Quantitative Aptitude PYQs

Shortcut Trick

Apply the BODMAS rule by grouping addition and subtraction for faster calculation.

⇒ 91 + 32 − 28 + (65 × 2) − (42 + 19)

⇒ (123 − 28) + 130 − 61

⇒ 95 + 130 − 61 = 164

∴ The correct answer is 164.

Alternate Method

Given: 91 + 32 − 28 + 65 × 2 − (42 + 19)

Formula Used: VBODMAS Rule (Vinculum, Brackets, Of, Division, Multiplication, Addition, Subtraction)

⇒ 91 + 32 − 28 + 65 × 2 − 61 (Solving the bracket: 42 + 19 = 61)

⇒ 91 + 32 − 28 + 130 − 61 (Solving the multiplication: 65 × 2 = 130)

⇒ 123 − 28 + 130 − 61 (Performing addition: 91 + 32 = 123)

⇒ 95 + 130 − 61 (Performing subtraction: 123 − 28 = 95)

⇒ 225 − 61 (Performing addition: 95 + 130 = 225)

⇒ 164 (Performing final subtraction: 225 − 61 = 164)

∴ The correct answer is 164.

Additional Information

BODMAS Rule

It is a mnemonic to help remember the order of operations: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Sign Rules

When performing operations, remember that (+) × (−) = (−) and (−) × (−) = (+). In addition, the sign of the number with the larger absolute value is kept.

Priority of Addition and Subtraction

Addition and subtraction have the same priority level and are generally solved from left to right in an expression.

Shortcut Trick

Solve the bracket and of operations first: (33 ÷ 3) = 11 and 18 of 1 = 18.

The expression becomes: 79 + 36 − 44 + 18 + 11 + 27.

Group all positive numbers: 79 + 36 + 18 + 11 + 27 = 171.

Subtract the remaining value: 171 − 44 = 127.

∴ The correct answer is 127.

Alternate Method

Given: 79 + 36 − 44 + 18 of 1 + (33 ÷ 3) + 27

To solve this, apply the BODMAS rule (Bracket, Of, Division, Multiplication, Addition, Subtraction).

⇒ 79 + 36 − 44 + 18 of 1 + 11 + 27

⇒ 79 + 36 − 44 + 18 + 11 + 27

⇒ 115 − 44 + 18 + 11 + 27

⇒ 71 + 18 + 11 + 27

⇒ 89 + 11 + 27

⇒ 100 + 27

⇒ 127

∴ The correct answer is 127.

Additional Information

BODMAS Rule

It defines the order of operations: Brackets, Of, Division, Multiplication, Addition, and then Subtraction.

Of Operator

The word 'of' acts as a multiplication operator but it must be calculated before any division or multiplication signs.

Bracket Sequence

Solve in the order: small brackets ( ), then curly brackets { }, and finally square brackets [ ].

Shortcut Trick

Product of two numbers = HCF × LCM

Other number = (18 × 540) ÷ 90

Other number = 18 × (540 ÷ 90) = 18 × 6

Other number = 108

∴ The correct answer is 108.

Alternate Method

Given:

HCF of two numbers = 18

LCM of two numbers = 540

First number (a) = 90

Formula Used:

Product of two numbers (a × b) = HCF(a, b) × LCM(a, b)

Calculations:

⇒ Let the other number be b.

⇒ 90 × b = 18 × 540

⇒ 90 × b = 9720

⇒ b = 9720 ÷ 90

⇒ b = 108

∴ The correct answer is 108.

Additional Information

HCF and LCM Property

For any two positive integers 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).

HCF as a Factor

The Highest Common Factor (HCF) of a set of numbers is always a divisor or factor of their Least Common Multiple (LCM).

Co-prime Numbers

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

Shortcut Trick

Use the algebraic identity: a⁴ − b⁴ = (a² − b²)(a² + b²).

The given expression simplifies to: a² − b².

Here, a = 0.7 and b = 0.3.

Calculation: (0.7)² − (0.3)² = 0.49 − 0.09 = 0.40.

∴ The correct answer is 0.40.

Alternate Method

Given:

Expression: ((0.7)⁴ − (0.3)⁴) / ((0.7)² + (0.3)²)

Formula Used:

Difference of squares: (x² − y²) = (x − y)(x + y)

Applying this to a⁴ − b⁴ by treating x = a² and y = b²:

a⁴ − b⁴ = (a²)² − (b²)² = (a² − b²)(a² + b²)

Calculations:

⇒ Let a = 0.7 and b = 0.3

⇒ Expression = [(a² − b²)(a² + b²)] / (a² + b²)

⇒ Expression = a² − b²

⇒ (0.7)² − (0.3)²

⇒ 0.49 − 0.09

⇒ 0.40

∴ The correct answer is 0.40.

Additional Information

Algebraic Identities

Basic identities like (a + b)(a − b) = a² − b² are frequently used to simplify high-power polynomial expressions quickly.

Higher Power Reductions

For even powers, always look for the form (aⁿ)² − (bⁿ)² to factorize the numerator and cancel common terms in the denominator.

BODMAS Rule

Always prioritize Brackets and Exponents before performing division or subtraction in complex numerical simplifications.

Shortcut Trick

For an expression of the form 1/(a + √b) + 1/(a − √b), the result is always 2a / (a² − b).

Here, a = 4 and b = 2.

Result = (2 × 4) / (4² − 2) = 8 / (16 − 2) = 8 / 14 = 4/7.

∴ The correct answer is 4/7.

Alternate Method

Given Expression: 1 / (4 + √2) + 1 / (4 − √2)

To solve this, we take the Least Common Multiple (LCM) of the denominators.

⇒ [(4 − √2) + (4 + √2)] / [(4 + √2) × (4 − √2)]

⇒ In the numerator, −√2 and +√2 cancel each other out.

⇒ Numerator = 4 + 4 = 8

⇒ In the denominator, use the algebraic identity (a + b)(a − b) = a² − b².

⇒ Denominator = 4² − (√2)²

⇒ Denominator = 16 − 2 = 14

⇒ Result = 8 / 14

⇒ Simplify the fraction by dividing both numbers by 2.

⇒ Result = 4 / 7

∴ The correct answer is 4/7.

Additional Information

Rationalizing the Denominator

To remove a radical from the denominator, multiply both the numerator and denominator by the conjugate of the denominator. For (a + √b), the conjugate is (a − √b).

Algebraic Identities

The identity (a + b)(a − b) = a² − b² is fundamental in simplifying expressions involving square roots and surds.

Surds Addition and Subtraction

Only like surds (surds with the same number under the radical sign) can be added or subtracted directly, such as 3√2 + 5√2 = 8√2.

Shortcut Trick

Find the individual square roots using simple visual observation:

⇒ √12321 = 111 (using the standard palindromic pattern where 111² = 12321)

⇒ √108900 = 330 (since 33² = 1089 and the square root of two trailing zeros is one zero)

⇒ Substituting these values: √(330 + 111) = √441 = 21

∴ The correct answer is 21.

Alternate Method

Given:

Expression = √(√108900 + √12321)

Calculation:

First, calculate the value of the inner square roots:

⇒ Let x = √12321

⇒ x = 111

⇒ Let y = √108900

⇒ y = 330

Now, substitute the values of x and y back into the main expression:

⇒ Required value = √(y + x)

⇒ Required value = √(330 + 111)

⇒ Required value = √441

⇒ Required value = 21

∴ The correct answer is 21.

Additional Information

Square Root of Palindromic Numbers

For numbers consisting of digits that increase sequentially up to a peak and then decrease (e.g., 121, 12321, 1234321), the square root contains only 1s, where the count of 1s equals the peak digit. Thus, √12321 = 111.

Square Root of Numbers with Trailing Zeros

When finding the square root of a perfect square ending in an even number of zeros, the square root contains exactly half as many zeros at the end. For example, √108900 = √1089 × √100 = 33 × 10 = 330.

Simplification Hierarchy

In mathematical expressions involving nested radicals, always simplify the innermost operations first before proceeding outward to the main radical.

Shortcut Trick

To find the area of a path built outside a rectangle, use the formula:

Area = 2w(L + B + 2w)

Where L = 150 m, B = 60 m, and w = 2 m.

Area = 2 × 2 × (150 + 60 + 2 × 2) = 4 × (210 + 4) = 4 × 214 = 856 m².

∴ The correct answer is 856 m².

Alternate Method

Given:

⇒ Length of garden (L) = 150 m

⇒ Breadth of garden (B) = 60 m

⇒ Width of path (w) = 2 m

Calculations:

⇒ Since the path is outside, Outer Length = 150 + 2 + 2 = 154 m

⇒ Outer Breadth = 60 + 2 + 2 = 64 m

⇒ Area of Path = (Outer Length × Outer Breadth) − (Inner Length × Inner Breadth)

⇒ Area = (154 × 64) − (150 × 60)

⇒ Area = 9856 − 9000

⇒ Area = 856 m²

∴ The correct answer is 856 m².

Additional Information

Path Inside a Rectangle

If the path is built inside the rectangular garden, the formula becomes Area = 2w(L + B − 2w).

Central Cross Paths

Area of two paths crossing each other at the center of a rectangle is given by Area = w(L + B − w).

Circular Path

The area of a circular path around a circle with radius 'r' and width 'w' is π((r+w)² − r²) = πw(2r + w).

Shortcut Trick

The sum of the squares of the first n natural numbers is calculated using the direct formula: [n × (n + 1) × (2n + 1)] / 6.

For n = 30, the sum is (30 × 31 × 61) / 6.

On simplifying: 5 × 31 × 61 = 155 × 61 = 9455.

∴ The correct answer is 9455.

Alternate Method

Given: The series is the sum of squares of the first 30 natural numbers: 1² + 2² + 3² + ... + 30².

Formula Used: Sum (S_n) = [n(n + 1)(2n + 1)] / 6

⇒ Here, the total number of terms n = 30.

⇒ S₃₀ = [30 × (30 + 1) × (2 × 30 + 1)] / 6

⇒ S₃₀ = [30 × 31 × 61] / 6

⇒ S₃₀ = 5 × 31 × 61

⇒ S₃₀ = 155 × 61

⇒ S₃₀ = 9455

∴ The correct answer is 9455.

Additional Information

Sum of First n Natural Numbers

The sum of the first n natural numbers (1 + 2 + 3 + ... + n) is given by the formula: [n(n + 1)] / 2.

Sum of Cubes of First n Natural Numbers

The sum of the cubes of the first n natural numbers (1³ + 2³ + ... + n³) is given by: [[n(n + 1)] / 2]².

Sum of First n Odd Numbers

The sum of the first n odd natural numbers (1 + 3 + 5 + ... up to n terms) is equal to n².

Shortcut Trick

Identify the Pythagorean triplet for the right triangle formed by half-diagonals and the side (15 cm): 9, 12, 15.

Check sum of diagonals: (2 × 9) + (2 × 12) = 18 + 24 = 42 cm (Matches given condition).

Diagonals are 18 cm and 24 cm.

Area = (18 × 24) ÷ 2 = 216 cm².

∴ The correct answer is 216 cm².

Alternate Method

Given:

Sum of diagonals (d₁ + d₂) = 42 cm

Side of rhombus (s) = 15 cm

Formula Used:

1. Relationship between side and diagonals: 4s² = d₁² + d₂²

2. Area of rhombus = (d₁ × d₂) ÷ 2

Calculations:

⇒ d₁² + d₂² = 4 × s²

⇒ d₁² + d₂² = 4 × 15² = 4 × 225 = 900

⇒ We know: (d₁ + d₂)² = d₁² + d₂² + 2d₁d₂

⇒ 42² = 900 + 2d₁d₂

⇒ 1764 = 900 + 2d₁d₂

⇒ 2d₁d₂ = 1764 − 900 = 864

⇒ d₁d₂ = 864 ÷ 2 = 432

⇒ Area = (d₁ × d₂) ÷ 2 = 432 ÷ 2 = 216 cm²

∴ The correct answer is 216 cm².

Additional Information

Properties of a Rhombus

A rhombus is a quadrilateral with all four sides equal. Its diagonals bisect each other at 90°.

Area and Perimeter

Area = ½ × d₁ × d₂ or Base × Height. Perimeter = 4 × side.

Diagonal and Side Relation

The side s and diagonals d₁, d₂ are related by s² = (d₁/2)² + (d₂/2)² based on the Pythagorean theorem.

Shortcut Trick

New Average = New Weight + (Initial Students × Reduction)

⇒ New Average = 40 + (39 × 0.5)

⇒ New Average = 40 + 19.5

⇒ New Average = 59.5 kg

∴ The correct answer is 59.5 kg.

Alternate Method

Given:

Initial students = 39

Weight of new student = 40 kg

Reduction in average = 0.5 kg

Formula Used:

Average = Total Sum of Observations ÷ Number of Observations

Calculations:

⇒ Let the initial average weight of the 39 students be x kg.

⇒ Total weight of 39 students = 39 × x = 39x

⇒ Total weight of 40 students (including the new student) = 39x + 40

⇒ New average weight = x − 0.5

⇒ New Average = (Total weight of 40 students) ÷ 40

⇒ x − 0.5 = (39x + 40) ÷ 40

⇒ 40 × (x − 0.5) = 39x + 40

⇒ 40x − 20 = 39x + 40

⇒ 40x − 39x = 40 + 20

⇒ x = 60 kg (Old Average)

⇒ New Average weight = 60 − 0.5 = 59.5 kg

∴ The correct answer is 59.5 kg.

Additional Information

Net Deviation Method

The sum of deviations of all observations from their average is always zero. New member weight − Old average = Total members × Change in average.

Weighted Average Formula

Weighted Average = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is the count and x is the individual average/value.

Inclusion/Exclusion Rule

When a new person joins, New Value = New Average + Initial Number × (Change in Average). If the average decreases, the change is considered positive in this specific adjustment format.

Shortcut Trick

The Average (p) of a set of numbers must always satisfy the condition: Smallest Number ≤ p ≤ Largest Number.

Given the Smallest number is 22, the Average (p) can be any value greater than or equal to 22, and the Largest number can be any value greater than or equal to 22.

Statement I claims p is exactly 35, which is not mandatory.

Statement II claims the Largest number is exactly 50, which is also not mandatory.

Since both are specific cases and not universal rules, neither is always correct.

∴ The correct answer is Neither I nor II.

Alternate Method

Given: Smallest number = 22, Average = p.

⇒ Let the set of numbers be {22, 23}. Here, Average (p) = 22.5 and Largest = 23. (Both I and II are False)

⇒ Let the set of numbers be {22, 48}. Here, Average (p) = 35 and Largest = 48. (Statement II is False)

⇒ Let the set of numbers be {22, 22, 106}. Here, Average (p) = 50 and Largest = 106. (Statement I is False)

⇒ Since we can find instances where these statements do not hold, they are not always correct.

∴ The correct answer is Neither I nor II.

Additional Information

Range Property of Average

The average of any data set must lie strictly between (or equal to) the minimum and maximum values of the set: x_min ≤ Average ≤ x_max.

Sum of Deviations

The algebraic sum of the deviations of a set of observations from their Arithmetic Mean is always zero: Σ(x_i − p) = 0.

Effect of Change in Origin

If a constant k is added to or subtracted from every term in a set, the new average will be p ± k.

Shortcut Trick

If the ratio of two numbers is a : b and their Highest Common Factor (HCF) is H, then the numbers are (a × H) and (b × H).

Given Ratio = 7 : 9 and HCF = 3.

Numbers = (7 × 3) and (9 × 3) = 21 and 27.

∴ The correct answer is 21, 27.

Alternate Method

Given: Ratio of numbers = 7 : 9 and their HCF = 3.

Formula Used: Number = Ratio Unit × HCF

⇒ Let the two numbers be 7x and 9x.

⇒ The HCF of 7x and 9x is x (since 7 and 9 are co-prime numbers).

⇒ According to the question, HCF = 3.

⇒ Therefore, x = 3.

⇒ First number = 7 × 3 = 21.

⇒ Second number = 9 × 3 = 27.

∴ The correct answer is 21, 27.

Additional Information

Fundamental Relation

For any two numbers, the product of the numbers is equal to the product of their HCF and LCM. Formula: HCF × LCM = n₁ × n₂.

LCM from HCF and Ratio

If the ratio of two numbers is x : y and their HCF is H, then their Least Common Multiple (LCM) is given by LCM = x × y × H.

Co-prime Numbers

Two numbers are called co-prime if their Highest Common Factor is 1. In a simplified ratio a : b, a and b are always co-prime.

Shortcut Trick

Direct relation: 16% of total mixture = 64 kg

Divide by 16 to find 1%: 1% of mixture = 64 ÷ 16 = 4 kg

Multiply by 100 for 100%: Total mixture = 4 × 100 = 400 kg

∴ The correct answer is 400 kg.

Alternate Method

Given: Weight of Element A = 64 kg, Percentage of Element A in the mixture = 16%

⇒ Let the total weight of the mixture be X kg.

⇒ 16% of X = 64

⇒ (16 ÷ 100) × X = 64

⇒ X = (64 × 100) ÷ 16

⇒ X = 4 × 100

⇒ X = 400 kg

∴ The correct answer is 400 kg.

Additional Information

Mass Percentage Formula

Concentration of a substance = (Mass of substance ÷ Total mass of mixture) × 100

Ratio Method for Mixtures

16% can be written as 4:25 (Part:Total). If 4 units represent 64 kg, then 1 unit is 16 kg, making 25 units equal to 400 kg.

Finding Total from Percentage Change

Total Value = Given Value ÷ Percentage Fraction (e.g., Total = 64 ÷ 0.16 = 400 kg).

Shortcut Trick

The ratio of Cost Price (CP) to Selling Price (SP) is 5 : 2.

Loss units = CP − SP = 5 − 2 = 3 units.

Given, 3 units = ₹300 → 1 unit = ₹100.

Cost Price = 5 units = 5 × ₹100 = ₹500.

∴ The correct answer is ₹500.

Alternate Method

Given:

Ratio of CP to SP = 5 : 2

Total Loss = ₹300

Formula Used:

Loss = Cost Price − Selling Price

Calculations:

⇒ Let the Cost Price be 5x and the Selling Price be 2x.

⇒ Loss = 5x − 2x

⇒ 300 = 3x

⇒ x = 300 ÷ 3

⇒ x = 100

⇒ Cost Price of the article = 5x = 5 × 100 = ₹500.

∴ The correct answer is ₹500.

Additional Information

Loss Percentage

It is the loss incurred per 100 units of cost price. Formula: Loss % = (Loss ÷ CP) × 100.

Profit and Loss Condition

If CP > SP, there is a Loss. If SP > CP, there is a Profit. If CP = SP, there is no profit and no loss.

Selling Price for Loss Case

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

Shortcut Trick

Given Cost Price (CP) : Selling Price (SP) = 10 : 13

Directly take CP = 10 units and SP = 13 units.

Profit = 13 − 10 = 3 units.

Required Ratio (Profit : CP) = 3 : 10.

∴ The correct answer is 3 : 10.

Alternate Method

Given:

Ratio of Cost Price to Selling Price = 10 : 13

Formula Used:

Profit = Selling Price (SP) − Cost Price (CP)

Calculations:

⇒ Let Cost Price (CP) = 10x

⇒ Selling Price (SP) = 13x

⇒ Profit = 13x − 10x = 3x

⇒ Ratio = Profit ÷ Cost Price

⇒ Ratio = 3x ÷ 10x = 3 : 10

∴ The correct answer is 3 : 10.

Additional Information

Profit Percentage

Profit percentage is always calculated on the Cost Price: Profit% = (Profit ÷ CP) × 100. In this case, Profit% = (3 ÷ 10) × 100 = 30%.

Loss Condition

When CP is greater than SP, a loss occurs: Loss = CP − SP. Loss Percentage = (Loss ÷ CP) × 100.

Marked Price and Discount

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

Shortcut Trick

Convert values to a common unit: Actual time = 500 min and Estimated time = 520 min.

Absolute Error = 520 − 500 = 20 min.

Error percentage = (Error ÷ Actual Value) × 100.

Error % = (20 ÷ 500) × 100 = 4%.

∴ The correct answer is 4 percent.

Alternate Method

Given:

Actual time interval = 8 hr 20 min

Estimated time interval = 8 hr 40 min

Formula Used:

Error Percentage = (Absolute Error ÷ Actual Value) × 100

Calculations:

⇒ Actual time in minutes = (8 × 60) + 20 = 480 + 20 = 500 min

⇒ Estimated time in minutes = (8 × 60) + 40 = 480 + 40 = 520 min

⇒ Absolute Error = 520 − 500 = 20 min

⇒ Error Percentage = (20 ÷ 500) × 100

⇒ Error Percentage = 2000 ÷ 500 = 4%

∴ The correct answer is 4 percent.

Additional Information

Absolute Error

It is the numerical difference between the measured value and the actual value: |Measured − Actual|.

Relative Error

It is the ratio of the absolute error to the actual value. It expresses how large the error is relative to the total size.

Percentage Change

The formula for percentage increase or decrease is always calculated on the original (base) value: (Change ÷ Original Value) × 100.

Shortcut Trick

In fraction-based problems, the Total Work is represented by the denominator.

If 1 unit (out of 5) takes 36 min to empty,

Then 5 units (complete tank) will take 36 × 5 = 180 min.

∴ The correct answer is 180 min.

Alternate Method

Given:

Part of the tank emptied = 1/5

Time taken to empty this part = 36 minutes

Formula Used:

Total Time = (Time Taken) ÷ (Part Emptied)

Calculations:

⇒ Let the total capacity of the tank be 1 unit.

⇒ Time taken to empty 1/5 part = 36 min

⇒ Time taken to empty 1 unit = 36 ÷ (1/5)

⇒ Total Time = 36 × 5

⇒ Total Time = 180 min

∴ The correct answer is 180 min.

Additional Information

Pipes and Cisterns Rule

If a pipe can empty a tank in x hours, then the part emptied in 1 hour is 1/x.

Unitary Method in Fractions

If a/b part of a work is done in T time, then the whole work is done in T × (b/a) time.

Efficiency and Time

The Efficiency of a pipe is inversely proportional to the Time taken to complete the work (filling or emptying).

Shortcut Trick

Total work (LCM of 15, 25, 50) = 150 units.

Combined efficiency of A, B, and C = 10 + 6 + 3 = 19 units/day.

Work completed by all three in 5 days = 19 × 5 = 95 units.

Remaining work = 150 − 95 = 55 units.

Time taken by C to finish = 55 ÷ 3 = 55/3 days.

∴ The correct answer is 55/3 days.

Alternate Method

Given: Time taken by A = 15 days, B = 25 days, C = 50 days, and time they worked together = 5 days.

Formula Used: Efficiency = Total Work ÷ Time taken.

⇒ Total efficiency of (A + B + C) = 10 + 6 + 3 = 19 units/day.

⇒ Work done by all three in 5 days = 19 × 5 = 95 units.

⇒ Remaining work to be completed = Total Work − Work Done = 150 − 95 = 55 units.

⇒ Time taken by C alone to complete the remaining work = 55 ÷ 3 days.

∴ The correct answer is 55/3 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken. If the ratio of time taken by two people is x : y, then the ratio of their efficiency is y :

x. Pipes and Cisterns

The same LCM method applies. Inlet pipes have positive efficiency, while outlet (leakage) pipes have negative efficiency.

Work and Wages

Wages are always distributed in the ratio of the actual work done by each person. If they work for the same duration, wages are distributed in the ratio of their efficiencies.

Shortcut Trick

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

Speed = 108 × (5/18)

Speed = 6 × 5 = 30 m/s

∴ The correct answer is 30 m/s.

Alternate Method

Given:

Speed of the car = 108 km/hr

Formula Used:

⇒ 1 kilometer = 1000 meters

⇒ 1 hour = 60 minutes = 3600 seconds

⇒ Conversion factor = 1000/3600 = 5/18

Calculations:

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

⇒ Speed = 108 × (5/18)

⇒ Speed = 108 ÷ 18 × 5

⇒ Speed = 6 × 5

⇒ Speed = 30 m/s

∴ The correct answer is 30 m/s.

Additional Information

Speed Conversion Rule

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

Average Speed Formula

When a body covers equal distances at speeds x and y, the average speed is (2xy)/(x + y).

Relative Speed

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

Shortcut Trick

Convert speed to m/s: 54 × 5/18 = 15 m/s.

Ratio of Train Length : Platform Length = 1 : 2.

Total units of distance = 1 + 2 = 3 units.

Total distance = 15 m/s × 18 sec = 270 m.

1 unit (Train Length) = 270 ÷ 3 = 90 m.

∴ The correct answer is 90 m.

Alternate Method

Given: Speed = 54 km/hr, Time = 18 sec, Platform = 2 × Train Length.

Formula: Total Distance = Length of Train + Length of Platform = Speed × Time.

⇒ Speed in m/s = 54 × (5/18) = 15 m/s

⇒ Let Length of the train be L.

⇒ Length of the platform = 2L.

⇒ Total Distance = L + 2L = 3L.

⇒ 3L = 15 × 18

⇒ 3L = 270

⇒ L = 270 ÷ 3 = 90

∴ The correct answer is 90 m.

Additional Information

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 Concepts

When a train crosses a pole or stationary man, distance = Length of train. When it crosses a bridge or platform, distance = Length of train + Length of bridge/platform.

Relative Speed

For two objects moving in the same direction, relative speed = S₁ − S₂. For opposite directions, relative speed = S₁ + S₂.