Quantitative Aptitude PYQs

Shortcut Trick

Apply the Unit Digit Method to identify the result quickly.

Sum of unit digits: 3 + 2 − 1 + (1 + 1) − 8 + 3

⇒ (3 + 2 + 2 + 3) − (1 + 8) = 10 − 9 = 1

Since the unit digit of the final value is 1, the answer is 121.

∴ The correct answer is 121.

Alternate Method

Given: 73 + 92 − 61 + (41 + 21) − 78 + 33

Follow the BODMAS rule (Bracket, Of, Division, Multiplication, Addition, Subtraction) to simplify the expression:

⇒ 73 + 92 − 61 + 62 − 78 + 33

⇒ (73 + 92 + 62 + 33) − (61 + 78)

⇒ 260 − 139

⇒ 121

∴ The correct answer is 121.

Additional Information

BODMAS Rule

The rule defines the order of operations: Brackets, Of, Division, Multiplication, Addition, and Subtraction. Always solve terms inside brackets first.

Grouping Method

For faster calculation, group all positive terms together and all negative terms together before performing the final subtraction.

Unit Digit Technique

This shortcut helps eliminate incorrect options by focusing only on the last digit of each number in the arithmetic operation.

Shortcut Trick

Apply VBODMAS: solve brackets and 'of' operations first.

Calculate (36 ÷ 6) = 6 and 42 of 1 = 42.

Group positive and negative terms: (79 + 29 + 42 + 13) − (6 + 24).

Final sum: 163 − 30 = 133.

∴ The correct answer is 133.

Alternate Method

Given: 79 + 29 − (36 ÷ 6) + 42 of 1 + 13 − 24

Formula Used: Simplify the expression using the BODMAS rule (Bracket, Of, Division, Multiplication, Addition, Subtraction).

⇒ 79 + 29 − 6 + 42 of 1 + 13 − 24 (Solving Bracket first)

⇒ 79 + 29 − 6 + 42 + 13 − 24 (Solving 'of': 42 × 1)

⇒ 108 − 6 + 42 + 13 − 24 (Addition from left to right)

⇒ 102 + 42 + 13 − 24

⇒ 144 + 13 − 24

⇒ 157 − 24

⇒ 133

∴ The correct answer is 133.

Additional Information

BODMAS Rule

This acronym defines the priority of operations: Bracket (), Of (order/powers), Division (÷), Multiplication (×), Addition (+), and Subtraction (−).

Priority of Of Operator

In mathematical expressions, 'of' is a multiplication operation that must be completed before any division or standard multiplication.

Negative Number Rule

When multiple subtractions are present, they can be combined as: −a − b = −(a + b).

Shortcut Trick

Product of two numbers = LCM × HCF

42 × Second Number = 168 × 14

Second Number = (168 × 14) ÷ 42

Second Number = 168 ÷ 3 = 56

∴ The correct answer is 56.

Alternate Method

Given:

⇒ Least Common Multiple (LCM) = 168

⇒ Highest Common Factor (HCF) = 14

⇒ First Number (n₁) = 42

Formula Used:

Product of two numbers = LCM × HCF

n₁ × n₂ = LCM(n₁, n₂) × HCF(n₁, n₂)

Calculations:

⇒ Let the other number be x.

⇒ 42 × x = 168 × 14

⇒ x = (168 × 14) ÷ 42

⇒ x = 2352 ÷ 42

⇒ x = 56

∴ The correct answer is 56.

Additional Information

HCF Property

The Highest Common Factor of two numbers always divides their LCM exactly.

Co-prime Numbers

If two numbers are co-prime, their HCF is 1 and their LCM is equal to the product of the two numbers.

Product Rule Constraint

The formula Product of Numbers = LCM × HCF is strictly valid only for two numbers and does not apply directly to three or more numbers.

Shortcut Trick

Identify the expression as the algebraic identity: a³ − 3a²b + 3ab² − b³ = (a − b)³.

Here, a = 4.2 and b = 0.2.

The expression simplifies to (4.2 − 0.2)³.

⇒ 4³ = 64.

∴ The correct answer is 64.

Alternate Method

Given Expression: (4.2)² × 4.2 − 3 × 0.2 × (4.2)² + 3 × 4.2 × (0.2)² − (0.2)³

Formula Used: (a − b)³ = a³ − 3a²b + 3ab² − b³

⇒ Rewrite the terms to match the identity: (4.2)³ − 3(4.2)²(0.2) + 3(4.2)(0.2)² − (0.2)³

⇒ Let a = 4.2 and b = 0.2.

⇒ Expression = (4.2 − 0.2)³

⇒ Expression = (4.0)³

⇒ 4 × 4 × 4 = 64

∴ The correct answer is 64.

Additional Information

Cube of Sum Identity

(a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b).

Difference of Two Cubes

a³ − b³ = (a − b)(a² + ab + b²).

Sum of Two Cubes

a³ + b³ = (a + b)(a² − ab + b²).

Shortcut Trick

Identify perfect square roots: √484 = 22, √121 = 11, and √64 = 8.

Substitute in equation: x = (22 ÷ 11) + 8

Solve for x: x = 2 + 8 = 10

Calculate x²: 10 × 10 = 100

∴ The correct answer is 100.

Alternate Method

Given: x = √484 ÷ √121 + √64

Formula Used: BODMAS (Division before Addition) and Square root values.

⇒ √484 = 22

⇒ √121 = 11

⇒ √64 = 8

⇒ x = 22 ÷ 11 + 8

⇒ x = 2 + 8

⇒ x = 10

⇒ x² = 10²

⇒ x² = 100

∴ The correct answer is 100.

Additional Information

Perfect Squares and Roots

Memorizing squares up to 30 helps in fast calculation (e.g., 20² = 400, 22² = 484, 25² = 625).

BODMAS Rule

Always follow the sequence: Brackets, Orders (Powers/Roots), Division, Multiplication, Addition, and Subtraction.

Square Root Estimation

For √484, since the last digit is 4, the root ends in 2 or 8. Since 20² = 400 and 30² = 900, the root is 22.

Shortcut Trick

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

Here, a = 5 and b = √3.

Sum of squares = 2 × (5² + (√3)²) = 2 × (25 + 3) = 56.

Required value = 56 − 40 = 16.

∴ The correct answer is 16.

Alternate Method

Given Expression: (5 + √3)² + (5 − √3)² − 40

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

⇒ Expansion of (5 + √3)² = 5² + (√3)² + 2 × 5 × √3 = 25 + 3 + 10√3 = 28 + 10√3

⇒ Expansion of (5 − √3)² = 5² + (√3)² − 2 × 5 × √3 = 25 + 3 − 10√3 = 28 − 10√3

⇒ Total = (28 + 10√3) + (28 − 10√3) − 40

⇒ Total = 28 + 28 + 10√3 − 10√3 − 40

⇒ Total = 56 − 40

⇒ 16

∴ The correct answer is 16.

Additional Information

Standard Algebraic Identities

The identity (a + b)² + (a − b)² = 2(a² + b²) is a quick way to sum the squares of binomials with identical terms.

Subtraction of Binomial Squares

Similarly, (a + b)² − (a − b)² = 4ab is used when the terms are being subtracted instead of added.

Difference of Two Squares

The product of (a + b) and (a − b) always results in a² − b², which is useful for rationalizing denominators.

Shortcut Trick

Check Divisibility by 8 first (last three digits) then Divisibility by 3 (sum of digits).

For 5448: Last three digits 448 are divisible by 8 (448 ÷ 8 = 56).

Sum of digits 5+4+4+8 = 21, which is divisible by 3 (21 ÷ 3 = 7).

Both conditions are satisfied for the number 5448.

∴ The correct answer is 5448.

Alternate Method

Given: Numbers (1) 5448, (2) 5240, (3) 5560, (4) 6170.

Rules: A number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 8 if the last three digits are divisible by 8.

⇒ Option 1: 5448

⇒ Sum = 5 + 4 + 4 + 8 = 21 (Divisible by 3 because 21 ÷ 3 = 7)

⇒ Last 3 digits = 448 (Divisible by 8 because 448 ÷ 8 = 56)

⇒ Option 2: 5240

⇒ Sum = 5 + 2 + 4 + 0 = 11 (Not divisible by 3)

⇒ Option 3: 5560

⇒ Sum = 5 + 5 + 6 + 0 = 16 (Not divisible by 3)

⇒ Option 4: 6170

⇒ Sum = 6 + 1 + 7 + 0 = 14 (Not divisible by 3)

∴ The correct answer is 5448.

Additional Information

Divisibility by 4

A number is divisible by 4 if the number formed by its last two digits is divisible by 4 or ends in 00.

Divisibility by 9

A number is divisible by 9 if the sum of all its digits is divisible by 9.

Divisibility by 11

The difference between the sum of digits at odd places and even places must be either 0 or a multiple of 11.

Shortcut Trick

For a path built outside a rectangular garden:

Area of path = 2w × (L + B + 2w)

Where L = 180 m, B = 70 m, and w = 4 m.

Area = 2 × 4 × (180 + 70 + 2 × 4) = 8 × (250 + 8)

Area = 8 × 258 = 2064 m².

∴ The correct answer is 2064 m².

Alternate Method

Given:

Length (L) of the rectangular garden = 180 m

Breadth (B) of the garden = 70 m

Width of the path = 4 m

Formula Used:

Area of path = (Area including path) − (Area of garden)

Calculations:

⇒ Outer Length = 180 + 4 + 4 = 188 m

⇒ Outer Breadth = 70 + 4 + 4 = 78 m

⇒ Inner Area (Garden) = 180 × 70 = 12600 m²

⇒ Outer Area (Total) = 188 × 78 = 14664 m²

⇒ Area of path = 14664 − 12600

⇒ Area of path = 2064 m²

∴ The correct answer is 2064 m².

Additional Information

Path Inside a Rectangle

If the path is built inside the rectangular field, the area is calculated as Area = 2w × (L + B − 2w).

Cross-roads Area

Area of two paths of width 'w' crossing each other at right angles in the middle of a field = w × (L + B − w).

Perimeter including Path

The perimeter of the outer boundary after adding a path of width 'w' is 2 × (L + B + 4w).

Shortcut Trick

Ratio of 1^st number to 2^nd number = 1 : 2

Sum of two numbers = Average × 2 = 66 × 2 = 132

Total ratio parts = 1 + 2 = 3 parts

3 parts = 132 ⇒ 1 part = 132 ÷ 3 = 44

∴ The correct answer is 44.

Alternate Method

Given:

Average of two numbers = 66

First number = 1/2 × Second number

Formula Used:

Average = (Sum of observations) ÷ (Number of observations)

Calculations:

Let the second number be 2x.

⇒ First number = (1/2) × 2x = x

⇒ Average = (x + 2x) ÷ 2 = 66

⇒ 3x ÷ 2 = 66

⇒ 3x = 66 × 2

⇒ 3x = 132

⇒ x = 132 ÷ 3

⇒ x = 44

∴ The correct answer is 44.

Additional Information

Average of n Numbers

The average is the central value of a set of data, calculated as Sum of all values ÷ Total count of values.

Sum and Average Relation

Total Sum = Average × Number of observations. This is used to find individual values when ratios are given.

Ratio to Value Conversion

If a total S is divided in ratio a:b, then the first part is [a ÷ (a+b)] × S.

Shortcut Trick

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

Half of the given diagonal = 30 ÷ 2 = 15 cm.

Using Pythagorean triplet (15, 20, 25), the other half-diagonal is 20 cm.

Total second diagonal = 20 × 2 = 40 cm.

Area = (1/2) × 30 × 40 = 600 cm².

∴ The correct answer is 600 cm².

Alternate Method

Given:

Perimeter of Rhombus = 100 cm

One diagonal (d₁) = 30 cm

Formula Used:

Side (s) = Perimeter ÷ 4

Relation: s² = (d₁/2)² + (d₂/2)²

Area = (1/2) × d₁ × d₂

Calculations:

⇒ Side = 100 ÷ 4 = 25 cm

⇒ Half of d₁ = 30 ÷ 2 = 15 cm

⇒ In right-angled triangle AOB (where O is intersection),

⇒ Side² = (d₁/2)² + (d₂/2)²

⇒ 25² = 15² + (d₂/2)²

⇒ 625 = 225 + (d₂/2)²

⇒ (d₂/2)² = 625 − 225 = 400

⇒ d₂/2 = √400 = 20 cm

⇒ Second diagonal (d₂) = 2 × 20 = 40 cm

⇒ Area of Rhombus = (1/2) × 30 × 40

⇒ Area = 600 cm²

∴ The correct answer is 600 cm².

Additional Information

Properties of Rhombus Diagonals

The diagonals of a rhombus always bisect each other at right angles (90°).

Relationship Formula

The relationship between the side and diagonals is given by: 4 × Side² = d₁² + d₂².

General Area Formula

Since a rhombus is also a parallelogram, its area can also be calculated as Base × Height.

Shortcut Trick

Assume base average is 23 for all 14 numbers.

The remaining 9 numbers are 37 − 23 = 14 higher than the base.

Total surplus = 9 × 14 = 126.

Distribute this surplus over all 14 numbers: 126 ÷ 14 = 9.

Value of m = Base + Increase = 23 + 9 = 32.

∴ The correct answer is 32.

Alternate Method

Given:

Total count of numbers = 14

Average of 14 numbers = m

Average of first 5 numbers = 23

Average of remaining 9 numbers = 37

Formula Used:

Average = Total Sum ÷ Count of numbers

Total Sum = (n₁ × Avg₁) + (n₂ × Avg₂)

Calculations:

⇒ Sum of the first 5 numbers = 5 × 23 = 115

⇒ Count of remaining numbers = 14 − 5 = 9

⇒ Sum of the remaining 9 numbers = 9 × 37 = 333

⇒ Total sum of 14 numbers = 115 + 333 = 448

⇒ Average of 14 numbers (m) = 448 ÷ 14

⇒ m = 32

∴ The correct answer is 32.

Additional Information

Weighted Average Formula

When two groups are combined, the average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Net Deviation Rule

In any set of numbers, the sum of deviations of all values from the arithmetic mean is always zero.

Adding/Removing Elements

New Average = Old Average + (Change in Sum ÷ New Count).

Shortcut Trick

The side of the largest square tile is the HCF of the floor dimensions.

Side = HCF(525, 700) = 175 cm.

Number of tiles along length = 700 ÷ 175 = 4.

Number of tiles along breadth = 525 ÷ 175 = 3.

Total tiles required = 4 × 3 = 12.

∴ The correct answer is 12.

Alternate Method

Given: Dimensions of the rectangular floor are 700 cm and 525 cm.

Formula: Number of tiles = (Area of floor) ÷ (Area of one square tile), where tile side = HCF(Length, Breadth).

⇒ Prime factors of 700 = 2² × 5² × 7

⇒ Prime factors of 525 = 3 × 5² × 7

⇒ HCF (Side of tile) = 5² × 7 = 25 × 7 = 175 cm.

⇒ Area of floor = 700 × 525 = 3,67,500 cm².

⇒ Area of one square tile = 175 × 175 = 30,625 cm².

⇒ Number of tiles = 3,67,500 ÷ 30,625 = 12.

∴ The correct answer is 12.

Additional Information

HCF (Highest Common Factor)

HCF is the largest positive integer that divides each of the integers without a remainder. For tiling problems, the HCF of the dimensions gives the largest possible square tile size.

Area of Rectangle and Square

Area of rectangle = Length × Breadth. Area of square = side². The number of objects required to cover an area is always Total Area ÷ Object Area.

Minimal Tiling Rule

To minimize the number of tiles used to cover a fixed rectangular area, the size of each tile must be maximized, which is why the HCF is calculated.

Shortcut Trick

Convert 60% to a fraction: 60/100 = 3/5.

Total fraction of the number = 2/5 × 3/5 = 6/25.

Given: 6 units = 420.

1 unit = 420 ÷ 6 = 70.

The total number (25 units) = 25 × 70 = 1750.

∴ The correct answer is 1750.

Alternate Method

Given:

Fraction part = 2/5

Percentage part = 60%

Value of the result = 420

Calculation:

Let the number be x.

⇒ (2/5) × 60% × x = 420

⇒ (2/5) × (60/100) × x = 420

⇒ (2/5) × (3/5) × x = 420

⇒ (6/25) × x = 420

⇒ x = (420 × 25) ÷ 6

⇒ x = 70 × 25

⇒ x = 1750

∴ The correct answer is 1750.

Additional Information

Percentage to Fraction Conversion

To convert a percentage to a fraction, divide by 100. Example: 20% = 20/100 = 1/5; 40% = 2/5; 60% = 3/5; 80% = 4/5.

Concept of "Of" in Mathematics

The word "of" represents the multiplication operation (×) in BODMAS/VBODMAS rules for simplification.

Successive Fractions

When taking a fraction of another fraction, the net multiplier is the product of the individual fractions: (a/b) of (c/d) = (a×c)/(b×d).

Shortcut Trick

Loss Percentage = 30% = 3/10 (where 3 units is Loss and 10 units is Cost Price).

The difference between CP and SP when there is a loss is the Loss itself.

Given: 3 units = ₹720 ⇒ 1 unit = ₹240.

Selling Price (SP) = CP − Loss = 10 units − 3 units = 7 units.

SP = 7 × 240 = ₹1680.

∴ The correct answer is ₹1680.

Alternate Method

Given:

Difference (CP − SP) = ₹720

Loss Percentage = 30%

Formula Used:

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

Loss % = (Loss ÷ CP) × 100

Calculations:

⇒ Loss = ₹720

⇒ 30 = (720 ÷ CP) × 100

⇒ CP = (720 × 100) ÷ 30

⇒ CP = 24 × 100 = ₹2400

⇒ SP = CP − Loss

⇒ SP = 2400 − 720

⇒ SP = ₹1680

∴ The correct answer is ₹1680.

Additional Information

Profit and Loss Relation

When Selling Price (SP) is greater than Cost Price (CP), there is a profit: Profit = SP − CP. When CP > SP, there is a loss: Loss = CP − SP.

Percentage Calculation Base

Profit % and Loss % are always calculated on the Cost Price (CP) unless specifically mentioned otherwise in the question.

Direct SP Formula

SP = CP × (100 − Loss%) ÷ 100 for loss cases, and SP = CP × (100 + Profit%) ÷ 100 for profit cases.

Shortcut Trick

Profit percentage of 50% can be written as the fraction 1/2.

In this fraction, the denominator represents the Cost Price (CP) and the numerator represents the Profit.

Let CP = 2 units and Profit = 1 unit.

Selling Price (SP) = CP + Profit = 2 + 1 = 3 units.

The required Ratio of SP to CP is 3 : 2.

∴ The correct answer is 3 : 2.

Alternate Method

Given:

Profit percentage = 50%

Formula Used:

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

Calculations:

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

⇒ Profit = 50% of 100 = 50 units.

⇒ Selling Price (SP) = CP + Profit

⇒ SP = 100 + 50 = 150 units.

⇒ Required Ratio = SP : CP

⇒ Ratio = 150 : 100

⇒ Dividing both sides by 50, we get 3 : 2.

∴ The correct answer is 3 : 2.

Additional Information

Profit and Loss Formulas

Profit = Selling Price − Cost Price. Loss = Cost Price − Selling Price. Both are always calculated on the Cost Price unless stated otherwise.

Marked Price and Selling Price

Selling Price = Marked Price − Discount. Discount % = (Discount ÷ Marked Price) × 100.

Successive Discount

Two successive discounts of d₁% and d₂% are equivalent to a single discount of (d₁ + d₂ − d₁d₂/100)%.

Shortcut Trick

Ratio of (Sum : First number) = 35 : 20.

Second number = Sum − First number = 35 − 20 = 15 units.

Required Percentage = (15 ÷ 20) × 100 = 75%.

∴ The correct answer is 75%.

Alternate Method

Given:

Sum of two numbers = (35 ÷ 20) × First number

Calculations:

⇒ Let the First number be 20x.

⇒ Sum = (35 ÷ 20) × 20x = 35x.

⇒ Let the Second number be y.

⇒ 20x + y = 35x

⇒ y = 35x − 20x = 15x

⇒ Required Percentage = (Second number ÷ First number) × 100

⇒ Percentage = (15x ÷ 20x) × 100

⇒ Percentage = (3 ÷ 4) × 100 = 75%

∴ The correct answer is 75%.

Additional Information

Base Value Identification

In percentage questions, the word 'of' or 'than' identifies the base value (denominator). Here, it is the First number.

Fraction to Percentage Conversion

Common fractions: 1/4 = 25%, 1/2 = 50%, 3/4 = 75%, and 1/5 = 20%.

Percentage Change Formula

Percentage = (Part ÷ Whole) × 100. If the Part is the difference, it represents the percentage increase/decrease.

Shortcut Trick

Total time to empty the tank = 20 hr.

Work done in 1 hour = 1/20 of the tank.

Work done in 5 hours = 5 × (1/20) = 1/4.

∴ The correct answer is 1/4.

Alternate Method

Given: Time taken by pipe to empty the tank = 20 hr.

Formula: Part of tank emptied = (Time given) ÷ (Total time to empty)

⇒ Total Capacity of the tank = 20 units (LCM of 20).

⇒ Efficiency of the pipe = 1 unit per hour.

⇒ Amount of water emptied in 5 hours = Efficiency × Time = 1 × 5 = 5 units.

⇒ Part of the tank emptied = (Emptied units) ÷ (Total Capacity) = 5 ÷ 20 = 1/4.

∴ The correct answer is 1/4.

Additional Information

Pipes and Cisterns Basics

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

Combined Work Formula

If Pipe A fills in x hours and Pipe B empties in y hours, the net work in 1 hour is (1/x − 1/y).

Work and Time Relationship

Total Work = Efficiency × Time. Efficiency is inversely proportional to the time taken to complete the work.

Shortcut Trick

Let Total Work be LCM of 12 and 18 = 36 units.

Efficiency of (P+Q) = 36 ÷ 12 = 3 units/day.

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

Total Efficiency (P+Q+R) = 3 + 2 = 5 units/day.

Work to be completed = 1/3 × 36 = 12 units.

Time = 12 ÷ 5 = 2.4 days.

∴ The correct answer is 2.4 days.

Alternate Method

Given: P and Q together = 12 days, R alone = 18 days.

Formula: Total Work = Efficiency × Time

⇒ Combined efficiency of P, Q, and R = 3 + 2 = 5 units/day.

⇒ Total work to be completed = 1/3 × 36 units.

⇒ Required work = 12 units.

⇒ Time taken = Required work ÷ Total efficiency.

⇒ Time taken = 12 ÷ 5 = 2.4 days.

∴ The correct answer is 2.4 days.

Additional Information

Work and Efficiency Relation

Work is directly proportional to Efficiency when time is constant. Efficiency = 1 ÷ Time.

Combined Time Formula

If A takes 'x' days and B takes 'y' days, then together they take (x × y) ÷ (x + y) days to finish the work.

Fractional Work

If a person completes 1/n part of a work in 't' days, the total time taken to complete the full work is n × t days.

Shortcut Trick

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

Speed in m/s = 90 × (5/18)

⇒ 5 × 5 = 25 m/s

∴ The correct answer is 25 m/s.

Alternate Method

Given:

Speed of the car = 90 km/hr

Formula Used:

1 kilometer (km) = 1000 meters (m)

1 hour (hr) = 3600 seconds (s)

Calculations:

⇒ Speed = 90 × (1000 m / 3600 s)

⇒ Speed = 90 × (10 / 36) m/s

⇒ Speed = 90 × (5 / 18) m/s

⇒ Speed = 5 × 5 m/s

⇒ Speed = 25 m/s

∴ The correct answer is 25 m/s.

Additional Information

m/s to km/hr Conversion

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

Average Speed

Average Speed = Total Distance ÷ Total Time. 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 is the difference of their speeds (u − v). In opposite directions, it is the sum (u + v).

Shortcut Trick

Relative Speed = Total Distance ÷ Meeting Time

Relative Speed = 390 ÷ 3 = 130 km/hr

Since they move in opposite directions, Relative Speed = Speed of G + Speed of H

130 = 80 + Speed of H

Speed of H = 130 − 80 = 50 km/hr

∴ The correct answer is 50 km/hr.

Alternate Method

Given:

Distance (D) = 390 km

Time (t) = 3 hr

Speed of G (S_G) = 80 km/hr

Let the Speed of H be v km/hr.

When two objects move towards each other, their relative speed is the sum of their individual speeds.

⇒ Relative Speed = S_G + S_H

⇒ Relative Speed = (80 + v) km/hr

Using the formula: Distance = Relative Speed × Time

⇒ 390 = (80 + v) × 3

⇒ 80 + v = 390 ÷ 3

⇒ 80 + v = 130

⇒ v = 130 − 80

⇒ v = 50 km/hr

∴ The correct answer is 50 km/hr.

Additional Information

Relative Speed in Opposite Directions

When two objects move towards each other or away from each other in opposite directions, their relative speed is S₁ + S₂.

Relative Speed in Same Direction

When two objects move in the same direction, their relative speed is the difference between their speeds |S₁ − S₂|.

Distance, Speed, and Time

The fundamental relationship is Distance = Speed × Time. For two moving objects, use the relative speed in place of speed.