Quantitative Aptitude PYQs

Shortcut Trick

The age difference between two people remains constant throughout their lives.

Ratio 5 years ago (Priya : Meera) = 3 : 2 → Difference in ratio units = 3 − 2 = 1 unit.

Given, 1 unit = 10 years. ∴ Meera's age 5 years ago = 2 × 10 = 20 years.

Meera's present age = 20 + 5 = 25 years.

∴ The correct answer is 25 years.

Alternate Method

Given:

Ratio of Priya and Meera (5 years ago) = 3 : 2

Difference in their present ages = 10 years

Calculations:

⇒ Let ages 5 years ago be 3x and 2x.

⇒ Present age of Priya = 3x + 5

⇒ Present age of Meera = 2x + 5

⇒ Difference = (3x + 5) − (2x + 5) = 10

⇒ x = 10

⇒ Meera's present age = 2(10) + 5

⇒ Meera's present age = 20 + 5 = 25 years

∴ The correct answer is 25 years.

Additional Information

Constant Difference Rule

The difference between the ages of two people is always constant, whether it is 10 years ago, present, or 10 years in the future.

Ratio Scaling

If the difference in ratio units is 'd' and the actual difference is 'D', then 1 ratio unit = D/d. This multiplier is used to find all related age values.

Shortcut Trick

Convert bases to 2: 16 = 2⁴ and 64 = 2⁶.

Equate the exponents: 3y + x = 4 and 2y − x = 6.

Add both equations to eliminate x: 5y = 10.

Solve for y: y = 2.

∴ The correct answer is 2.

Alternate Method

Given:

2^{3y + x} = 16

2^{2y − x} = 64

Formula Used:

If a^m = aⁿ, then m = n (when a > 0 and a ≠ 1).

Calculations:

⇒ 2^{3y + x} = 2⁴

⇒ 3y + x = 4 ---(Equation 1)

⇒ 2^{2y − x} = 2⁶

⇒ 2y − x = 6 ---(Equation 2)

⇒ Adding (Equation 1) and (Equation 2):

⇒ (3y + x) + (2y − x) = 4 + 6

⇒ 5y = 10

⇒ y = 10 ÷ 5

⇒ y = 2

∴ The correct answer is 2.

Additional Information

Laws of Indices

When the bases are equal, their powers can be equated. Also, a^m × aⁿ = a^m+n and (a^m)ⁿ = a^m×n.

Linear Equations in Two Variables

To find the value of one variable, use the elimination method by adding or subtracting equations to remove the other variable.

Powers of 2

Recognizing standard powers like 2⁴ = 16, 2⁵ = 32, and 2⁶ = 64 is essential for quick calculation in exponential equations.

Shortcut Trick

Ratio of speeds S_A : S_B = 100 : 150 = 2 : 3.

Using the meeting point formula: S_A / S_B = √(T_B / T_A).

2 / 3 = √(p / 90.9)

Squaring both sides: 4 / 9 = p / 90.9

p = 4 × 10.1 = 40.4 hours.

∴ The correct answer is 40.4 hours.

Alternate Method

Given:

Time taken by A after meeting (T_A) = 90.9 hours

Time taken by B after meeting (T_B) = p hours

Speed of B (S_B) = 1.5 × Speed of A (S_A)

Formula Used:

For two objects starting simultaneously from opposite ends and meeting at a point, the ratio of their speeds is inversely proportional to the square root of the time taken to reach the opposite destinations after meeting: S_A / S_B = √(T_B / T_A)

Calculations:

⇒ S_A / S_B = 1 / 1.5 = 1 / (3/2) = 2/3

⇒ 2/3 = √(p / 90.9)

⇒ (2/3)² = p / 90.9

⇒ 4 / 9 = p / 90.9

⇒ p = (4 × 90.9) ÷ 9

⇒ p = 4 × 10.1

⇒ p = 40.4

∴ The correct answer is 40.4 hours.

Additional Information

Meeting Point Time

The time when they meet (T_M) is given by the geometric mean of the remaining times: T_M = √(T_A × T_B).

Relative Speed

When two objects move towards each other, their relative speed is the sum of their individual speeds (S_A + S_B).

Speed-Time Relationship

For a constant distance, speed is inversely proportional to time (S ∝ 1/T).

Shortcut Trick

Let Total Work (LCM of 10 and 15) = 30 units.

Efficiency of M = 3; Efficiency of N = 2.

Work done by M + N in 5 minutes = (3 + 2) × 5 = 25 units.

Remaining work = 30 − 25 = 5 units, filled by N in 5 ÷ 2 = 2.5 minutes.

Total time = 5 + 2.5 = 7.5 minutes.

∴ The correct answer is 7.5 min.

Alternate Method

Given: Time for Pipe M = 10 min, Time for Pipe N = 15 min.

Calculations:

⇒ Both pipes M and N opened for first 5 minutes.

⇒ Work done in 5 min = (Efficiency of M + Efficiency of N) × 5

⇒ Work done = (3 + 2) × 5 = 25 units

⇒ Remaining work = Total Work − Completed Work = 30 − 25 = 5 units

⇒ Time taken by Pipe N to fill remaining part = 5 ÷ 2 = 2.5 minutes

⇒ Total time required = 5 min + 2.5 min = 7.5 minutes

∴ The correct answer is 7.5 min.

Additional Information

Pipes & Cisterns Efficiency

The rate of filling is inversely proportional to the time taken. If a pipe fills a tank in 'x' hours, its 1-hour work is 1/x.

Inlet and Outlet Pipes

Inlet pipe efficiency is taken as positive (+), while outlet (leakage) pipe efficiency is taken as negative (−).

Total Work Concept

Total work is the LCM of individual times taken by the pipes to fill or empty the tank independently.

Shortcut Trick

Bina SP (Shyam CP) = 81% of Bina CP = ₹4,860.

Bina CP = 4,860 ÷ 0.81 = ₹6,000.

Shyam SP (Bina 17% Profit Price) = 117% of 6,000 = ₹7,020.

Shyam Profit = Shyam SP − Shyam CP = 7,020 − 4,860 = ₹2,160.

∴ The correct answer is ₹2,160.

Alternate Method

Given → Bina SP = ₹4,860, Bina Loss = 19%, Target Profit = 17%. Formulas: CP = SP × 100 ÷ (100 − Loss%), New SP = CP × (100 + Profit%) ÷ 100.

⇒ Cost Price (CP) of article for Bina = 4,860 × 100 ÷ (100 − 19) = 4,860 × 100 ÷ 81 = ₹6,000. ⇒ Selling Price (SP) for Shyam (Bina 17% profit price) = 6,000 × 117 ÷ 100 = ₹7,020. ⇒ Cost Price (CP) for Shyam = ₹4,860. ⇒ Profit for Shyam = 7,020 − 4,860 = ₹2,160.

∴ The correct answer is ₹2,160.

Additional Information

Chain Sale Calculations

In chain sales, the Selling Price for one person becomes the Cost Price for the next person in the sequence.

Profit and Loss Relation

SP = CP × (100 ± Gain/Loss)% ÷ 100. Always calculate the profit or loss percentage on the Cost Price unless stated otherwise.

Shortcut Trick

Ratio of Cost Price (CP) to Marked Price (MP) is given by (100 − Discount%) : (100 + Profit%).

CP : MP = (100 − 20) : (100 + 10) = 80 : 110 = 8 : 11.

Given MP = 11 units = 880, so 1 unit = 880 ÷ 11 = 80.

CP = 8 units = 8 × 80 = 640.

∴ The correct answer is ₹ 640.

Alternate Method

Given: Marked Price (MP) = ₹ 880, Discount = 20%, Profit = 10%.

Formulas: SP = MP × (100 − Discount%) ÷ 100 | CP = SP × 100 ÷ (100 + Profit%).

⇒ Selling Price (SP) = 880 × (100 − 20) ÷ 100 = 880 × 0.8 = ₹ 704.

⇒ Cost Price (CP) = 704 × 100 ÷ (100 + 10) = 704 × 100 ÷ 110.

⇒ CP = 7040 ÷ 11 = ₹ 640.

∴ The correct answer is ₹ 640.

Additional Information

Marked Price & Discount

The Marked Price is the price listed on the label. Discount is always calculated on the MP to arrive at the Selling Price (SP).

Relation between CP and MP

When both profit percentage and discount percentage are known, use the direct relation CP/MP = (100 − D%)/(100 + P%).

Shortcut Trick

Apply the Empirical Formula: Mode = 3 × Median − 2 × Mean.

46.2 = 3 × 40 − 2 × Mean

46.2 = 120 − 2 × Mean

2 × Mean = 120 − 46.2 = 73.8

Mean = 36.9

∴ The correct answer is 36.9.

Alternate Method

Given:

Mode = 46.2

Median = 40

Formula Used:

The relationship between the three measures of central tendency is given by: Mode = 3(Median) − 2(Mean)

Calculations:

⇒ 46.2 = 3 × 40 − 2 × Mean

⇒ 46.2 = 120 − 2 × Mean

⇒ 2 × Mean = 120 − 46.2

⇒ 2 × Mean = 73.8

⇒ Mean = 73.8 ÷ 2

⇒ Mean = 36.9

∴ The correct answer is 36.9.

Additional Information

Empirical Formula

It is used for moderately skewed distributions where the relationship is Mode = 3 Median − 2 Mean.

Symmetric Distribution

In a perfectly symmetrical distribution (like the Normal Distribution), Mean = Median = Mode.

Mean Calculation

Mean can also be calculated as Mean = Median − (Mode − Median) ÷ 2, which is a derivation of the same formula.

Skewness Relation

If Mean > Median > Mode, the data is positively skewed; if Mode > Median > Mean, the data is negatively skewed.

Shortcut Trick

Using Alligation method: Difference between Type A (30%) and Mean (42%) is 12.

Difference between Type B (50%) and Mean (42%) is 8.

Ratio of A : B = 8 : 12 = 2 : 3.

Total parts = 2 + 3 = 5 parts = 20 kg → 1 part = 4 kg.

Type A = 2 × 4 = 8 kg and Type B = 3 × 4 = 12 kg.

∴ The correct answer is 8 kg of A, 12 kg of B.

Alternate Method

Given: Total mixture = 20 kg, Nitrogen required = 42%. Nitrogen in A = 30%, Nitrogen in B = 50%.

⇒ Ratio of quantity of A to B = 8 : 12 = 2 : 3.

⇒ Let the quantities be 2x and 3x.

⇒ 2x + 3x = 20 kg

⇒ 5x = 20 kg ⇒ x = 4 kg.

⇒ Quantity of Fertiliser A = 2 × 4 = 8 kg.

⇒ Quantity of Fertiliser B = 3 × 4 = 12 kg.

∴ The correct answer is 8 kg of A, 12 kg of B.

Additional Information

Alligation Rule

It states that (Quantity of Cheaper / Quantity of Dearer) = (CP of Dearer − Mean Price) / (Mean Price − CP of Cheaper).

Weighted Average Formula

The final concentration C = (n₁c₁ + n₂c₂) / (n₁ + n₂), where n is quantity and c is concentration.

Mixture Applications

This method is widely used in profit/loss, average speed, simple interest rates, and solution concentrations.

Shortcut Trick

Product of H.C.F. and L.C.M. = Product of the two numbers.

Other number = (18 × 1296) ÷ 162

Other number = 1296 ÷ 9 = 144

∴ The correct answer is 144.

Alternate Method

Given:

H.C.F. = 18

L.C.M. = 1296

First Number (A) = 162

Formula Used:

H.C.F. × L.C.M. = A × B

Calculations:

⇒ 18 × 1296 = 162 × B

⇒ B = (18 × 1296) ÷ 162

⇒ B = 23328 ÷ 162

⇒ B = 144

∴ The correct answer is 144.

Additional Information

H.C.F. and L.C.M. Property

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. This property holds true only for two numbers.

Co-prime Numbers

If two numbers are co-prime, their H.C.F. is always 1 and their L.C.M. is equal to the product of the two numbers.

H.C.F. of Fractions

H.C.F. = (H.C.F. of Numerators) ÷ (L.C.M. of Denominators).

L.C.M. of Fractions

L.C.M. = (L.C.M. of Numerators) ÷ (H.C.F. of Denominators).

Shortcut Trick

Surface Area of Sphere = 4πr² = 5544 cm².

r² = 5544 ÷ (4 × 22/7) = 5544 × 7 ÷ 88 = 441.

r = √441 = 21 cm.

Diameter = 2r = 2 × 21 = 42 cm.

∴ The correct answer is 42 cm.

Alternate Method

Given: Surface Area of sphere = 5544 cm², π = 22/7.

Formula Used: Surface Area of Sphere = 4πr² and Diameter (d) = 2r.

⇒ 4 × (22/7) × r² = 5544

⇒ r² = (5544 × 7) ÷ (4 × 22)

⇒ r² = (5544 × 7) ÷ 88

⇒ r² = 63 × 7 = 441

⇒ r = √441 = 21 cm

⇒ Diameter = 2 × 21 = 42 cm

∴ The correct answer is 42 cm.

Additional Information

Volume of a Sphere

The total space occupied by a sphere is given by Volume = (4/3)πr³.

Hemisphere Properties

For a hemisphere of radius r, Curved Surface Area = 2πr² and Total Surface Area = 3πr².

Relation between Radius and Diameter

The diameter is the longest chord of a sphere passing through the center, where d = 2r.

Shortcut Trick

When A finishes the race in 36 seconds, B is still running and takes 9 more seconds (45 − 36) to finish.

Distance B covers in those 9 seconds is the distance by which A beats B.

Distance = (Total Distance ÷ B's Time) × Time Difference

Distance = (100 ÷ 45) × 9 = 20 m.

∴ The correct answer is 20 metres.

Alternate Method

Given:

Distance = 100 m

Time taken by A = 36 seconds

Time taken by B = 45 seconds

Formula Used:

Speed = Distance ÷ Time

Distance = Speed × Time

Calculations:

⇒ Speed of B = 100 ÷ 45 m/s

⇒ Speed of B = 20/9 m/s

⇒ Distance covered by B in 36 seconds = (20/9) × 36

⇒ Distance covered by B = 20 × 4 = 80 m

⇒ Distance by which A beats B = Total Distance − Distance covered by B in 36 seconds

⇒ Beating distance = 100 − 80 = 20 m

∴ The correct answer is 20 metres.

Additional Information

Beating by Distance

If A beats B by x metres, it means when A reaches the finish line, B is x metres behind.

Beating by Time

If A beats B by t seconds, it means B takes t more seconds than A to complete the same distance.

Relative Speed in Races

In a linear race starting at the same time, the ratio of distances covered is equal to the ratio of their speeds: D₁/D₂ = S₁/S₂.

Shortcut Trick

Ratio of Expenditure to Savings = 275 : 100 = 11 : 4

Net change in Income = [(11 × −12) + (4 × 40.5)] ÷ (11 + 4)

Net change = [−132 + 162] ÷ 15 = 30 ÷ 15 = 2%

∴ The correct answer is 2%.

Alternate Method

Given: Expenditure is 175% more than Savings.

Formula Used: Income = Expenditure + Savings

⇒ Let Savings = 100 units

⇒ Expenditure = 100 + 175 = 275 units

⇒ Initial Income = 275 + 100 = 375 units

⇒ New Expenditure (12% decrease) = 275 − (12% of 275) = 275 − 33 = 242 units

⇒ New Savings (40.5% increase) = 100 + 40.5 = 140.5 units

⇒ New Income = 242 + 140.5 = 382.5 units

⇒ Increase in Income = 382.5 − 375 = 7.5 units

⇒ Percentage Increase = (7.5 ÷ 375) × 100 = 2%

∴ The correct answer is 2%.

Additional Information

Percentage Change Rule

Income change can be calculated using the weighted average of expenditure and savings changes: %ΔI = (%ΔE × E + %ΔS × S) ÷ (E + S).

Base Value Concept

Always identify the base correctly; "175% more than x" means the new value is 2.75x, whereas "175% of x" means 1.75x.

Shortcut Trick

Effective Compound Interest rate for 2 years at 10% = 10 + 10 + (10 × 10)/100 = 21%.

Total Amount = 100% (Principal) + 21% (Interest) = 121% of Principal.

Amount = 1.21 × 3,300 = 3,993.

∴ The correct answer is ₹ 3,993.

Alternate Method

Given:

Principal (P) = ₹ 3,300

Rate of Interest (R) = 10% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 3300(1 + 10/100)²

⇒ A = 3300(1 + 1/10)²

⇒ A = 3300(11/10)²

⇒ A = 3300 × (121/100)

⇒ A = 33 × 121

⇒ A = 3,993

∴ The correct answer is ₹ 3,993.

Additional Information

Effective Rate for Successive Periods

The effective interest rate for two successive periods with rates r₁ and r₂ is given by [r₁ + r₂ + (r₁ × r₂)/100]%.

Compound Interest Formula

Compound Interest (CI) = Amount − Principal = P[(1 + R/100)ⁿ − 1].

Difference between CI and SI

For 2 years, the difference between CI and SI = P(R/100)².

Shortcut Trick

Compare the amounts: ₹ 13,600 is exactly double of ₹ 6,800.

Since Price and Quantity are directly proportional, the number of stands will also double.

⇒ Number of stands = 34 × 2 = 68

∴ The correct answer is 68 mobile stands.

Alternate Method

Given: Cost Price of 34 mobile stands = ₹ 6,800. New Total Amount = ₹ 13,600.

Formula Used: Unitary Method (Value of 1 item = Total Cost ÷ Total Quantity)

⇒ Cost of 1 mobile stand = 6,800 ÷ 34

⇒ Cost of 1 mobile stand = ₹ 200

⇒ Number of stands for ₹ 13,600 = 13,600 ÷ 200

⇒ Number of stands = 68

∴ The correct answer is 68 mobile stands.

Additional Information

Direct Proportion

When two quantities increase or decrease together at the same rate, they are in direct proportion (x/y = k).

Unitary Method

A technique used to find the value of a single unit first, and then multiply by the desired quantity to find the final value.

Inverse Proportion

When one quantity increases while the other decreases (e.g., speed and time), the product remains constant (x × y = k).

Shortcut Trick

Use the successive discount formula: Single Discount = a + b − (a × b) ÷ 100.

Here, a = 2% and b = 21%.

Equivalent Discount = 2 + 21 − (2 × 21) ÷ 100.

Equivalent Discount = 23 − 42 ÷ 100 = 23 − 0.42 = 22.58%.

∴ The correct answer is 22.58%.

Alternate Method

Given: First discount = 2%, Second discount = 21%.

Formula: Single equivalent discount = [1 − (1 − d₁/100) × (1 − d₂/100)] × 100

⇒ Let the Marked Price (MP) be 100.

⇒ Price after 1^st discount = 100 − 2% of 100 = 98.

⇒ Price after 2^nd discount = 98 − 21% of 98 = 98 − 20.58 = 77.42.

⇒ Total Discount = 100 − 77.42 = 22.58.

⇒ Discount percentage = (22.58 ÷ 100) × 100 = 22.58%.

∴ The correct answer is 22.58%.

Additional Information

Successive Discounts

When two or more discounts are applied one after another on the Marked Price, they are called successive discounts. The total discount is always less than the sum of individual discount rates.

Effective Discount Rate

For three successive discounts a%, b%, and c%, the net discount can be calculated by applying the formula twice or using the multiplier method: 1 − [(100−a)/100 × (100−b)/100 × (100−c)/100].

Shortcut Trick

Use identities: sec² A = 1 + tan² A and cosec² A = 1 + cot² A.

Substitute these into the given equation: (1 + tan² A) + (1 + cot² A) = 13.

⇒ 2 + tan² A + cot² A = 13.

⇒ tan² A + cot² A = 13 − 2 = 11.

∴ The correct answer is 11.

Alternate Method

Given: sec² A + cosec² A = 13

Formula Used: 1 + tan² θ = sec² θ | 1 + cot² θ = cosec² θ

⇒ sec² A + cosec² A = 13

⇒ (1 + tan² A) + (1 + cot² A) = 13

⇒ 1 + 1 + tan² A + cot² A = 13

⇒ 2 + tan² A + cot² A = 13

⇒ tan² A + cot² A = 13 − 2

⇒ tan² A + cot² A = 11

∴ The correct answer is 11.

Additional Information

Trigonometric Identities

Fundamental identities include sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = cosec² θ.

Reciprocal Relations

tan θ = 1 / cot θ and sec θ = 1 / cos θ, which are useful for simplifying algebraic expressions in trigonometry.

Product Form of sec and cosec

Note that sec² A + cosec² A is also equal to sec² A × cosec² A.

Shortcut Trick

Outer radius R = 5 cm, Inner radius r = 3 cm, Length h = 21 cm.

Total Surface Area = 2π(R + r)(h + R − r)

⇒ 2π(5 + 3)(21 + 5 − 3) = 2π × 8 × 23

⇒ 16 × 23 × π = 368π cm².

∴ The correct answer is 368π cm².

Alternate Method

Given:

Outer Diameter = 10 cm ⇒ Outer Radius (R) = 5 cm

Inner Diameter = 6 cm ⇒ Inner Radius (r) = 3 cm

Length of pipe (h) = 21 cm

Formula Used:

Total Surface Area (TSA) of a hollow cylinder = CSA_outer + CSA_inner + 2 × Area of circular rings

TSA = 2πRh + 2πrh + 2π(R² − r²)

Calculations:

⇒ TSA = 2πh(R + r) + 2π(R + r)(R − r)

⇒ TSA = 2π(R + r) [h + R − r]

⇒ TSA = 2π(5 + 3) [21 + 5 − 3]

⇒ TSA = 2π(8) [23]

⇒ TSA = 16π × 23

⇒ TSA = 368π cm²

∴ The correct answer is 368π cm².

Additional Information

Volume of Hollow Cylinder

The space occupied by the material is Volume = π(R² − r²)h.

Curved Surface Area

The total curved area (inner + outer) is 2πh(R + r).

Area of Cross-section

The area of the base ring is π(R² − r²).

Shortcut Trick

Combined efficiency (P + Q) = 1/3 and P efficiency = 1/12.

Q efficiency = 1/3 − 1/12 = 3/12 = 1/4 cistern per hour.

Time for Q to fill 1/4 cistern = (1/4 cistern) ÷ (1/4 cistern/hour) = 1 hour.

∴ The correct answer is 1 hour.

Alternate Method

Given: Time (P + Q) = 3 hours, Time (P) = 12 hours.

Formula Used: Total Work = LCM of time; Efficiency = Total Work ÷ Time.

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

⇒ Efficiency of Q = 4 − 1 = 3 units/hour

⇒ Required work = 1/4 of total work = 1/4 × 12 = 3 units

⇒ Time taken by Q = Required work ÷ Efficiency of Q = 3 ÷ 3 = 1 hour

∴ The correct answer is 1 hour.

Additional Information

Pipes & Cisterns Efficiency

Efficiency is inversely proportional to time. If a pipe fills a tank in 'x' hours, its 1-hour work is 1/x.

Inlet and Outlet Pipes

Inlet pipes have positive efficiency (+), while outlet pipes (leaks) have negative efficiency (−).

Combined Work Formula

If A takes 'a' days and B takes 'b' days, together they take (a × b) ÷ (a + b) days.

Shortcut Trick

Expenditure = 75% → Savings = 100% − 75% = 25%

25% of Income = ₹ 10,000

Income (100%) = 10,000 × 4 = ₹ 40,000

∴ The correct answer is ₹ 40,000.

Alternate Method

Given:

Monthly Expenditure = 75% of Monthly Income

Monthly Savings = ₹ 10,000

Formula Used:

Income = Expenditure + Savings

Calculations:

⇒ Let the total Monthly Income be 100 units.

⇒ Expenditure = 75% of 100 = 75 units.

⇒ Savings = 100 − 75 = 25 units.

⇒ According to the question, 25 units = ₹ 10,000.

⇒ 1 unit = 10,000 ÷ 25 = ₹ 400.

⇒ Total Monthly Income = 100 × 400 = ₹ 40,000.

∴ The correct answer is ₹ 40,000.

Additional Information

Percentage Change

To find the percentage of a value: (Value ÷ Base) × 100. In income problems, Income is always the base (100%).

Ratio Representation

75% can be written as 3/4. This means for every ₹ 4 of Income, ₹ 3 is Expenditure and ₹ 1 is Savings.

Successive Percentage

If income increases by x% and expenditure increases by y%, the new savings are calculated using the Net Change formula or component-wise calculation.

Shortcut Trick

Let x = p − q, y = q − r, and z = r − p.

Observe that their sum is x + y + z = (p − q) + (q − r) + (r − p) = 0.

According to the algebraic property, if x + y + z = 0, then x³ + y³ + z³ = 3xyz.

The numerator becomes 3(p − q)(q − r)(r − p), which when divided by the denominator (p − q)(q − r)(r − p), equals 3.

∴ The correct answer is 3.

Alternate Method

Given: The expression to simplify is ((p − q)³ + (q − r)³ + (r − p)³) ÷ ((p − q)(q − r)(r − p)).

Formula Used: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

⇒ Let a = p − q, b = q − r, and c = r − p.

⇒ First, calculate a + b + c = (p − q) + (q − r) + (r − p) = 0.

⇒ Substituting a + b + c = 0 into the algebraic identity:

⇒ a³ + b³ + c³ − 3abc = 0 × (a² + b² + c² − ab − bc − ca) = 0.

⇒ Therefore, a³ + b³ + c³ = 3abc.

⇒ The given expression simplifies to [3(p − q)(q − r)(r − p)] ÷ [(p − q)(q − r)(r − p)].

⇒ By cancelling common terms, the result is 3.

∴ The correct answer is 3.

Additional Information

Algebraic Identity for Three Variables

The formula a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) is a fundamental polynomial identity used in factorization.

Conditional Cube Identity

A special case arises when a + b + c = 0; in this scenario, the sum of cubes a³ + b³ + c³ is always equal to 3abc.

Square of a Trinomial

Relatedly, the square of a trinomial is given by (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).