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Shortcut Trick

Total Work = LCM(12, 15, 20, 30) = 60 units. Efficiencies: A=5, B=4, C=3, D=2.

Subtract work of person leaving after starting: 60 − (D × 2) = 60 − 4 = 56 units (D is removed).

Add work of persons leaving before completion: 56 + (C × 3) + (B × 2) = 56 + 9 + 8 = 73 units.

Total time = Adjusted Work ÷ (Sum of efficiencies of A, B, and C) = 73 ÷ 12 = 6¹/₁₂ days.

Converting fraction to hours: ¹/₁₂ × 24 = 2 hours.

∴ The correct answer is 6 days 2 hr.

Alternate Method

Given: Time taken by A = 12 days, B = 15 days, C = 20 days, and D = 30 days.

Formula Used: Total Work = Efficiency × Time | Efficiency = Total Work ÷ Time

⇒ Let the total time to complete the work be x days.

⇒ D works for 2 days, C works for (x − 3) days, B works for (x − 2) days, and A works for all x days.

⇒ Total work equation: (5 × x) + 4 × (x − 2) + 3 × (x − 3) + (2 × 2) = 60

⇒ 5x + 4x − 8 + 3x − 9 + 4 = 60

⇒ 12x − 13 = 60

⇒ 12x = 73

⇒ x = 73 ÷ 12 = 6¹/₁₂ days

⇒ 6 days + (¹/₁₂ × 24) hours = 6 days 2 hours.

∴ The correct answer is 6 days 2 hr.

Additional Information

Leaving Before vs. After Completion

If a person leaves n days after starting, subtract their work and exclude them. If they leave n days before completion, add their hypothetical work and keep them in the divisor.

Work and Wages

Wages are always distributed in the ratio of the actual work done by each person, which is (Efficiency × Time Spent).

Pipes and Cisterns Analogy

Similar to Time and Work, but inlet pipes have positive efficiency and outlet/leak pipes have negative efficiency.

Shortcut Trick

Value of one-third of the number = 107 − 87 = 20.

Original Number = 20 × 3 = 60.

Sum of digits of the number = 6 + 0 = 6.

∴ The correct answer is 6.

Alternate Method

Given:

Result after increasing one-third of a number by 87 = 107

Formula Used:

Linear Equation: (1/n) × x + c = R (where x is the original number)

Calculations:

Let the original number be x.

⇒ (1/3) × x + 87 = 107

⇒ x/3 = 107 − 87

⇒ x/3 = 20

⇒ x = 20 × 3

⇒ x = 60

Now, calculate the sum of the digits of the number 60:

⇒ Sum = 6 + 0 = 6

∴ The correct answer is 6.

Additional Information

Linear Equations

A linear equation in one variable is an equation which can be written in the form ax + b = 0, where a and b are constants and x is the variable.

Sum of Digits Property

The sum of the digits is often used in divisibility rules. For example, a number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9 respectively.

Fraction Manipulation

To find the whole from a part, multiply the value of the part by the reciprocal of the fraction (e.g., if 1/n = V, then Total = V × n).

Shortcut Trick

In Compound Interest, if a sum becomes n times in t years, it becomes n^k times in k × t years.

Given: Sum becomes 3 times in 8 years.

Target time = 24 years ⇒ k = 24 ÷ 8 = 3.

New Amount = Principal × 3³ ⇒ 8000 × 27 = ₹216000.

∴ The correct answer is ₹216000.

Alternate Method

Given:

Principal (P) = ₹8000

Time (t) = 8 years, Amount (A) = 3P

Total Time (T) = 24 years

Formula Used:

A = P(1 + r/100)ⁿ

Calculations:

⇒ 3P = P(1 + r/100)⁸

⇒ (1 + r/100)⁸ = 3

⇒ For 24 years, Amount (A) = P(1 + r/100)²⁴

⇒ A = 8000 × [(1 + r/100)⁸]³

⇒ A = 8000 × 3³

⇒ A = 8000 × 27

⇒ A = ₹216000

∴ The correct answer is ₹216000.

Additional Information

CI Multiplication Property

If a sum grows by a factor of m in t years, it will grow by a factor of m^k in k × t years.

Rule of 72

To estimate the time required to double an investment, divide 72 by the annual interest rate r (Time ≈ 72 ÷ r).

Difference between CI and SI for 2 Years

The difference D between CI and SI for 2 years is given by D = P × (r/100)².

Shortcut Trick

Convert fractions to decimals directly: 7/2 = 3.5 and 3/4 = 0.75.

Group numbers for easier addition: (3.5 + 32.5) + (1.245 + 0.75).

36 + 1.995 = 37.995.

∴ The correct answer is 37.995.

Alternate Method

Given Expression: 7/2 + 32.5 + 1.245 + 3/4

Method: Convert all terms into decimal form and perform vertical addition.

⇒ Convert 7/2: 7 ÷ 2 = 3.5

⇒ Convert 3/4: 3 ÷ 2² = 0.75

⇒ Substitute the values: 3.5 + 32.5 + 1.245 + 0.75

⇒ Aligning by decimal point:

   03.500

  32.500

   01.245

• 00.750

−−−−−−−

  37.995

∴ The correct answer is 37.995.

Additional Information

Decimal Addition Rule

Always align the decimal points vertically. If numbers have different lengths, add trailing zeros (e.g., 3.5 becomes 3.500) to avoid errors in place value.

Common Fraction to Decimal Conversions

Memorize standard conversions for speed: 1/2 = 0.5, 1/4 = 0.25, 3/4 = 0.75, 1/5 = 0.2, and 1/8 = 0.125.

Simplification Priority (BODMAS)

In expressions with mixed operators, solve fractions (division) and multiplication before final addition or subtraction.

Shortcut Trick

Total work = LCM of 5 and 50 = 50 units.

Efficiency of A = 50 ÷ 5 = 10 units/day; Efficiency of B = 50 ÷ 50 = 1 unit/day.

Work done by B in 10 days = 10 × 1 = 10 units.

Remaining work = 50 − 10 = 40 units.

Days taken by A = 40 ÷ 10 = 4 days.

∴ The correct answer is 4 days.

Alternate Method

Given:

Time taken by A = 5 days

Time taken by B = 50 days

Duration B worked alone = 10 days

Formula Used:

Work = Efficiency × Time

Efficiency = Total Work ÷ Time

Calculations:

⇒ Total work (LCM of 5, 50) = 50 units

⇒ Work completed by B in 10 days = Efficiency of B × Time

⇒ Work done by B = 1 × 10 = 10 units

⇒ Remaining work = Total Work − Work done by B

⇒ Remaining work = 50 − 10 = 40 units

⇒ Days taken by A alone = Remaining work ÷ Efficiency of A

⇒ Days taken by A = 40 ÷ 10 = 4 days

∴ The correct answer is 4 days.

Additional Information

Efficiency-Time Relationship

The efficiency of a person is inversely proportional to the time taken to complete a work: Efficiency ∝ 1/Time.

Combined Work Rule

If two people work together, their combined efficiency is the sum of their individual efficiencies: Eff_Total = Eff_A + Eff_B.

LCM Method for Total Work

Always assume the total work as the Least Common Multiple (LCM) of the number of days given to simplify calculations into whole numbers.

Shortcut Trick

Total work remains constant: M₁ × D₁ = M₂ × D₂

Product of workers and days in first case = Product in second case.

15 × 10 = M₂ × 6

150 = 6 × M₂ ⇒ M₂ = 25

∴ The correct answer is 25 workers.

Alternate Method

Given:

Initial workers (M₁) = 15

Initial days (D₁) = 10

Required days (D₂) = 6

Formula Used:

M₁ × D₁ = M₂ × D₂

Calculations:

⇒ 15 × 10 = M₂ × 6

⇒ 150 = 6 × M₂

⇒ M₂ = 150 ÷ 6

⇒ M₂ = 25

∴ The correct answer is 25 workers.

Additional Information

Chain Rule Formula

The general formula for work problems is (M₁ × D₁ × H₁) ÷ W₁ = (M₂ × D₂ × H₂) ÷ W₂, where H is hours and W is work.

Inverse Proportion

For a fixed amount of work, the number of workers is inversely proportional to the time taken. If time decreases, workers must increase.

Work and Efficiency

Total Work = Total Efficiency × Time. If efficiency of each worker is 1, then Total Work = Number of Workers × Time.

Shortcut Trick

Identify the standard algebraic identity: (az − by)² + (bz + ay)² = (a² + b²)(z² + y²).

Substitute a = 5 and b = 12: (5² + 12²)(z² + y²) = (25 + 144)(z² + y²) = 169z² + 169y².

Subtract the remaining term: 169z² + 169y² − 144z² = 25z² + 169y².

∴ The correct answer is 25z² + 169y².

Alternate Method

Given Expression: (5z − 12y)² + (12z + 5y)² − 144z²

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

⇒ [(5z)² + (12y)² − 2(5z)(12y)] + [(12z)² + (5y)² + 2(12z)(5y)] − 144z²

⇒ [25z² + 144y² − 120zy] + [144z² + 25y² + 120zy] − 144z²

⇒ 25z² + 144z² − 144z² + 144y² + 25y² − 120zy + 120zy

⇒ 25z² + 169y² + 0

⇒ 25z² + 169y²

∴ The correct answer is 25z² + 169y².

Additional Information

Algebraic Identities

The square of a binomial sum is (a + b)² = a² + 2ab + b², while the square of a binomial difference is (a − b)² = a² − 2ab + b².

Brahmagupta-Fibonacci Identity

The product of two sums of squares is itself a sum of squares: (a² + b²)(c² + d²) = (ac − bd)² + (ad + bc)².

Pythagorean Triplet

In this problem, the coefficients 5, 12, and 13 form a Pythagorean triplet where 5² + 12² = 13² = 169.

Shortcut Trick

Net percentage increase = x + y + (xy/100)

Successive increase of 25% and 20% = 25 + 20 + (25 × 20 / 100) = 50%

New number of girls = 36 + 50% of 36 = 36 + 18 = 54

Difference = 54 − 46 = 8

∴ The correct answer is 8.

Alternate Method

Given:

Initial number of boys = 46

Initial number of girls = 36

Calculations:

⇒ Number of girls after 25% increase = 36 × (1 + 25/100) = 36 × 1.25 = 45

⇒ Number of girls after subsequent 20% increase = 45 × (1 + 20/100) = 45 × 1.2 = 54

⇒ Difference between final girls and boys = 54 − 46 = 8

∴ The correct answer is 8.

Additional Information

Successive Percentage Change

When a value is changed by a% and then by b%, the effective change is (a + b + ab/100)%. This is also known as the compounding effect.

Fractional Equivalence

To solve quickly, use fractions: a 25% increase is multiplying by 5/4, and a 20% increase is multiplying by 6/5. Total multiplier = 5/4 × 6/5 = 3/2 (or 1.5).

Percentage Difference

Percentage difference = (Difference / Reference Value) × 100. Always identify the base (denominator) correctly based on the phrasing of the question.

Shortcut Trick

Substitute a standard angle θ = 45° to quickly evaluate the expression.

At θ = 45°: tan 45° = 1, cot 45° = 1, sec 45° = √2, and cosec 45° = √2.

Expression = 4 + (1)2 + (1)2 − (√2)2 x (√2)2

Expression = 4 + 1 + 1 − 2 x 2 = 6 − 4 = 2

∴ The correct answer is 2.

Alternate Method

Given: The trigonometric expression is 2 + tan²θ + cot²θ − sec²θ cosec²θ.

Formula Used:

1. 1 + tan²θ = sec²θ ⇒ tan²θ = sec²θ − 1
2. 1 + cot²θ = cosec²θ ⇒ cot²θ = cosec²θ − 1
3. sec²θ + cosec²θ = sec²θ × cosec²θ

Calculations:

⇒ 4 + tan2θ + cot2θ − sec2θ cosec2θ

⇒ 4 + (sec2θ − 1) + (cosec2θ − 1) − sec2θ cosec2θ

⇒ 4 + sec2θ + cosec2θ − 2 − sec2θ cosec2θ

⇒ 2 + sec2θ + cosec2θ − sec2θ cosec2θ

⇒ 2 + (1/cos2θ + 1/sin2θ) − sec2θ cosec2θ

⇒ 2 + [(sin2θ + cos2θ) / (sin2θ cos2θ)] − sec2θ cosec2θ

⇒ 2 + [1 / (sin2θ cos2θ)] − sec2θ cosec2θ

⇒ 2 + sec2θ cosec2θ − sec2θ cosec2θ = 2

∴ The correct answer is 2.

Additional Information

Sum-to-Product Identity

For secant and cosecant, the sum of their squares is equal to the product of their squares: sec²θ + cosec²θ = sec²θ × cosec²θ.

Tangent and Cotangent Relation

The expression tan θ + cot θ can be simplified as 1/(sin θ cos θ), which is also equal to sec θ cosec θ.

Pythagorean Identities

Always remember sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ as they are the building blocks for such simplifications.

Shortcut Trick

Solve the expression by segmenting it into smaller independent parts using VBODMAS.

Outer segments: 8 × 8 + 6 × 6 = 64 + 36 = 100.

Inner parts: 6 × {2 × 6} − {2 + 24} = 72 − 26 = 46.

Combine and divide: 100 + 46 ÷ 2 = 100 + 23 = 123.

∴ The correct answer is 123.

Alternate Method

Given Expression: 8 × 8 + 6 × 6 + [6 × {4 ÷ 2 × (3 of 2)} − {36 ÷ (6 × 3) + (8 × 3)}] ÷ 2

We use the BODMAS rule to simplify the expression step by step.

⇒ 8 × 8 + 6 × 6 + [6 × {4 ÷ 2 × 6} − {36 ÷ 18 + 24}] ÷ 2

⇒ 64 + 36 + [6 × {2 × 6} − {2 + 24}] ÷ 2

⇒ 100 + [6 × 12 − 26] ÷ 2

⇒ 100 + [72 − 26] ÷ 2

⇒ 100 + 46 ÷ 2

⇒ 100 + 23 = 123

∴ The correct answer is 123.

Additional Information

BODMAS Hierarchy

The order of operations is Brackets (), Of (powers/roots), Division (÷), Multiplication (×), Addition (+), and Subtraction (−). If a vinculum (bar) is present, solve it before brackets.

Bracket Sequence

Always solve from the innermost bracket outwards: start with Parentheses ( ), then Braces { }, and finally Square brackets [ ].

Of vs Multiplication

The 'Of' operator mathematically represents multiplication but holds a higher precedence than both Division and standard Multiplication in the sequence.