Quantitative Aptitude PYQs

Shortcut Trick

Solve the 'Of' operator first, then group positive and negative terms for faster calculation.

18 of 2 = 36

Positive terms: 88 + 21 + 28 + 36 + 17 = 190

Negative terms: −34 − 13 = −47

190 − 47 = 143

∴ The correct answer is 143.

Alternate Method

Given: 88 + 21 − 34 + 28 + (18 of 2) + 17 − 13

Formula Used: Apply BODMAS Rule (Bracket, Of, Division, Multiplication, Addition, Subtraction).

⇒ 88 + 21 − 34 + 28 + (18 × 2) + 17 − 13

⇒ 88 + 21 − 34 + 28 + 36 + 17 − 13

⇒ (88 + 21 + 28 + 36 + 17) − (34 + 13)

⇒ 190 − 47

⇒ 143

∴ The correct answer is 143.

Additional Information

BODMAS Rule

The rule defines the order of precedence: Brackets (), Of (orders), Division (÷), Multiplication (×), Addition (+), and Subtraction (−).

Precedence of 'Of'

The 'Of' operator is treated as multiplication but must be performed before division and standard multiplication.

Grouping Property

When multiple additions and subtractions are present, grouping like-signed numbers helps reduce calculation errors.

Shortcut Trick

Check the product of HCF and LCM: 80 × 3600 = 288,000.

Check options where the product is 288,000 and difference is 320.

Option 4: 720 − 400 = 320 (Matches difference).

Option 4: 720 × 400 = 288,000 (Matches HCF × LCM product).

∴ The correct answer is 720 and 400.

Alternate Method

Given: HCF = 80, LCM = 3600, Difference = 320

Formula Used: Product of numbers = HCF × LCM | Numbers = HCF × a and HCF × b (where a, b are co-primes)

⇒ Let the numbers be 80a and 80b.

⇒ LCM = 80 × a × b = 3600

⇒ ab = 3600 ÷ 80 = 45

⇒ Difference = 80a − 80b = 320

⇒ 80(a − b) = 320

⇒ a − b = 320 ÷ 80 = 4

⇒ We need two co-prime numbers whose product is 45 and difference is 4.

⇒ The factors of 45 are (9, 5) since 9 × 5 = 45 and 9 − 5 = 4.

⇒ First number = 80 × 9 = 720

⇒ Second number = 80 × 5 = 400

∴ The correct answer is 720 and 400.

Additional Information

Fundamental Property of HCF and LCM

For any two numbers x and y, the product of the numbers is always equal to the product of their HCF and LCM (x × y = HCF × LCM).

Relation with Difference

The HCF of two numbers is always a factor of their difference, their sum, and their LCM.

Co-prime Factors

If HCF is h, the numbers can be expressed as ha and hb, where a and b are co-prime (HCF of a and b is 1).

Shortcut Trick

Separate the whole numbers and the fractions from the mixed terms.

Sum of integers = 3 + 33 + 333 + 3333 = 3702.

Sum of fractional parts = 1/2 + 1/2 + 1/2 + 1/2 = 2.

Total value = 3702 + 2 = 3704.

∴ The correct answer is 3704.

Alternate Method

Given: 3(1/2) + 33(1/2) + 333(1/2) + 3333(1/2)

Formula Used: A(b/c) = A + b/c

⇒ (3 + 1/2) + (33 + 1/2) + (333 + 1/2) + (3333 + 1/2)

⇒ (3 + 33 + 333 + 3333) + (1/2 + 1/2 + 1/2 + 1/2)

⇒ 3702 + (1 + 1 + 1 + 1)/2

⇒ 3702 + 4/2

⇒ 3702 + 2 = 3704

∴ The correct answer is 3704.

Additional Information

Mixed Fraction Addition

A mixed fraction consists of a whole number and a proper fraction. When adding multiple mixed fractions, it is mathematically simpler to sum the integer parts and the fractional parts independently.

BODMAS Rule

For simplification, always follow the BODMAS order (Brackets, Orders, Division, Multiplication, Addition, and Subtraction) to ensure accurate results in multi-step expressions.

Fraction Simplification

Fractions with the same denominator can be added by simply summing their numerators while keeping the denominator constant: a/c + b/c = (a + b)/c.

Shortcut Trick

Use the Algebraic Identity: (a + b)² − (a − b)² = 4ab.

In the given problem, a = 7.5 and b = 2.5.

The expression is in the form: (4ab) ÷ (ab).

The term (ab) cancels out, leaving only the constant 4.

∴ The correct answer is 4.

Alternate Method

Given: Expression = ((7.5 + 2.5)² − (7.5 − 2.5)²) ÷ (7.5 × 2.5)

Method: Perform direct addition and subtraction inside the brackets first.

⇒ (10)² − (5)² ÷ (7.5 × 2.5)

⇒ (100 − 25) ÷ 18.75

⇒ 75 ÷ 18.75

⇒ 4

∴ The correct answer is 4.

Additional Information

Square of Sum and Difference

The identity (a + b)² + (a − b)² = 2(a² + b²) is another frequently used variation in simplification.

Difference of Squares

The formula a² − b² = (a + b)(a − b) can also be applied to solve the numerator by treating (a+b) and (a-b) as two separate terms.

BODMAS Rule

Always solve Brackets first, followed by Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Shortcut Trick

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

Here, let a = √5 and b = √3.

Expression = 2 × [(√5)² + (√3)²]

Expression = 2 × (5 + 3) = 2 × 8 = 16.

∴ The correct answer is 16.

Alternate Method

Given: (√5 + √3)² + (√5 − √3)²

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

⇒ [(√5)² + (√3)² + 2 × √5 × √3] + [(√5)² + (√3)² − 2 × √5 × √3]

⇒ [5 + 3 + 2√15] + [5 + 3 − 2√15]

⇒ 8 + 2√15 + 8 − 2√15

⇒ 16

∴ The correct answer is 16.

Additional Information

Algebraic Difference Identity

When subtracting the squares of sums and differences, the result is (a + b)² − (a − b)² = 4ab.

Square Root Property

For any non-negative number x, the square of its square root is the number itself: (√x)² = x.

Difference of Squares

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

Shortcut Trick

For any three consecutive even numbers, the average of the numbers is always equal to the middle number.

Middle Number = Sum ÷ 3 = 24 ÷ 3 = 8

Since the numbers are consecutive even numbers, the set is: (8 − 2), 8, and (8 + 2) ⇒ 6, 8, 10

Product of the numbers = 6 × 8 × 10 = 480

∴ The correct answer is 480.

Alternate Method

Given: Sum of three consecutive even numbers = 24

Formula Used: Let the three consecutive even numbers be x, x + 2, and x + 4.

Calculations:

⇒ x + (x + 2) + (x + 4) = 24

⇒ 3x + 6 = 24

⇒ 3x = 24 − 6

⇒ 3x = 18

⇒ x = 18 ÷ 3 = 6

⇒ First number = 6

⇒ Second number = 6 + 2 = 8

⇒ Third number = 6 + 4 = 10

⇒ Product = 6 × 8 × 10 = 480

∴ The correct answer is 480.

Additional Information

Consecutive Even Numbers

These are even integers that follow each other in order, where each number is 2 greater than the previous one (e.g., n, n + 2, n + 4).

Arithmetic Mean Property

For any Arithmetic Progression (AP) with an odd number of terms, the Average is always the exact middle term.

Sum of First n Even Numbers

The sum of the first n even natural numbers is given by the formula S_n = n(n + 1).

Shortcut Trick

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

Here, a = 5 and b = √3.

⇒ 4 × 5 × √3 − 4√3

⇒ 20√3 − 4√3 = 16√3

∴ The correct answer is 16√3.

Alternate Method

Given: (5 + √3)² − (5 − √3)² − 4√3

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

⇒ [(5)² + 2 × 5 × √3 + (√3)²] − [(5)² − 2 × 5 × √3 + (√3)²] − 4√3

⇒ [25 + 10√3 + 3] − [25 − 10√3 + 3] − 4√3

⇒ [28 + 10√3] − [28 − 10√3] − 4√3

⇒ 28 + 10√3 − 28 + 10√3 − 4√3

⇒ 20√3 − 4√3

⇒ 16√3

∴ The correct answer is 16√3.

Additional Information

Difference of Squares Identity

The identity (a + b)² − (a − b)² = 4ab is a derived form of the basic square identities, extremely useful in simplification.

Sum of Squares Identity

Similarly, (a + b)² + (a − b)² = 2(a² + b²), which helps in solving expressions involving sum of squares.

Operations on Like Surds

Irrational numbers with the same radical part are called like surds and can be added or subtracted directly: x√y ± z√y = (x ± z)√y.

Shortcut Trick

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

Here, a = 290 and b = 210.

Expression = 4 × 290 × 210.

Rearrange as (2 × 290) × (2 × 210) = 580 × 420.

∴ The correct answer is 580 × 420.

Alternate Method

Given: (290 + 210)² − (290 − 210)²

Formula Used: (a + b)² − (a − b)² = 4ab

⇒ Let a = 290 and b = 210.

⇒ Applying the identity: (290 + 210)² − (290 − 210)² = 4 × 290 × 210.

⇒ Split the factor 4 into 2 × 2.

⇒ (2 × 290) × (2 × 210).

⇒ 580 × 420.

∴ The correct answer is 580 × 420.

Additional Information

Sum of Squares Identity

(a + b)² + (a − b)² = 2(a² + b²). This is used when the middle sign is positive.

Difference of Two Squares

a² − b² = (a − b)(a + b). This is a fundamental identity for factorisation.

Standard Expansion

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

Shortcut Trick

Identify the standard Pythagorean Triplet ratio: 3 : 4 : 5.

Scale the triplet by a factor of 2: (3 × 2) : (4 × 2) : (5 × 2) = 6 : 8 : 10.

Given that the Length is 6 cm and the Diagonal (hypotenuse) is 10 cm.

The missing side, which is the Breadth, must correspond to the middle value of the triplet.

∴ The correct answer is 8 cm.

Alternate Method

Given:

Length (l) of the rectangle = 6 cm

Diagonal (d) of the rectangle = 10 cm

Formula Used:

In a Rectangle, the diagonal forms a right-angled triangle with the length and breadth. According to Pythagoras Theorem:

Diagonal² = Length² + Breadth²

Calculations:

⇒ 10² = 6² + Breadth²

⇒ 100 = 36 + Breadth²

⇒ Breadth² = 100 − 36

⇒ Breadth² = 64

⇒ Breadth = √64

⇒ Breadth = 8 cm

∴ The correct answer is 8 cm.

Additional Information

Area of Rectangle

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

Perimeter of Rectangle

The total boundary length is calculated as 2 × (Length + Breadth).

Pythagorean Triplets

Sets of three integers that satisfy a² + b² = c², such as (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Properties of Rectangle

All internal angles are 90°, and diagonals are equal and bisect each other.

Shortcut Trick

Check the options for divisibility by both 8 and 9 (factors of 72).

For 9936: Sum of digits = 9+9+3+6 = 27 (divisible by 9).

Last 3 digits 936 ÷ 8 = 117 (divisible by 8).

Other options: 9986 (sum 32 ✗), 9972 (972 ÷ 8 ✗), 9948 (sum 30 ✗).

∴ The correct answer is 9936.

Alternate Method

Given: The largest 4-digit number is 9999 and the divisor is 72.

Formula: Largest divisible number = Largest number − Remainder

⇒ Divide 9999 by 72 to find the remainder.

⇒ 9999 ÷ 72

⇒ 72 × 138 = 9936

⇒ Remainder = 9999 − 9936 = 63

⇒ Required number = 9999 − 63

⇒ 9936

∴ The correct answer is 9936.

Additional Information

Divisibility Rule of 72

A number is divisible by 72 if it is divisible by both 8 and 9, as they are co-prime factors.

Divisibility by 8 and 9

A number is divisible by 8 if its last three digits are divisible by 8, and divisible by 9 if the sum of its digits is a multiple of 9.

Largest N-digit Number Divisibility

To find the largest n-digit number divisible by x, divide the largest n-digit number by x and subtract the remainder from the original number.

Shortcut Trick

Total sum of ages of E, F, and G = 50 × 3 = 150 years.

Sum of ages of E and F = 96 years.

Age of G = Total sum − Sum of (E + F) = 150 − 96 = 54 years.

∴ The correct answer is 54 years.

Alternate Method

Given:

Sum of ages of E and F = 96 years

Average age of E, F, and G = 50 years

Formula Used:

Average = (Sum of observations) ÷ (Total number of observations)

Calculations:

⇒ (E + F + G) ÷ 3 = 50

⇒ E + F + G = 50 × 3

⇒ E + F + G = 150

⇒ 96 + G = 150

⇒ G = 150 − 96

⇒ G = 54

∴ The correct answer is 54 years.

Additional Information

Sum from Average

The total sum of n items can be calculated by multiplying the Average by n. Formula: Sum = Average × n.

Finding a Missing Value

To find the value of a missing observation, subtract the sum of the known observations from the total calculated sum.

Weighted Average

If two groups have averages A₁ and A₂ and sizes n₁ and n₂, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Shortcut Trick

If 50% of students passed, then the remaining 50% must have failed.

Since the percentage of Passed and Failed students is equal (both 50%), their numerical counts must also be equal.

Given Failed students = 20000, it directly implies Passed students = 20000.

∴ The correct answer is 20000.

Alternate Method

Given:

Percentage of passed students = 50%

Total number of failed students = 20000

Formula:

Total Students (100%) = Passed Students + Failed Students

Calculations:

⇒ Percentage of failed students = 100% − 50% = 50%

⇒ Let the total number of appeared students be x

⇒ 50% of x = 20000

⇒ (50 ÷ 100) × x = 20000

⇒ 0.5 × x = 20000

⇒ x = 20000 ÷ 0.5 = 40000

⇒ Number of passed students = 50% of 40000

⇒ Passed students = 0.5 × 40000 = 20000

∴ The correct answer is 20000.

Additional Information

Basic Percentage Concept

Percentage represents a part per hundred. Value = (Percentage ÷ 100) × Total.

Complementary Percentages

If a set is divided into two categories (e.g., Pass/Fail), the sum of their percentages always equals 100%.

Fraction Equivalents

Common percentages like 50% can be treated as fractions (1/2) to simplify calculations quickly: Value = Total × 1/2.

Shortcut Trick

Ratio of Cost Price (CP) to Selling Price (SP) = 5000 : 4000 = 5 : 4.

Loss unit = 5 − 4 = 1 unit.

Loss % = (Loss ÷ CP) × 100.

Loss % = (1 ÷ 5) × 100 = 20%.

∴ The correct answer is 20%.

Alternate Method

Given: The Cost Price (CP) of the article is ₹5000 and the Selling Price (SP) is ₹4000.

Formula Used: Loss % = [(Cost Price − Selling Price) ÷ Cost Price] × 100

⇒ Loss = CP − SP

⇒ Loss = ₹5000 − ₹4000 = ₹1000

⇒ Loss percentage = (Loss ÷ CP) × 100

⇒ Loss % = (1000 ÷ 5000) × 100

⇒ Loss % = (1 ÷ 5) × 100

⇒ Loss % = 20%

∴ The correct answer is 20%.

Additional Information

Profit Percentage

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

Relationship between CP and SP

Cost Price (CP) = [100 ÷ (100 − Loss%)] × SP, or Selling Price (SP) = CP × [(100 − Loss%) ÷ 100].

Marked Price and Discount

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

Shortcut Trick

Gap between selling prices = 1600 − 1200 = 400.

Ratio of Loss : Profit = 1 : 19 ⇒ Total 20 units = 400.

1 unit (Loss) = 20 ⇒ Cost Price = 1200 + 20 = ₹1220.

New Selling Price = 1220 × 130% = ₹1586.

∴ The correct answer is ₹1586.

Alternate Method

Given:

Selling Price 1 (SP₁) = ₹1600

Selling Price 2 (SP₂) = ₹1200

Profit at SP₁ = 19 × Loss at SP₂

Formula Used:

Profit = Selling Price − Cost Price

Loss = Cost Price − Selling Price

Calculations:

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

⇒ According to the condition: (1600 − x) = 19 × (x − 1200)

⇒ 1600 − x = 19x − 22800

⇒ 19x + x = 1600 + 22800

⇒ 20x = 24400

⇒ x = 24400 ÷ 20 = ₹1220

⇒ Desired Profit Percentage = 30%

⇒ New Selling Price = 1220 × (1 + 30 ÷ 100)

⇒ New Selling Price = 1220 × 1.3 = ₹1586

∴ The correct answer is ₹1586.

Additional Information

Profit and Loss Basics

Profit % = (Profit ÷ CP) × 100 and Loss % = (Loss ÷ CP) × 100. Both are always calculated on Cost Price.

Effective SP Formula

When a profit of P% is required, the Selling Price is given by: SP = CP × (100 + P) ÷ 100.

Relationship Shortcut

If Profit at SP₁ is n times Loss at SP₂, then CP = (SP₁ + n × SP₂) ÷ (n + 1).

Shortcut Trick

Apply the Successive Percentage formula: Net change = a + b + (ab/100).

Here, a = +40% (increase) and b = −25% (decrease).

Net % = 40 − 25 + [(40 × −25) / 100]

Net % = 15 − 10 = +5%.

A positive sign indicates a net increase in salary.

∴ The correct answer is 5% increase.

Alternate Method

Given:

Initial salary increase = 40%

Subsequent salary reduction = 25%

Calculations:

⇒ Let the initial Salary be 100 units.

⇒ Salary after 40% increase = 100 + (40/100 × 100) = 100 + 40 = 140 units.

⇒ Now, the salary is reduced by 25% on the new value (140).

⇒ Reduction amount = 25% of 140 = (25/100) × 140 = 1/4 × 140 = 35 units.

⇒ Final Salary = 140 − 35 = 105 units.

⇒ Net change = Final Salary − Initial Salary = 105 − 100 = 5.

⇒ Net percentage change = (5 / 100) × 100 = 5%.

∴ The correct answer is 5% increase.

Additional Information

Multiplying Factor (MF) Method

Final Value = Initial Value × MF₁ × MF₂. For a 40% increase, MF₁ = 1.40. For a 25% decrease, MF₂ = 0.75. Net MF = 1.40 × 0.75 = 1.05 (i.e., 5% increase).

Base Value 100 Rule

In percentage change problems, assuming 100 as the base value simplifies calculations because any change is directly equivalent to the percentage value.

Successive Change Application

This logic is widely used in Profit and Loss (successive discounts), Compound Interest, and Population Growth problems.

Shortcut Trick

Remaining fraction after Day 1 = 1 − 2/3 = 1/3.

Remaining fraction after Day 2 = 1/3 × (1 − 1/4) = 1/3 × 3/4 = 1/4.

Given that 1/4 of the Total work = 20 units.

Total work = 20 × 4 = 80 units.

∴ The correct answer is 80 units.

Alternate Method

Given:

Work done on first day = 2/3 part

Work done on second day = 1/4 of the remaining work

Final remaining work = 20 units

Formula Used:

Total Work = Work Done + Remaining Work

Calculations:

⇒ Let the Total work be 1 unit.

⇒ Work done on Day 1 = 2/3

⇒ Remaining work after Day 1 = 1 − 2/3 = 1/3

⇒ Work done on Day 2 = 1/4 × 1/3 = 1/12

⇒ Total work done in 2 days = 2/3 + 1/12

⇒ Total work done = (8 + 1) / 12 = 9/12 = 3/4

⇒ Final remaining work = 1 − 3/4 = 1/4

⇒ According to the question, 1/4 part = 20 units

⇒ 1 part (Total Work) = 20 × 4 = 80 units

∴ The correct answer is 80 units.

Additional Information

Successive Remaining Fractions

If fractions f₁, f₂... are removed successively from a remainder, the final remainder is Total × (1 − f₁) × (1 − f₂)...

LCM Method for Work

To avoid fractions, assume the Total Work as the LCM of all denominators (e.g., LCM of 3 and 4 is 12).

Unitary Method in Work

If '1/n' part of work is 'k' units, then the complete work (1 unit) is 'n × k' units.

Shortcut Trick

Use the combined work formula: Time = (X × Y) ÷ (X + Y)

Where X = 12 days (for P and Q together) and Y = 18 days (for R alone).

Total time for P, Q, and R = (12 × 18) ÷ (12 + 18)

Total time = 216 ÷ 30 = 7.2 days.

∴ The correct answer is 7.2 days.

Alternate Method

Given:

Time taken by P and Q together = 12 days

Time taken by R alone = 18 days

Formula Used:

Total Work = Efficiency × Time

Efficiency = Total Work ÷ Time

⇒ Total work is LCM of 12 and 18 = 36 units.

⇒ Combined efficiency of P, Q, and R = (Efficiency of P + Q) + (Efficiency of R)

⇒ Total Efficiency = 3 + 2 = 5 units/day

⇒ Time taken by P, Q, and R = 36 ÷ 5

⇒ Time taken = 7.2 days

∴ The correct answer is 7.2 days.

Additional Information

Efficiency and Time Relation

The time taken to complete a work is inversely proportional to the efficiency of the worker (Time = 1/Efficiency).

Total Work Assumption

Assuming Total Work as the LCM of given time periods makes calculations easier by providing integer values for efficiency.

Combined Rate Rule

If multiple people work together, their individual efficiencies (work done per unit time) are added to find the total work rate.

Shortcut Trick

Total distance to be covered = 250 + 350 = 600 m

Relative speed (Opposite direction) = Total distance ÷ Time = 600 ÷ 60 = 10 m/s

Since speeds are equal, speed of each train = 10 ÷ 2 = 5 m/s

Convert m/s to km/hr: 5 × (18/5) = 18 km/hr

∴ The correct answer is 18 km/hr.

Alternate Method

Given:

Length of Train A = 250 m

Length of Train B = 350 m

Crossing time = 60 sec

Speed of Train A = Speed of Train B = S

Formula Used:

Relative Speed (Opposite Direction) = Speed₁ + Speed₂

Distance = Relative Speed × Time

Calculations:

⇒ Total Distance = 250 + 350 = 600 m

⇒ Relative Speed = S + S = 2S

⇒ 2S = 600 ÷ 60

⇒ 2S = 10 m/s

⇒ S = 5 m/s

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

⇒ Speed = 5 × (18/5) = 18 km/hr

∴ The correct answer is 18 km/hr.

Additional Information

Relative Speed in Opposite Direction

When two objects move in opposite directions, their relative speed is the sum of their individual speeds: V_rel = V₁ + V₂.

Relative Speed in Same Direction

When two objects move in the same direction, their relative speed is the difference between their speeds: V_rel = |V₁ − V₂|.

Unit Conversion

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

Crossing Stationary Objects

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

Shortcut Trick

The speed per hour forms an Arithmetic Progression (AP): 22, 30, 38, ...

Distance after n hours = (n/2) × [2 × Initial Speed + (n − 1) × Increment]

Check options: For n = 6 ⇒ (6/2) × [2(22) + (6 − 1)8] = 3 × [44 + 40] = 3 × 84 = 252 km.

Since the sum matches the total distance, the time taken is 6 hr.

∴ The correct answer is 6 hr.

Alternate Method

Given: Initial speed (a) = 22 km/hr, Increment (d) = 8 km/hr, Total distance (S_n) = 252 km.

Formula Used: Sum of AP, S_n = (n/2)[2a + (n − 1)d]

⇒ 252 = (n/2) × [2 × 22 + (n − 1) × 8]

⇒ 252 = (n/2) × [44 + 8n − 8]

⇒ 252 = (n/2) × [36 + 8n]

⇒ 252 = n × (18 + 4n)

⇒ 252 = 18n + 4n²

⇒ 4n² + 18n − 252 = 0

⇒ 2n² + 9n − 126 = 0

⇒ 2n² + 21n − 12n − 126 = 0

⇒ n(2n + 21) − 6(2n + 21) = 0

⇒ (n − 6)(2n + 21) = 0

⇒ n = 6 (Since time cannot be negative)

∴ The correct answer is 6 hr.

Additional Information

Arithmetic Progression (AP) Sum

The sum of the first n terms of an AP is given by S_n = (n/2)[2a + (n − 1)d], where a is the first term and d is the common difference.

General Term of AP

The n^th term of an AP is calculated as T_n = a + (n − 1)d. In this question, it represents the speed in the n^th hour.

Quadratic Formula

For an equation ax² + bx + c = 0, the roots are found using x = [−b ± √(b² − 4ac)] / 2a.

Shortcut Trick

Discount 30% = 3/10 → MP : SP = 10 : 7

Discount 20% = 1/5 → MP : SP = 5 : 4

Multiply corresponding parts: (10 × 5) : (7 × 4) = 50 : 28

On simplifying, 50 : 28 ÷ 2 = 25 : 14

∴ The correct answer is 25 : 14.

Alternate Method

Given: First discount = 30%, Second discount = 20%

Formula: Net Discount = d₁ + d₂ − (d₁ × d₂)/100

⇒ Let Marked Price = 100 units

⇒ After 30% discount, Price = 100 − (30% of 100) = 100 − 30 = 70 units

⇒ After 20% discount on remaining, Selling Price = 70 − (20% of 70) = 70 − 14 = 56 units

⇒ Ratio of MP to SP = 100 : 56

⇒ Dividing both sides by 4, we get 25 : 14

∴ The correct answer is 25 : 14.

Additional Information

Successive Discount Formula

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

Relation Between MP and SP

Selling Price = Marked Price × (1 − d₁/100) × (1 − d₂/100), where d represents discount percentages.

Markup and Discount

Discount is always calculated on the Marked Price (MP), while Profit or Loss is calculated on the Cost Price (CP).