Search Previous Year Questions

Given:

((23/8))¹² × ((8/23))¹⁵ × ((23/8))⁹ = ((8/23))^{8m + 8}

Formula used:

(a/b)ⁿ = (b/a)⁻ⁿ

a^x × a^y × a^z = a^{(x + y + z)}

If a^p = a^q, then p = q

Calculations:

The given equation is:

((23/8))¹² × ((8/23))¹⁵ × ((23/8))⁹ = ((8/23))^{8m + 8}

Convert all terms to a common base (8/23):

⇒ ((8/23))⁻¹² × ((8/23))¹⁵ × ((8/23))⁻⁹ = ((8/23))^{8m + 8}

Combine the exponents on the left side:

⇒ ((8/23))^{(−12 + 15 − 9)} = ((8/23))^{8m + 8}

⇒ ((8/23))⁻⁶ = ((8/23))^{8m + 8}

Compare the exponents of the same base:

⇒ −6 = 8m + 8

⇒ 8m = −6 − 8

⇒ 8m = −14

⇒ m = −14/8

⇒ m = −7/4

∴ The correct answer is option (4).

Shortcut Trick

Total Work (LCM of 12 and 33) = 132 units.

Efficiency of A = 11, B = 4. Combined = 15 units/day.

Work done by A and B in 4 days = 15 × 4 = 60 units.

Remaining Work = 132 − 60 = 72 units.

Time taken by B to complete remaining work = 72 ÷ 4 = 18 days.

∴ The correct answer is 18 days.

Alternate Method

Given:

Time taken by A = 12 days

Time taken by B = 33 days

Time for which both worked = 4 days

Formula Used:

Total Work = LCM of time taken | Efficiency = Work ÷ Time

Calculations:

⇒ Total Work = LCM(12, 33) = 132 units.

⇒ Efficiency of A = 132 ÷ 12 = 11 units/day.

⇒ Efficiency of B = 132 ÷ 33 = 4 units/day.

⇒ Total work done by (A + B) in 4 days = (11 + 4) × 4 = 60 units.

⇒ Remaining work = 132 − 60 = 72 units.

⇒ Time taken by B for remaining work = 72 ÷ 4 = 18 days.

∴ The correct answer is 18 days.

Additional Information

Work and Efficiency Relationship

The Efficiency of a person is inversely proportional to the Time taken to complete the work. Efficiency = Work ÷ Time.

Total Work by LCM Method

For multiple workers, taking the LCM of their individual time periods as the Total Work simplifies fractional calculations into whole numbers.

Combined Work Equation

If multiple people work together, their Total Efficiency is the sum of their individual efficiencies: E_total = E₁ + E₂ + ... + E_n.

Shortcut Trick

Use the effective price factor method: Profit Factor = (1 + Markup%) × (1 − Discount%) × (True Weight ÷ Faulty Weight).

Net Selling Price Factor = 1.25 × 0.80 × (1000 ÷ 800) = 1 × 1.25 = 1.25.

Effective Selling Price per kg = 1.25 × ₹100 = ₹125.

Profit per kg = ₹125 − ₹100 = ₹25.

∴ The correct answer is ₹25.

Alternate Method

Given:

Cost Price (CP) of 1 kg = ₹100

Markup = 25% | Discount = 20%

Actual Weight sold as 1 kg = 800 g

Formula Used:

Marked Price (MP) = CP × (100 + Markup%) ÷ 100

Selling Price (SP) = MP × (100 − Discount%) ÷ 100

Profit = Final Revenue − Initial Investment

Calculations:

⇒ Marked Price (MP) per kg = 100 + (25% of 100) = ₹125

⇒ Selling Price (SP) for a claimed kg = 125 − (20% of 125) = 125 − 25 = ₹100

⇒ In reality, the Merchant receives ₹100 for selling only 800 g of tea.

⇒ Total Revenue from selling 1000 g = (100 ÷ 800) × 1000 = ₹125

⇒ Profit per kg = Total Revenue − Total Cost Price = 125 − 100 = ₹25

∴ The correct answer is ₹25.

Additional Information

Dishonest Dealer Profit

When a dealer uses a faulty weight, Profit% = [Error ÷ (True Value − Error)] × 100.

Successive Percentage Change

The net change for markup and discount is given by [M − D − (MD ÷ 100)]%.

Relation between CP and MP

MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%). This is used to find markup directly when profit and discount are known.

Shortcut Trick

The mean proportional of two numbers a and b is calculated as √(a × b).

Here, a = 4 and b = 64.

Value of x = √(4 × 64) = √256 = 16.

∴ The correct answer is 16.

Alternate Method

Given:

Length of the rectangular garden = 16 m

Width of the rectangular garden = x m

Numbers for mean proportional = 4 m and 64 m

Formula Used:

If x is the mean proportional between a and b, then a : x :: x : b

⇒ a/x = x/b

⇒ x² = a × b

Calculations:

⇒ x² = 4 × 64

⇒ x² = 256

⇒ x = √256

⇒ x = 16

∴ The correct answer is 16.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is the square root of their product: x = √(a × b).

Third Proportional

If a, b, and c are in continued proportion (a : b :: b : c), then c is the third proportional: c = b² ÷ a.

Fourth Proportional

If four numbers a, b, c, and d are in proportion (a : b :: c : d), then d is the fourth proportional: d = (b × c) ÷ a.

Shortcut Trick

Ratio of products = (p₁ × p₂ × p₃) ÷ (p₂ × p₃ × p₄) = 105 ÷ 385.

Common factors p₂ and p₃ cancel out, leaving p₁ ÷ p₄ = 105 ÷ 385.

Simplifying the fraction: 105 ÷ 385 = 21 ÷ 77 = 3 ÷ 11.

Since the numbers are prime numbers listed in order, we have p₁ = 3 and p₄ = 11.

The largest prime number is p₄ = 11.

∴ The correct answer is 11.

Alternate Method

Given:

Product of the first three prime numbers = 105

Product of the last three prime numbers = 385

Calculations:

⇒ Let the four prime numbers be p₁, p₂, p₃, and p₄.

⇒ Step 1: Factorise 105 into prime numbers ⇒ 105 = 3 × 5 × 7.

⇒ Step 2: Factorise 385 into prime numbers ⇒ 385 = 5 × 7 × 11.

⇒ Step 3: By comparing the two sets of factors, the common primes are 5 and 7.

⇒ The unique prime from the first set is 3, and the unique prime from the last set is 11.

⇒ The sequence of four prime numbers is 3, 5, 7, 11.

⇒ The largest prime number is 11.

∴ The correct answer is 11.

Additional Information

Prime Numbers

A natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself (e.g., 2, 3, 5, 7, 11...).

Fundamental Theorem of Arithmetic

Every composite number can be expressed as a unique product of prime numbers, regardless of the order of factors.

Co-prime Numbers

Two numbers are called co-prime if their Highest Common Factor (HCF) is 1. All distinct prime numbers are co-prime to each other.

Shortcut Trick

The ratio of sugar to flour is 7 : 11.

Given sugar = 154 g ⇒ 7 units = 154 g.

1 unit = 154 ÷ 7 = 22 g.

Flour required = 11 units = 11 × 22 = 242 g.

∴ The correct answer is 242 g.

Alternate Method

Given:

Sugar : Flour = 7 : 11

Amount of sugar = 154 g

Let the amount of sugar be 7x and the amount of flour be 11x.

⇒ 7x = 154

⇒ x = 154 ÷ 7

⇒ x = 22

⇒ Quantity of flour = 11x

⇒ 11 × 22 = 242 g

∴ The correct answer is 242 g.

Additional Information

Direct Proportion

When two quantities increase or decrease in the same ratio, they are in direct proportion. Formula: x₁/y₁ = x₂/y₂.

Property of Ratios

Multiplying or dividing both terms of a ratio by the same non-zero number does not change the ratio (a : b = ka : kb).

Unitary Method

Finding the value of a single unit first and then multiplying it by the required number of units to find the total value.

Shortcut Trick

Divide the whole equation sin θ + cos θ = √2 cos θ by cos θ.

(sin θ ÷ cos θ) + (cos θ ÷ cos θ) = √2 (cos θ ÷ cos θ)

tan θ + 1 = √2

tan θ = √2 − 1

∴ The correct answer is √2 − 1.

Alternate Method

Given: sin θ + cos θ = √2 cos θ

Formula Used: tan θ = sin θ ÷ cos θ

Calculations:

⇒ sin θ + cos θ = √2 cos θ

⇒ sin θ = √2 cos θ − cos θ

⇒ sin θ = cos θ × (√2 − 1)

⇒ sin θ ÷ cos θ = √2 − 1

⇒ tan θ = √2 − 1

∴ The correct answer is √2 − 1.

Additional Information

Trigonometric Ratio Definition

The tan θ ratio represents the slope of the hypotenuse and is calculated as Opposite ÷ Adjacent.

Relationship between Ratios

For any angle θ, tan θ = sin θ ÷ cos θ and cot θ = cos θ ÷ sin θ.

Complementary Angles

In a right triangle, the ratios of complementary angles are related such that tan(90° − θ) = cot θ.

Shortcut Trick

Ratio of Boys : Girls : Teachers = 25 : 55 : (100 − 25 − 55) = 25 : 55 : 20 = 5 : 11 : 4.

Use Weighted Average method: Average = Σ(Ratio × Value) ÷ ΣRatio

Average weight = (5 × 50 + 11 × 55 + 4 × 65) ÷ (5 + 11 + 4) = (250 + 605 + 260) ÷ 20

Average weight = 1115 ÷ 20 = 55.75 kg.

∴ The correct answer is 55.75 kg.

Alternate Method

Given:

Percentage of Boys = 25% | Average weight = 50 kg

Percentage of Girls = 55% | Average weight = 55 kg

Percentage of Teachers = 100% − (25% + 55%) = 20% | Average weight = 65 kg

Calculations:

⇒ Let the total number of people in the group = 100.

⇒ Total weight of Boys = 25 × 50 = 1250 kg.

⇒ Total weight of Girls = 55 × 55 = 3025 kg.

⇒ Total weight of Teachers = 20 × 65 = 1300 kg.

⇒ Sum of all weights = 1250 + 3025 + 1300 = 5575 kg.

⇒ Average weight of the group = Total weight ÷ Total number

⇒ Average weight = 5575 ÷ 100 = 55.75 kg.

∴ The correct answer is 55.75 kg.

Additional Information

Weighted Average Formula

When a group is divided into subgroups, the overall average is given by (n₁w₁ + n₂w₂ + ... + n_kw_k) ÷ (n₁ + n₂ + ... + n_k).

Combined Average Shortcut

Instead of using actual numbers, you can use the simplest ratio of the quantities to calculate the combined average weight without changing the final result.

Deviation Method

Assume a base average (e.g., 55) and calculate net deviation as Σ(Ratio × Difference from base) ÷ Total Ratio to find the adjustment to the assumed average.

Shortcut Trick

Combined efficiency of Pipe A and Pipe B = 1/20 + 1/30 = 5/60 = 1/12 part per min.

Time taken to fill the full tank = 12 min.

Time taken to fill 2/3 of the tank = 12 × 2/3 = 8 min.

∴ The correct answer is 8 min.

Alternate Method

Given:

Time taken by Pipe A = 20 min

Time taken by Pipe B = 30 min

Formula Used:

Efficiency = Total Work ÷ Time

Calculations:

⇒ Total capacity (LCM of 20, 30) = 60 units

⇒ Target volume to fill = 2/3 of total capacity = 2/3 × 60 = 40 units

⇒ Combined efficiency of (A + B) = 3 + 2 = 5 units/min

⇒ Time required to fill 40 units = 40 ÷ 5 = 8 min

∴ The correct answer is 8 min.

Additional Information

Pipes and Cisterns Logic

Filling pipes are considered to have positive efficiency, while emptying pipes (leaks) are considered to have negative efficiency.

Net Efficiency

The net work done per unit time is the algebraic sum of the efficiencies of all pipes operating simultaneously.

Time and Work Relation

Total Work = Efficiency × Time. If work is constant, Time is inversely proportional to Efficiency.

Shortcut Trick

The Area of any Quadrilateral whose diagonals intersect at right angles (90°) is given by the formula:

Area = 1/2 × d₁ × d₂

Given diagonals are 12 m and 16 m.

⇒ Area = 1/2 × 12 × 16 = 6 × 16 = 96 m².

∴ The correct answer is 96 m².

Alternate Method

Given:

Length of first diagonal (d₁) = 12 m

Length of second diagonal (d₂) = 16 m

Angle of intersection = 90°

Formula Used:

Area of an orthodiagonal quadrilateral = (1/2) × Product of diagonals

Calculations:

⇒ Area = 1/2 × d₁ × d₂

⇒ Area = 1/2 × 12 × 16

⇒ Area = 6 × 16

⇒ Area = 96 m²

∴ The correct answer is 96 m².

Additional Information

Orthodiagonal Quadrilateral

Any quadrilateral where diagonals cross at right angles is called orthodiagonal. Common examples include the rhombus, square, and kite.

Area of a Rhombus

Since a rhombus is a special type of orthodiagonal quadrilateral, its area is also 1/2 × d₁ × d₂, where d₁ and d₂ are the perpendicular diagonals.

Sides and Diagonals Relationship

In an orthodiagonal quadrilateral, the sum of the squares of opposite sides is equal: a² + c² = b² + d².

Area of a General Quadrilateral

If the angle θ between diagonals is not 90°, Area = 1/2 × d₁ × d₂ × sin(θ).