Quantitative Aptitude PYQs

Shortcut Trick

Teacher's age = New Average + (Initial count × Change in average)

Increase in average = 18 − 17 = 1 year

Teacher's age = 18 + (24 × 1) = 18 + 24

Teacher's age = 42 years

∴ The correct answer is 42 years.

Alternate Method

Given:

Initial count = 24 students

Initial average age = 17 years

New count (including teacher) = 25 people

New average age = 18 years

Formula Used:

Age of Teacher = (Total age of 25 people) − (Total age of 24 students)

⇒ Total age of 24 students = 24 × 17 = 408 years

⇒ Total age of 25 people = 25 × 18 = 450 years

⇒ Age of the teacher = 450 − 408

⇒ Age of the teacher = 42 years

∴ The correct answer is 42 years.

Additional Information

Net Deviation Method

The sum of deviations from the average is always zero. If the average increases by 1 for 25 people, the total deviation is +25, which must be balanced by the new person relative to the old average.

Exclusion Case

When a person leaves: Age of left person = Old Average + (Final count × Decrease in Average).

Weighted Average Formula

Combined Average = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is count and x is the individual average.

Shortcut Trick

For two lines to be parallel, the ratio of the coefficients of x and y must be equal.

(k − 4) / 2 = 2 / 1

⇒ k − 4 = 4

⇒ k = 8

∴ The correct answer is 8.

Alternate Method

Given:

Equation 1: (k − 4)x + 2y = 6

Equation 2: 2x + y = 4

Formula Used:

For two linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be parallel, the condition is:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Calculations:

⇒ Comparing the given equations with the standard form:

⇒ a₁ = k − 4, b₁ = 2, c₁ = −6

⇒ a₂ = 2, b₂ = 1, c₂ = −4

⇒ For the lines to be parallel, we set the ratios of the coefficients equal:

⇒ (k − 4) / 2 = 2 / 1

⇒ k − 4 = 2 × 2

⇒ k − 4 = 4

⇒ k = 4 + 4

⇒ k = 8

⇒ Verification of the constant ratio: c₁/c₂ = −6/−4 = 1.5. Since 2 ≠ 1.5, the lines are strictly parallel and not coincident.

∴ The correct answer is 8.

Additional Information

Coincident Lines

Lines are coincident if a₁/a₂ = b₁/b₂ = c₁/c₂, resulting in infinitely many solutions as the lines lie on top of each other.

Intersecting Lines

Lines intersect at exactly one point if a₁/a₂ ≠ b₁/b₂, which provides a unique solution for the system of equations.

Perpendicular Lines Condition

Two lines are perpendicular if the product of their slopes is −1 (m₁ × m₂ = −1) or if a₁a₂ + b₁b₂ = 0.

Shortcut Trick

Identify the base value (A's marks) and the comparing value (B's marks).

Required Percentage = (B's marks ÷ A's marks) × 100

⇒ (50 ÷ 75) × 100 = (2 ÷ 3) × 100 = 66.666...

∴ The correct answer is 66.67%.

Alternate Method

Given:

Marks obtained by B = 50

Marks obtained by A = 75

Maximum marks = 80

Formula Used:

Percentage = (Value ÷ Base Value) × 100

Calculations:

⇒ Required Percentage = (50 ÷ 75) × 100

⇒ Dividing both numbers by 25: (2 ÷ 3) × 100

⇒ 200 ÷ 3 = 66.6666...

⇒ Rounding off to two decimal places = 66.67

∴ The correct answer is 66.67%.

Additional Information

Percentage Change

To find how much more or less A is than B: Percentage Difference = [(A − B) ÷ B] × 100.

Base Value Identification

In phrases like 'X is what % of Y', the value following 'of' is always the denominator (base).

Rounding Rules

If the third decimal digit is 5 or greater, increment the second decimal digit by 1.

Shortcut Trick
Let the capacity of the tank be the LCM of 7 and 56 = 56 units.
Efficiency of filling pipe = 56 ÷ 7 = +8 units/minute.
Efficiency of emptying pipe = 56 ÷ 56 = −1 unit/minute.
Combined efficiency of both pipes = 8 − 1 = 7 units/minute.
Time taken to fill one-half of the tank (28 units) = 28 ÷ 7 = 4 minutes.
∴ The correct answer is 4 minutes.Alternate Method
Given:
Time taken by filling pipe A = 7 minutes
Time taken by emptying pipe B = 56 minutes

Person/Pipe
Days/Time (mins)
Total Work (LCM)
Efficiency

Pipe A (Fill)
7
56 units
+8

Pipe B (Empty)
56
−1

Calculations:
⇒ Total capacity of the tank = 56 units
⇒ One-half capacity of the tank = 56 ÷ 2 = 28 units
⇒ Net efficiency of both pipes working together = 8 − 1 = 7 units/minute
⇒ Time required to fill 28 units = 28 ÷ 7 = 4 minutes
∴ The correct answer is 4 minutes.Additional Information
Pipes and Cisterns Concept
In pipes and cisterns, an inlet pipe does positive work (+ efficiency) while an outlet or leak does negative work (− efficiency).
Combined Work Formula
If one pipe fills a tank in X minutes and another empties it in Y minutes (where Y > X), the net part filled in 1 minute is (1/X − 1/Y).

Shortcut Trick

The relation between Cost Price (CP) and Marked Price (MP) or Listed Price is given by the ratio:

CP : MP = (100 − Discount%) : (100 + Profit%)

CP : MP = (100 − 77) : (100 + 49)

CP : MP = 23 : 149

∴ The correct answer is 23 : 149.

Alternate Method

Given → Discount = 77%, Profit = 49%.

Formulas: SP = MP × (100 − D%) ÷ 100 | SP = CP × (100 + P%) ÷ 100

⇒ Let Listed Price (MP) be 100.

⇒ Selling Price (SP) = 100 − 77% of 100 = 23.

⇒ Since Profit is 49%, SP = CP × 1.49.

⇒ 23 = CP × (149 ÷ 100) ⇒ CP = 2300 ÷ 149.

⇒ Ratio CP : MP = (2300 ÷ 149) : 100.

⇒ Ratio CP : MP = 2300 : 14900 = 23 : 149.

∴ The correct answer is 23 : 149.

Additional Information

Marked Price & Discount

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

CP-MP Relation

This direct ratio (100−D):(100+P) is a derivation of equating SP from both CP and MP sides. It is the most efficient way to solve markup or listed price problems.

Shortcut Trick

If x + 1/x = k, then x² + 1/x² = k² − 2.

Given: 2x + 2/x = 3 ⇒ 2(x + 1/x) = 3 ⇒ x + 1/x = 3/2

Here, k = 3/2.

Result = (3/2)² − 2 = 9/4 − 2 = 1/4.

∴ The correct answer is 1/4.

Alternate Method

Given: 2x + (2/x) = 3

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

⇒ 2(x + 1/x) = 3

⇒ x + 1/x = 3/2

Squaring both sides:

⇒ (x + 1/x)² = (3/2)²

⇒ x² + (1/x)² + 2 × x × (1/x) = 9/4

⇒ x² + 1/x² + 2 = 9/4

⇒ x² + 1/x² = 9/4 − 2

⇒ x² + 1/x² = (9 − 8)/4 = 1/4

∴ The correct answer is 1/4.

Additional Information

Algebraic Identities for Reciprocals

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k. This is derived from the expansion of (a + b)³.

Negative Reciprocal Square

If x − 1/x = k, then x² + 1/x² = k² + 2. Note that the sign changes because (x − 1/x)² = x² + 1/x² − 2.

Higher Powers

To find x⁴ + 1/x⁴, square the result of x² + 1/x² and subtract 2 again. Let x² + 1/x² = m, then x⁴ + 1/x⁴ = m² − 2.

Shortcut Trick

To find the Highest Common Factor (HCF), identify all common prime factors.

For each common factor, take the minimum exponent (lowest power) present in both factorizations.

Multiply these together to get the HCF.

∴ The correct answer is Prime Factorization Model: Find the prime factorization of both numbers, then multiply the common prime factors with the lowest powers.

Alternate Method

Given: Two numbers for which the HCF needs to be calculated.

Definition: Highest Common Factor (HCF) or Greatest Common Divisor (GCD) is the largest positive integer that divides each of the integers.

⇒ Let numbers be N₁ and N₂.

⇒ Perform Prime Factorization: N₁ = p₁^a × p₂^b ... and N₂ = p₁^x × p₂^y ...

⇒ HCF = p₁^min(a,x) × p₂^min(b,y) ...

⇒ This model ensures the resulting number is the largest divisor shared by both.

∴ The correct answer is Prime Factorization Model: Find the prime factorization of both numbers, then multiply the common prime factors with the lowest powers.

Additional Information

Euclidean Algorithm

A systematic method for computing the HCF of two numbers by repeatedly replacing the larger number with the remainder of its division by the smaller number until the remainder is zero.

Relationship between HCF and LCM

For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b.

Least Common Multiple (LCM)

Found by taking the product of the highest powers of all prime factors present in the numbers.

Shortcut Trick

The sum of the exterior angles of any convex polygon is always 360°.

For a regular pentagon (5 sides):

Each exterior angle = 360° ÷ 5 = 72°.

Since the interior and exterior angles form a linear pair:

Each interior angle = 180° − 72° = 108°.

∴ The correct answer is 108°.

Alternate Method

Given:

Number of sides (n) of a regular pentagon = 5.

Formula Used:

Measure of each interior angle of a regular polygon = [(n − 2) × 180°] ÷ n

Calculations:

⇒ Sum of interior angles = (5 − 2) × 180°

⇒ Sum of interior angles = 3 × 180° = 540°

⇒ Measure of each interior angle = 540° ÷ 5

⇒ Measure of each interior angle = 108°

∴ The correct answer is 108°.

Additional Information

Regular Polygon Properties

Shortcut Trick

Total deduction = 40 + 83 + 68 = ₹191

Remaining amount = 13403 − 191 = ₹13212

Ratio of remaining parts = 21 : 11 : 4 (Total units = 36)

Difference between B and C = (11 − 4) + (83 − 68) = 7 units + ₹15

1 unit = 13212 ÷ 36 = ₹367

Difference = (7 × 367) + 15 = 2569 + 15 = ₹2584

∴ The correct answer is ₹2584.

Alternate Method

Given:

Total amount = ₹13,403

Deductions: S = ₹40, B = ₹83, C = ₹68

Remaining Ratio S : B : C = 21 : 11 : 4

Calculations:

⇒ Sum of remaining parts = 13403 − (40 + 83 + 68)

⇒ Sum = 13403 − 191 = ₹13212

⇒ Total ratio units = 21 + 11 + 4 = 36 units

⇒ 36 units = 13212

⇒ 1 unit = 13212 ÷ 36 = 367

⇒ Share of B (Original) = (11 × 367) + 83 = 4037 + 83 = ₹4120

⇒ Share of C (Original) = (4 × 367) + 68 = 1468 + 68 = ₹1536

⇒ Difference = 4120 − 1536 = ₹2584

∴ The correct answer is ₹2584.

Additional Information

Ratio and Proportion

When constant values are deducted from shares, subtract the total deduction from the total sum before applying the ratio division.

Calculation of Individual Shares

If Amount A is divided in ratio x:y, then part 1 = [x / (x+y)] × A and part 2 = [y / (x+y)] × A.

Difference Rule

The difference between original shares is the sum of the difference in ratio parts and the difference in their respective constant deductions.

Shortcut Trick

Speed ratio = 3 : 4 → Time ratio = 4 : 3 (since distance is constant).

Difference in time ratio parts = 4 − 3 = 1 unit.

Given: 1 unit = 11 minutes.

Usual time (smaller ratio part) = 3 units = 3 × 11 = 33 minutes.

∴ The correct answer is 33 min.

Alternate Method

Given: New Speed = 3/4 of Usual Speed | Delay = 11 minutes.

Formula Used: Speed × Time = Distance | Time ∝ 1/Speed (when Distance is constant).

⇒ Let usual speed be S and usual time be T.

⇒ New speed = (3/4)S

⇒ New time = T + 11

⇒ Since Distance = Speed × Time:

⇒ S × T = (3/4)S × (T + 11)

⇒ T = (3/4)(T + 11)

⇒ 4T = 3T + 33

⇒ T = 33 minutes.

∴ The correct answer is 33 min.

Additional Information

Inverse Proportion of Speed and Time

When the distance is constant, the ratio of speeds is inversely proportional to the ratio of time taken. If speed ratio is a:b, then time ratio is b:a.

Late/Early Concept Formula

Usual Time = (Late time × n) / (1 − n) where speed becomes 'n' times the original speed. Here, Usual Time = (11 × 3/4) / (1 − 3/4) = (33/4) / (1/4) = 33 min.

Average Speed

If a person covers the same distance at two different speeds x and y, the average speed is 2xy / (x + y).

Shortcut Trick

Find the LCM of 15, 25, and 8.

LCM (15, 25, 8) = 600.

The number to be added = LCM − Given Number (if the number is just below the LCM).

Number to be added = 600 − 517 = 83.

∴ The correct answer is 83.

Alternate Method

Given: Number = 517, Divisors = 15, 25, 8.

Step 1: Calculate the Least Common Multiple (LCM) of divisors.

⇒ 15 = 3 × 5

⇒ 25 = 5²

⇒ 8 = 2³

⇒ LCM = 2³ × 3 × 5² = 8 × 3 × 25 = 600.

Step 2: Find the difference between the LCM and the given number.

⇒ Required number = 600 − 517

⇒ Required number = 83.

∴ The correct answer is 83.

Additional Information

LCM (Least Common Multiple)

The smallest positive integer that is divisible by each of the given numbers. It is used to find a common scale for different values.

Divisibility Rule for 15

A number is divisible by 15 if it is divisible by both 3 (sum of digits is multiple of 3) and 5 (ends in 0 or 5).

Divisibility Rule for 8

A number is divisible by 8 if its last three digits are divisible by 8.

Shortcut Trick

Ratio of Grandfather (GF) to Grandson (GS): Now = 5 : 1, after 10 years = 3 :

1. Equating the ratio difference: (5−1) = 4; (3−1) = 2. Multiply second ratio by 2 → 6 : 2.

Change in ratio units (6−5 or 2−1) = 1 unit. Given 1 unit = 10 years.

GF current age = 5 × 10 = 50 years. Cousin age = 50 + 5 = 55 years.

Cousin 3 years ago = 55 − 3 = 52 years.

∴ The correct answer is 52 years.

Alternate Method

Given: Ratio of Grandfather to Grandson is 5 :

1. After 10 years, ratio becomes 3 :
2. Cousin is 5 years older than Grandfather.

Formula Used: Age Equation method.

⇒ Let Grandson's present age be x and Grandfather's age be 5x.

⇒ After 10 years: (5x + 10) = 3 × (x + 10)

⇒ 5x + 10 = 3x + 30

⇒ 2x = 20 ⇒ x = 10

⇒ Grandfather's present age = 5 × 10 = 50 years.

⇒ Cousin's present age = 50 + 5 = 55 years.

⇒ Cousin's age 3 years ago = 55 − 3 = 52 years.

∴ The correct answer is 52 years.

Additional Information

Ratio Difference Method

When the time gap is the same for two people, the difference in their ratio parts must be made equal to maintain a constant age gap.

Linear Equation in Ages

Always define the present age as the base variable (x) to simplify past or future age expressions.

Constant Age Difference

The difference between the ages of two individuals remains constant regardless of the number of years passed or preceding.

Shortcut Trick

Incomes of Mohan and David are in ratio 5 : 9.

Expenditure ratio is 2 : 4, which simplifies to 1 : 2.

Cross-multiplication method: (5 × 2 − 9 × 1) units = ₹48,000 × (2 − 1).

1 unit = ₹48,000.

Mohan's income = 5 units = 5 × ₹48,000 = ₹2,40,000.

∴ The correct answer is ₹2,40,000.

Alternate Method

Given: Ratio of Income = 5 : 9, Savings = ₹48,000 each, Ratio of Expenditure = 2 : 4.

Formula Used: Income − Savings = Expenditure

⇒ Let Mohan's income be 5x and David's income be 9x.

⇒ (5x − 48000) / (9x − 48000) = 2 / 4

⇒ (5x − 48000) / (9x − 48000) = 1 / 2

⇒ 2 × (5x − 48000) = 1 × (9x − 48000)

⇒ 10x − 96000 = 9x − 48000

⇒ 10x − 9x = 96000 − 48000

⇒ x = 48000

⇒ Mohan's income = 5 × 48000 = ₹2,40,000.

∴ The correct answer is ₹2,40,000.

Additional Information

Income-Expenditure Relationship

Basic identity: Income = Expenditure + Savings. For two people, use (I₁ − S₁)/(I₂ − S₂) = E₁/E₂.

Ratio Balancing

If savings are equal, ensure the difference between income ratio parts and expenditure ratio parts is equal by multiplying ratios with appropriate factors.

Proportion Rule

If a/b = c/d, then a × d = b × c. This cross-multiplication is the most efficient way to solve linear ratio equations.

Shortcut Trick

Break the numbers into products of perfect squares: √50 = 5√2 and √8 = 2√2.

Combine the like terms by adding the coefficients: 5√2 + 2√2 = 7√2.

To match the options, convert back to a pure surd: 7√2 = √(7² × 2) = √(49 × 2) = √98.

∴ The correct answer is √98.

Alternate Method

Given: √50 + √8

Formula: √(a² × b) = a√b

⇒ √(25 × 2) + √(4 × 2)

⇒ 5√2 + 2√2

⇒ (5 + 2)√2

⇒ 7√2

⇒ √(7² × 2)

⇒ √(49 × 2)

⇒ √98

∴ The correct answer is √98.

Additional Information

Addition of Surds

Only like surds (surds with the same radicand) can be added or subtracted: x√a ± y√a = (x ± y)√a.

Mixed Surd to Pure Surd

A mixed surd k√x can be expressed as a pure surd by squaring the coefficient and moving it inside: √(k² × x).

Multiplication Rule

For any non-negative numbers a and b, the square root of their product is the product of their square roots: √(a × b) = √a × √b.

Shortcut Trick

Discount of 20% is equal to the extra amount of ₹6,925.

Ratio of Discount to Selling Price = 20 : (100 − 20) = 20 : 80 = 1 : 4.

Since 1 unit = ₹6,925, then 4 units = 4 × 6,925.

Ajeet paid = ₹27,700.

∴ The correct answer is ₹27,700.

Alternate Method

Given → Discount Percentage = 20% | Extra Amount (Discount Value) = ₹6,925.
Formulas: Discount = MP − SP | Discount% = (Discount ÷ MP) × 100 | SP = MP × (100 − D%) ÷ 100.

⇒ Let the Marked Price (MP) be x.
⇒ 20% of x = 6,925.
⇒ (20 ÷ 100) × x = 6,925 ⇒ x = 6,925 × 5 = ₹34,625.
⇒ Selling Price (SP) = MP − Discount = 34,625 − 6,925.
⇒ SP = ₹27,700.

∴ The correct answer is ₹27,700.

Additional Information

Marked Price & Discount

The Marked Price is the price listed on the product. Discount is always calculated on the Marked Price to arrive at the Selling Price.

Percentage Increase/Decrease

If a value is decreased by r%, the new value is (100 − r)% of the original value.

Shortcut Trick

Total hours for A = 12 × 6 = 72 hours

Total hours for B = 18 × 8 = 144 hours

Combined efficiency (LCM = 144): A(2) + B(1) = 3 units/hour

Total time needed = 144 ÷ 3 = 48 hours

Days required working 64/9 hours per day = 48 ÷ (64/9) = 6.75 days

∴ The correct answer is 6(3/4) days.

Alternate Method

Given: A takes 12 days at 6 hrs/day. B takes 18 days at 8 hrs/day. Combined work rate is 7(1/9) hrs/day.

Formula Used: Total Work = Days × Hours/day × Efficiency

⇒ Total efficiency of A + B = 2 + 1 = 3 units/hr

⇒ Total hours required = 144 ÷ 3 = 48 hours

⇒ Hours per day = 7(1/9) = 64/9 hours

⇒ Number of days = 48 ÷ (64/9) = 48 × (9/64)

⇒ Number of days = 3 × (9/4) = 27/4 = 6(3/4) days

∴ The correct answer is 6(3/4) days.

Additional Information

Man-Hour-Day Formula

The relationship between work, men, days, and hours is given by (M₁×D₁×H₁)/W₁ = (M₂×D₂×H₂)/W₂.

Efficiency and Time

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

Shortcut Trick

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

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

Required Amount = 121% of ₹3,200 = 1.21 × 3200 = ₹3,872.

∴ The correct answer is ₹3,872.

Alternate Method

Given:

Principal (P) = ₹3,200

Rate (R) = 10% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 3200 × (1 + 10/100)²

⇒ A = 3200 × (110/100)²

⇒ A = 3200 × (11/10) × (11/10)

⇒ A = 3200 × (121/100)

⇒ A = 32 × 121

⇒ A = ₹3,872

∴ The correct answer is ₹3,872.

Additional Information

Compound Interest Formula

Interest is calculated on the principal and the accumulated interest of previous periods: CI = P[(1 + R/100)ⁿ − 1].

Difference between CI and SI for 2 Years

The difference between compound interest and simple interest for two years is given by: Difference = P(R/100)².

Effective Rate of Interest

For two successive cycles of R% interest, the effective rate is [2R + (R²/100)]%.

Shortcut Trick

Quantity of milk remains constant.

Initial Milk % = 100 − 12 = 88% | Final Milk % = 100 − 20 = 80%.

New Total Mixture = (Initial Quantity × Initial Milk %) ÷ Final Milk %

New Total = (50 × 88) ÷ 80 = 55 litres.

Water added = 55 − 50 = 5 litres.

∴ The correct answer is 5 litres.

Alternate Method

Given:

Initial Volume = 50 litres | Water % = 12%

⇒ Initial Water = 12% of 50 = (12/100) × 50 = 6 litres.

⇒ Initial Milk = 50 − 6 = 44 litres.

⇒ Let x litres of water be added.

⇒ New Total Volume = 50 + x

⇒ New Water Content = 6 + x

⇒ According to question: (6 + x) / (50 + x) = 20 / 100

⇒ (6 + x) / (50 + x) = 1 / 5

⇒ 5 × (6 + x) = 1 × (50 + x)

⇒ 30 + 5x = 50 + x

⇒ 4x = 20

⇒ x = 5 litres.

∴ The correct answer is 5 litres.

Additional Information

Rule of Constant Quantity

When only one component of a mixture is added or removed, the quantity of the other component remains constant throughout the process.

Alligation Method

This can be solved using Alligation by taking Water %: (12% initial) mixed with (100% pure water) to get (20% final concentration).

Mixture Ratio Change

If initial ratio is a:b and x quantity of 'b' is added, make the 'a' parts equal in both ratios to find the change in 'b' units.

Shortcut Trick

Directly apply the Surface Area formula: Area = 4 × π × (Radius)².

Square the Radius: 30² = 900.

Multiply by 4: 4 × 900 = 3600.

Keep the answer in terms of π as per the options.

∴ The correct answer is 3600π cm².

Alternate Method

Given:

Radius of the Sphere (r) = 30 cm

Formula Used:

Surface Area of a Sphere = 4πr²

Calculations:

⇒ Surface Area = 4 × π × (30)²

⇒ Surface Area = 4 × π × (30 × 30)

⇒ Surface Area = 4 × π × 900

⇒ Surface Area = 3600π cm²

∴ The correct answer is 3600π cm².

Additional Information

Volume of a Sphere

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

Surface Area of Hemisphere

For a hemisphere, the Curved Surface Area (CSA) is 2πr² and the Total Surface Area (TSA) is 3πr².

Ratio of Surface Area to Volume

For any sphere, the ratio of its Surface Area to its Volume is 3/r.

Shortcut Trick

Total Capacity (LCM of 14, 28) = 28 units.

Efficiency of (P + Q) = 28 ÷ 14 = 2 units/hour.

Efficiency of P = 28 ÷ 28 = 1 unit/hour.

Efficiency of Q = 2 − 1 = 1 unit/hour.

Time for Q to fill 1/4^th cistern = (28 ÷ 4) ÷ 1 = 7 hours.

∴ The correct answer is 7 hours.

Alternate Method

Given:

Time taken by P and Q = 14 hours

Time taken by P alone = 28 hours

Formula Used:

Total Work = Efficiency × Time

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

⇒ Efficiency of Q = 2 − 1 = 1 unit/hour

⇒ Total work to be done by Q = 1/4 × Total Capacity

⇒ Work for Q = 28 ÷ 4 = 7 units

⇒ Time taken by Q = 7 ÷ 1 = 7 hours

∴ The correct answer is 7 hours.

Additional Information

Pipes and Cisterns Logic

Filling pipes have positive efficiency, while drainage or leak pipes have negative efficiency.

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a specific task (E ∝ 1/T).

Combined Work Formula

If two pipes fill a tank in x and y hours respectively, they together fill it in (x × y) ÷ (x + y) hours.