ Price after 10% discount = 93 × 0.90 = 83.7.
⇒ Price after 15% discount = 83.7 × 0.85 = 71.145.
⇒ Total Discount = 100 − 71.145 = 28.855%.
⇒ Approximate value = 28.85%.

∴ The correct answer is 28.85%.

Additional Information

Successive Discount Formula

For two discounts a% and b%, the single equivalent discount is (a + b − ab/100)%.

Marked Price & Discount

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

Shortcut Trick

Total Work (LCM of 15, 18, 27) = 270 units.

Efficiencies: A = 18, B = 15, C = −10.

First 5 hr (A+B): (18 + 15) × 5 = 165 units.

Next 12 hr (B+C): (15 − 10) × 12 = 60 units.

Remaining: 270 − (165 + 60) = 45 units.

Time for B alone: 45 ÷ 15 = 3 hours.

∴ The correct answer is 3 hr.

Alternate Method

⇒ Work done by Pipe A and Pipe B in 5 hours = (18 + 15) × 5 = 33 × 5 = 165 units.

⇒ Work done by Pipe B and Pipe C in 12 hours = (15 − 10) × 12 = 5 × 12 = 60 units.

⇒ Total work done so far = 165 + 60 = 225 units.

⇒ Remaining work = Total capacity − Work done = 270 − 225 = 45 units.

⇒ Time taken by Pipe B alone = Remaining work ÷ Efficiency of B = 45 ÷ 15 = 3 hours.

∴ The correct answer is 3 hr.

Additional Information

Inlet and Outlet Pipes

Inlet pipes add water (positive efficiency) while outlet pipes remove water (negative efficiency). Net efficiency is the algebraic sum.

Work and Time Relationship

Total Work = Time taken × Efficiency. For cisterns, capacity is the LCM of individual filling/emptying times.

Shortcut Trick

To find the greatest possible scale to measure the dimensions exactly, find the HCF (Highest Common Factor) of the given dimensions.

Given dimensions: 16 m, 20 m, 24 m.

HCF(16, 20, 24) = 4 m.

∴ The correct answer is 4 m.

Alternate Method

Given:

Dimensions of the room = 16 m, 20 m, 24 m

Formula:

Length of the greatest possible scale = HCF (Highest Common Factor) of the dimensions

Calculations:

⇒ Prime factorization of 16 = 2 × 2 × 2 × 2 = 2⁴

⇒ Prime factorization of 20 = 2 × 2 × 5 = 2² × 5

⇒ Prime factorization of 24 = 2 × 2 × 2 × 3 = 2³ × 3

⇒ HCF(16, 20, 24) = 2² = 4

⇒ Greatest scale length = 4 m

∴ The correct answer is 4 m.

Additional Information

Highest Common Factor (HCF)

The HCF of two or more numbers is the greatest number that divides each of them exactly without leaving a remainder.

Least Common Multiple (LCM)

The LCM of two or more numbers is the smallest number that is a multiple of all the given numbers.

HCF and LCM Relation

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.

Shortcut Trick

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

Here, a = 3x and b = 5y, so 4ab = 4 × 3x × 5y = 60xy.

The numerator becomes (3x − 5y)², which is identical to the denominator.

Expression = (3x − 5y)² ÷ (3x − 5y)² = 1.

∴ The correct answer is 1.

Alternate Method

Given: [(3x + 5y)² − 60xy] ÷ (3x − 5y)²

Formula Used: (a + b)² = a² + 2ab + b²

⇒ Expand the numerator: (3x)² + 2(3x)(5y) + (5y)² − 60xy

⇒ Numerator = 9x² + 30xy + 25y² − 60xy = 9x² − 30xy + 25y²

⇒ Expand the denominator: (3x − 5y)² = (3x)² − 2(3x)(5y) + (5y)²

⇒ Denominator = 9x² − 30xy + 25y²

⇒ Division = (9x² − 30xy + 25y²) ÷ (9x² − 30xy + 25y²)

⇒ 1

∴ The correct answer is 1.

Additional Information

Standard Algebraic Identities

(a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b² are basic expansions.

Relationship Identity

(a + b)² − (a − b)² = 4ab is often used to simplify complex polynomial divisions.

Alternative Relationship

(a + b)² + (a − b)² = 2(a² + b²) is another useful variant in simplification.

Shortcut Trick

Test the given options to satisfy both Malati and Rani conditions.

For Option (4): Rose = ₹120, Jasmine = ₹90.

Condition 1: (5 × 120) + (9 × 90) = 600 + 810 = ₹1410 (Satisfied).

Condition 2: (6 × 120) + (5 × 90) = 720 + 450 = ₹1170 (Satisfied).

∴ The correct answer is 120; 90.

Alternate Method

Let the price of 1 Rose plant be ₹ R and 1 Jasmine plant be ₹ J.

Given:

Case 1 (Malati): 5R + 9J = 1410 → (Eq. 1)

Case 2 (Rani): 6R + 5J = 1170 → (Eq. 2)

Multiply (Eq. 1) by 5 and (Eq. 2) by 9 to eliminate J:

⇒ 25R + 45J = 7050

⇒ 54R + 45J = 10530

Subtract the first resulting equation from the second:

⇒ (54R − 25R) = 10530 − 7050

⇒ 29R = 3480

⇒ R = 3480 ÷ 29 = 120

Substitute R = 120 in (Eq. 2):

⇒ 6(120) + 5J = 1170

⇒ 720 + 5J = 1170

⇒ 5J = 1170 − 720

⇒ 5J = 450

⇒ J = 90

∴ The correct answer is 120; 90.

Additional Information

Linear Equations in Two Variables

A pair of equations with two variables (like x and y) can be solved using Elimination or Substitution methods to find unique values.

Elimination Method

Multiply equations by suitable constants so that the coefficients of one variable become equal, allowing its removal through subtraction.

Verification

Always verify the calculated values by substituting them back into the original equations to ensure they satisfy the given conditions.

Shortcut Trick

Ratio of cost of TV to Washing Machine (WM) = 100 : 75 = 4 : 3.

Total initial cost for 4 WM and 5 TV = (4 × 3) + (5 × 4) = 12 + 20 = 32 units.

Change in WM cost = +87% of 12 = 10.44; Change in TV cost = −45% of 20 = −9.

Net change = 10.44 − 9 = 1.44 units.

Percentage change = (1.44 ÷ 32) × 100 = 4.5%.

∴ The correct answer is Increase by 4.5%.

Alternate Method

Formulas: Total Cost = (Price × Quantity) | Percentage Change = (New − Old) ÷ Old × 100

⇒ Let the cost of TV be 100. Then cost of Washing Machine = 75.

⇒ Total Initial Cost = (4 × 75) + (5 × 100) = 300 + 500 = 800.

⇒ New WM cost = 75 + (87% of 75) = 75 + 65.25 = 140.25.

⇒ New TV cost = 100 − (45% of 100) = 55.

⇒ Total New Cost = (4 × 140.25) + (5 × 55) = 561 + 275 = 836.

⇒ Percentage Change = [(836 − 800) ÷ 800] × 100 = (36 ÷ 800) × 100 = 4.5%.

∴ The correct answer is Increase by 4.5%.

Additional Information

Percentage Increase and Decrease

To increase a value by x%, multiply it by (100 + x) ÷ 100. To decrease by x%, multiply by (100 − x) ÷ 100.

Weighted Average Concept

Net % change = (w₁c₁ + w₂c₂) ÷ (w₁ + w₂), where w is initial weight and c is % change.

Shortcut Trick

Total Distance = Sum of lengths = 232 + 248 = 480 m

Relative Speed (Opposite) = 51 + 57 = 108 km/hr

Relative Speed in m/s = 108 × (5/18) = 30 m/s

Time = Distance ÷ Relative Speed = 480 ÷ 30 = 16 seconds

∴ The correct answer is 16 sec.

Alternate Method

Given:

Length of first train (L₁) = 232 m

Length of second train (L₂) = 248 m

Speed of first train (S₁) = 51 km/hr

Speed of second train (S₂) = 57 km/hr

Formula Used:

Time = (Total Length) ÷ (Relative Speed)

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

Calculations:

⇒ Total distance to be covered = L₁ + L₂ = 232 + 248 = 480 m

⇒ Relative speed in opposite direction = S₁ + S₂ = 51 + 57 = 108 km/hr

⇒ Relative speed in m/s = 108 × (5/18) = 6 × 5 = 30 m/s

⇒ Time taken = 480 ÷ 30 = 16 seconds

∴ The correct answer is 16 sec.

Additional Information

Relative Speed Concepts

When two objects move in opposite directions, relative speed is (S₁ + S₂). When they move in the same direction, relative speed is |S₁ − S₂|.

Unit Conversion

Always ensure units are consistent. Multiply km/hr by 5/18 to get m/s; multiply m/s by 18/5 to get km/hr.

Train Crossing a Point Object

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

Shortcut Trick

0.36 = (3/5)², 0.216 = (3/5)³, 0.1296 = (3/5)⁴

LHS = [(3/5)²]³ × [(3/5)³]³ ÷ [(3/5)⁴]³

= (3/5)⁶ × (3/5)⁹ ÷ (3/5)¹²

= (3/5)³

RHS = (60/100)^x+1 = (3/5)^x+1

⇒ (3/5)³ = (3/5)^x+1

⇒ x + 1 = 3 ⇒ x = 2

∴ Answer = 2

Alternate Method

Given:

(0.36)³ × (0.216)³ ÷ (0.1296)³ = (60/100)^x+1

Formula used:

a^m × aⁿ ÷ a^p = a^m+n−p

Calculation:

⇒ 0.36 = (3/5)²

⇒ 0.216 = (3/5)³

⇒ 0.1296 = (3/5)⁴

⇒ LHS = [(3/5)²]³ × [(3/5)³]³ ÷ [(3/5)⁴]³

⇒ = (3/5)⁶ × (3/5)⁹ ÷ (3/5)¹²

⇒ = (3/5)⁶⁺⁹⁻¹² = (3/5)³

⇒ RHS = (60/100)^x+1 = (3/5)^x+1

⇒ Equate powers:

⇒ x + 1 = 3

⇒ x = 2

∴ The correct answer is 2.

Additional Information

Exponent Rule

Add/subtract powers for same base

Base Conversion

Convert decimals into fractional powers

Equality Rule

If bases are same, equate exponents

Key Insight

Simplify into single base before solving

Shortcut Trick

Required percentage = (Marks of B / Marks of A) × 100

Required percentage = (50 / 75) × 100 = (2 / 3) × 100

2/3 of 100% is approximately 66.67%.

∴ The correct answer is 66.67%.

Alternate Method

Given:

Marks obtained by B = 50

Marks obtained by A = 75

Formula Used:

Percentage = (Value / Base Value) × 100

Calculations:

⇒ Required Percentage = (Marks of B / Marks of A) × 100

⇒ Required Percentage = (50 / 75) × 100

⇒ Required Percentage = (2 / 3) × 100

⇒ Required Percentage = 66.666...

⇒ Rounding off to two decimal places, we get 66.67%.

∴ The correct answer is 66.67%.

Additional Information

Percentage Change

Percentage change is calculated as [(New Value − Old Value) / Old Value] × 100.

Percentage of Total

To find what percentage a score is of the maximum marks, use: (Obtained Marks / Maximum Marks) × 100.

Fraction to Percentage Conversion

Common fractions like 1/3 ≈ 33.33%, 2/3 ≈ 66.67%, and 1/4 = 25% are useful for quick calculations.

Shortcut Trick

Relative Speed in same direction = 101 − 65 = 36 kmph.

Convert Speed to m/s = 36 × (5/18) = 10 m/s.

Total distance (2L) = Speed × Time = 10 × 66 = 660 m.

Length of one train (L) = 660 ÷ 2 = 330 m.

∴ The correct answer is 330 m.

Alternate Method

Given: Speed of Rajdhani = 101 kmph, Speed of Garib Rath = 65 kmph, Time = 66 seconds, Lengths are equal (L).

Formula Used: Relative Speed (same direction) = S₁ − S₂; Distance = Speed × Time.

⇒ Relative Speed = 101 − 65 = 36 kmph

⇒ Speed in m/s = 36 × (5/18) = 10 m/s

⇒ Total Distance = 2 × L

⇒ 2L = 10 × 66

⇒ 2L = 660 m

⇒ L = 660 ÷ 2 = 330 m

∴ The correct answer is 330 m.

Additional Information

Relative Speed Concepts

When moving in the same direction, relative speed is S₁ − S₂. When moving in the opposite direction, it is S₁ + S₂.

Unit Conversion

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

Train Crossing Objects

When a train crosses another train, the total distance covered is the sum of their lengths (L₁ + L₂).

Shortcut Trick

Ratio of Numbers: N1 : N2 = 7 : 14 ⇒ 1 : 2

Ratio of Numbers: N2 : N3 = 15 : 15 ⇒ 1 : 1 ⇒ 2 : 2

Combined Ratio N1 : N2 : N3 = 1 : 2 : 2

Total units = 1 + 2 + 2 = 5 units → 180

Value of 2 units (Second number) = (180 ÷ 5) × 2 = 36 × 2 = 72

∴ The correct answer is 72.

Alternate Method

Given: Sum of numbers = 180; N2 : N1 = 14 : 7; N3 : N2 = 15 : 15

⇒ Simplify First Ratio: N1 : N2 = 7 : 14 = 1 : 2

⇒ Simplify Second Ratio: N2 : N3 = 15 : 15 = 1 : 1

⇒ To combine, make N2 same: N2 : N3 = (1×2) : (1×2) = 2 : 2

⇒ Combined Ratio N1 : N2 : N3 = 1 : 2 : 2

⇒ Let numbers be 1x, 2x, and 2x

⇒ 1x + 2x + 2x = 180

⇒ 5x = 180

⇒ x = 180 ÷ 5 = 36

⇒ Second number (N2) = 2x = 2 × 36 = 72

∴ The correct answer is 72.

Additional Information

Compound Ratio

When two or more ratios are multiplied together, the result is the compound ratio. For a:b and c:d, it is (ac):(bd).

Ratio Balancing

To combine A:B and B:C, multiply both ratios by suitable factors to make the common term B equal in both.

Proportion Rule

If a:b = c:d, then product of extremes (a × d) is always equal to the product of means (b × c).

Shortcut Trick

The Empirical Relationship states: Mode − Median = 2/3 (Mode − Mean).

Given: Mode − Mean = 27.3.

Required: Mode − Median.

⇒ Mode − Median = 2/3 × 27.3

⇒ Mode − Median = 2 × 9.1 = 18.2.

∴ The correct answer is 18.2.

Alternate Method

Given:

⇒ Difference between Mode and Mean = 27.3

⇒ Mode − Mean = 27.3

Formula Used:

⇒ Empirical Formula: Mode = 3 × Median − 2 × Mean

Calculations:

⇒ From the given data: Mean = Mode − 27.3

⇒ Substitute Mean in the empirical formula:

⇒ Mode = 3 × Median − 2 × (Mode − 27.3)

⇒ Mode = 3 × Median − 2 × Mode + 54.6

⇒ Mode + 2 × Mode − 3 × Median = 54.6

⇒ 3 × Mode − 3 × Median = 54.6

⇒ 3 × (Mode − Median) = 54.6

⇒ Mode − Median = 54.6 ÷ 3

⇒ Mode − Median = 18.2

∴ The correct answer is 18.2.

Additional Information

Empirical Relationship

For a moderately skewed distribution, the relationship is: Mode = 3 Median − 2 Mean. It helps find one measure when the other two are known.

Symmetric Distribution

In a perfectly symmetrical distribution, the values of Mean, Median, and Mode are all equal (Mean = Median = Mode).

Karl Pearson Coefficient of Skewness

Skewness is calculated as (Mean − Mode) ÷ Standard Deviation or 3(Mean − Median) ÷ Standard Deviation.

Relationship of Differences

The distance between Mean and Mode is 3 times the distance between Mean and Median: (Mean − Mode) = 3(Mean − Median).

Shortcut Trick

For an angle of elevation of 45°, the height of the building and its base distance are always equal.

In a 45°−45°−90° triangle, the ratio of sides is 1 : 1 : √2.

Since Height = 20 m, then Distance = 20 m.

∴ The correct answer is 20 m.

Alternate Method

Given:

Height of the building (AB) = 20 m

Angle of elevation (θ) = 45°

Formula Used:

tan θ = Perpendicular ÷ Base

Calculations:

⇒ tan 45° = AB ÷ BC

⇒ 1 = 20 ÷ BC

⇒ BC = 20 m

∴ The correct answer is 20 m.

Additional Information

Trigonometric Ratios for Standard Angles

For 30°, tan 30° = 1/√3. For 60°, tan 60° = √3. These ratios help find heights and distances quickly.

Angle of Elevation vs Depression

The angle of elevation is measured upward from the horizontal, while the angle of depression is measured downward. Both are numerically equal due to alternate interior angles.

Shortcut Trick

Using the inverse proportionality rule: Total Work = M₁ × D₁ = M₂ × D₂.

Initial men = 102, Final men = 102 + 68 = 170.

Days ratio = Inverse of Men ratio = 170 : 102 = 5 : 3.

5 units = 30 weeks → 1 unit = 6 weeks → 3 units = 18 weeks.

∴ The correct answer is 18 weeks.

Alternate Method

Given:

Initial number of men (M₁) = 102

Initial duration (D₁) = 30 weeks

Additional men = 68

Formula Used:

M₁ × D₁ = M₂ × D₂

Calculations:

⇒ M₂ = 102 + 68 = 170 men

⇒ 102 × 30 = 170 × D₂

⇒ 3060 = 170 × D₂

⇒ D₂ = 3060 ÷ 170

⇒ D₂ = 306 ÷ 17

⇒ D₂ = 18 weeks

∴ The correct answer is 18 weeks.

Additional Information

Chain Rule

The general formula is (M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂, where M is men, D is days, H is hours, and W is work.

Inverse Variation

In provision problems, the number of men is inversely proportional to the number of days the food lasts (M ∝ 1/D).

Constant Product

When the total amount of food (provisions) is constant, the product of the number of people and the time must remain the same.