Quantitative Aptitude PYQs

Shortcut Trick

Total time = 1 + 1.5 = 2.5 hours.

Total distance = Average speed × Total time = 55 × 2.5 = 137.5 km.

Distance covered in 1^st part = 70 × 1 = 70 km.

Distance covered in 2^nd part = 137.5 − 70 = 67.5 km.

Speed p = 67.5 ÷ 1.5 = 45 km/hr.

∴ The correct answer is 45 km/hr.

Alternate Method

Given:

Speed 1 = 70 km/hr, Time 1 = 1 hour

Speed 2 = p km/hr, Time 2 = 1.5 hours

Average Speed = 55 km/hr

Formula Used:

Average Speed = Total Distance ÷ Total Time

Distance = Speed × Time

Calculations:

⇒ Total Distance = (70 × 1) + (p × 1.5)

⇒ Total Time = 1 + 1.5 = 2.5 hours

⇒ Average Speed = (70 + 1.5p) ÷ 2.5

⇒ 55 = (70 + 1.5p) ÷ 2.5

⇒ 55 × 2.5 = 70 + 1.5p

⇒ 137.5 = 70 + 1.5p

⇒ 1.5p = 137.5 − 70

⇒ 1.5p = 67.5

⇒ p = 67.5 ÷ 1.5 = 45

∴ The correct answer is 45 km/hr.

Additional Information

Average Speed Formula

For different speeds v₁, v₂ over different time intervals t₁, t₂: Avg Speed = (v₁t₁ + v₂t₂) ÷ (t₁ + t₂).

Harmonic Mean Case

If distances are equal, Average Speed = 2v₁v₂ ÷ (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.

Shortcut Trick

Given cot θ = 14 / 3 = Base / Perpendicular.

Let Base (B) = 14 and Perpendicular (P) = 3.

Hypotenuse (H) = √(14² + 3²) = √(196 + 9) = √205.

Since cos θ = Base / Hypotenuse = 14 / √205.

∴ The correct answer is 14/√205.

Alternate Method

Given: cot θ = 14/3

Formula Used:

⇒ cot θ = Base / Perpendicular

⇒ cos θ = Base / Hypotenuse

⇒ Hypotenuse² = Perpendicular² + Base²

Calculations:

⇒ Base (B) = 14, Perpendicular (P) = 3

⇒ Hypotenuse (H) = √(14² + 3²)

⇒ H = √(196 + 9) = √205

⇒ cos θ = B / H = 14 / √205

∴ The correct answer is 14/√205.

Additional Information

Trigonometric Identities

Relation between functions: cosec²θ − cot²θ = 1 and sec²θ − tan²θ = 1.

Reciprocal Relations

cot θ = 1/tan θ and cos θ = 1/sec θ.

Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: H² = P² + B².

Shortcut Trick

Effective Multiplier = 0.55 × 0.40 × 0.45 × 1.50

Effective Multiplier = 0.22 × 0.675 = 0.1485

Percentage Change = (0.1485 − 1) × 100 = −85.15%

∴ The correct answer is Decreases by 85.15%.

Alternate Method

Given:

First Month Change = 45% Decrease

Second Month Change = 60% Decrease

Third Month Change = 55% Decrease

Fourth Month Change = 50% Increase

Let the Original Price be 100.

⇒ Price after 1^st month = 100 × (100 − 45)/100 = 55

⇒ Price after 2^nd month = 55 × (100 − 60)/100 = 55 × 0.4 = 22

⇒ Price after 3^rd month = 22 × (100 − 55)/100 = 22 × 0.45 = 9.9

⇒ Price after 4^th month = 9.9 × (100 + 50)/100 = 9.9 × 1.5 = 14.85

⇒ Total Percentage Change = [(Final Price − Original Price) ÷ Original Price] × 100

⇒ Percentage Change = [(14.85 − 100) ÷ 100] × 100 = −85.15%

∴ The correct answer is Decreases by 85.15%.

Additional Information

Successive Percentage Formula

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

Multiplier Method

Final Value = Initial Value × (1 ± r₁/100) × (1 ± r₂/100) × ... × (1 ± r_n/100).

Percentage Decrease

If the final value is less than the initial value, the percentage change is negative, indicating a decrease.

Shortcut Trick

Effective Cost Price (CP) = 200 + 10% of 200 = 200 + 20 = ��220.

To earn 25% profit, use the multiplier: Selling Price (SP) = CP × 1.25.

SP = 220 × (5 ÷ 4) = 55 × 5 = ₹275.

∴ The correct answer is ₹275.

Alternate Method

Given: CP = ₹200, Overhead = 10% of CP, Profit% = 25%.

Formulas: Total CP = CP + Overhead | SP = Total CP × (100 + P%) ÷ 100

⇒ Overhead Expenses = 10% of 200 = (10 ÷ 100) × 200 = ₹20.

⇒ Total Cost Price = 200 + 20 = ₹220.

⇒ Selling Price = 220 × (100 + 25) ÷ 100.

⇒ SP = 220 × (125 ÷ 100) = 220 × (5 ÷ 4).

⇒ SP = 55 × 5 = ₹275.

∴ The correct answer is ₹275.

Additional Information

Overhead Expenses

Any extra expenditure incurred on goods like freight, transportation, repair, or labor is added to the initial Cost Price to calculate the Total Cost Price.

Profit Calculation

Profit or Loss is always calculated on the Total Cost Price (CP + Overheads), not just the buying price. Profit = SP − Total CP.

Shortcut Trick

Required percentage = (Value being compared ÷ Base value) × 100

Compare B's marks (50) with A's marks (75) as the base.

Required % = (50 ÷ 75) × 100 = 2/3 × 100 = 66.666...

∴ The correct answer is 66.67%.

Alternate Method

Given:

Marks obtained by A = 75

Marks obtained by B = 50

Maximum marks = 95 (Not required for this calculation as the base is A's marks)

Formula Used:

Percentage = (B ÷ A) × 100

Calculations:

⇒ Required Percentage = (50 ÷ 75) × 100

⇒ Required Percentage = (2 ÷ 3) × 100

⇒ Required Percentage = 200 ÷ 3

⇒ Required Percentage = 66.6666...

⇒ Rounding off to two decimal places = 66.67%

∴ The correct answer is 66.67%.

Additional Information

Base Value Identification

In percentage problems, the value following the word 'of' or 'than' usually acts as the denominator (base).

Percentage Change

Percentage increase/decrease = (Difference ÷ Original Value) × 100.

Fraction to Percentage Conversion

1/3 = 33.33%, 2/3 = 66.67%, 1/4 = 25%, 1/5 = 20%.

Shortcut Trick

Fourth Proportional of 1, 3, 6 → (3 × 6) ÷ 1 = 18.

Mean Proportional of 12, 3 → √(12 × 3) = √36 = 6.

Required Ratio → 18 : 6 = 3 :

1. ∴ The correct answer is 3 : 1.

Alternate Method

Given: Numbers for Fourth Proportional (1, 3, 6) and Mean Proportional (12, 3).

Formula Used: Fourth Proportional of a, b, c is (b × c) / a. Mean Proportional of a, b is √(a × b).

⇒ Fourth Proportional = (3 × 6) ÷ 1 = 18.

⇒ Mean Proportional = √(12 × 3) = √36 = 6.

⇒ Required Ratio = 18 : 6.

⇒ Dividing both sides by 6, we get 3 :

1. ∴ The correct answer is 3 : 1.

Additional Information

Third Proportional

The third proportional of two numbers a and b is defined as c such that a : b :: b : c, which gives c = b² / a.

Continued Proportion

Three numbers a, b, c are said to be in continued proportion if the ratio of the first to the second is equal to the ratio of the second to the third (a/b = b/c).

Invertendo and Alternendo

If a : b = c : d, then Invertendo states b : a = d : c and Alternendo states a : c = b : d.

Shortcut Trick

Separate the integers and decimals for faster mental calculation.

⇒ (36 − 30) + (300.30 − 0.26)

⇒ 6 + 300.04

⇒ 306.04

∴ The correct answer is 306.04.

Alternate Method

Given:

36 − 30 + 300.3 − 0.26

Formula Used:

Group all positive values and negative values together to simplify addition and subtraction.

Calculations:

⇒ Sum of positive values = 36 + 300.3 = 336.3

⇒ Sum of negative values = −30 − 0.26 = −30.26

⇒ Total = 336.3 − 30.26

⇒ Add a trailing zero to match decimal places: 336.30 − 30.26

⇒ Total = 306.04

∴ The correct answer is 306.04.

Additional Information

BODMAS Rule

It specifies the correct order of mathematical operations: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction. Addition and Subtraction hold the same priority and can be grouped or calculated strictly from left to right.

Decimal Addition and Subtraction

Always align numbers exactly by their decimal point. Insert trailing zeros to the right of the decimal so all values have the exact same number of decimal places before computing.

Checking with Decimal Digits

In decimal simplifications with distinct options, evaluating only the final decimal digits (e.g., .30 − .26 = .04) can frequently help you pinpoint the correct option immediately without having to complete the full arithmetic sequence.

Shortcut Trick

Let the Total Capacity of the cistern be the LCM of 12 and 14 = 84 units.

Efficiency of (P + Q) = 84 ÷ 12 = 7 units/hour.

Efficiency of P = 84 ÷ 14 = 6 units/hour.

Efficiency of Q = 7 − 6 = 1 unit/hour.

Time for Q to fill 1/4^th cistern = (84 ÷ 4) ÷ 1 = 21 hours.

∴ The correct answer is 21 hours.

Alternate Method

Given:

Time taken by P and Q = 12 hours

Time taken by P alone = 14 hours

Formula Used:

Total Work = Time × Efficiency

Calculations:

⇒ Efficiency of Q = Efficiency of (P + Q) − Efficiency of P

⇒ Efficiency of Q = 7 − 6 = 1 unit/hour

⇒ One-fourth of the cistern capacity = 84 ÷ 4 = 21 units

⇒ Time taken by Q alone = 21 ÷ 1 = 21 hours

∴ The correct answer is 21 hours.

Additional Information

Pipes & Cisterns Efficiency

If a pipe fills a tank in 'x' hours and another empties it in 'y' hours, the net work done in one hour is (1/x − 1/y).

Work & Wages

Wages are always distributed in the ratio of the efficiency or the work done by the individuals, not the time taken.

Shortcut Trick

Assume an estimated mean of 60 for the given observations.

Calculate deviations: +35, −15, +25, −15, −9, +25, +24, +6, −13.

Sum of deviations = 35 − 15 + 25 − 15 − 9 + 25 + 24 + 6 − 13 = 63.

Average deviation = 63 ÷ 9 = 7.

Actual Mean = Assumed Mean + Average Deviation = 60 + 7 = 67.

∴ The correct answer is 67.

Alternate Method

Given: Observations are 95, 45, 85, 45, 51, 85, 84, 66, and 47.

Formula Used: Arithmetic Mean = (Sum of all observations) ÷ (Total number of observations)

Calculations:

⇒ Sum = 95 + 45 + 85 + 45 + 51 + 85 + 84 + 66 + 47

⇒ Sum = 603

⇒ Total count (n) = 9

⇒ Arithmetic Mean = 603 ÷ 9

⇒ Arithmetic Mean = 67

∴ The correct answer is 67.

Additional Information

Properties of Arithmetic Mean

The sum of the deviations of items from their arithmetic mean is always zero: Σ(x − x̄) = 0.

Empirical Relationship

For moderately skewed distributions: Mode = 3 × Median − 2 × Mean.

Weighted Arithmetic Mean

When observations have different importance, use Mean = Σ(w × x) ÷ Σw, where w is the weight of each observation.

Shortcut Trick

Ratio 10 years ago (F:S) = 3 : 1 (Difference = 2)

Ratio 10 years hence (F:S) = 2 : 1 (Difference = 1)

To equalize difference, multiply second ratio by 2 → 4 : 2

Increase in units = 4 − 3 = 1 unit. Total time gap = 10 + 10 = 20 years.

∴ 1 unit = 20 years. Son's age 10 years ago = 1 unit = 20 years.

Present age of son = 20 + 10 = 30 years.

∴ The correct answer is 30 years.

Alternate Method

Given: Ratio 10 years ago = 3:1, Ratio 10 years later = 2:1.

Formula: Age after n years = Age + n.

⇒ Let Son's age 10 years ago be x and Father's age be 3x.

⇒ Present ages: Son = x + 10, Father = 3x + 10.

⇒ Ages after 10 years: Son = x + 20, Father = 3x + 20.

⇒ According to question: (3x + 20) = 2 × (x + 20)

⇒ 3x + 20 = 2x + 40

⇒ x = 20

⇒ Son's present age = x + 10 = 20 + 10 = 30 years.

∴ The correct answer is 30 years.

Additional Information

Ratio Difference Method

When the time interval added to two people is the same, the numerical difference between their ratio parts must be equalized to solve directly.

Age Constant Property

The difference between the ages of two persons remains constant throughout their lives, regardless of the number of years passed.

Shortcut Trick

The perimeter of a triangle formed by joining the mid-points of the sides of a given triangle is always half of the perimeter of the original triangle.

Perimeter of triangle ABC = 5 + 6 + 7 = 18 cm.

Required Perimeter = 18 ÷ 2 = 9 cm.

∴ The correct answer is 9 cm.

Alternate Method

Given: Sides of ΔABC are AB = 5 cm, BC = 6 cm, and CA = 7 cm.

According to the Mid-point Theorem, the segment joining the mid-points of two sides of a triangle is parallel to the third side and half its length.

Let D, E, and F be the mid-points of sides AB, BC, and CA respectively.

⇒ Side DE = CA ÷ 2 = 7 ÷ 2 = 3.5 cm

⇒ Side EF = AB ÷ 2 = 5 ÷ 2 = 2.5 cm

⇒ Side FD = BC ÷ 2 = 6 ÷ 2 = 3.0 cm

⇒ Perimeter of ΔDEF = DE + EF + FD

⇒ Perimeter = 3.5 + 2.5 + 3.0 = 9 cm

∴ The correct answer is 9 cm.

Additional Information

Mid-point Theorem

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Area of Mid-point Triangle

The area of the triangle formed by joining the mid-points is exactly 1/4^th of the area of the original triangle ABC.

Similarity Property

The triangle formed by joining the mid-points is similar to the original triangle with a side ratio of 1:2.

Shortcut Trick

Let Total Work (LCM of 18, 12, 36) = 36 units.

Efficiency of Pipe A = 36 ÷ 18 = 2 units/hr.

Efficiency of Pipe B = 36 ÷ 12 = 3 units/hr.

Efficiency of Leakage = 36 ÷ 36 = −1 unit/hr.

Combined Efficiency = 2 + 3 − 1 = 4 units/hr.

Time to fill 1/3^rd capacity = (36 × 1/3) ÷ 4 = 12 ÷ 4 = 3 hours.

∴ The correct answer is 3 hr.

Alternate Method

Given: Filling time for Pipe A = 18 hr, Pipe B = 12 hr, Emptying time for Leakage = 36 hr.

Calculations:

⇒ Total Efficiency when all are active = 2 + 3 − 1 = 4 units/hr

⇒ Capacity to be filled = 1/3 × 36 = 12 units

⇒ Required Time = 12 ÷ 4 = 3 hr

∴ The correct answer is 3 hr.

Additional Information

Pipes and Cisterns Efficiency

Inlet pipes have positive efficiency while outlet pipes or leakages have negative efficiency in calculations.

Combined Work Formula

Total Time = (Total Work to be done) ÷ (Sum of individual efficiencies).

LCM Method

Total work is assumed as the Least Common Multiple of individual time periods to avoid calculations with fractions.

Shortcut Trick

Convert all numbers into same base

8 = 2³ ⇒ 2^3m × 2³ = 2⁹ ⇒ 3m + 3 = 9 ⇒ m = 2

9 = 3² ⇒ 3²ⁿ × 3² = 3⁶ ⇒ 2n + 2 = 6 ⇒ n = 2

m/n = 2/2 = 1

∴ Answer = 1

Alternate Method

Given:

2^3m × 8 = 2⁹ and 3²ⁿ × 9 = 3⁶

Formula used:

a^x × a^y = a^{(x + y)}

Calculation:

⇒ 2^3m × 2³ = 2⁹

⇒ 2^{(3m + 3)} = 2⁹

⇒ 3m + 3 = 9

⇒ 3m = 6 ⇒ m = 2

⇒ 3²ⁿ × 3² = 3⁶

⇒ 3^{(2n + 2)} = 3⁶

⇒ 2n + 2 = 6

⇒ 2n = 4 ⇒ n = 2

⇒ m/n = 2/2 = 1

∴ The correct answer is 1.

Additional Information

Laws of Exponents

a^x × a^y = a^{(x + y)}, (a^x)^y = a^(xy)

Equal Bases Rule

If bases are equal, then powers must be equal.

Power Conversion

Convert numbers like 8 = 2³ and 9 = 3² to simplify equations.

Shortcut Trick

The Volume of a Cuboid is calculated by the product of its three dimensions.

Volume = 34 × 10 × 26

Perform multiplication: (34 × 26) × 10 = 884 × 10 = 8840 m³.

∴ The correct answer is 8840 m³.

Alternate Method

Given:

Length (L) of the tank = 34 m

Breadth (B) of the tank = 10 m

Height (H) of the tank = 26 m

Formula Used:

Volume of a Cuboid = L × B × H

Calculations:

⇒ Volume = 34 × 10 × 26

⇒ Volume = 340 × 26

⇒ Volume = 8840 m³

∴ The correct answer is 8840 m³.

Additional Information

Total Surface Area of Cuboid

The total surface area is the sum of the areas of all 6 faces: TSA = 2(LB + BH + HL).

Diagonal of a Cuboid

The length of the longest rod that can be placed inside a cuboid is given by √(L² + B² + H²).

Volume of a Cube

For a cube where all sides are equal (L = B = H = a), the volume is simply a³.

Shortcut Trick

Time = Distance ÷ Speed

Time = 60 ÷ 12 = 5

∴ The correct answer is 5 hr.

Alternate Method

Given:

Total Distance = 60 km

Speed of the cyclist = 12 km/hr

Formula Used:

Time = Distance / Speed

Calculations:

⇒ Time taken = 60 / 12

⇒ Time taken = 5 hr

∴ The correct answer is 5 hr.

Additional Information

Basic Speed Formula

Speed is defined as the distance covered per unit time. Speed = Distance ÷ Time.

Unit Conversion

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

Average Speed

Average Speed = (Total Distance) ÷ (Total Time taken).

Shortcut Trick

In a 30°−60°−90° triangle, the ratio of sides opposite to these angles is 1 : √3 : 2.

The Hypotenuse (Ladder) corresponds to 2 units → 10 m.

The Height (Opposite to 60°) corresponds to √3 units.

⇒ Value of 1 unit = 10 ÷ 2 = 5 m.

⇒ Height of the wall = 5 × √3 = 5√3 m.

∴ The correct answer is 5√3 m.

Alternate Method

Given:

Length of the Ladder (Hypotenuse, AC) = 10 m

Angle of elevation (θ) = 60°

Formula Used:

sin θ = Perpendicular / Hypotenuse

Calculations:

⇒ sin 60° = Height of wall / Length of ladder

⇒ √3 / 2 = Height / 10

⇒ Height = 10 × (√3 / 2)

⇒ Height = 5√3 m

∴ The correct answer is 5√3 m.

Additional Information

Trigonometric Ratios

For a right-angled triangle, sin θ = P/H, cos θ = B/H, and tan θ = P/B.

Standard Values

sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2.

Pythagoras Theorem

In any right-angled triangle: (Hypotenuse)² = (Base)² + (Perpendicular)².

Shortcut Trick

To find the greatest volume container, we need to calculate the HCF (Highest Common Factor) of the given quantities.

Given volumes: 300, 280, and 180 litres.

Common factor 10: 10 × (30, 28, 18)

Common factor 2: 10 × 2 × (15, 14, 9)

HCF = 10 × 2 = 20.

∴ The correct answer is 20 litres.

Alternate Method

Given: Volumes are 300 litres, 280 litres, and 180 litres.

Formula Used: Required container volume = HCF of (300, 280, 180).

⇒ Prime Factorization of 300 = 2² × 3 × 5²

⇒ Prime Factorization of 280 = 2³ × 5 × 7

⇒ Prime Factorization of 180 = 2² × 3² × 5

⇒ HCF = 2^min(2,3,2) × 5^min(2,1,1)

⇒ HCF = 2² × 5¹ = 4 × 5 = 20.

∴ The correct answer is 20 litres.

Additional Information

HCF (Highest Common Factor)

It is the largest positive integer that divides each of the integers without a remainder. Used for maximum capacity or equal distribution problems.

LCM (Least Common Multiple)

It is the smallest positive integer that is divisible by each of the given numbers. Used for simultaneous events or recurring interval problems.

Relation between HCF and LCM

For two numbers a and b, HCF(a, b) × LCM(a, b) = a × b.

Shortcut Trick

Net work done in 1 hour = (1/Filling Time) − (1/Emptying Time)

Combined efficiency = 1/9 − 1/12 = (4 − 3)/36 = 1/36

Time taken = 36 hours

∴ The correct answer is 36 hr.

Alternate Method

Given:

Time taken by Filling Tap = 9 hours

Time taken by Emptying Hole = 12 hours

Calculations:

⇒ Net efficiency when both are open = 4 + (−3) = 1 unit/hr

⇒ Total time required to fill the empty cistern = Total Work ÷ Net Efficiency

⇒ Time = 36 ÷ 1 = 36 hr

∴ The correct answer is 36 hr.

Additional Information

Pipes and Cisterns Logic

Efficiency of a filling pipe is taken as positive, while the efficiency of an outlet/leakage is taken as negative.

Direct Formula

If pipe A fills in x hours and pipe B empties in y hours, combined time T = (x × y) ÷ (y − x) provided y > x.

Shortcut Trick

In any data set, the Mode is the value that has the highest frequency (occurs most often).

By scanning the given values, we count the frequency of each number:

Number 47 appears 5 times, while 51 and 43 appear only 3 times each.

Since 47 has the maximum repetition, it is the mode.

∴ The correct answer is 47.

Alternate Method

Given: Data set = 47, 51, 51, 48, 53, 54, 47, 52, 51, 46, 43, 47, 52, 40, 47, 41, 43, 43, 46, 47

Formula Used: Mode = Observation with the highest frequency.

To find the mode, we list the frequency of all distinct observations:

⇒ Number 40: 1 time

⇒ Number 41: 1 time

⇒ Number 43: 3 times (43, 43, 43)

⇒ Number 46: 2 times (46, 46)

⇒ Number 47: 5 times (47, 47, 47, 47, 47)

⇒ Number 48: 1 time

⇒ Number 51: 3 times (51, 51, 51)

⇒ Number 52: 2 times (52, 52)

⇒ Number 53: 1 time

⇒ Number 54: 1 time

⇒ The maximum frequency is 5.

⇒ The value corresponding to frequency 5 is 47.

∴ The correct answer is 47.

Additional Information

Empirical Relationship

For a moderately skewed distribution, the relationship between measures of central tendency is: Mode = 3 × Median − 2 × Mean.

Types of Mode

A data set with one mode is Unimodal, two modes is Bimodal, and more than two is Multimodal.

Mean and Median

Mean is the average of all values (Sum ÷ Count), and Median is the middle value when data is arranged in ascending order.

Shortcut Trick

Fourth proportional to 20, 15, 8 ⇒ (15 × 8) ÷ 20 = 6.

This is equal to the third proportional to A and 30.

Third proportional formula ⇒ 30² ÷ A = 6.

A = 900 ÷ 6 = 150.

∴ The correct answer is 150.

Alternate Method

Given: First set (20, 15, 8) and Second set (A, 30).

Formula Used: Fourth Proportional (x) of a, b, c is x = (b × c) / a. Third Proportional (c) of a, b is c = b² / a.

⇒ Fourth proportional (x) = (15 × 8) ÷ 20

⇒ x = 120 ÷ 20 = 6

⇒ Third proportional to A and 30 is also 6.

⇒ 6 = (30 × 30) ÷ A

⇒ 6 × A = 900

⇒ A = 900 ÷ 6 = 150

∴ The correct answer is 150.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is √(a × b).

Continued Proportion

Three numbers a, b, c are in continued proportion if a/b = b/c, which implies b² = ac.

Fourth Proportional Rule

If a : b :: c : d, then d is the fourth proportional and Product of Extremes = Product of Means (ad = bc).