Quantitative Aptitude PYQs

Shortcut Trick

Apply the BODMAS rule sequentially: Of → Division → Addition/Subtraction.

Calculate priority operations: 28 of 2 = 56 and 46 ÷ 2 = 23.

Combine all positive terms: 77 + 56 + 23 + 29 = 185.

Combine all negative terms: −33 − 17 = −50.

Net result: 185 − 50 = 135.

∴ The correct answer is 135.

Alternate Method

Given: 77 − 33 + 28 of 2 + 46 ÷ 2 − 17 + 29

Formula Used: BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction)

⇒ 77 − 33 + (28 × 2) + 46 ÷ 2 − 17 + 29

⇒ 77 − 33 + 56 + (46 ÷ 2) − 17 + 29

⇒ 77 − 33 + 56 + 23 − 17 + 29

⇒ 44 + 56 + 23 − 17 + 29

⇒ 100 + 23 − 17 + 29

⇒ 123 − 17 + 29

⇒ 106 + 29 = 135

∴ The correct answer is 135.

Additional Information

BODMAS / VBODMAS Rule

This rule defines the order of priority: V (Vinculum/Bar), B (Brackets), O (Of), D (Division), M (Multiplication), A (Addition), and S (Subtraction).

The 'Of' Operator

The word 'of' serves as a multiplication operator but must be calculated before any division or standard multiplication in the expression.

Group Calculation Trick

To avoid sign errors in long expressions, sum all positive numbers and negative numbers separately before performing the final subtraction.

Shortcut Trick

Perform the decimal divisions directly by balancing the number of decimal places.

First term: 0.08 ÷ 0.4 = 0.8 ÷ 4 = 0.2

Second term: 0.024 ÷ 0.6 = 0.24 ÷ 6 = 0.04

Difference = 0.2 − 0.04 = 0.16

∴ The correct answer is 0.16.

Alternate Method

Given Expression: (0.08 / 0.4) − (0.024 / 0.6)

According to the BODMAS rule, division operations must be completed before subtraction.

⇒ Term 1 = 0.08 ÷ 0.4 = (8 / 100) ÷ (4 / 10)

⇒ Term 1 = (8 / 100) × (10 / 4) = 80 / 400 = 1 / 5 = 0.2

⇒ Term 2 = 0.024 ÷ 0.6 = (24 / 1000) ÷ (6 / 10)

⇒ Term 2 = (24 / 1000) × (10 / 6) = 240 / 6000 = 4 / 100 = 0.04

⇒ Final Value = 0.2 − 0.04 = 0.16

∴ The correct answer is 0.16.

Additional Information

BODMAS Rule

In mathematical calculations, always follow the sequence: Brackets, Orders, Division, Multiplication, Addition, and Subtraction.

Decimal Conversion

To simplify division with decimals, multiply both numerator and denominator by 10ⁿ to convert them into whole numbers.

Subtraction of Decimals

While subtracting decimals like 0.2 and 0.04, ensure the decimal points are aligned vertically by adding a placeholder zero (e.g., 0.20 − 0.04).

Shortcut Trick

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

Identify the values: a = 7.75 and b = 2.75.

Calculate the difference: (a − b) = 7.75 − 2.75 = 5.

The final result is (5)² = 25.

∴ The correct answer is 25.

Alternate Method

Given:

Expression: (7.75 + 2.75)² − 4 × 7.75 × 2.75

Formula Used:

(a + b)² = a² + 2ab + b²

Calculations:

⇒ (7.75 + 2.75)² − 4 × 7.75 × 2.75

⇒ (10.5)² − (4 × 21.3125)

⇒ 110.25 − 85.25

⇒ 25

∴ The correct answer is 25.

Additional Information

Algebraic Identity - Square of Sum

The expression (a + b)² expands to a² + 2ab + b².

Algebraic Identity - Square of Difference

The expression (a − b)² expands to a² − 2ab + b².

Relationship between Sum and Difference Squares

The identities are related as (a + b)² − (a − b)² = 4ab and (a + b)² + (a − b)² = 2(a² + b²).

Shortcut Trick

In square root calculations, if the decimal point shifts by 2n places inside the root, it shifts by n places in the result.

Movement inside √ : 210.25 → 0.00021025 (6 places to the left).

Movement in result : 14.5 → (6 ÷ 2 = 3 places to the left).

Final Value : 14.5 → 0.0145.

∴ The correct answer is 0.0145.

Alternate Method

Given: √210.25 = 14.5

We need to find the value of √0.00021025.

⇒ 0.00021025 = 210.25 ÷ 1,000,000

⇒ 0.00021025 = 210.25 ÷ 10⁶

Taking square root on both sides:

⇒ √0.00021025 = √(210.25 ÷ 10⁶)

⇒ √0.00021025 = √210.25 ÷ √10⁶

⇒ √0.00021025 = 14.5 ÷ 10³

⇒ √0.00021025 = 14.5 ÷ 1000

⇒ √0.00021025 = 0.0145

∴ The correct answer is 0.0145.

Additional Information

Square Root of Powers of 10

The square root of 10²ⁿ is always 10ⁿ. For example, √100 = 10 and √10,000 = 100.

Decimal Rule in Square Roots

For every two decimal places inside a square root, there is one decimal place in the square root's value (e.g., √0.01 = 0.1).

Squares of Numbers Ending in 5

To square a number like 14.5, multiply 14 by 15 (210) and append 0.25 to get 210.25.

Shortcut Trick

List all Prime Numbers between 20 and 40: 23, 29, 31, 37.

Group them for faster calculation: (23 + 37) + (29 + 31).

Sum = 60 + 60 = 120.

∴ The correct answer is 120.

Alternate Method

Given: Range for prime numbers is 20 to 40.

A Prime Number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Step-by-step Calculation:

⇒ Odd numbers between 20 and 40: 21, 23, 25, 27, 29, 31, 33, 35, 37, 39.

⇒ Eliminate non-primes: 21 (3×7), 25 (5×5), 27 (3×9), 33 (3×11), 35 (5×7), 39 (3×13).

⇒ Remaining primes: 23, 29, 31, 37.

⇒ Total Sum = 23 + 29 + 31 + 37

⇒ Total Sum = 52 + 68

⇒ Total Sum = 120

∴ The correct answer is 120.

Additional Information

Prime Numbers Definition

A Prime Number has only two factors: 1 and the number itself. The smallest prime number is 2, and it is the only even prime.

Composite Numbers

Numbers with more than two factors are called Composite Numbers. 1 is neither prime nor composite.

Primes up to 100

There are exactly 25 prime numbers between 1 and 100. The sum of prime numbers from 1 to 50 is 328.

Shortcut Trick

Identify the algebraic identity: a³ + b³ + 3ab(a + b) = (a + b)³.

In the given expression, let a = 15 and b = 5, which means a + b = 20.

The expression becomes: (15 + 5)³ ⇒ 20³.

Result = 20 × 20 × 20 = 8000.

∴ The correct answer is 8000.

Alternate Method

Given: (15)³ + (5)³ + 3 × 15 × 5 × 20

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

⇒ Let a = 15 and b = 5.

⇒ The sum of a and b is 15 + 5 = 20.

⇒ Substituting these into the formula: 15³ + 5³ + 3 × 15 × 5 × (15 + 5).

⇒ This is exactly the same as (15 + 5)³.

⇒ (20)³ = 20 × 20 × 20 = 8000.

∴ The correct answer is 8000.

Additional Information

Cube of Difference Identity

The identity for the cube of a difference is (a − b)³ = a³ − b³ − 3ab(a − b).

Sum of Cubes Factorization

The expression a³ + b³ can be factored as (a + b)(a² − ab + b²).

Difference of Cubes Factorization

The expression a³ − b³ can be factored as (a − b)(a² + ab + b²).

Shortcut Trick

In a rectangle, the perimeter is twice the sum of its length and breadth.

Sum of (Length + Breadth) = Perimeter ÷ 2

Given Perimeter = 24 cm

∴ Sum = 24 ÷ 2 = 12 cm

∴ The correct answer is 12 cm.

Alternate Method

Given:

Perimeter of rectangle = 24 cm

Length (L) = 5 cm

Formula Used:

Perimeter = 2 × (Length + Breadth)

Calculations:

⇒ 24 = 2 × (5 + Breadth)

⇒ 24 ÷ 2 = 5 + Breadth

⇒ 12 = 5 + Breadth

⇒ Breadth = 12 − 5 = 7 cm

⇒ Sum of length and breadth = 5 + 7 = 12 cm

∴ The correct answer is 12 cm.

Additional Information

Area of Rectangle

The total space occupied by the rectangle is calculated as Length × Breadth.

Diagonal of Rectangle

The distance between opposite corners is given by √(Length² + Breadth²).

Perimeter of Square

For a square where all sides (a) are equal, the perimeter is 4 × a.

Relationship of Perimeter and Sides

The semi-perimeter of a rectangle is exactly equal to the sum of its adjacent sides (L + B).

∴ The correct answer is 12 cm.

Shortcut Trick

Solve the nested radical expression mentally from the innermost root outwards.

√49 = 7

√(93 + 7) = √100 = 10

√(1671 + 10) = √1681 = 41

∴ The correct answer is 41.

Alternate Method

Given:

x = √(1671 + √(93 + √49))

Formula Used:

For any non-negative number a, √a² = a

Calculations:

⇒ x = √(1671 + √(93 + 7))

⇒ x = √(1671 + √100)

⇒ x = √(1671 + 10)

⇒ x = √1681

⇒ x = √(41²)

⇒ x = 41

∴ The correct answer is 41.

Additional Information

Perfect Square Identification

To find √1681, observe that 40² = 1600. Since 1681 ends in 1, the square root must end in either 1 or 9. Thus, 41 is the most likely candidate.

BODMAS in Radicals

In nested radicals, the order of operations dictates that the innermost grouping (the smallest radical) must be simplified before moving to the outer layers.

Unit Digit Rule

Numbers ending in 1, 4, 5, 6, or 9 can be perfect squares. A number ending in 1 will have a square root ending in 1 or 9.

Shortcut Trick

A number is divisible by 6 if it is divisible by both 2 and 3.

84P6 ends in 6 (even), so it is divisible by 2.

Sum of digits = 8 + 4 + P + 6 = 18 + P must be a multiple of 3.

Checking options: 18 + 0 = 18, 18 + 3 = 21, 18 + 9 = 27 are all divisible by 3.

However, 18 + 7 = 25 is not divisible by 3.

∴ The correct answer is 7.

Alternate Method

Given: The number 84P6 is divisible by 6.

Formula: For a number to be divisible by 6, it must satisfy the divisibility rules of both 2 and 3.

⇒ Divisibility by 2: The last digit of 84P6 is 6, which is an even number. Thus, the number is divisible by 2 for any digit value of P.

⇒ Divisibility by 3: The sum of the digits must be divisible by 3.

⇒ Sum = 8 + 4 + P + 6 = 18 + P.

⇒ For (18 + P) to be divisible by 3, P must be 0, 3, 6, or 9 (since 18 is already divisible by 3).

⇒ Testing Option 1: If P = 0, Sum = 18 + 0 = 18 (Divisible by 3).

⇒ Testing Option 2: If P = 3, Sum = 18 + 3 = 21 (Divisible by 3).

⇒ Testing Option 3: If P = 9, Sum = 18 + 9 = 27 (Divisible by 3).

⇒ Testing Option 4: If P = 7, Sum = 18 + 7 = 25 (NOT divisible by 3).

∴ The correct answer is 7.

Additional Information

Divisibility Rule of 3

A number is divisible by 3 if the sum of all its digits is divisible by 3.

Divisibility Rule of 4

A number is divisible by 4 if the last two digits of the number are divisible by 4 or are both zero.

Divisibility Rule of 9

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

Shortcut Trick

Identify the net change in the total sum: New value − Old value = 10 − 30 = −20.

Calculate the change in average: Net change ÷ Total numbers = −20 ÷ 10 = −2.

New average = Old average + Change in average = X + (−2) = X − 2.

∴ The correct answer is X − 2.

Alternate Method

Given:

Total numbers = 10

Initial average = X

Old number = 30

New number = 10

Formula used:

Sum = Average × Count

New Average = (Total Sum − Old Number + New Number) ÷ Count

Calculations:

⇒ Total initial sum = 10 × X = 10X

⇒ New sum = 10X − 30 + 10

⇒ New sum = 10X − 20

⇒ New average = (10X − 20) ÷ 10

⇒ New average = 10X/10 − 20/10

⇒ New average = X − 2

∴ The correct answer is X − 2.

Additional Information

General Average Formula

The average (Arithmetic Mean) is defined as Sum of observations ÷ Total number of observations.

Effect of Replacement

When one value is replaced, the new average = Old average + (Difference ÷ n), where Difference = New value − Old value and n = total items.

Uniform Change Property

If every number in a set is increased or decreased by a constant value k, the average also increases or decreases by exactly k.

Shortcut Trick

To find the largest number that leaves the same remainder, calculate the differences between the given numbers.

Difference 1: 121 − 103 = 18

Difference 2: 139 − 121 = 18

Difference 3: 157 − 139 = 18

The required number is the HCF of these differences: HCF(18, 18, 18) = 18.

∴ The correct answer is 18.

Alternate Method

Given: The numbers are 103, 121, 139, and 157.

Formula Used: If numbers x, y, z, and w leave the same remainder when divided by a number N, then N is the Highest Common Factor (HCF) of |x − y|, |y − z|, and |z − w|.

Calculations:

⇒ First difference = 121 − 103 = 18

⇒ Second difference = 139 − 121 = 18

⇒ Third difference = 157 − 139 = 18

⇒ Required number = HCF(18, 18, 18)

⇒ Since all differences are 18, their HCF is 18.

⇒ Verification: 103÷18 (Rem=13), 121÷18 (Rem=13), 139÷18 (Rem=13), 157÷18 (Rem=13).

∴ The correct answer is 18.

Additional Information

HCF with Same Remainder

When the remainder is the same but not given, the largest divisor is the HCF of the differences between the numbers: HCF(|a−b|, |b−c|, |c−a|).

HCF with Given Remainder

If the largest number divides x, y, and z leaving remainders a, b, and c respectively, the number is HCF((x−a), (y−b), (z−c)).

Common Remainder Property

The common remainder 'r' can be calculated using any number 'n' from the set: r = n mod (HCF of differences).

Shortcut Trick

If the Cost Price (CP) and Selling Price (SP) are both increased or decreased by the same percentage, the ratio CP : SP remains constant.

Since the Profit or Loss percentage depends only on the ratio of CP and SP, it remains unchanged.

Initial Loss percentage = 10%.

Increase in CP and SP = 20% each → No change in the loss percentage.

∴ The correct answer is 10 percent.

Alternate Method

Given: Initial loss percentage = 10%, increase in both CP and SP = 20%.

Formula Used: Loss % = [(CP − SP) ÷ CP] × 100

⇒ Let the initial Cost Price (CP) = 100

⇒ Initial Selling Price (SP) = 100 − 10% of 100 = 90

⇒ New CP after 20% increase = 100 + (20% of 100) = 120

⇒ New SP after 20% increase = 90 + (20% of 90) = 90 + 18 = 108

⇒ New Loss = New CP − New SP = 120 − 108 = 12

⇒ New Loss percentage = (12 ÷ 120) × 100 = 10%

∴ The correct answer is 10 percent.

Additional Information

Profit and Loss Formulas

Profit % = [(SP − CP) ÷ CP] × 100 and Loss % = [(CP − SP) ÷ CP] × 100. These percentages are always calculated relative to the Cost Price.

Constant Percentage Change Property

If both CP and SP are multiplied by the same factor k (which happens during a fixed percentage increase or decrease), the ratio CP : SP remains the same, hence the Profit/Loss % is invariant.

Selling Price Calculation

When CP and Loss % are known, SP = CP × (100 − Loss %) ÷ 100.

Shortcut Trick

Convert percentage to fraction: 20% = 1/5.

If A is 1/5 less than B, then the numerator remains 1 and the denominator becomes (5 − 1) = 4.

B is 1/4 more than A.

1/4 × 100 = 25%.

∴ The correct answer is 25%.

Alternate Method

Given: A is 20% less than B.

Formula: Percentage more = [(Difference) ÷ (Lesser Value)] × 100

⇒ Let the value of B = 100.

⇒ Value of A = 100 − 20% of 100 = 80.

⇒ Difference between B and A = 100 − 80 = 20.

⇒ Percentage by which B is more than A = (Difference ÷ A) × 100.

⇒ Percentage more = (20 ÷ 80) × 100.

⇒ Percentage more = (1 ÷ 4) × 100 = 25%.

∴ The correct answer is 25%.

Additional Information

Percentage Base Change Formula

If A is x% less than B, then B is [x ÷ (100 − x)] × 100% more than A.

Fractional Equivalence

Common percentages to remember: 20% = 1/5, 25% = 1/4, 33.33% = 1/3, 50% = 1/2.

Successive Percentage Change

Net change for two successive changes of a% and b% is given by [a + b + (ab ÷ 100)]%.

Shortcut Trick

Loss % = [Loss ÷ (Selling Price + Loss)] × 100

⇒ Loss % = [500 ÷ (3500 + 500)] × 100

⇒ Loss % = (500 ÷ 4000) × 100 = 12.5%

∴ The correct answer is 12.5%.

Alternate Method

Given:

Selling Price (SP) = ₹3500

Loss = ₹500

Formula Used:

1. Cost Price (CP) = Selling Price + Loss
2. Loss Percentage = (Loss ÷ Cost Price) × 100

Calculations:

⇒ Cost Price (CP) = 3500 + 500 = ₹4000

⇒ Loss Percentage = (500 ÷ 4000) × 100

⇒ Loss Percentage = (1 ÷ 8) × 100

⇒ Loss Percentage = 12.5%

∴ The correct answer is 12.5%.

Additional Information

Profit Percentage

When Selling Price is greater than Cost Price, Profit % = [(SP − CP) ÷ CP] × 100.

Relation with Marked Price

Selling Price is also calculated as Marked Price (MP) − Discount.

Cost Price Base

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

Shortcut Trick

Use the Successive Percentage formula: Net Change = x + y + (xy ÷ 100)

Here, x = +25% (increase) and y = −40% (reduction).

Net Change = 25 − 40 + [25 × (−40) ÷ 100]

Net Change = −15 − (1000 ÷ 100) = −15 − 10 = −25%.

Negative sign indicates a decrease.

∴ The correct answer is 25 percent decrease.

Alternate Method

Given:

Initial increase = 25%

Subsequent reduction = 40%

Assume the initial salary of the worker is 100 units.

⇒ First Change (25% increase): 100 + (25% of 100)

⇒ New Salary = 100 + 25 = 125 units

⇒ Second Change (40% reduction): 125 − (40% of 125)

⇒ Final Salary = 125 − (0.4 × 125) = 125 − 50 = 75 units

⇒ Net change in salary = Initial − Final

⇒ Net change = 100 − 75 = 25 units

⇒ Net percentage change = (25 ÷ 100) × 100 = 25%

Since the final value is less than the initial value, it is a decrease.

∴ The correct answer is 25 percent decrease.

Additional Information

Successive Percentage Formula

For two successive changes of a% and b%, the effective change is given by [a + b + (ab/100)]%.

Fraction Method

25% increase = 5/4 factor; 40% decrease = 3/5 factor. Net factor = (5/4) × (3/5) = 3/4. A 3/4 factor represents a 1/4 or 25% decrease.

Equal Increase and Decrease

If a value is increased by x% and then decreased by x%, the net result is always a decrease of (x²/100)%.

Shortcut Trick

Work rate of R and S together = 1/24 + 1/36 = (3 + 2)/72 = 5/72 units/day.

Work completed in 8 days = 8 × (5/72) = 5/9 part.

Remaining part of work = 1 − 5/9 = 4/9.

∴ The correct answer is 4/9.

Alternate Method

Given:

Time taken by R = 24 days

Time taken by S = 36 days

Formula Used:

Efficiency = Total Work ÷ Time

Remaining Work = Total Work − Work Done

Calculations:

⇒ Total Efficiency of (R + S) = 3 + 2 = 5 units/day

⇒ Work done in 8 days = 5 × 8 = 40 units

⇒ Remaining Work = 72 − 40 = 32 units

⇒ Part of total work left = Remaining Work ÷ Total Work

⇒ Part left = 32/72 = 4/9

∴ The correct answer is 4/9.

Additional Information

Time and Work Relation

Efficiency is inversely proportional to time taken. If a person completes work in n days, their 1-day work is 1/n.

Work and Wages

Wages are distributed in the ratio of the work done or in the ratio of efficiencies when the time period is the same.

Combined Efficiency

If A takes 'a' days and B takes 'b' days, together they take (a × b) / (a + b) days to complete the work.

Shortcut Trick

Apply the Chain Rule formula: M₁D₁H₁ = M₂D₂H₂

30 × 40 × 9 = M₂ × 60 × 6

10800 = M₂ × 360

M₂ = 10800 ÷ 360 = 30

∴ The correct answer is 30 men.

Alternate Method

Given:

Initial Men (M₁) = 30

Initial Days (D₁) = 40

Initial Hours (H₁) = 9

Final Days (D₂) = 60

Final Hours (H₂) = 6

Formula:

Total Work done is the same in both cases, so:

M₁ × D₁ × H₁ = M₂ × D₂ × H₂

Calculations:

⇒ 30 × 40 × 9 = M₂ × 60 × 6

⇒ 1200 × 9 = M₂ × 360

⇒ 10800 = 360 × M₂

⇒ M₂ = 10800 ÷ 360

⇒ M₂ = 30

∴ The correct answer is 30 men.

Additional Information

General Chain Rule Formula

For varying work and efficiency, use (M₁ × D₁ × H₁ × E₁) ÷ W₁ = (M₂ × D₂ × H₂ × E₂) ÷ W₂.

Inverse Proportion in Time & Work

If Work is constant, the number of Men is inversely proportional to the Time taken (Days × Hours).

Man-Hours Concept

Total effort or work is often measured in Man-Hours, which is the product of number of men and total hours worked.

Shortcut Trick

Speed ratio of A to B = 10 : 20 = 1 : 2.

For a constant distance, the time taken is inversely proportional to the speed (Time ∝ 1/Speed).

Therefore, Time ratio = Inverse of Speed ratio = 2 :

1. ∴ The correct answer is 2 : 1.

Alternate Method

Given: Speed of A = 10 km/hr, Speed of B = 20 km/hr.

Formula Used: Time = Distance ÷ Speed

⇒ Let the constant distance covered by both be D km.

⇒ Time taken by A = T_A = D ÷ 10 hours.

⇒ Time taken by B = T_B = D ÷ 20 hours.

⇒ Ratio of time taken = T_A : T_B = (D ÷ 10) : (D ÷ 20).

⇒ Ratio = (1 ÷ 10) : (1 ÷ 20) = 20 : 10 = 2 :

1. ∴ The correct answer is 2 : 1.

Additional Information

Relationship between Speed, Time, and Distance

When distance is constant, Speed₁ × Time₁ = Speed₂ × Time₂. This means speed is inversely proportional to time.

Average Speed Formula

If a person covers the same distance at two different speeds x and y, the average speed for the whole journey is (2 × x × y) ÷ (x + y).

Unit Conversion

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

Shortcut Trick

Speed Ratio (Usual : New) = 100 : 60 = 5 : 3

Time Ratio (Usual : New) = 3 : 5 (Inverse of speed ratio as distance is constant)

Difference in time = 5 − 3 = 2 units

Given 2 units = 18 min ⇒ 1 unit = 9 min

Usual time = 3 units = 3 × 9 = 27 min

∴ The correct answer is 27 min.

Alternate Method

Given:

Reduced speed = 60% of Usual speed

Delay = 18 minutes

Formula Used:

Distance = Speed × Time

Let Usual speed = s and Usual time = t

⇒ Distance (d) = s × t

⇒ Reduced speed = 0.6s

⇒ New time = t + 18

⇒ Distance = 0.6s × (t + 18)

Since distance is the same:

⇒ st = 0.6s × (t + 18)

⇒ t = 0.6 × (t + 18)

⇒ t = 0.6t + 10.8

⇒ t − 0.6t = 10.8

⇒ 0.4t = 10.8

⇒ t = 10.8 ÷ 0.4

⇒ t = 27 min

∴ The correct answer is 27 min.

Additional Information

Inversely Proportional Relation

When distance is constant, Speed is inversely proportional to Time (S₁/S₂ = T₂/T₁).

Average Speed Formula

If a person travels from A to B at speed x and returns at speed y, then Average Speed = 2xy ÷ (x + y).

Relative Speed

For two objects moving in the same direction, Relative Speed = S₁ − S₂. In opposite directions, it is S₁ + S₂.

Shortcut Trick

Assume Marked Price (X) = 100.

Cost Price (CP) for the person = 100 − 40% = 60.

Selling Price (SP) = 100 + 10% = 110.

Profit = 110 − 60 = 50.

Profit % = (50 ÷ 60) × 100 = 83.33%.

∴ The correct answer is 83.33 percent.

Alternate Method

Given:

Marked Price (MP) = Rs. X

Discount = 40%

Final Selling Price (SP) = X + 10% of X

Calculations:

⇒ Purchase Price (CP) = X − 40% of X = 0.6X

⇒ Selling Price (SP) = X + 10% of X = 1.1X

⇒ Profit = SP − CP = 1.1X − 0.6X = 0.5X

⇒ Profit Percentage = (Profit ÷ CP) × 100

⇒ Profit % = (0.5X ÷ 0.6X) × 100

⇒ Profit % = (5 ÷ 6) × 100 = 83.33%

∴ The correct answer is 83.33 percent.

Additional Information

Profit Percentage

Profit % = [(SP − CP) ÷ CP] × 100. It is always calculated on the Cost Price unless specified otherwise.

Marked Price and Discount

Selling Price = Marked Price − Discount. Discount is usually calculated as a percentage of the Marked Price.

Markup Percentage

Markup % = [(MP − CP) ÷ CP] × 100. It represents how much the Marked Price is above the Cost Price.