Quantitative Aptitude PYQs

Shortcut Trick

Solve the expression using the BODMAS rule priority.

First, evaluate the bracket: (33 + 26) = 59.

Next, perform multiplication: 52 × 2 = 104.

Then, group addition and subtraction: (89 + 96 + 104) − (47 + 59).

Final calculation: 289 − 106 = 183.

∴ The correct answer is 183.

Alternate Method

Given:

89 − 47 + 96 − (33 + 26) + 52 × 2

Formula Used:

The BODMAS rule determines the order of operations: Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Calculations:

⇒ 89 − 47 + 96 − 59 + 52 × 2 (Solving bracket)

⇒ 89 − 47 + 96 − 59 + 104 (Solving multiplication)

⇒ 42 + 96 − 59 + 104 (Performing subtraction)

⇒ 138 − 59 + 104 (Performing addition)

⇒ 79 + 104 (Performing subtraction)

⇒ 183

∴ The correct answer is 183.

Additional Information

BODMAS Rule

It stands for Brackets (), Orders (powers/roots), Division (÷), Multiplication (×), Addition (+), and Subtraction (−). Operations must be performed in this specific sequence.

Unit Digit Check

For quick verification, use unit digits: 9 − 7 + 6 − (3 + 6) + (2 × 2) → 9 − 7 + 6 − 9 + 4 = 3. The final answer must end in 3.

Grouping Method

To avoid errors, group all positive terms and negative terms separately before the final subtraction: (Sum of Positives) − (Sum of Negatives).

Shortcut Trick

Solve using BODMAS rule: Bracket → Division → Multiplication → Addition → Subtraction.

Step 1: 87 + 22 − 6 + 29 − 13 + 18

Step 2: Group positive and negative terms: (87 + 22 + 29 + 18) − (6 + 13)

Step 3: 156 − 19 = 137

∴ The correct answer is 137.

Alternate Method

Given: 87 + 22 − (36 ÷ 6) + 29 × 1 − 13 + 18

Formula Used: BODMAS Rule

⇒ 87 + 22 − 6 + 29 × 1 − 13 + 18 (Solving Bracket and Division)

⇒ 87 + 22 − 6 + 29 − 13 + 18 (Solving Multiplication)

⇒ 109 − 6 + 29 − 13 + 18 (Solving Addition from left)

⇒ 103 + 29 − 13 + 18

⇒ 132 − 13 + 18

⇒ 119 + 18

⇒ 137

∴ The correct answer is 137.

Additional Information

BODMAS / VBODMAS Rule

It defines the order of operations: Vinculum (bar), Brackets, Of (orders/powers), Division, Multiplication, Addition, and Subtraction.

Left-to-Right Precedence

Addition and subtraction have the same priority level; they are typically solved from left to right to avoid confusion, though grouping positives and negatives is also a valid shortcut.

Multiplication by 1

Any number multiplied by 1 remains unchanged (Identity property of multiplication), which simplifies 29 × 1 to 29.

Shortcut Trick

To find the maximum number of students, calculate the HCF (Highest Common Factor) of 84 and 108.

The difference between 108 and 84 is 24. Factors of 24 are 24, 12, 8, 6, 4, 3, 2, 1.

Check the largest factor of the difference that divides both numbers: 12 × 7 = 84 and 12 × 9 = 108.

So, the maximum students = 12.

Items per student: 84 ÷ 12 = 7 chocolates and 108 ÷ 12 = 9 candies.

∴ The correct answer is 12 students; 7 chocolates and 9 candies.

Alternate Method

Given: Total chocolates = 84 and total candies = 108.

For equal distribution among the maximum number of students, we need the HCF of (84, 108).

⇒ Prime factorization of 84 = 2² × 3 × 7

⇒ Prime factorization of 108 = 2² × 3³

⇒ HCF = 2² × 3 = 4 × 3 = 12

⇒ Number of chocolates for each student = 84 ÷ 12 = 7

⇒ Number of candies for each student = 108 ÷ 12 = 9

∴ The correct answer is 12 students; 7 chocolates and 9 candies.

Additional Information

HCF (Highest Common Factor)

HCF is the largest number that divides two or more numbers without leaving a remainder. It is used for problems involving maximum grouping or equal distribution.

Relationship between HCF and LCM

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

HCF by Difference Method

The HCF of two numbers is always a factor of their difference. This is a quick way to narrow down potential HCF values in competitive exams.

Shortcut Trick

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

Here, let a = 12.35 and b = 7.65.

Substituting the values: (12.35 + 7.65)²

⇒ (20)² = 400

∴ The correct answer is 400.

Alternate Method

Given:

The mathematical expression to solve is: (12.35 − 7.65)² + 4 × 12.35 × 7.65

Formula Used:

The standard algebraic expansion formulas are:

1. (a − b)² = a² + b² − 2ab

2. (a + b)² = a² + b² + 2ab

Adding 4ab to the first equation gives: (a − b)² + 4ab = a² + b² − 2ab + 4ab = a² + b² + 2ab = (a + b)²

Calculations:

⇒ Let a = 12.35 and b = 7.65.

⇒ The given expression is in the form: (a − b)² + 4ab

⇒ According to the identity, (a − b)² + 4ab = (a + b)²

⇒ Expression = (12.35 + 7.65)²

⇒ Expression = (20.00)²

⇒ Expression = 400

∴ The correct answer is 400.

Additional Information

Standard Algebraic Identities

Familiarity with identities like (a+b)² − (a−b)² = 4ab and (a+b)² + (a−b)² = 2(a²+b²) helps in solving complex simplification problems quickly.

Sum and Difference Properties

When the sum (a+b) or difference (a−b) results in a base number (like 10, 20, 100), always look for expansion patterns to avoid direct multiplication of decimals.

Shortcut Trick

Apply the standard algebraic identity: (a + b)² + (a − b)² = 2(a² + b²).

Here, let a = √3 and b = √2.

Result = 2 × [(√3)² + (√2)²]

Result = 2 × (3 + 2) = 2 × 5 = 10.

∴ The correct answer is 10.

Alternate Method

Given:

(√3 + √2)² + (√3 − √2)²

Formula Used:

(a + b)² = a² + b² + 2ab

(a − b)² = a² + b² − 2ab

Calculations:

⇒ [(√3)² + (√2)² + 2 × √3 × √2] + [(√3)² + (√2)² − 2 × √3 × √2]

⇒ [3 + 2 + 2√6] + [3 + 2 − 2√6]

⇒ 5 + 2√6 + 5 − 2√6

⇒ 5 + 5 + 2√6 − 2√6

⇒ 10

∴ The correct answer is 10.

Additional Information

Square of Difference Identity

The difference between the squares of sum and difference is given by (a + b)² − (a − b)² = 4ab.

Difference of Squares

The product of sum and difference of two terms is (a + b)(a − b) = a² − b².

Square of a Binomial

A basic expansion formula: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².

Shortcut Trick

Recognize the algebraic identity a² − b² = (a + b)(a − b).

Here, a = 6.203 and b = 5.797.

Sum (a + b) = 12 and Difference (a − b) = 0.406.

The expression becomes (12 × 0.406) ÷ 0.812.

Since 0.812 = 2 × 0.406, the result is 12 ÷ 2 = 6.

∴ The correct answer is 6.

Alternate Method

Given:

Expression: ((6.203)² − (5.797)²) ÷ 0.812

Formula Used:

Algebraic identity: a² − b² = (a + b)(a − b)

Calculations:

⇒ [(6.203 + 5.797) × (6.203 − 5.797)] ÷ 0.812

⇒ [12.000 × 0.406] ÷ 0.812

⇒ 4.872 ÷ 0.812

⇒ 4872 ÷ 812

⇒ 6

∴ The correct answer is 6.

Additional Information

Difference of Squares

The identity a² − b² = (a + b)(a − b) is used to simplify the subtraction of two squared terms.

Square of a Sum

(a + b)² = a² + 2ab + b² is used for calculating the square of the sum of two numbers.

Square of a Difference

(a − b)² = a² − 2ab + b² is used for calculating the square of the difference between two numbers.

Shortcut Trick

Use the standard Pythagorean Triplet (3, 4, 5).

Multiply the triplet by 3: (3 × 3, 4 × 3, 5 × 3) = (9, 12, 15).

Given Length = 9 cm and Diagonal = 15 cm, the Width must be 12 cm.

Area = Length × Width = 9 × 12 = 108 cm².

∴ The correct answer is 108 cm².

Alternate Method

Given:

Length (l) = 9 cm

Diagonal (d) = 15 cm

Formula Used:

Diagonal² = Length² + Width²

Area = Length × Width

Calculations:

⇒ 15² = 9² + Width²

⇒ 225 = 81 + Width²

⇒ Width² = 225 − 81

⇒ Width² = 144

⇒ Width = √144 = 12 cm

⇒ Area of Rectangle = Length × Width

⇒ Area = 9 cm × 12 cm = 108 cm²

∴ The correct answer is 108 cm².

Additional Information

Pythagorean Triplet

A set of three integers satisfying a² + b² = c². Common examples are (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Perimeter of Rectangle

The distance around the outside of the rectangle, calculated as P = 2 × (length + width).

Properties of Rectangle

Opposite sides are equal and parallel, all four angles are 90°, and diagonals are equal and bisect each other.

Shortcut Trick

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

Here, let a = 412 and b = 388.

⇒ (412 + 388)² = (800)²

⇒ 640000

∴ The correct answer is 640000.

Alternate Method

Given: (412 − 388)² + 4 × 412 × 388

Formula Used: Expansion of (a − b)² and (a + b)²

⇒ (412 − 388) = 24

⇒ (24)² + 4 × 412 × 388

⇒ 576 + 639424

⇒ 640000

∴ The correct answer is 640000.

Additional Information

Square of Sum

The identity for the square of a sum is (a + b)² = a² + 2ab + b².

Square of Difference

The identity for the square of a difference is (a − b)² = a² − 2ab + b².

Relationship Identity

Combining the two provides the useful relation: (a + b)² − (a − b)² = 4ab.

Shortcut Trick

Let the Ratio of Length to Breadth be 3 :

1. Perimeter ratio = 2 × (3 + 1) = 8 units = 48 cm.

1 unit = 6 cm ⇒ Breadth = 6 cm and Length = 18 cm.

Diagonal = √(18² + 6²) = 6 × √(3² + 1²) = 6√10 cm.

∴ The correct answer is 6√10 cm.

Alternate Method

Given: Perimeter of Rectangle = 48 cm, Length (L) = 3 × Breadth (B).

Formula Used: Perimeter = 2(L + B) | Diagonal (d) = √(L² + B²)

⇒ 2(L + B) = 48

⇒ L + B = 24

⇒ Substitute L = 3B: 3B + B = 24

⇒ 4B = 24 ⇒ B = 6 cm

⇒ L = 3 × 6 = 18 cm

⇒ Diagonal (d) = √(18² + 6²)

⇒ d = √(324 + 36) = √360

⇒ d = √(36 × 10) = 6√10 cm

∴ The correct answer is 6√10 cm.

Additional Information

Area of a Rectangle

The total space occupied by a Rectangle is calculated as Area = Length × Breadth.

Pythagoras Theorem

For a Right-angled Triangle formed by the Diagonal, L² + B² = d².

Circumcircle of a Rectangle

The Radius of the circle passing through all four vertices of a Rectangle is half of its Diagonal (R = d/2).

Shortcut Trick

Ratio of Average of first 4 and last 4 numbers = 2 :

1. Since the count of numbers is the same (4 each), the Sum ratio is also 2 :
2. Total units for sum = 2 + 1 = 3 units.

Total Sum of 8 numbers = 8 × 57 = 456.

Sum of last 4 numbers (1 unit) = 456 ÷ 3 = 152.

∴ The correct answer is 152.

Alternate Method

Given: Average of 8 numbers = 57. Average of first 4 is double the average of the last 4.

Formula Used: Sum = Average × Number of observations.

⇒ Total Sum of all 8 numbers = 8 × 57 = 456.

⇒ Let the average of the last 4 numbers = x.

⇒ Average of the first 4 numbers = 2x.

⇒ Sum of first 4 numbers = 4 × 2x = 8x.

⇒ Sum of last 4 numbers = 4 × x = 4x.

⇒ Total sum = 8x + 4x = 12x.

⇒ 12x = 456.

⇒ x = 456 ÷ 12 = 38.

⇒ Sum of the last 4 numbers = 4x = 4 × 38 = 152.

∴ The correct answer is 152.

Additional Information

Weighted Average

When two groups are combined, the total average is calculated as A = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂).

Sum of Observations

The total sum is always the product of the Arithmetic Mean and the total number of items: Sum = Average × Count.

Equal Groups Property

If the number of items in two groups is equal, the Combined Average is simply the average of their individual averages: (a₁ + a₂) ÷ 2.

Shortcut Trick

For any set of numbers in an Arithmetic Progression (like consecutive natural numbers), the Average is always equal to the Arithmetic Mean of the first and last terms.

Average (A) = (Smallest + Largest) ÷ 2

By cross-multiplying:

Sum of Smallest and Largest = 2 × Average = 2A

∴ The correct answer is 2A.

Alternate Method

Given: Total numbers (n) = 17, Average of these consecutive numbers = A.

Formula Used: Average of an Arithmetic Progression (AP) = (a + l) ÷ 2, where 'a' is the first term and 'l' is the last term.

⇒ Let the 17 consecutive natural numbers be n₁, n₂, n₃, ..., n₁₇.

⇒ These numbers form an Arithmetic Progression with a common difference (d) = 1.

⇒ For any AP, the Sum of all terms = (n ÷ 2) × (First term + Last term).

⇒ Also, Average = Sum ÷ n.

⇒ Average (A) = [(n ÷ 2) × (n₁ + n₁₇)] ÷ n

⇒ A = (n₁ + n₁₇) ÷ 2

⇒ n₁ + n₁₇ = 2 × A = 2A

∴ The correct answer is 2A.

Additional Information

Middle Term Property

In an Arithmetic Progression with an odd number of terms (n), the Average is always the exact middle term, which is the ((n+1)/2)^th term.

Sum of First 'n' Natural Numbers

The sum of the first 'n' natural numbers is calculated using the formula: S_n = n(n + 1) ÷ 2.

Average of Even and Odd Numbers

The average of the first 'n' consecutive even numbers is (n + 1), while the average of the first 'n' consecutive odd numbers is simply n.

Shortcut Trick

To find the maximum number of boys, calculate the HCF of 210 and 294.

HCF(210, 294): 294 − 210 = 84. Factors of 84 are 42, 21, etc. Since 42 divides both, HCF = 42.

Number of chips per boy = Total chips ÷ HCF = 210 ÷ 42 = 5.

∴ The correct answer is 5 chips.

Alternate Method

Given: Total chips = 210, Total toffees = 294.

Formula Used: Maximum group size = HCF (Highest Common Factor); Items per person = Total items ÷ Group size.

⇒ Prime factorization of 210 = 2 × 3 × 5 × 7

⇒ Prime factorization of 294 = 2 × 3 × 7 × 7

⇒ HCF (Maximum number of boys) = 2 × 3 × 7 = 42

⇒ Chips received by each boy = Total chips ÷ Total boys

⇒ Chips per boy = 210 ÷ 42 = 5

∴ The correct answer is 5 chips.

Additional Information

Highest Common Factor (HCF)

HCF is the greatest number that divides two or more numbers without leaving a remainder. It is used to find the maximum possible size for equal distribution.

Least Common Multiple (LCM)

LCM is the smallest positive integer that is divisible by each of the given numbers. Relationship: Product of two numbers = HCF × LCM.

HCF by Difference Method

The HCF of two numbers is also a factor of their difference. This is a quick way to narrow down the HCF in exams.

Shortcut Trick

Let the total number be 100%.

Given that 30% of the number + 70 = 100%.

Subtracting percentages: 100% − 30% = 70%.

This implies that 70% of the number is equal to 70.

Calculating the full value: 100% = (70 / 70) × 100 = 100.

∴ The correct answer is 100.

Alternate Method

Given:

Percentage added = 30%

Constant value added = 70

Formula Used:

Number = [Value / (100% − Percentage)] × 100

Let the number be x.

⇒ (30 / 100) × x + 70 = x

⇒ 0.3x + 70 = x

⇒ 70 = x − 0.3x

⇒ 70 = 0.7x

⇒ x = 70 / 0.7

⇒ x = 100

∴ The correct answer is 100.

Additional Information

Percentage to Fraction

To convert a percentage to a fraction, divide the number by 100. Example: 30% = 30/100 = 3/10.

Percentage Change

The percentage change is calculated as [(Final Value − Initial Value) / Initial Value] × 100.

Base Value Rule

In most word problems, the 'number' or 'original quantity' is always treated as the 100% base for calculations.

Shortcut Trick

Profit of 30% implies that the Selling Price (SP) is 130% of the Cost Price (CP).

Ratio method: 30% = 3/10 ⇒ CP = 10 units, Profit = 3 units, SP = 13 units.

13 units = ₹7540

1 unit = 7540 ÷ 13 = 580

CP = 10 units = 580 × 10 = ₹5800

∴ The correct answer is ₹5800.

Alternate Method

Given:

Selling Price (SP) = ₹7540

Profit Percentage = 30%

Formula Used:

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

Calculations:

⇒ CP = [7540 × 100] ÷ [100 + 30]

⇒ CP = 754000 ÷ 130

⇒ CP = 75400 ÷ 13

⇒ CP = ₹5800

∴ The correct answer is ₹5800.

Additional Information

Basic Profit and Loss Formulas

Profit = SP − CP and Profit % = (Profit ÷ CP) × 100.

Selling Price Calculation

SP = CP × (100 + Profit %)/100 (if profit) or SP = CP × (100 − Loss %)/100 (if loss).

Marked Price and Discount

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

Shortcut Trick

The difference between Selling Price (SP) and Cost Price (CP) is the Profit.

Given Profit = 30% of CP = ₹120.

⇒ 30% → 120

⇒ 1% → 4

Selling Price = 100% (CP) + 30% (Profit) = 130%.

⇒ 130 × 4 = ₹520.

∴ The correct answer is ₹520.

Alternate Method

Given: Difference between CP and SP (Profit) = ₹120, Profit Percentage = 30%.

Formula Used: Profit = SP − CP; Profit % = (Profit ÷ CP) × 100; SP = CP + Profit.

⇒ Let the Cost Price (CP) be ₹x.

⇒ Profit = 30% of x = 120

⇒ (30 ÷ 100) × x = 120

⇒ x = (120 × 100) ÷ 30

⇒ x = 4 × 100 = ₹400

⇒ Selling Price (SP) = CP + Profit

⇒ SP = 400 + 120

⇒ SP = ₹520

∴ The correct answer is ₹520.

Additional Information

Profit and Loss Fundamentals

Profit occurs when SP > CP and Loss occurs when CP > SP. Profit/Loss is always calculated on the Cost Price unless specified otherwise.

Basic Formulas

Profit % = (Profit ÷ CP) × 100 | Loss % = (Loss ÷ CP) × 100.

Finding CP from SP

CP = [100 ÷ (100 + Profit %)] × SP (for profit) or CP = [100 ÷ (100 − Loss %)] × SP (for loss).

Shortcut Trick

Let the first number be 20 units.

The sum of the two numbers = 45/20 × 20 = 45 units.

The second number = Sum − First number = 45 − 20 = 25 units.

Required percentage = (Second number / First number) × 100

Required percentage = (25 / 20) × 100 = 125%.

∴ The correct answer is 125 percent.

Alternate Method

Given:

Let the first number be 'x' and the second number be 'y'.

The sum of the two numbers (x + y) = (45 / 20) × x

Formula Used:

Percentage = (Value / Base) × 100

Calculations:

⇒ x + y = (45 / 20)x

⇒ y = (45 / 20)x − x

⇒ y = (45x − 20x) / 20

⇒ y = 25x / 20

⇒ y / x = 25 / 20

⇒ Required Percentage = (y / x) × 100

⇒ Required Percentage = (25 / 20) × 100

⇒ Required Percentage = 1.25 × 100 = 125%

∴ The correct answer is 125 percent.

Additional Information

Percentage Representation

A fraction a/b can be converted to a percentage by multiplying it by 100. Percentage = (a/b) × 100%.

Ratio and Proportion

If the sum and one part are known in terms of units, the other part is the difference between the total sum and the known part.

Percentage of Base

To find what percent 'y' is of 'x', 'x' always acts as the denominator (base) in the calculation.

Shortcut Trick

Time taken is directly proportional to the volume of the tank to be emptied.

1/4 part → 8 min

Multiply both sides by 2 to get 1/2 part:

(1/4 × 2) part → 8 × 2 min

1/2 part → 16 min

∴ The correct answer is 16 min.

Alternate Method

Given:

Time to empty 1/4 part = 8 min

Formula Used:

Total Time = Time taken / Fraction of work done

Calculations:

⇒ Time to empty the full tank (1 part) = 8 × 4

⇒ Full tank time = 32 min

⇒ Time to empty 1/2 part of the tank = 32 × (1/2)

⇒ Time = 16 min

∴ The correct answer is 16 min.

Additional Information

Unitary Method for Fractions

If 1/n part of a task is completed in 't' time, then the full task (1 unit) is completed in (n × t) time.

Pipes and Cisterns Efficiency

Efficiency of an emptying pipe is considered negative when working together with an inlet pipe in total work calculations.

Time and Work Relation

Work Done = Rate of Work × Time. For a constant rate, Work and Time are in direct proportion.

Shortcut Trick

Given P + R = 5Q. Adding Q to both sides: P + Q + R = 6Q.

This means the Total Efficiency of P, Q, and R combined is 6 times the efficiency of Q alone.

Since Time is inversely proportional to Efficiency, Q will take 6 times as much time as the group.

Time taken by Q = 6 × 10 = 60 days.

∴ The correct answer is 60 days.

Alternate Method

Given: P, Q and R together complete work in 10 days.

Ratio of efficiencies: (Q + R) / P = 8 / 1 (Total units = 9) and (P + R) / Q = 5 / 1 (Total units = 6).

⇒ To equalize the total efficiency, take LCM(9, 6) = 18.

⇒ New Ratio 1 (multiply by 2): (Q + R) : P = 16 : 2.

⇒ New Ratio 2 (multiply by 3): (P + R) : Q = 15 : 3.

⇒ Thus, efficiency of P = 2, Q = 3, and Total = 18.

⇒ Total Work = Total Efficiency × Total Days = 18 × 10 = 180 units.

⇒ Time taken by Q alone = Total Work ÷ Efficiency of Q = 180 ÷ 3 = 60 days.

∴ The correct answer is 60 days.

Additional Information

Efficiency and Time Relationship

Efficiency (E) is inversely proportional to Time (T) for a constant amount of work: E₁ × T₁ = E₂ × T₂.

Combined Efficiency

When multiple people work together, their individual Efficiencies are added to find the Total Efficiency of the group.

Work-Wages Principle

Wages are generally distributed in the direct ratio of the Work Done or the ratio of Efficiencies (if time spent is same).

Shortcut Trick

When two trains cross in opposite direction, their Relative Speed is the sum of their individual speeds.

Relative Speed = (Length of Train A + Length of Train B) ÷ Time

Total Speed = (400 + 800) ÷ 80 = 1200 ÷ 80 = 15 m/s

∴ The correct answer is 15 m/s.

Alternate Method

Given:

Length of Train A (L_A) = 400 m

Length of Train B (L_B) = 800 m

Time taken to cross (t) = 80 sec

Formula Used:

Relative Speed (Opposite Direction) = (L_A + L_B) ÷ t

Calculations:

⇒ Total Distance to be covered = 400 + 800 = 1200 m

⇒ Relative Speed = 1200 ÷ 80

⇒ Relative Speed = 15 m/s

∴ The correct answer is 15 m/s.

Additional Information

Relative Speed in Same Direction

If two trains of speeds S_A and S_B run in the same direction, the relative speed is |S_A − S_B|.

Unit Conversion

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

Crossing a Stationary Point Object

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

Shortcut Trick

Relative Speed = Distance ÷ Time = 400 ÷ 8 = 50 m/s.

Ratio of Train speed to Man speed = 4 :

1. In opposite direction, relative units = 4 + 1 = 5 units.

5 units = 50 m/s ⇒ 1 unit = 10 m/s.

Speed of train = 4 units = 40 m/s = 40 × (18/5) = 144 km/hr.

∴ The correct answer is 144 km/hr.

Alternate Method

Given:

Length of Train (L) = 400 m

Time taken (t) = 8 seconds

Speed of train (V_t) = 4 × Speed of man (V_m)

Formula used:

Relative Speed = Total Distance ÷ Time

In opposite direction, Relative Speed = V₁ + V₂

Calculations:

⇒ Relative Speed = 400 ÷ 8

⇒ V_t + V_m = 50 m/s

⇒ Since V_t = 4V_m, substituting the value:

⇒ 4V_m + V_m = 50

⇒ 5V_m = 50 ⇒ V_m = 10 m/s

⇒ Speed of train (V_t) = 4 × 10 = 40 m/s

⇒ Converting to km/hr: 40 × (18 ÷ 5)

⇒ 8 × 18 = 144 km/hr

∴ The correct answer is 144 km/hr.

Additional Information

Relative Speed in Trains

When two objects move in opposite directions, Relative Speed = S₁ + S₂. When they move in the same direction, Relative Speed = |S₁ − S₂|.

Speed Unit Conversion

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

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.