Quantitative Aptitude PYQs

Shortcut Trick

Increase in average = (New score − Old average) ÷ Total number of innings

Increase = (266 − 77) ÷ 13

Increase = 189 ÷ 13 ≈ 14.538

Rounding 14.538 to the nearest whole number ⇒ 15.

∴ The correct answer is 15 runs.

Alternate Method

Given:

Initial number of innings = 12

Initial average runs = 77

Runs in the 13^th inning = 266

Formula Used:

Average = Total Runs ÷ Number of Innings

Calculations:

⇒ Total runs in 12 innings = 12 × 77 = 924

⇒ Total runs after 13 innings = 924 + 266 = 1190

⇒ New average = 1190 ÷ 13

⇒ New average ≈ 91.538

⇒ Increase in average = New average − Old average

⇒ Increase = 91.538 − 77 = 14.538

⇒ Rounded to the nearest whole number = 15

∴ The correct answer is 15 runs.

Additional Information

Fundamental Average Formula

Average = (Sum of all observations) ÷ (Total number of observations). To find the sum, multiply average by the number of items.

Inclusion Rule

When a new value is added, New Sum = (Old Average × Old Count) + New Value. The new average is then New Sum ÷ (Old Count + 1).

Rounding Rules

To round to the nearest whole number, if the decimal part is 0.5 or greater, round up; if less than 0.5, round down.

Shortcut Trick

Net discount = a + b − (a × b) ÷ 100 = 20 + 25 − (20 × 25) ÷ 100 = 45 − 5 = 40%.

Ratio of Marked Price (MP) to Cost Price (CP) = (100 + Gain%) ÷ (100 − Discount%).

MP ÷ CP = (100 + 25) ÷ (100 − 40) = 125 ÷ 60 = 25 ÷ 12.

Given CP = 12 units = ₹8,400 ⇒ 1 unit = ₹700.

MP = 25 units = 25 × 700 = ₹17,500.

∴ The correct answer is ₹17,500.

Alternate Method

Given: CP = ₹8,400, Gain% = 25%, Discounts = 20% and 25%.
Formulas: SP = CP × (100 + P%) ÷ 100 | MP = SP × 100 ÷ (100 − D1%) × 100 ÷ (100 − D2%)

⇒ Required SP = 8,400 × (125 ÷ 100) = 8,400 × 1.25 = ₹10,500.

⇒ Let Marked Price be MP.

⇒ SP = MP × (80 ÷ 100) × (75 ÷ 100) = 10,500.

⇒ MP × (4 ÷ 5) × (3 ÷ 4) = 10,500.

⇒ MP × (3 ÷ 5) = 10,500.

⇒ MP = (10,500 × 5) ÷ 3 = 3,500 × 5 = ₹17,500.

∴ The correct answer is ₹17,500.

Additional Information

Marked Price & Discount

The price listed on the label is MP. Discount is always calculated on MP. SP = MP − Discount.

Successive Discount Formula

If two discounts d₁ and d₂ are given, the equivalent single discount is (d₁ + d₂ − d₁d₂/100)%.

Shortcut Trick

3⁵ × 7⁵ = (3×7)⁵ = 21⁵

⇒ 21⁵ ÷ 21³ = 21²

⇒ = 441

∴ Answer = 441

Alternate Method

Given:

3⁵ × 7⁵ ÷ 21³

Concept Used:

Exponent laws: a^m × b^m = (ab)^m, a^m ÷ aⁿ = a^m−n

Calculation:

⇒ 3⁵ × 7⁵ = (3×7)⁵ = 21⁵

⇒ 21⁵ ÷ 21³ = 21⁵⁻³

⇒ = 21²

⇒ = 441

∴ The correct answer is 441.

Additional Information

Exponent Rule

a^m × b^m = (ab)^m

Division Rule

a^m ÷ aⁿ = a^m−n

Simplification Tip

Combine same powers before applying division.

Shortcut Trick

Required Percentage = (Value to be compared / Base value) × 100

Here, B’s marks are compared with A’s marks as the base.

Ratio of B : A = 50 : 75 = 2 : 3

Percentage = (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 = 115

Formula Used:

Percentage = (Part / Whole) × 100

Calculations:

⇒ Required Percentage = (Marks of B / Marks of A) × 100

⇒ Percentage = (50 / 75) × 100

⇒ Percentage = (2 / 3) × 100

⇒ Percentage = 200 / 3

⇒ Percentage ≈ 66.666...

⇒ Rounding off to two decimal places, we get 66.67%.

∴ The correct answer is 66.67%.

Additional Information

Percentage Comparison

When calculating what percent x is of y, the value following the word ‘of’ or ‘than’ is always taken as the denominator (base).

Fraction to Percentage Conversion

Common fractions like 1/3 = 33.33% and 2/3 = 66.67% are useful for speeding up calculations in competitive exams.

Base Value Identification

Ensure the correct base is chosen; here the maximum marks (115) is irrelevant because the comparison is specifically against A’s marks.

Shortcut Trick

Let the total valid votes be 100%.

Candidate 1 (Loser) = 28%

Candidate 2 (Winner) = 100% − 28% = 72%

Vote Difference = 72% − 28% = 44%

Given 44% → 13,200, then 1% → 300.

∴ The correct answer is 30,000 votes.

Alternate Method

Given:

Votes obtained by the losing candidate = 28%

Difference in votes (Margin) = 13,200

Calculations:

⇒ Let Total Valid Votes = V

⇒ Votes of Winner = (100 − 28)% of V = 72% of V

⇒ Difference = 72% of V − 28% of V = 44% of V

⇒ 44% of V = 13,200

⇒ (44/100) × V = 13,200

⇒ V = (13,200 × 100) ÷ 44

⇒ V = 300 × 100

⇒ V = 30,000

∴ The correct answer is 30,000 votes.

Additional Information

Percentage Change

Percentage difference between two values is calculated as |Value₁ − Value₂|/Base × 100.

Election Problems

Total Votes = Valid Votes + Invalid Votes. Winner's Share + Loser's Share = 100% of Valid Votes.

Base Change Property

If A is x% more than B, then B is [x/(100+x)] × 100% less than A.

Shortcut Trick

Assume an Assumed Mean (A) = 50.

Calculate deviations from 50: −25, +42, −2, −33, −12, +43, −13, +40, +23.

Sum of deviations = (−25 − 2 − 33 − 12 − 13) + (42 + 43 + 40 + 23) = −85 + 148 = +63.

Average deviation = 63 ÷ 9 = +7.

Actual Mean = Assumed Mean + Average Deviation = 50 + 7 = 57.

∴ The correct answer is 57.

Alternate Method

Given: Observations = 25, 92, 48, 17, 38, 93, 37, 90, 73

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

⇒ Total number of observations (n) = 9

⇒ Sum of observations = 25 + 92 + 48 + 17 + 38 + 93 + 37 + 90 + 73

⇒ Sum of observations = 513

⇒ Arithmetic Mean = 513 ÷ 9

⇒ Arithmetic Mean = 57

∴ The correct answer is 57.

Additional Information

Sum of Deviations

The algebraic sum of the deviations of a set of observations from their arithmetic mean is always zero: Σ(x_i − Mean) = 0.

Combined Mean

If two groups have sizes n₁, n₂ and means x₁, x₂, their combined mean is (n₁x₁ + n₂x₂) ÷ (n₁ + n₂).

Effect of Arithmetic Operations

If each observation is increased, decreased, multiplied, or divided by a constant k, the mean also changes by the same constant k.

Shortcut Trick

Relative Speed (Opposite Direction) = 60 + 90 = 150 km/hr

Convert Speed to m/s = 150 × (5/18) = 125/3 m/s

Total Distance = Length X + Length Y = 150 + 250 = 400 m

Time Taken = 400 ÷ (125/3) = 400 × 3 ÷ 125 = 9.6 sec

∴ The correct answer is 9.6 sec.

Alternate Method

Given:

Length of Train X (L₁) = 150 m

Length of Train Y (L₂) = 250 m

Speed of Train X (S₁) = 60 km/hr

Speed of Train Y (S₂) = 90 km/hr

Formula Used:

Time = Total Distance ÷ Relative Speed

Relative Speed (Opposite) = S₁ + S₂

Conversion: 1 km/hr = 5/18 m/sec

Calculations:

⇒ Total Distance = 150 + 250 = 400 m

⇒ Relative Speed = 60 + 90 = 150 km/hr

⇒ Relative Speed in m/s = 150 × (5/18) = 750/18 = 125/3 m/s

⇒ Time = 400 ÷ (125/3)

⇒ Time = (400 × 3) ÷ 125

⇒ Time = 1200 ÷ 125 = 9.6 sec

∴ The correct answer is 9.6 sec.

Additional Information

Relative Speed Concept

When two objects move in opposite directions, their speeds are added (S₁ + S₂). If they move in the same direction, their speeds are subtracted (|S₁ − S₂|).

Unit Conversion

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

Train Crossing Objects

When a train crosses a stationary pole/man, distance = length of train. When it crosses another train/platform, distance = sum of both lengths.

Shortcut Trick

Total days from 8 February to 22 April 2023: 20 (Feb) + 31 (Mar) + 22 (Apr) = 73 days.

In simple interest problems, 73 days is exactly 1/5 of a standard year (365 days).

Interest for 1 year = 7% of 2000 = ₹140.

Interest for 1/5 year = 140 ÷ 5 = ₹28.

∴ The correct answer is ₹28.

Alternate Method

Given:

Principal (P) = ₹2,000

Rate of Interest (R) = 7% per annum

Time Period: 8 February 2023 to 22 April 2023

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

Calculations:

⇒ Days in February (after 8^th) = 28 − 8 = 20 days

⇒ Days in March = 31 days

⇒ Days in April = 22 days

⇒ Total Days = 20 + 31 + 22 = 73 days

⇒ Time (T) in years = 73 ÷ 365 = 1/5 years

⇒ SI = (2000 × 7 × 1/5) ÷ 100

⇒ SI = (20 × 7) ÷ 5

⇒ SI = 4 × 7 = 28

∴ The correct answer is ₹28.

Additional Information

Counting Days for Interest

While calculating the time period, the day on which money is borrowed is generally excluded, while the day on which money is returned is included.

Fractional Year Constants

Common exam values: 73 days = 1/5 year, 146 days = 2/5 year, 219 days = 3/5 year, and 292 days = 4/5 year (based on a 365-day year).

Amount Formula

Total Amount (A) = Principal (P) + Simple Interest (SI). In this case, A = 2000 + 28 = ₹2028.

Shortcut Trick

Let the original price paid by A be x.

Successive multiplier: x × (120/100) × (75/100) × (140/100) = 252.

x × (6/5) × (3/4) × (7/5) = 252.

x × (126/100) = 252 ⇒ x = (252 × 100) ÷ 126 = 200.

∴ The correct answer is ₹200.

Alternate Method

Given: A's Profit = 20%, B's Loss = 25%, C's Profit = 40%, D's Payment = ₹252.
Formulas: Final Price = Initial Price × (100 ± P/L₁)% × (100 ± P/L₂)% ...

⇒ Let A's cost price be 100 units.
⇒ B's cost price = 100 × 120/100 = 120 units.
⇒ C's cost price = 120 × 75/100 = 90 units.
⇒ D's cost price = 90 × 140/100 = 126 units.
⇒ Given 126 units = ₹252.
⇒ 1 unit = 252 ÷ 126 = 2.
⇒ 100 units = 100 × 2 = 200.

∴ The correct answer is ₹200.

Additional Information

Chain Rule in Profit and Loss

When an article is sold through multiple people, the Selling Price of one becomes the Cost Price of the next. The final value is calculated using the formula: Final Price = Initial Price × (1 ± r₁/100) × (1 ± r₂/100)...

Successive Percentage Change

Net change for two successive updates of x% and y% is given by [x + y + (xy/100)]%. This can also be applied to successive profit/loss scenarios.

Shortcut Trick

If (x + 6) is the HCF, then x = −6 must be a root of both expressions.

Substitute x = −6 into the second expression: (−6)² + k(−6) + 12 = 0.

36 − 6k + 12 = 0 ⇒ 48 = 6k.

k = 8.

∴ The correct answer is 8.

Alternate Method

Given: HCF is (x + 6). Expressions are P(x) = x² + 10x + 24 and Q(x) = x² + kx + 12.

Formula: If (x − a) is a factor of a polynomial f(x), then f(a) = 0.

⇒ x + 6 = 0

⇒ x = −6

⇒ Since (x + 6) is the Highest Common Factor, it is a factor of x² + kx + 12.

⇒ (−6)² + k(×−6) + 12 = 0

⇒ 36 − 6k + 12 = 0

⇒ 48 − 6k = 0

⇒ 6k = 48

⇒ k = 48 ÷ 6

⇒ k = 8

∴ The correct answer is 8.

Additional Information

Factor Theorem

A polynomial f(x) has a factor (x − a) if and only if f(a) = 0. This is used to find unknown coefficients when the HCF or a factor is known.

Relationship between HCF and LCM

For two polynomials P(x) and Q(x), the product of the polynomials is equal to the product of their HCF and LCM: P(x) × Q(x) = HCF × LCM.

Quadratic Roots

For a quadratic equation ax² + bx + c = 0, the sum of roots is −b/a and the product of roots is c/a.

Shortcut Trick

Check odd numbers between 368 and 376: 369, 371, 373, 375.

369: Sum of digits (3+6+9 = 18) → Divisible by 9.

371: 371 = 7 × 53 → Divisible by 7.

375: Ends in 5 → Divisible by 5.

373: Not divisible by 2, 3, 5, 7, 11, 13, 17, 19 → Prime.

∴ The correct answer is 1.

Alternate Method

Given: Range between 368 and 376.

Numbers in range: 369, 370, 371, 372, 373, 374, 375.

⇒ 370, 372, 374 are even (composite).

⇒ 369 = 3 × 123 (composite).

⇒ 371 = 7 × 53 (composite).

⇒ 375 = 5 × 75 (composite).

⇒ 373: √373 ≈ 19.3. Checking divisibility by 2, 3, 5, 7, 11, 13, 17, 19 yields no divisors.

∴ The correct answer is 1.

Additional Information

Definition of Prime Number

A natural number greater than 1 that has no positive divisors other than 1 and itself.

Testing for Primality

To check if N is prime, test divisibility by all primes less than or equal to √N.

Prime Numbers between 1 and 100

There are exactly 25 prime numbers between 1 and 100.

Shortcut Trick

Product of two numbers = [(Sum + Difference) ÷ 2] × [(Sum − Difference) ÷ 2]

Given Sum = 86 and Difference = 12.

First number = (86 + 12) ÷ 2 = 49

Second number = (86 − 12) ÷ 2 = 37

Product = 49 × 37 = 1813

∴ The correct answer is 1813.

Alternate Method

Let the two numbers be x and y (where x > y).

Given:

x + y = 86

x − y = 12

Calculations:

⇒ Adding the two equations: (x + y) + (x − y) = 86 + 12

⇒ 2x = 98

⇒ x = 98 ÷ 2 = 49

⇒ Substituting x = 49 in the first equation: 49 + y = 86

⇒ y = 86 − 49 = 37

⇒ Product of the numbers = x × y = 49 × 37

⇒ Product = 1813

∴ The correct answer is 1813.

Additional Information

Algebraic Identity for Product

The product of two numbers can be found using the identity: 4xy = (x + y)² − (x − y)².

Sum and Difference Rule

If the sum (S) and difference (D) of two numbers are known, the larger number is (S + D) ÷ 2 and the smaller number is (S − D) ÷ 2.

Difference of Squares

The difference of the squares of two numbers is equal to the product of their sum and their difference: x² − y² = (x + y)(x − y).

Shortcut Trick

Combine the ratios P : Q (17 : 5) and Q : R (6 : 12 or 1 : 2).

Ratio P : Q : R → (17×6) : (5×6) : (5×12) ⇒ 102 : 30 : 60 ⇒ 51 : 15 : 30.

Total units = 51 + 15 + 30 = 96 units.

Share of Q = (15 / 96) × 28,800 = 15 × 300 = ₹4,500.

∴ The correct answer is ₹4,500.

Alternate Method

Given:

Ratio P : Q = 17 : 5

Ratio Q : R = 6 : 12 = 1 : 2

Total Profit = ₹28,800

Calculations:

⇒ To equalize Q, multiply (P : Q) by 6 and (Q : R) by 5.

⇒ P : Q = (17 × 6) : (5 × 6) = 102 : 30

⇒ Q : R = (6 × 5) : (12 × 5) = 30 : 60

⇒ Combined Ratio P : Q : R = 102 : 30 : 60

⇒ Simplified Ratio P : Q : R = 51 : 15 : 30 (dividing by 2)

⇒ Sum of ratio parts = 51 + 15 + 30 = 96

⇒ Share of Q = (15 / 96) × 28,800

⇒ Share of Q = 15 × 300

⇒ Share of Q = ₹4,500

∴ The correct answer is ₹4,500.

Additional Information

Compound Ratio

When two or more ratios are multiplied together, the result is the compound ratio. For a : b and c : d, the compound ratio is (ac) : (bd).

Profit Distribution in Partnership

Profit is distributed in the ratio of (Capital × Time). If time is constant, profit is shared in the ratio of investment.

Balancing Ratios

To combine ratios like A : B and B : C, find the LCM of the common term B to create a unified ratio A : B : C.

Shortcut Trick

Discount = 75% ⇒ Selling Price (SP) = 100% − 75% = 25% of Marked Price (MP).

Given 25% of MP = 69,356.

Ratio of MP : SP = 100 : 25 = 4 :

1. 1 unit = 69,356 ⇒ 4 units = 69,356 × 4 = 2,77,424.

∴ The correct answer is ₹2,77,424.

Alternate Method

Given: Selling Price (SP) = ₹69,356, Discount % = 75%.

Formula: SP = MP × (100 − Discount %) ÷ 100

⇒ Let Marked Price be x.

⇒ 69,356 = x × (100 − 75) ÷ 100

⇒ 69,356 = x × 25 ÷ 100

⇒ 69,356 = x × 1 ÷ 4

⇒ x = 69,356 × 4 = 2,77,424

∴ The correct answer is ₹2,77,424.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP). SP = MP − Discount.

Successive Discount

If two discounts a% and b% are given, total discount = a + b − (a × b ÷ 100).

Shortcut Trick

Total ratio parts = 8 + 25 + 15 = 48 units.

Difference between Suman and Neha = 15 − 8 = 7 units.

Using ratio-unitary method: 48 units = ₹44,544.

Value of 1 unit = 44,544 ÷ 48 = ₹928.

Required difference = 7 × 928 = ₹6,496.

∴ The correct answer is ₹6,496.

Alternate Method

Given → Investment Ratio (S : T : N) = 8 : 25 : 15, Total Profit = ₹44,544.

Formula → Profit Share ∝ Investment × Time. Since time is constant, Profit Ratio = Investment Ratio.

⇒ Total Profit Share = 8x + 25x + 15x = 48x.

⇒ 48x = 44,544.

⇒ x = 44,544 ÷ 48 = 928.

⇒ Difference (Neha − Suman) = 15x − 8x = 7x.

⇒ Difference = 7 × 928 = 6,496.

∴ The correct answer is ₹6,496.

Additional Information

Partnership Basics

When multiple partners invest for the same time period, the profit is distributed solely based on the ratio of their investments.

Ratio Distribution

To find an individual share, use the formula: Individual Share = (Individual Ratio / Sum of Ratios) × Total Profit.

Shortcut Trick

Average rate for the whole week = 2200 ÷ 120 = 55/3 per km.

Using Alligation: (Taxi rate: 20) and (Car rate: 10) with mean (55/3).

Distance Ratio (Taxi : Car) = (55/3 − 10) : (20 − 55/3) = 25/3 : 5/3 = 5 :

1. Total units = 5 + 1 = 6 units → 120 km ⇒ 1 unit = 20 km.

Taxi distance = 5 × 20 = 100 km.

∴ The correct answer is 100 km.

Alternate Method

Given: Rate for taxi = ₹20/km, Rate for car = ₹10/km, Total distance = 120 km, Total claim = ₹2,200.

Let the distance traveled by taxi be x km.

Then, the distance traveled by own car = (120 − x) km.

⇒ Total Claim = (Rate × Distance)_Taxi + (Rate × Distance)_Car

⇒ 2200 = 20 × x + 10 × (120 − x)

⇒ 2200 = 20x + 1200 − 10x

⇒ 2200 − 1200 = 10x

⇒ 1000 = 10x

⇒ x = 100 km.

∴ The correct answer is 100 km.

Additional Information

Weighted Average Formula

The average value of a mixture is given by A = (n₁v₁ + n₂v₂) ÷ (n₁ + n₂), where n represents quantity and v represents the rate.

Rule of Alligation

It is a shorter way to find the ratio in which two ingredients at given prices must be mixed to produce a mixture at a given price: (Quantity of Cheaper) ÷ (Quantity of Dearer) = (CP of Dearer − Mean Price) ÷ (Mean Price − CP of Cheaper).

Linear Equations in One Variable

Problems involving two unknown quantities related by a total sum can always be simplified to a single variable equation by substituting (Total − x) for the second quantity.

Shortcut Trick

Check the odd numbers between 376 and 384: 377, 379, 381, 383.

377 is 13 × 29. 381 is 3 × 127.

379 and 383 are prime numbers.

∴ The correct answer is 2.

Alternate Method

Given: Range between 376 and 384.

Formula Used: A prime number has only two factors: 1 and itself. Use divisibility rules for 2, 3, 5, 7, 11, 13, 17, 19.

⇒ List numbers: 377, 378, 379, 380, 381, 382, 383.

⇒ Eliminate even numbers: 378, 380, 382.

⇒ Check 377: 377 ÷ 13 = 29 (Not prime).

⇒ Check 379: Not divisible by 2, 3, 5, 7, 11, 13, 17, 19 (Is prime).

⇒ Check 381: Sum of digits (3+8+1) = 12, divisible by 3 (Not prime).

⇒ Check 383: Not divisible by 2, 3, 5, 7, 11, 13, 17, 19 (Is prime).

∴ The correct answer is 2.

Additional Information

Prime Testing Rule

To check if N is prime, test divisibility by all primes less than or equal to √N. For 384, √384 ≈ 19.6.

Co-prime Numbers

Two numbers are co-prime if their Highest Common Factor (HCF) is 1, regardless of whether the numbers themselves are prime.

Divisibility by 3

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

Shortcut Trick

LCM × HCF = Product of two numbers

LCM × 3 = 324

LCM = 324 ÷ 3 = 108

∴ The correct answer is 108.

Alternate Method

Given:

Product of two numbers = 324

HCF = 3

Formula Used:

LCM × HCF = n₁ × n₂ (where n₁ and n₂ are the two numbers)

Calculations:

⇒ LCM × 3 = 324

⇒ LCM = 324 ÷ 3

⇒ LCM = 108

∴ The correct answer is 108.

Additional Information

HCF (Highest Common Factor)

The greatest number that divides two or more numbers without leaving a remainder.

LCM (Least Common Multiple)

The smallest non-zero common multiple of two or more numbers.

Co-prime Numbers

Two numbers are co-prime if their HCF is 1. For co-prime numbers, LCM = Product of the numbers.

Shortcut Trick

Let the total work be LCM of 3, 6, and 6 = 6 units.

Combined efficiency of A, B, and C = (6 ÷ 3) + (6 ÷ 6) + (6 ÷ 6) = 2 + 1 + 1 = 4 units/day.

Twice the work = 2 × 6 = 12 units.

Required time = 12 ÷ 4 = 3 days.

∴ The correct answer is 3 days.

Alternate Method

Given:

Time taken by A = 3 days

Time taken by B = 6 days

Time taken by C = 6 days

Formula Used:

Efficiency = Total Work ÷ Time Taken

Time Taken = Total Work ÷ Efficiency

Calculations:

⇒ Total efficiency of A + B + C = 2 + 1 + 1 = 4 units/day

⇒ New Total Work = 2 × 6 = 12 units

⇒ Time taken to complete twice the work = 12 ÷ 4 = 3 days

∴ The correct answer is 3 days.

Additional Information

LCM Method in Time and Work

Using the LCM of individual times as the total work helps avoid complex fraction calculations and makes the determination of efficiencies much faster.

Work and Wages Relation

Wages are always distributed in proportion to the total work done by each individual, which is also the ratio of their efficiencies if they work for the same amount of time.

Shortcut Trick

Remaining work in man-days = 24 men × (12 − 5 days) = 168 man-days.

Remaining men = 24 − 6 = 18 men.

Extra days needed = 168 ÷ 18 = 28/3 = 9(1/3) days.

∴ The correct answer is 9(1/3) days.

Alternate Method

Given: Total men = 24, Total days = 12, Work done for 5 days, Men left = 6.

Formula Used: Total Work = Men × Days

⇒ Total work = 24 × 12 = 288 units

⇒ Work completed in 5 days = 24 × 5 = 120 units

⇒ Remaining work = 288 − 120 = 168 units

⇒ Remaining men = 24 − 6 = 18 men

⇒ Days required by remaining men = 168 ÷ 18 = 28/3 days

⇒ Days = 9(1/3) days

∴ The correct answer is 9(1/3) days.

Additional Information

Chain Rule Formula

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

Work and Efficiency

Efficiency is inversely proportional to the time taken to complete the work when the total work is constant.

Man-Days Concept

Total work is often expressed as the product of the number of people and the time they work, known as man-hours or man-days.