Quantitative Aptitude PYQs

Shortcut Trick

Successive multiplier = (1 − 0.12) × (1 + 0.20) = 0.88 × 1.2 = 1.056.

Let Cost Price of A be x.

x × 1.056 = 1,122.

x = 1,122 ÷ 1.056 = 1,062.50.

∴ The correct answer is ₹1,062.50.

Alternate Method

Given: Loss % (A to B) = 12%, Profit % (B to C) = 20%, Amount paid by C = ₹1,122.

Formula: CP = SP ÷ [(1 ± R₁/100) × (1 ± R₂/100)]

⇒ Let CP for A = x

⇒ SP for A (which is CP for B) = x × (100 − 12) / 100 = 0.88x

⇒ SP for B (which is CP for C) = 0.88x × (100 + 20) / 100

⇒ 1,122 = 0.88x × 1.20

⇒ 1,122 = 1.056x

⇒ x = 1,122 / 1.056

⇒ x = 1,062.50

∴ The correct answer is ₹1,062.50.

Additional Information

Successive Percentage Change

Net percentage change for two successive changes of a% and b% is given by [a + b + (ab/100)]%.

Cost Price Formula

To find CP when SP and Profit% are given: CP = [SP × 100] ÷ [100 + Profit%]. For loss, replace + with −.

Ratio Method

Convert percentages to fractions: 12% loss = 22/25 multiplier; 20% profit = 6/5 multiplier. Initial CP = Final SP × (25/22) × (5/6).

Shortcut Trick

For quarterly compounding, Rate = 16% ÷ 4 = 4% per quarter.

Time = 6 months = 2 quarters.

Effective interest rate for 2 quarters = 4 + 4 + (4 × 4)/100 = 8.16%.

CI = 5625 × 8.16% = 459.

∴ The correct answer is ₹459.

Alternate Method

Given:

Principal (P) = ₹5,625

Rate (R) = 16% per annum

Time (T) = 6 months

Formula Used: Amount (A) = P(1 + r/100)ⁿ; CI = A − P

Since interest is compounded quarterly:

⇒ Quarterly Rate (r) = 16% ÷ 4 = 4%

⇒ Number of quarters (n) = 6 months ÷ 3 months = 2

⇒ A = 5625 × (1 + 4/100)²

⇒ A = 5625 × (26/25) × (26/25)

⇒ A = 5625 × (676/625)

⇒ A = 9 × 676 = 6084

⇒ CI = 6084 − 5625 = 459

∴ The correct answer is ₹459.

Additional Information

Quarterly Compounding

When interest is compounded quarterly, the annual rate is divided by 4 and the time in years is multiplied by 4.

Effective Rate of Interest

For two periods with rate r, the effective rate is r + r + (r × r)/100.

Compound Interest for Half-Yearly

If compounded half-yearly, the rate is halved (R/2) and the time is doubled (2T).

Shortcut Trick

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

Let a = 6pq and b = 14q

Required result = 4 × (6pq) × (14q)

⇒ 24 × 14 × p × q × q = 336pq²

∴ The correct answer is 336pq².

Alternate Method

Given: (6pq + 14q)² − (6pq − 14q)²

Formula used: x² − y² = (x + y)(x − y)

⇒ [(6pq + 14q) + (6pq − 14q)] × [(6pq + 14q) − (6pq − 14q)]

⇒ [6pq + 14q + 6pq − 14q] × [6pq + 14q − 6pq + 14q]

⇒ [12pq] × [28q]

⇒ 12 × 28 × p × q × q

⇒ 336pq²

∴ The correct answer is 336pq².

Additional Information

Algebraic Identities of Squares

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

Sum and Difference of Squares

(a + b)² + (a − b)² = 2(a² + b²) while (a + b)² − (a − b)² = 4ab.

Factorization Identity

a² − b² = (a + b)(a − b). This is useful for simplifying expressions involving difference of squares.

Shortcut Trick

Let original fraction be N/D.

New fraction = (125% of N) / (91% of D) = 5/7

N/D = (5/7) × (91/125)

N/D = (1/1) × (13/25) = 13/25

Denominator − Numerator = 25 − 13 = 12.

∴ The correct answer is The difference between denominator and numerator is 12.

Alternate Method

Given:

Numerator increase = 25%

Denominator decrease = 9%

Resulting fraction = 5/7

Let the original fraction be x/y.

⇒ According to the question:

⇒ [x × (100 + 25)/100] / [y × (100 − 9)/100] = 5/7

⇒ (x × 125/100) / (y × 91/100) = 5/7

⇒ 125x / 91y = 5/7

⇒ x/y = (5 × 91) / (7 × 125)

⇒ x/y = 455 / 875

⇒ x/y = 13 / 25

⇒ Original Numerator = 13, Original Denominator = 25

⇒ Checking

Statements:

⇒ Denominator − Numerator = 25 − 13 = 12

∴ The correct answer is The difference between denominator and numerator is 12.

Additional Information

Percentage Change in Fractions

To find the original fraction given a percentage change, use the formula: Original × (New% Numerator / New% Denominator) = Final Fraction.

Proper and Improper Fractions

A fraction is proper if the numerator is less than the denominator (e.g., 13/25), and improper if the numerator is greater than or equal to the denominator.

Ratio Comparison

If the numerator increases and the denominator decreases, the value of the fraction always increases.

Shortcut Trick

New distance is 3 times the old distance (26.4 ÷ 8.8 = 3).

New speed is 2 times the old speed.

Since Time = Distance ÷ Speed, the new time factor = 3 ÷ 2 = 1.5.

Initial time = 1 hour 20 minutes = 80 minutes.

New time = 80 × 1.5 = 120 minutes = 2 hours.

∴ The correct answer is 2 hours.

Alternate Method

Given: Distance₁ = 8.8 km, Time₁ = 1 hour 20 min, Distance₂ = 26.4 km, Speed₂ = 2 × Speed₁.

Formula used: Speed = Distance ÷ Time | Time = Distance ÷ Speed

⇒ Time₁ = 1 + 20/60 = 1 + 1/3 = 4/3 hours

⇒ Speed₁ = 8.8 ÷ (4/3) = (8.8 × 3) ÷ 4 = 2.2 × 3 = 6.6 kmph

⇒ Speed₂ = 2 × 6.6 = 13.2 kmph

⇒ Time₂ = Distance₂ ÷ Speed₂

⇒ Time₂ = 26.4 ÷ 13.2 = 2 hours

∴ The correct answer is 2 hours.

Additional Information

Speed-Time-Distance Proportionality

When speed is constant, Distance ∝ Time. When distance is constant, Speed ∝ 1/Time. When time is constant, Distance ∝ Speed.

Average Speed

Average Speed = Total Distance ÷ Total Time. For two equal distances with speeds x and y, Average Speed = 2xy ÷ (x + y).

Unit Conversion

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

Shortcut Trick

Solve the innermost bracket expressions first: (36 ÷ 6) = 6 and (9 ÷ 3) = 3.

Simplify the terms inside the braces: {6 + 3 − 3} = 6.

Multiply the result by the outer fraction: 1/6 × 6 = 1.

Add the final constant term: 1 + 18 = 19.

∴ The correct answer is 19.

Alternate Method

Given expression: [1/6 × {(36 ÷ 6) + (9 ÷ 3) − 3} + 18]

Method: Apply BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction).

⇒ [1/6 × {6 + 3 − 3} + 18]

⇒ [1/6 × {9 − 3} + 18]

⇒ [1/6 × 6 + 18]

⇒ 1 + 18

⇒ 19

∴ The correct answer is 19.

Additional Information

BODMAS Rule

It defines the order of operations: Brackets (), Orders (roots/powers), Division (÷), Multiplication (×), Addition (+), and Subtraction (−).

Types of Brackets

Solve brackets in the order: Parentheses ( ), then Curly Braces { }, and finally Square Brackets [ ].

Division Shortcut

Division is the same as multiplication by the reciprocal. For example, a ÷ b = a × (1/b).

Shortcut Trick

Total current age of 2 members = Average × Number of members = 35 × 2 = 70 years.

In 11 years, each member's age increases by 11.

Total increase in sum for 2 members = 2 × 11 = 22 years.

Total age after 11 years = Current sum + Total increase = 70 + 22 = 92 years.

∴ The correct answer is 92 years.

Alternate Method

Given:

Ratio of ages = 6:1

Average age = 35 years

Formula used: Average = Sum of observations ÷ Total number of observations

Let the current ages of the two members be 6x and x respectively.

⇒ Average = (6x + x) ÷ 2 = 35

⇒ 7x ÷ 2 = 35

⇒ 7x = 70

⇒ x = 10

⇒ Present ages: Member 1 = 6 × 10 = 60 years | Member 2 = 1 × 10 = 10 years

⇒ Age of Member 1 after 11 years = 60 + 11 = 71 years

⇒ Age of Member 2 after 11 years = 10 + 11 = 21 years

⇒ Total age after 11 years = 71 + 21 = 92 years

∴ The correct answer is 92 years.

Additional Information

Sum and Average Relation

Total sum of ages is always equal to the Average multiplied by the total number of individuals (Sum = Avg × n).

Future Sum of Ages

If there are 'n' people, the total sum of their ages after 't' years will increase by (n × t).

Ratio and Sum

If the ratio of two quantities is a:b, then their sum can be represented as k(a + b), where k is a constant factor.

Shortcut Trick

Area of a general quadrilateral = 1/2 × Diagonal × (Sum of perpendiculars)

Diagonal (d) = 10 m, Perpendiculars h₁ = 12 m, h₂ = 13 m

Area = 1/2 × 10 × (12 + 13)

Area = 5 × 25 = 125 m²

∴ The correct answer is 125 m².

Alternate Method

Given:

Length of diagonal (d) = 10 m

Perpendicular offsets (h₁ and h₂) = 12 m and 13 m

Formula Used:

Area of Quadrilateral = Area of Triangle₁ + Area of Triangle₂

Area = (1/2 × base × height₁) + (1/2 × base × height₂)

⇒ Area = 1/2 × d × (h₁ + h₂)

⇒ Area = 1/2 × 10 × (12 + 13)

⇒ Area = 5 × 25

⇒ Area = 125 m²

∴ The correct answer is 125 m².

Additional Information

Area of Rhombus

When both diagonals are known, Area = 1/2 × d₁ × d₂.

Area of Trapezium

Area = 1/2 × (sum of parallel sides) × h, where h is the distance between them.

Area of Parallelogram

Area = base × height. Also, if two adjacent sides a, b and the angle θ between them are given, Area = ab sinθ.

Area of Cyclic Quadrilateral

Area = √[(s−a)(s−b)(s−c)(s−d)], where s is the semi-perimeter (a+b+c+d)/2.

Shortcut Trick

Total capacity of cistern = LCM (8, 20) = 40 units

Efficiency of (P + Q) = 40 ÷ 8 = 5 units/hr

Efficiency of P = 40 ÷ 20 = 2 units/hr

Efficiency of Q = (P + Q) − P = 5 − 2 = 3 units/hr

3/4^th of capacity = 40 × 3/4 = 30 units

Time taken by Q = 30 ÷ 3 = 10 hours

∴ The correct answer is 10.

Alternate Method

Given:

Time taken by (P + Q) = 8 hours

Time taken by P = 20 hours

Formula Used:

Total Work = Efficiency × Time

⇒ Efficiency of P = 40 ÷ 20 = 2

⇒ Combined Efficiency of P and Q = 40 ÷ 8 = 5

⇒ Efficiency of Q = 5 − 2 = 3

⇒ Total work to be done = 3/4 of the cistern capacity

⇒ Required Work = 40 × (3/4) = 30 units

⇒ Time taken by Q = Required Work ÷ Efficiency of Q

⇒ Time taken by Q = 30 ÷ 3 = 10 hours

∴ The correct answer is 10.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken. If P takes x hours and Q takes y hours, their efficiencies are 1/x and 1/y respectively.

Pipes and Cisterns Basic Formula

If an inlet pipe fills in A hours and another fills in B hours, they together fill in (A × B) ÷ (A + B) hours.

Negative Work (Outlet/Leakage)

If a pipe empties the tank, its efficiency is taken as negative in the total efficiency calculation.

Shortcut Trick

LCM of (20, 24, 30) = 120 units (Total Work).

Efficiencies: (X+Y) = 6, (Y+Z) = 5, (Z+X) = 4.

Combined efficiency 2(X+Y+Z) = 6 + 5 + 4 = 15.

Efficiency (X+Y+Z) = 15 ÷ 2 = 7.5.

Days taken = 120 ÷ 7.5 = 16 days.

∴ The correct answer is 16 days.

Alternate Method

Given:

Time taken by (X + Y) = 20 days

Time taken by (Y + Z) = 24 days

Time taken by (Z + X) = 30 days

Formula used: Total Work = Time × Efficiency

Calculations:

⇒ Total efficiency of (X+Y) + (Y+Z) + (Z+X) = 6 + 5 + 4

⇒ 2(X + Y + Z) efficiency = 15 units/day

⇒ (X + Y + Z) efficiency = 15 ÷ 2 = 7.5 units/day

⇒ Time taken by (X + Y + Z) together = Total Work ÷ Combined Efficiency

⇒ Time = 120 ÷ 7.5

⇒ Time = 16 days

∴ The correct answer is 16 days.

Additional Information

Pipes and Cisterns

This concept is identical to Time and Work, where an inlet pipe has positive efficiency and an outlet pipe (leak) has negative efficiency.

Work and Wages

Wages are always distributed in the ratio of the work actually done by each person (which is efficiency × time).

Combined Efficiency Rule

When individuals work in pairs (A+B, B+C, C+A), the sum of their efficiencies is twice the efficiency of all three working together: 2(A+B+C) = ΣEfficiencies.

Shortcut Trick

Apply the standard algebraic identity: (a − b)² = a² − 2ab + b².

Substituting a = x and b = 2:

(x − 2)² = (x)² − 2 × (x) × (2) + (2)²

= x² − 4x + 4

∴ The correct answer is x² − 4x + 4.

Alternate Method

Given: (x − 2)²

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

⇒ (x − 2)² = (x − 2) × (x − 2)

⇒ x(x − 2) − 2(x − 2)

⇒ x² − 2x − 2x + 4

⇒ x² − 4x + 4

∴ The correct answer is x² − 4x + 4.

Additional Information

Square of a Sum

The algebraic identity for the square of a sum is (a + b)² = a² + 2ab + b².

Difference of Squares

The identity for the difference of two squares is a² − b² = (a − b)(a + b).

Square of a Trinomial

For three terms, the identity is (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Shortcut Trick

LCM of 30 and 40 minutes = 120 minutes = 2 hours.

Total duration from 8 am to 2 pm = 6 hours.

Number of times = (Total Duration ÷ LCM interval) + 1

Number of times = (6 ÷ 2) + 1 = 3 + 1 = 4 times.

∴ The correct answer is 4.

Alternate Method

Given:

Bell 1 interval = 30 minutes

Bell 2 interval = 40 minutes

Starting time = 8:00 am

Closing time = 2:00 pm

Formula Used:

To find the simultaneous occurrence of cyclic events, calculate the Least Common Multiple (LCM) of their time intervals.

⇒ LCM of 30 and 40:

⇒ 30 = 2 × 3 × 5

⇒ 40 = 2 × 2 × 2 × 5

⇒ LCM = 2³ × 3 × 5 = 120 minutes

⇒ Interval in hours = 120 ÷ 60 = 2 hours

⇒ Total duration = 2:00 pm − 8:00 am = 6 hours

⇒ Number of intervals = 6 ÷ 2 = 3

⇒ Total times the bells ring together = (Number of intervals + 1) for the initial 8 am ring.

⇒ Total = 3 + 1 = 4

∴ The correct answer is 4.

Additional Information

LCM for Cyclic Events

When multiple objects start at the same time and repeat at fixed intervals, they will coincide again at a time which is the LCM of their individual intervals.

Total Occurrences Formula

Total Occurrences = (Total Time Period ÷ LCM of intervals) + 1. The +1 accounts for the first time they ring together at the starting point.

Unit Consistency

Always ensure the time intervals and the total duration are in the same units (either all in minutes or all in hours) before performing division.

Shortcut Trick

Net filling rate = (1/6) − (1/24) = 3/24 = 1/8 tank per minute.

Time to fill the full tank = 8 minutes.

Time to fill one-half of the tank = 8 × (1/2) = 4 minutes.

∴ The correct answer is 4.

Alternate Method

Given:

Time taken by Pipe A to fill = 6 minutes

Time taken by Pipe B to empty = 24 minutes

Formula used: Efficiency = Total Work ÷ Time

⇒ Total capacity of the tank = LCM(6, 24) = 24 units

⇒ Efficiency of Pipe A = 24 ÷ 6 = +4 units/min

⇒ Efficiency of Pipe B = 24 ÷ 24 = −1 units/min

⇒ Net efficiency of (A + B) = 4 − 1 = 3 units/min

⇒ One-half of the tank capacity = 24 ÷ 2 = 12 units

⇒ Time required to fill 12 units = 12 ÷ 3 = 4 minutes

∴ The correct answer is 4.

Additional Information

Negative Work in Pipes

When a pipe empties a tank, its efficiency or rate of work is taken as negative (e.g., −1/T).

Pipes & Cisterns Basic Formula

If pipe A fills in x hours and pipe B empties in y hours, net work in 1 hour = (1/x) − (1/y).

LCM Method for Multiple Pipes

Total capacity is assumed as the LCM of individual times. Efficiency = Capacity ÷ Time.

Shortcut Trick

Ratio of A : B (Present) = 5 : 1 (Difference = 4 units)

Ratio of A : B (5 yrs ago) = 6 : 1 (Difference = 5 units)

Equalize the difference by multiplying ratios by 5 and 4 respectively:

Present ratio = 25 : 5 | Past ratio = 24 : 4

Difference in B's ratio parts = 5 − 4 = 1 unit → 5 years

Present age of B = 5 units × 5 = 25 years

∴ The correct answer is 25 years.

Alternate Method

Given:

Present ratio of A and B = 5 : 1

Ratio 5 years ago = 6 : 1

Let the present age of A be 5x and B be x.

According to the question, five years ago:

⇒ (5x − 5) ÷ (x − 5) = 6 ÷ 1

⇒ 1 × (5x − 5) = 6 × (x − 5)

⇒ 5x − 5 = 6x − 30

⇒ 6x − 5x = 30 − 5

⇒ x = 25

Present age of B = x = 25 years.

∴ The correct answer is 25 years.

Additional Information

Constant Difference Rule

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

Future and Past Age

To find age 'n' years ago, subtract n from present age (x − n). To find age 'n' years hence, add n to present age (x + n).

Ratio Balancing

When solving age ratios, multiplying the ratios to make the numerical difference between parts equal to the time gap simplifies the calculation.

Shortcut Trick

A number is divisible by 45 if it is divisible by both 5 and 9.

Divisibility by 5: Last digit must be 0 or 5. Only (2) 20475 and (4) 555555 qualify.

Divisibility by 9: Sum of digits must be divisible by 9.

For 20475, sum = 2+0+4+7+5 = 18 (divisible by 9).

For 555555, sum = 5×6 = 30 (not divisible by 9).

∴ The correct answer is 20475.

Alternate Method

Given:

Numbers: 306999, 20475, 25436, 555555

Formula Used:

Divisibility rule of 45: A number N is divisible by 45 if N is divisible by 5 and 9 simultaneously (since 5 × 9 = 45 and HCF(5, 9) = 1).

Calculations:

⇒ Option (1) 306999: Last digit is 9. Not divisible by 5.

⇒ Option (3) 25436: Last digit is 6. Not divisible by 5.

⇒ Option (4) 555555: Last digit is 5. Sum = 5+5+5+5+5+5 = 30. 30 is not divisible by 9.

⇒ Option (2) 20475: Last digit is 5. Divisible by 5.

⇒ Sum of digits of 20475 = 2 + 0 + 4 + 7 + 5 = 18.

⇒ Since 18 is divisible by 9, 20475 is divisible by 9.

⇒ As 20475 is divisible by both 5 and 9, it must be divisible by 45.

∴ The correct answer is 20475.

Additional Information

Divisibility by 3

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

Divisibility by 4

A number is divisible by 4 if its last two digits form a number divisible by 4.

Divisibility by 11

A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is 0 or a multiple of 11.

Shortcut Trick

The average of the first n odd natural numbers is always n.

Given value of n = 14.

∴ Average = 14.

∴ The correct answer is 14.

Alternate Method

Given:

Number of terms (n) = 14

Formula Used:

Sum of first n odd natural numbers = n²

Average = Sum of observations ÷ Total number of observations

⇒ First 14 odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27

⇒ Sum = 14²

⇒ Sum = 196

⇒ Average = 196 ÷ 14

⇒ Average = 14

∴ The correct answer is 14.

Additional Information

Average of first n natural numbers

The average is given by the formula (n + 1) ÷ 2.

Average of first n even natural numbers

The average of the first n even natural numbers is (n + 1).

Average of Arithmetic Progression (AP)

For any sequence in AP, Average = (First term + Last term) ÷ 2.

Shortcut Trick

Remaining Distance = 3/4 × 40 = 30 km

Remaining Time = 2/3 × 5 = 10/3 hours

Required Speed = Remaining Distance ÷ Remaining Time

Speed = 30 ÷ (10/3) = (30 × 3) / 10 = 9 km/h

∴ The correct answer is 9.

Alternate Method

Given:

Total Distance = 40 km

Total Time = 5 hours

Formula Used:

Speed = Distance / Time

⇒ Total Distance = 40 km

⇒ Covered Distance = 40 × (1/4) = 10 km

⇒ Remaining Distance = 40 − 10 = 30 km

⇒ Total Time = 5 hours

⇒ Covered Time = 5 × (1/3) = 5/3 hours

⇒ Remaining Time = 5 − 5/3 = 10/3 hours

⇒ Required Speed = Remaining Distance ÷ Remaining Time

⇒ Required Speed = 30 ÷ (10/3)

⇒ Required Speed = (30 × 3) / 10

⇒ Required Speed = 9 km/h

∴ The correct answer is 9.

Additional Information

Average Speed

It is the ratio of total distance covered to the total time taken. Average Speed = Total Distance ÷ Total Time.

Relative Speed

When moving in the same direction, speed = S₁ − S₂. When moving in opposite directions, speed = S₁ + S₂.

Unit Conversion

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

Shortcut Trick

Volume of metal in a hollow pipe = π × (R² − r²) × h

Difference of squares (R² − r²) = (7 + 5) × (7 − 5) = 12 × 2 = 24.

Volume = π × 24 × 14

Volume = 336π cm³.

∴ The correct answer is 336π cm³.

Alternate Method

Given:

Outer radius (R) = 7 cm

Inner radius (r) = 5 cm

Length/Height (h) = 14 cm

Formula for the volume of metal in a hollow cylinder:

Volume = External Volume − Internal Volume

⇒ Volume = πR²h − πr²h

⇒ Volume = πh(R² − r²)

⇒ Volume = π × 14 × (7² − 5²)

⇒ Volume = π × 14 × (49 − 25)

⇒ Volume = π × 14 × 24

⇒ Volume = 336π cm³

∴ The correct answer is 336π cm³.

Additional Information

Volume of a Solid Cylinder

The volume of a solid cylinder with radius r and height h is given by V = πr²h.

Curved Surface Area (CSA) of Hollow Pipe

The total curved surface area (inner + outer) is 2πRh + 2πrh = 2πh(R + r).

Total Surface Area (TSA) of Hollow Pipe

TSA = CSA + 2 × Area of Base Rings = 2πh(R + r) + 2π(R² − r²) = 2π(R + r)(h + R − r).

Shortcut Trick

Efficiency Ratio (A : B) = 100 : 125 = 4 : 5

Total efficiency = 4 + 5 = 9 units

Total work = 9 units/hr × 8 hr = 72 units

Time for Pipe A = 72 ÷ 4 = 18 hours

Capacity = 18 × 60 min × 5 litres/min = 5400 litres

∴ The correct answer is 5400 litres.

Alternate Method

Given: Time (A+B) = 8 hours, Speed of A = 5 L/min

Efficiency of B = 125% of A ⇒ B = 1.25A

⇒ Let Efficiency of A = 4 units/hr and B = 5 units/hr

⇒ Total Efficiency (A + B) = 4 + 5 = 9 units/hr

⇒ Total Work = 9 × 8 = 72 units

⇒ Time taken by Pipe A = 72 ÷ 4 = 18 hours

⇒ Rate of Pipe A = 5 litres/min × 60 mins/hr = 300 litres/hr

⇒ Capacity of Cistern = Rate × Time = 300 × 18 = 5400 litres

∴ The correct answer is 5400 litres.

Additional Information

Efficiency and Time Relation

Efficiency is inversely proportional to the time taken. If the ratio of efficiencies is x:y, the ratio of time taken will be y:x.

Cistern Capacity Formula

Total Capacity = Rate of flow (litres/unit time) × Total time taken to fill the tank.

Pipes and Cisterns Work

Net work done = (Sum of work by inlet pipes) − (Sum of work by outlet pipes).

Shortcut Trick

Total Distance = 120 + 180 = 300 km

Total Time = (120 ÷ 50) + (180 ÷ 80) = 2.4 + 2.25 = 4.65 hours

Average Speed = Total Distance ÷ Total Time

Average Speed = 300 ÷ 4.65 ≈ 64.52 km/h

∴ The correct answer is 64.52 km/h.

Alternate Method

Given:

Distance 1 (d₁) = 120 km, Speed 1 (s₁) = 50 km/h

Distance 2 (d₂) = 180 km, Speed 2 (s₂) = 80 km/h

Formula Used:

Average Speed = Total Distance ÷ Total Time

Time = Distance ÷ Speed

⇒ Time taken for first part (t₁) = 120 ÷ 50 = 2.4 hours

⇒ Time taken for second part (t₂) = 180 ÷ 80 = 2.25 hours

⇒ Total Distance = 120 + 180 = 300 km

⇒ Total Time = 2.4 + 2.25 = 4.65 hours

⇒ Average Speed = 300 ÷ 4.65

⇒ Average Speed = 64.516129... ≈ 64.52 km/h

∴ The correct answer is 64.52 km/h.

Additional Information

Average Speed for Equal Distances

When a journey covers two equal distances at speeds s₁ and s₂, Average Speed = (2 × s₁ × s₂) ÷ (s₁ + s₂).

Average Speed for Equal Time Intervals

If a person travels for equal time intervals at speeds s₁ and s₂, Average Speed = (s₁ + s₂) ÷ 2.

Unit Conversion Shortcut

To convert speed from km/h to m/s, multiply by 5/18. To convert from m/s to km/h, multiply by 18/5.