Quantitative Aptitude PYQs

Shortcut Trick

Walking time = 40 ÷ 5 = 8 hours.

Running time = 30 ÷ 10 = 3 hours.

Ratio of walking time to running time = 8 : 3.

∴ The correct answer is 8 : 3.

Alternate Method

Given:

Walking distance = 40 km, Walking speed = 5 km/hr.

Running distance = 30 km, Running speed = 10 km/hr.

Formula Used:

Time = Distance ÷ Speed

⇒ Walking time (T_w) = 40 ÷ 5 = 8 hours

⇒ Running time (T_r) = 30 ÷ 10 = 3 hours

⇒ Ratio = T_w : T_r = 8 : 3

∴ The correct answer is 8 : 3.

Additional Information

Average Speed

The average speed for the entire journey is calculated as Total Distance ÷ Total Time. Here, (40 + 30) ÷ (8 + 3) = 70 ÷ 11 km/hr.

Speed Conversion

To convert speed from km/hr to m/s, multiply the value by 5/18. For example, 10 km/hr = 10 × (5/18) m/s.

Inverse Proportion

When distance is constant, the time taken is inversely proportional to the speed (T ∝ 1/S).

Shortcut Trick

Let the difference between Savings and Expenditure be 1 unit.

Then, Expenditure = 2 × 1 = 2 units.

Savings = Expenditure + Difference = 2 + 1 = 3 units.

Total Salary = Expenditure + Savings = 2 + 3 = 5 units.

Expenditure % = (2 ÷ 5) × 100 = 40%.

∴ The correct answer is 40%.

Alternate Method

Given:

Expenditure (E) = 2 × (Savings (V) − Expenditure (E))

Formula Used:

Salary = Expenditure + Savings

Expenditure Percentage = (Expenditure ÷ Salary) × 100

⇒ E = 2V − 2E

⇒ 3E = 2V

⇒ V = (3 ÷ 2)E = 1.5E

⇒ Salary = E + V = E + 1.5E = 2.5E

⇒ Expenditure Percentage = (E ÷ 2.5E) × 100

⇒ Expenditure Percentage = (100 ÷ 2.5) = 40%

∴ The correct answer is 40%.

Additional Information

Income Expenditure Relation

The basic formula for income problems is: Salary = Expenditure + Savings.

Percentage Calculation

Percentage of any component is calculated as: (Component ÷ Total Salary) × 100.

Ratio and Proportion

When components are given in linear equations, converting them into Ratios (e.g., E:V = 2:3) simplifies calculations.

Given:

0.375 + 3/8 − 0.125

Formula used:

Decimal conversion: 3/8 = 0.375

Calculations:

0.375 + 3/8 − 0.125

⇒ 0.375 + 0.375 − 0.125

⇒ 0.750 − 0.125

⇒ 0.625

∴ The correct answer is option (2).

Shortcut Trick

For the number 12A916B to be divisible by 24, it must be divisible by both 8 and 3.

Divisibility by 8: Last three digits 16B must be divisible by 8. Possible values for B are 0 or 8. To maximize (A + B), let B = 8.

Divisibility by 3: Sum of digits (1 + 2 + A + 9 + 1 + 6 + 8 = 27 + A) must be divisible by 3.

Since 27 is already divisible by 3, A can be 0, 3, 6, or 9. To maximize (A + B), let A = 9.

Maximum value of (A + B) = 9 + 8 = 17.

∴ The correct answer is 17.

Alternate Method

Given: The 7-digit number is 12A916B and it is divisible by 24.

For divisibility by 24, the number must satisfy divisibility rules for 3 and 8 simultaneously.

Step 1: Apply divisibility rule of 8.

⇒ The last three digits 16B must be divisible by 8.

⇒ 160 + B must be divisible by 8.

⇒ Since 160 is divisible by 8, B must be 0 or 8.

⇒ For maximum (A + B), we choose B = 8.

Step 2: Apply divisibility rule of 3.

⇒ Sum of digits = 1 + 2 + A + 9 + 1 + 6 + B

⇒ Sum = 19 + A + B

⇒ Substitute B = 8: Sum = 19 + A + 8 = 27 + A

⇒ For (27 + A) to be divisible by 3, A can be 0, 3, 6, or 9.

⇒ For maximum value, we choose A = 9.

Step 3: Calculate the maximum value.

⇒ Maximum (A + B) = 9 + 8 = 17.

∴ The correct answer is 17.

Additional Information

Divisibility Rule of 8

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

Divisibility Rule of 3

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

Divisibility of Composite Numbers

To check divisibility by a composite number like 24, we check divisibility by its co-prime factors (3 and 8 in this case, because 3 × 8 = 24 and HCF(3, 8) = 1).

Shortcut Trick

Convert the red marble ratios into percentages to compare the proportion directly.

Bag A: 2 ÷ (2 + 3) = 2/5 = 40%

Bag B: 4 ÷ (4 + 1) = 4/5 = 80%

Bag C: 3 ÷ (3 + 2) = 3/5 = 60%

Comparing the values: 80% > 60% > 40%.

∴ The correct answer is Bag B.

Alternate Method

Given:

Ratio of red marbles to blue marbles in:

Bag A = 2 : 3

Bag B = 4 : 1

Bag C = 3 : 2

Formula Used:

Proportion of red marbles = Red Marbles ÷ (Red Marbles + Blue Marbles)

Calculations:

⇒ Red proportion in Bag A = 2 ÷ (2 + 3) = 2/5 = 0.40

⇒ Red proportion in Bag B = 4 ÷ (4 + 1) = 4/5 = 0.80

⇒ Red proportion in Bag C = 3 ÷ (3 + 2) = 3/5 = 0.60

⇒ Comparing the fractions: 0.80 > 0.60 > 0.40

⇒ Bag B has the highest proportion of red marbles at 0.80 (or 80%).

∴ The correct answer is Bag B.

Additional Information

Ratio to Proportion Conversion

To find the proportion of a specific component in a ratio x : y, use the formula x ÷ (x + y).

Comparing Fractions

Fractions can be compared by converting them to a common denominator or by converting them into decimal or percentage forms.

Percentage Calculation

The percentage of a component is (Value ÷ Total) × 100. This is useful for standardized comparison across different total quantities.

Shortcut Trick

Net interest percentage = Rate × Time = 5.5% × 5.5 = 30.25%

Simple Interest = 30.25% of ₹55,000

SI = (30.25 ÷ 100) × 55,000 = 0.3025 × 55,000 = 16,637.5

∴ The correct answer is ₹16,637.5.

Alternate Method

Given:

Principal (P) = ₹55,000

Rate (R) = 5.5% per annum

Time (T) = 5.5 years

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ SI = (55,000 × 5.5 × 5.5) ÷ 100

⇒ SI = 550 × 5.5 × 5.5

⇒ SI = 550 × 30.25

⇒ SI = 16,637.5

∴ The correct answer is ₹16,637.5.

Additional Information

Amount Formula

The total Amount (A) to be paid back after the time period is the sum of the Principal and the Simple Interest: A = P + SI.

Simple Interest vs Compound Interest

In Simple Interest, the interest is calculated only on the initial principal, whereas in Compound Interest, it is calculated on the principal plus any accumulated interest.

Rate of Interest p.a.

The Rate of Interest is usually expressed as a percentage per annum (p.a.). If the time is in months, it must be divided by 12 to convert it to years.

Shortcut Trick

Net filling rate of cistern = 1/4 + 1/20 − 1/20 = 1/4 cistern/hr.

Target work is to fill one-fourth of the empty cistern.

Time taken = Work ÷ Net Rate = (1/4) ÷ (1/4) = 1 hour.

∴ The correct answer is 1 hr.

Alternate Method

Given:

Time taken by Tap 1 (Filling) = 4 hr

Time taken by Tap 2 (Filling) = 20 hr

Time taken by Tap 3 (Emptying) = 20 hr

Calculations:

⇒ Total Capacity of cistern = LCM(4, 20, 20) = 20 units

⇒ Target to fill = 1/4 × 20 = 5 units

⇒ Combined Efficiency of all taps = 5 + 1 − 1 = 5 units/hr

⇒ Time required = Target Work ÷ Combined Efficiency

⇒ Time required = 5 ÷ 5 = 1 hour

∴ The correct answer is 1 hr.

Additional Information

Negative Efficiency

In Pipes and Cisterns, the efficiency of an outlet/emptying tap is always taken as negative while the efficiency of an inlet/filling tap is positive.

Work-Time Relation

Total Work = Efficiency × Time. For fractional work, Time = (Total Work × Fraction) ÷ Combined Efficiency.

Combined Rate Formula

If two taps fill in X and Y hours and one empties in Z hours, the net rate is (1/X + 1/Y − 1/Z) parts per hour.

Shortcut Trick

Find the LCM of 6, 10, and 14, which is 210.

Divide the largest 4-digit number (9999) by 210; the remainder is 129.

Largest multiple = 9999 − 129 = 9870.

Add the required remainder: 9870 + 3 = 9873.

∴ The correct answer is 9873.

Alternate Method

Given: Divisors = 6, 10, 14 and Remainder = 3.

Formula Used: Required Number = (LCM(Divisors) × k) + Remainder

⇒ First, find the prime factors of the divisors:

⇒ 6 = 2 × 3

⇒ 10 = 2 × 5

⇒ 14 = 2 × 7

⇒ LCM = 2 × 3 × 5 × 7 = 210

⇒ The largest 4-digit number is 9999.

⇒ 9999 ÷ 210 ⇒ Quotient = 47, Remainder = 129.

⇒ Largest multiple of 210 = 9999 − 129 = 9870.

⇒ Required number = 9870 + 3 = 9873.

∴ The correct answer is 9873.

Additional Information

General Form of Numbers

Any number that leaves a common remainder r when divided by x, y, z is written in the form [LCM(x, y, z) × k] + r.

LCM Calculation

LCM is calculated by taking the highest power of all prime factors involved in the prime factorization of the given numbers.

Divisibility Verification

To verify, subtract the remainder from the result; the value 9870 is exactly divisible by 6, 10, and 14.

Shortcut Trick

For a Building of height H and a Tower of height h, if the angles of depression to the top and bottom are θ₁ and θ₂:

Height of Tower h = H × [1 − (tan θ₁ / tan θ₂)]

Given: H = 20 m, θ₁ = 30°, θ₂ = 60°

h = 20 × [1 − (tan 30° / tan 60°)]

h = 20 × [1 − (1/√3 / √3)] = 20 × [1 − 1/3] = 20 × 2/3 = 13.33 m

∴ The correct answer is 13.33 m.

Alternate Method

Given:

Height of the Building (AB) = 20 m

Angle of depression of the Tower top (C) = 30°

Angle of depression of the Tower bottom (D) = 60°

Let the Height of the Tower be h and the distance between the building and the tower be x.

⇒ In right ▵ABD, tan 60° = AB / BD

⇒ √3 = 20 / x

⇒ x = 20 / √3

⇒ Draw a horizontal line from C to the building meeting it at E. Then, AE = 20 − h and CE = x.

⇒ In right ▵ACE, tan 30° = AE / CE

⇒ 1 / √3 = (20 − h) / (20 / √3)

⇒ 1 / √3 = (20 − h) × √3 / 20

⇒ 20 = 3 × (20 − h)

⇒ 20 = 60 − 3h

⇒ 3h = 40

⇒ h = 40 / 3 = 13.33 m

∴ The correct answer is 13.33 m.

Additional Information

Angle of Depression

The angle between the horizontal line and the line of sight when an observer looks at an object situated at a lower level.

Trigonometric Ratios

For any right-angled triangle, tan θ = Perpendicular / Base. Common values: tan 30° = 1/√3 and tan 60° = √3.

Relationship between H and h

If angles of depression from height H to the top and bottom of another object of height h are α and β respectively, then h = H(1 − tan α / tan β).

Calculation:

Arrange the terms in ascending order based on their coefficients: x/5, x/4, x/3, 2x/5, 2x/3, 3x/4, x.

Equate the median-representing term 2x/5 to the given median value:

2x/5 = 8.4

x = (8.4 × 5) ÷ 2 = 21

∴ The correct answer is 21.

Given:

30 + [{-9 × (26 − 11 + 3)}]

Formula used:

The BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction) is used to simplify the expression.

Calculations:

30 + [{-9 × (26 − 11 + 3)}]

⇒ 30 + [{-9 × (15 + 3)}]

⇒ 30 + [{-9 × 18}]

⇒ 30 + (−162)

⇒ 30 − 162

⇒ −132

∴ The correct answer is option (4).

Shortcut Trick

Sum of ratio squares = 3² + 5² + 7² = 9 + 25 + 49 = 83 parts.

Value of 1 part² = 2075 ÷ 83 = 25.

Value of 1 part = √25 = 5.

Middle number (5 parts) = 5 × 5 = 25.

∴ The correct answer is 25.

Alternate Method

Given:

Ratio of three positive numbers = 3 : 5 : 7

Sum of their squares = 2075

Calculations:

⇒ Let the three positive numbers be 3x, 5x, and 7x.

⇒ Sum of their squares = (3x)² + (5x)² + (7x)² = 2075

⇒ 9x² + 25x² + 49x² = 2075

⇒ 83x² = 2075

⇒ x² = 2075 ÷ 83

⇒ x² = 25

⇒ x = √25 = 5 (Since numbers are positive)

⇒ The middle number = 5x = 5 × 5 = 25

∴ The correct answer is 25.

Additional Information

Ratio and Proportion Basics

A ratio compares two quantities. If a : b = c : d, then a × d = b × c. To find actual values from a ratio, always multiply by a common constant x.

Duplicate Ratio

The duplicate ratio of a : b is a² : b². When squares of numbers in a ratio are given, the ratio of those squares is the duplicate ratio.

Mean Proportional

The mean proportional between two numbers a and b is √(a × b). This is a common property used in geometric progression and ratio problems.

Shortcut Trick

Ratio (8 years ago) G : R = 4 : 1 (Difference = 3 units)

Ratio (6 years later) G : R = 2 : 1 (Difference = 1 unit)

Make Difference equal by multiplying second ratio by 3 → 6 : 3

Change in units = 6 − 4 = 2 units | Total years gap = 8 + 6 = 14 years

2 units = 14 years ⇒ 1 unit = 7 years | G (8 years ago) = 4 × 7 = 28 years

∴ The correct answer is 36 years.

Alternate Method

Given:

Age of Mr. Gupta (8 years ago) = 4 × age of Rohit (8 years ago)

Age of Mr. Gupta (after 6 years) = 2 × age of Rohit (after 6 years)

Let Rohit’s age 8 years ago be x.

⇒ Mr. Gupta’s age 8 years ago = 4x

⇒ Present ages: Rohit = x + 8, Mr. Gupta = 4x + 8

⇒ Ages after 6 years: Rohit = x + 8 + 6 = x + 14, Mr. Gupta = 4x + 8 + 6 = 4x + 14

According to the question:

⇒ 4x + 14 = 2 × (x + 14)

⇒ 4x + 14 = 2x + 28

⇒ 4x − 2x = 28 − 14

⇒ 2x = 14

⇒ x = 7

⇒ Present Age of Mr. Gupta = 4x + 8 = 4 × 7 + 8 = 28 + 8 = 36 years

∴ The correct answer is 36 years.

Additional Information

Constant Age Difference

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

Ratio Balancing Method

To solve age problems quickly, ensure the difference between ratio parts is equal for both time periods by multiplying with suitable constants.

Linear Equation Formulation

Always define variables for the past or future to simplify the present age expression: Present = Past + t or Present = Future − t.

Shortcut Trick

Net discount for Option 2 = 20 + 15 − (20 × 15) ÷ 100 = 32%

Discount for Option 1 = 30%

Difference in discount percentage = 32% − 30% = 2%

Amount difference = 2% of ₹8,000 = ₹160

Since 32% > 30%, Option 2 provides the greater discount.

∴ The correct answer is Option 2 and by ₹160.

Alternate Method

Given:

Marked Price (MP) = ₹8,000

Option 1: Single discount = 30%

Option 2: Successive discounts = 20% and 15%

Calculations:

⇒ Discount in Option 1 = 30% of 8,000 = (30 ÷ 100) × 8,000 = ₹2,400

⇒ Successive discount formula = x + y − (xy ÷ 100)

⇒ Effective discount % for Option 2 = 20 + 15 − (20 × 15 ÷ 100) = 35 − 3 = 32%

⇒ Discount in Option 2 = 32% of 8,000 = (32 ÷ 100) × 8,000 = ₹2,560

⇒ Difference in discount amounts = ₹2,560 − ₹2,400 = ₹160

⇒ Comparing both, Option 2 offers a discount that is ₹160 more than Option 1.

∴ The correct answer is Option 2 and by ₹160.

Additional Information

Successive Discount Formula

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

Marked Price and Selling Price

Discount is always calculated on the Marked Price (MP). The price after deduction is the Selling Price (SP), where SP = MP × (100 − Discount%)/100.

Three Successive Discounts

If three discounts a%, b%, and c% are given, first find the equivalent of a and b, then use that result with c using the same formula.

Shortcut Trick

Value of 4 extra ₹500 notes = 4 × 500 = ₹2,000.

Remaining Total Amount = 24,800 − 2,000 = ₹22,800.

Combined value of one ₹100 note and one ₹500 note = 100 + 500 = ₹600.

Number of ₹100 notes = 22,800 ÷ 600 = 38.

∴ The correct answer is 38.

Alternate Method

Given: Total Amount = ₹24,800; ₹500 notes = 4 + ₹100 notes.

⇒ Let the number of ₹100 notes be x.

⇒ Number of ₹500 notes = x + 4.

⇒ Total Value = (x × 100) + ((x + 4) × 500) = 24,800.

⇒ 100x + 500x + 2,000 = 24,800.

⇒ 600x + 2,000 = 24,800.

⇒ 600x = 24,800 − 2,000 = 22,800.

⇒ x = 22,800 ÷ 600 = 38.

∴ The correct answer is 38.

Additional Information

Currency Denomination Principle

The Total Value of currency is calculated by the sum of (Value of one note × Number of notes) for each denomination.

Linear Equations in One Variable

A Linear Equation involves solving for an unknown variable by balancing constants and variables on either side of the equals sign.

Unitary Method Adjustment

In problems where one quantity is more than another, subtracting the extra value first allows the remaining amount to be divided by the sum of unit values.

Shortcut Trick

Number of sides (n) = (2 × Area) ÷ (Side × Apothem)

n = (2 × 247.5) ÷ (11 × 9)

n = 495 ÷ 99

n = 5

∴ The correct answer is 5.

Alternate Method

Given:

Side length (s) = 11 cm

Area (A) = 247.5 cm²

Perpendicular distance from centre to side (apothem, a) = 9 cm

Formula Used:

Area of a regular polygon = (1/2) × Perimeter × Apothem

Where Perimeter (P) = Number of sides (n) × Side length (s)

Calculations:

⇒ Area = (1/2) × (n × s) × a

⇒ 247.5 = (1/2) × (n × 11) × 9

⇒ 247.5 = (99/2) × n

⇒ 247.5 = 49.5 × n

⇒ n = 247.5 ÷ 49.5

⇒ n = 5

∴ The correct answer is 5.

Additional Information

Interior Angle of a Regular Polygon

The measure of each interior angle is given by [(n − 2) × 180°] ÷ n, where n is the number of sides.

Sum of Exterior Angles

For any convex polygon, the sum of all exterior angles is always 360°.

Relation between Side and Apothem

For a regular polygon, the side length (s) and apothem (a) are related by: s = 2 × a × tan(180°/n).

Shortcut Trick

Discount = 10630 − 9248.10 = 1381.90

Discount % = (Discount ÷ Marked Price) × 100

Discount % = (1381.90 ÷ 10630) × 100 = 13%

∴ The correct answer is 13%.

Alternate Method

Given:

Marked Price (MP) of the article = ₹10,630

Selling Price (SP) of the article = ₹9248.10

Formula Used:

Discount = Marked Price − Selling Price

Discount Percentage = (Discount ÷ Marked Price) × 100

Calculations:

⇒ Discount = 10630 − 9248.10

⇒ Discount = 1381.90

⇒ Discount % = (1381.90 ÷ 10630) × 100

⇒ Discount % = 138190 ÷ 10630

⇒ Discount % = 13%

∴ The correct answer is 13%.

Additional Information

Marked Price (List Price)

The price at which an article is listed or labeled by the retailer or manufacturer is called its Marked Price.

Successive Discounts

If two discounts d₁% and d₂% are given, the single equivalent discount is [d₁ + d₂ − (d₁ × d₂) ÷ 100]%.

Relationship between CP and MP

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

Shortcut Trick

Since the height H remains constant, the Volume of a cylinder is directly proportional to the square of its radius R (V ∝ R²).

Original Radius = R ⇒ Original Volume ∝ R².

New Radius = R/2 ⇒ Volume of one new cylinder ∝ (R/2)² = R²/4.

Total Volume of 3 new cylinders = 3 × (R²/4) = 3/4 of the original volume.

Unused fraction of Volume = 1 − 3/4 = 1/4.

∴ The correct answer is 1/4.

Alternate Method

Given: Original radius = R, height = H. New radius = R/2, height = H, and number of new cylinders = 3.

Formula Used: Volume of a Cylinder = π × r² × h

⇒ Volume of original cylinder (V₁) = πR²H

⇒ Volume of one new cylinder (v) = π × (R/2)² × H = π × (R²/4) × H = πR²H/4

⇒ Total volume of 3 new cylinders (V₂) = 3 × (πR²H/4) = 3πR²H/4

⇒ Unused Volume = V₁ − V₂ = πR²H − 3πR²H/4

⇒ Unused Volume = (4πR²H − 3πR²H) / 4 = πR²H/4

⇒ Fraction of original volume remaining unused = (Unused Volume) ÷ (Original Volume)

⇒ Fraction = (πR²H/4) ÷ (πR²H) = 1/4

∴ The correct answer is 1/4.

Additional Information

Volume of a Cylinder

The amount of space occupied by a cylinder is given by V = πr²h, where r is the radius and h is the height.

Scaling Effect on Volume

If the radius is changed by a factor of 'k' and the height remains constant, the volume changes by a factor of k².

Total Surface Area of Cylinder

The total surface area including the top and bottom bases is given by 2πr(h + r).

Shortcut Trick

Identify the direct algebraic relationship: (Difference)² = (Sum)² − 4 × (Product).

Substitute the given values: (Difference)² = 88² − 4 × 1927

(Difference)² = 7744 − 7708 = 36

Difference = √36 = 6

∴ The correct answer is 6.

Alternate Method

Given:

Sum of two numbers (x + y) = 88

Product of two numbers (xy) = 1927

Formula Used:

(x − y)² = (x + y)² − 4xy

Calculations:

⇒ (x − y)² = (88)² − 4 × (1927)

⇒ Square of 88 = (90 − 2)² = 8100 − 360 + 4 = 7744

⇒ 4 × 1927 = 7708

⇒ (x − y)² = 7744 − 7708

⇒ (x − y)² = 36

⇒ x − y = √36

⇒ x − y = 6

∴ The correct answer is 6.

Additional Information

Algebraic Identity

The relation (x + y)² − (x − y)² = 4xy is a fundamental identity used to find the difference when sum and product are known.

Quadratic Roots Property

For a quadratic equation t² − (Sum)t + (Product) = 0, the roots are the two numbers. The difference between roots is given by √D/a where D = b² − 4ac.

Squaring Shortcut

To square numbers near 100 like 88: (100 − 12)² = 10000 − 2400 + 144 = 7744.

Shortcut Trick

Let Total Work (Capacity) = LCM(10, 15, 30) = 30 units.

Efficiencies: Pipe A = +3, Pipe B = +2, Pipe C = −1. Combined = +4 units/hr.

Work done by all three in 6 hr = 4 × 6 = 24 units.

Remaining Work = 30 − 24 = 6 units.

Efficiency of (B + C) = 2 − 1 = 1 unit/hr. Additional time = 6 ÷ 1 = 6 hr.

Total time = 6 (initial) + 6 (extra) = 12 hr.

∴ The correct answer is 12 hr.

Alternate Method

Given: Pipe A fills in 10 hr, Pipe B fills in 15 hr, and Pipe C empties in 30 hr.

Total Work is the LCM of 10, 15, and 30.

⇒ Combined Efficiency (A + B + C) = 3 + 2 − 1 = 4 units/hr

⇒ Work completed by all pipes in first 6 hours = 4 × 6 = 24 units

⇒ Remaining capacity to be filled = 30 − 24 = 6 units

⇒ Since Pipe A is closed, remaining pipes (B and C) fill the rest.

⇒ Combined Efficiency of (B + C) = 2 − 1 = 1 unit/hr

⇒ Time taken to fill remaining 6 units = 6 ÷ 1 = 6 hr

⇒ Total Time taken to fill the cistern = Initial time + Extra time = 6 hr + 6 hr = 12 hr

∴ The correct answer is 12 hr.

Additional Information

Inlet and Outlet Pipes

The rate of work of an inlet pipe (filling) is taken as positive, while that of an outlet pipe (emptying) is taken as negative efficiency.

Net Work Done

If multiple pipes work together, the net efficiency is the algebraic sum of their individual efficiencies (e.g., Net E = E₁ + E₂ − E₃).

Time and Efficiency Relation

Total Capacity = Efficiency × Time. If the net efficiency is negative, the cistern will eventually be emptied rather than filled.