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Shortcut Trick

The equation 4sinθ + 1/sinθ = 4 is in the form x + 1/x = 2 where x = 2sinθ.

This standard form implies x = 1, which means 2sinθ = 1 or sinθ = 1/2.

For the range 0° < θ < 90°, the value of θ is 30°.

Substitute θ = 30° in the required expression: sin(3 × 30°) + cos(3 × 30°).

sin 90° + cos 90° = 1 + 0 = 1.

∴ The correct answer is 1.

Alternate Method

Given: 4sinθ + cosecθ = 4, where 0° < θ < 90°

Formula Used: cosecθ = 1/sinθ and (a − b)² = a² − 2ab + b²

Calculations:

⇒ 4sinθ + 1/sinθ = 4

⇒ (4sin²θ + 1) / sinθ = 4

⇒ 4sin²θ + 1 = 4sinθ

⇒ 4sin²θ − 4sinθ + 1 = 0

⇒ (2sinθ − 1)² = 0

⇒ 2sinθ − 1 = 0

⇒ sinθ = 1/2

⇒ Since sin 30° = 1/2 and 0° < θ < 90°, then θ = 30°.

⇒ Required value = sin 3θ + cos 3θ

⇒ sin(3 × 30°) + cos(3 × 30°)

⇒ sin 90° + cos 90°

⇒ 1 + 0 = 1

∴ The correct answer is 1.

Additional Information

Reciprocal Identities

The trigonometric functions sinθ and cosecθ are reciprocals, meaning sinθ × cosecθ = 1.

Standard Trigonometric Values

Key values include sin 30° = 1/2, sin 90° = 1, and cos 90° = 0.

Complementary Angles

For any angle A, sin A = cos(90° − A) and cos A = sin(90° − A).

Shortcut Trick

To find when the bells ring together again, calculate the Least Common Multiple (LCM) of the two time intervals.

Find the common factor: 1764 = 252 × 7 and 2268 = 252 × 9.

LCM = 252 × 7 × 9 = 252 × 63 = 15876 seconds.

Convert 15876 seconds: 15876 ÷ 3600 = 4 hours, remainder 1476; 1476 ÷ 60 = 24 minutes, remainder 36 seconds.

∴ The correct answer is 4 hr 24 min 36 sec.

Alternate Method

Given:

Interval of Bell 1 = 1,764 seconds

Interval of Bell 2 = 2,268 seconds

Formula Used:

Time to ring together again = LCM (Interval 1, Interval 2)

Calculations:

⇒ Prime Factorization of 1764 = 2² × 3² × 7²

⇒ Prime Factorization of 2268 = 2² × 3⁴ × 7¹

⇒ LCM = 2² × 3⁴ × 7²

⇒ LCM = 4 × 81 × 49 = 15876 seconds

⇒ To convert seconds to hours: 15876 ÷ 3600 = 4 hours (14400 seconds)

⇒ Remaining seconds = 15876 − 14400 = 1476 seconds

⇒ To convert remaining seconds to minutes: 1476 ÷ 60 = 24 minutes (14400 seconds)

⇒ Remaining seconds = 1476 − 1440 = 36 seconds

⇒ Total Time = 4 hr 24 min 36 sec

∴ The correct answer is 4 hr 24 min 36 sec.

Additional Information

LCM in Periodic Events

For problems involving bells, traffic lights, or runners on a circular track starting together, the next meeting time is always the LCM of their individual time periods.

Time Conversion Rules

1 hour = 60 minutes = 3600 seconds. To convert seconds to larger units, divide by 3600 for hours and divide the remainder by 60 for minutes.

HCF and LCM Relation

The product of two numbers is equal to the product of their HCF and LCM: Number₁ × Number₂ = HCF × LCM.

Shortcut Trick

If the ratio parts are pairwise coprime (meaning no two parts share a common factor), the LCM is simply the HCF × Product of Ratio Parts.

Ratio parts 4, 13, and 9 are pairwise coprime.

⇒ 6084 = HCF × (4 × 13 × 9)

⇒ 6084 = HCF × 468

⇒ HCF = 6084 ÷ 468 = 13

∴ The correct answer is 13.

Alternate Method

Given: Ratio of numbers = 4 : 13 : 9 and Least Common Multiple (LCM) = 6084.

Let the Highest Common Factor (HCF) be x.

Therefore, the three numbers are 4x, 13x, and 9x.

To find the LCM of these numbers, we look at the common and non-common factors:

⇒ LCM(4x, 13x, 9x) = x × LCM(4, 13, 9)

Since 4, 13, and 9 have no common factors between them (pairwise coprime):

⇒ LCM = x × (4 × 13 × 9) = 468x

According to the question:

⇒ 468x = 6084

⇒ x = 6084 ÷ 468

⇒ x = 13

∴ The correct answer is 13.

Additional Information

HCF and LCM Relation

For two numbers a and b, HCF(a, b) × LCM(a, b) = a × b. This formula is only strictly valid for exactly two numbers.

Co-prime Numbers

Two numbers are co-prime if their HCF is 1. The LCM of a set of co-prime numbers is the product of those numbers.

LCM of Fractions

LCM of fractions = (LCM of Numerators) ÷ (HCF of Denominators). Similarly, HCF = (HCF of Numerators) ÷ (LCM of Denominators).

Shortcut Trick

Work done by Pipes X, Y, and Z in 3 hr = 3/8 part.

Remaining work = 1 − 3/8 = 5/8 part.

X and Y fill 5/8 part in 10 hr ⇒ full tank in 10 × 8/5 = 16 hr.

Efficiency of pipe Z = 1/8 − 1/16 = 1/16.

Time taken by pipe Z alone = 16 hr.

∴ The correct answer is 16 hr.

Alternate Method

Given:

Time taken by (X+Y+Z) = 8 hr

Work done by (X+Y+Z) in 3 hr = 3/8

Remaining work = 1 − 3/8 = 5/8

Time taken by (X+Y) to fill 5/8 part = 10 hr

⇒ Time for (X+Y) to fill full tank = 10 × 8/5 = 16 hr

Calculations:

⇒ Efficiency of pipe Z = Efficiency of (X+Y+Z) − Efficiency of (X+Y)

⇒ Efficiency of pipe Z = 2 − 1 = 1

⇒ Time taken by pipe Z alone = Total Work ÷ Efficiency of Z

⇒ Time = 16 ÷ 1 = 16 hr

∴ The correct answer is 16 hr.

Additional Information

Work, Time, and Efficiency Relationship

The total work is the product of efficiency and time (Work = Efficiency × Time). Efficiency is the amount of work done in a unit of time.

Combined Rate Rule

When multiple pipes work together, their combined efficiency is the sum of their individual efficiencies: E_total = E₁ + E₂ + E₃.

Inlet vs. Outlet Pipes

Inlet pipes (filling) have positive efficiency, while outlet pipes (emptying/leakage) have negative efficiency in calculation.

Shortcut Trick

Perimeter of 100 m ⇒ Side = 100 ÷ 4 = 25 m.

Given one diagonal is 14 m ⇒ half-diagonal = 14 ÷ 2 = 7 m.

Using the Pythagorean triplet (7, 24, 25), the other half-diagonal is 24 m.

Length of the other diagonal = 24 × 2 = 48 m.

∴ The correct answer is 48 m.

Alternate Method

Given: Perimeter of the rhombus = 100 m, Diagonal 1 (d₁) = 14 m.

Formula Used: Side (s) = Perimeter ÷ 4; (d₁/2)² + (d₂/2)² = s²

⇒ Side of the rhombus (s) = 100 ÷ 4 = 25 m

⇒ Half of first diagonal = 14 ÷ 2 = 7 m

⇒ Let the other diagonal be d₂.

⇒ (d₂/2)² = 25² − 7²

⇒ (d₂/2)² = 625 − 49

⇒ (d₂/2)² = 576

⇒ d₂/2 = √576

⇒ d₂/2 = 24 m

⇒ d₂ = 24 × 2 = 48 m

∴ The correct answer is 48 m.

Additional Information

Area of a Rhombus

The area of a rhombus can be calculated using its diagonals: Area = 1/2 × d₁ × d₂.

Side-Diagonal Relationship

The relationship between the side (s) and diagonals is 4s² = d₁² + d₂², which is derived from the Pythagoras theorem.

Diagonals Property

In a rhombus, all sides are equal, and the diagonals bisect each other at right angles (90°).

Shortcut Trick

Since Tap 2 (fills in 37 hr) and Tap 3 (empties in 37 hr) have the same rate, they cancel each other out completely.

The net filling depends only on Tap 1, which fills the full cistern in 2 hr.

To fill half of the cistern, the time taken is simply half of the total time for Tap 1.

Required Time = 2 hr ÷ 2 = 1 hr.

∴ The correct answer is 1 hr.

Alternate Method

Given:

Time to fill by Tap A = 2 hr

Time to fill by Tap B = 37 hr

Time to empty by Tap C = 37 hr

Efficiency Table:

Calculations:

⇒ Net Efficiency (A + B + C) = 37 + 2 − 2 = 37 units/hr

⇒ Total capacity of cistern = 74 units

⇒ Work required to fill half the cistern = 74 ÷ 2 = 37 units

⇒ Time taken = Work ÷ Efficiency = 37 ÷ 37 = 1 hr

∴ The correct answer is 1 hr.

Additional Information

Inlet and Outlet Pipes

Pipes that fill the tank are called Inlets (positive efficiency), while those that empty it are Outlets (negative efficiency).

LCM Method

Assuming Total Work as the Least Common Multiple (LCM) of all given time periods simplifies efficiency calculations significantly.

Combined Rate Formula

If Tap 1 fills in 'x' and Tap 2 empties in 'x', the net rate is zero, meaning they don't change the tank level when working together.

Shortcut Trick

Total cost for 12 months = 150 × 12 = ₹1800.

Savings = Discount % of Total Cost.

Savings = 15% of 1800 = 15 × 18 = ₹270.

∴ The correct answer is ₹270.

Alternate Method

Given:

Monthly charge = ₹150

Duration = 12 months

Discount rate = 15%

Calculation:

⇒ Total cost without discount = 150 × 12

⇒ Total cost = ₹1800

⇒ Total savings = 15% of 1800

⇒ Savings = (15 ÷ 100) × 1800

⇒ Savings = 15 × 18 = ₹270

∴ The correct answer is ₹270.

Additional Information

Percentage Discount Formula

Discount Amount = (Discount % ÷ 100) × Marked Price. It represents the reduction in the original price.

Net Payable Amount

The Final Price (Selling Price) = Original Price − Discount Amount, or Final Price = Original Price × (100 − Discount%) ÷ 100.

Successive Discounts

When two discounts a% and b% are applied sequentially, the equivalent single discount is given by (a + b − ab/100)%.

Shortcut Trick

Identify factors of the product 170 that add up to the sum 27.

170 = 17 × 10

Checking the sum: 17 + 10 = 27 (Satisfied)

Required positive difference: 17 − 10 = 7

∴ The correct answer is 7.

Alternate Method

Let the two positive numbers be x and y.

Given: x + y = 27 and xy = 170

Formula Used: (x − y)² = (x + y)² − 4xy

⇒ (x − y)² = (27)² − 4 × 170

⇒ (x − y)² = 729 − 680

⇒ (x − y)² = 49

⇒ x − y = √49

⇒ x − y = 7

∴ The correct answer is 7.

Additional Information

Algebraic Identity Relation

The relation between sum, difference, and product of two numbers is given by: (a + b)² − (a − b)² = 4ab.

Quadratic Equation Formation

If the sum (S) and product (P) of two numbers are known, the numbers are the roots of the quadratic equation: t² − St + P = 0.

Sum of Squares

The sum of the squares of the numbers can be calculated as: x² + y² = (x + y)² − 2xy. In this case, 27² − 2(170) = 729 − 340 = 389.

Shortcut Trick

Apply the successive percentage change formula: Net Change = x + y + (xy ÷ 100)

For depreciation, take x = −10 and y = −10.

Net change = −10 − 10 + [(−10 × −10) ÷ 100]

Net change = −20 + 1 = −19%

The negative sign indicates a total depreciation of 19%.

∴ The correct answer is 19%.

Alternate Method

Given:

Initial rate of depreciation = 10% per year

Time period = 2 years

Formula Used:

Value after n years = P × (1 − r ÷ 100)ⁿ, where P is the initial value and r is the rate.

⇒ Let the initial value of the machine be 100 units.

⇒ Value after 1^st year = 100 − (10% of 100) = 100 − 10 = 90 units.

⇒ Value after 2^nd year = 90 − (10% of 90) = 90 − 9 = 81 units.

⇒ Total depreciation = Initial Value − Final Value = 100 − 81 = 19 units.

⇒ Total percentage depreciation = (Total Depreciation ÷ Initial Value) × 100

⇒ Total percentage depreciation = (19 ÷ 100) × 100 = 19%.

∴ The correct answer is 19%.

Additional Information

Successive Percentage Change

When a value changes by a% and then by b% successively, the net percentage change is given by [a + b + (ab ÷ 100)]%.

Compound Depreciation

It is similar to compound interest but with a negative rate. The value after n years is V = P(1 − r/100)ⁿ.

Appreciation

If the value of an asset increases over time, it is called appreciation, calculated as V = P(1 + r/100)ⁿ.

Shortcut Trick

For a fixed perimeter, the area of a triangle is maximized when it is an equilateral triangle.

Side of the equilateral triangle = Perimeter ÷ 3 = 18 ÷ 3 = 6 cm.

Maximum Area = (√3 ÷ 4) × side²

Maximum Area = (√3 ÷ 4) × 6² = (√3 ÷ 4) × 36 = 9√3 cm².

∴ The correct answer is 9√3 cm².

Alternate Method

Given: Perimeter (a + b + c) = 18 cm

Formula Used: Heron's Formula Area = √[s(s − a)(s − b)(s − c)], where s = (a + b + c) ÷ 2

⇒ Semi-perimeter (s) = 18 ÷ 2 = 9 cm

⇒ For maximum area, a = b = c = 18 ÷ 3 = 6 cm

⇒ Area = √[9 × (9 − 6) × (9 − 6) × (9 − 6)]

⇒ Area = √[9 × 3 × 3 × 3]

⇒ Area = √[243] = √[81 × 3] = 9√3 cm²

∴ The correct answer is 9√3 cm².

Additional Information

Equilateral Triangle Properties

All sides and angles (60°) are equal. Height h = (√3 ÷ 2) × side and Area = (√3 ÷ 4) × side².

In-radius and Circum-radius

For an equilateral triangle with side 'a', in-radius (r) = a ÷ (2√3) and circum-radius (R) = a ÷ √3.

Maximum Area of Polygons

Among all polygons with 'n' sides and a fixed perimeter, the regular polygon (like a square for n=4) always has the maximum area.