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Shortcut Trick

Use the standard trigonometric identity: tan θ + cot θ = 2 cosec 2θ.

Given the value of the expression tan θ + cot θ = b.

Equating both, we get b = 2 cosec 2θ.

Solving for the required term: cosec 2θ = b/2.

∴ The correct answer is b/2.

Alternate Method

Given: sec θ + cosec θ = a and tan θ + cot θ = b

Formula Used: tan θ = sin θ / cos θ, cot θ = cos θ / sin θ, and sin 2θ = 2 sin θ cos θ

Calculations:

⇒ tan θ + cot θ = b

⇒ (sin θ / cos θ) + (cos θ / sin θ) = b

⇒ (sin² θ + cos² θ) / (sin θ cos θ) = b

⇒ 1 / (sin θ cos θ) = b [Since sin² θ + cos² θ = 1]

⇒ sin θ cos θ = 1/b

⇒ sin 2θ = 2 sin θ cos θ = 2 × (1/b) = 2/b

⇒ cosec 2θ = 1 / sin 2θ = 1 / (2/b) = b/2

∴ The correct answer is b/2.

Additional Information

Double Angle Identities

The primary double angle formulas are sin 2θ = 2 sin θ cos θ and cos 2θ = cos² θ − sin² θ = 2 cos² θ − 1 = 1 − 2 sin² θ.

Reciprocal Relations

The sum tan θ + cot θ is always equal to sec θ × cosec θ, which further simplifies to 2 cosec 2θ.

Pythagorean Identities

Fundamental identities include sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = cosec² θ.

Shortcut Trick

Since the frequency (20) is identical for all 5 class intervals, the data is perfectly uniform and symmetric.

In a symmetric distribution with equal frequencies, the Median lies exactly at the midpoint of the entire range.

Total Range = 54 to 64.

Median = (54 + 64) ÷ 2 = 59.

∴ The correct answer is ₹59.

Alternate Method

Given: Total workers (N) = 100. Classes and frequencies (f) are provided.

Formula Used: Median = l + [((N ÷ 2) − cf) ÷ f] × h

Cumulative Frequencies (CF):

⇒ 54-56: 20

⇒ 56-58: 20 + 20 = 40

⇒ 58-60: 40 + 20 = 60 (Median class since 60 ≥ 50)

Median Calculation:

⇒ l (Lower limit) = 58

⇒ N/2 = 100 ÷ 2 = 50

⇒ cf (Previous CF) = 40

⇒ f (Frequency of median class) = 20

⇒ h (Class height) = 2

⇒ Median = 58 + [(50 − 40) ÷ 20] × 2

⇒ Median = 58 + [10 ÷ 20] × 2

⇒ Median = 58 + 0.5 × 2 = 58 + 1

⇒ Median = 59

∴ The correct answer is ₹59.

Additional Information

Empirical Relationship

For a moderately skewed distribution: Mode = 3 × Median − 2 × Mean.

Median for Grouped Data

The median is the value that divides the total frequency into two equal halves (N/2).

Class Interval (h)

Class width or height (h) is the difference between the upper and lower limits of a class (e.g., 56 − 54 = 2).

Shortcut Trick

The given expression follows the algebraic identity: (a³ − b³) ÷ (a² + ab + b²) = (a − b).

In this problem, a = 1.4 and b = 1.2.

Applying the trick: 1.4 − 1.2 = 0.2.

∴ The correct answer is 0.2.

Alternate Method

Given:

Numerator = (1.4 × 1.4 × 1.4 − 1.2 × 1.2 × 1.2)

Denominator = (1.4 × 1.4 + 1.4 × 1.2 + 1.2 × 1.2)

Formula Used:

The difference of two cubes formula is: a³ − b³ = (a − b)(a² + ab + b²)

⇒ Let a = 1.4 and b = 1.2.

⇒ Numerator = a³ − b³ = (a − b)(a² + ab + b²)

⇒ Denominator = (a² + ab + b²)

⇒ Expression = [(a − b)(a² + ab + b²)] ÷ (a² + ab + b²)

⇒ After cancelling the common term (a² + ab + b²), we get: a − b

⇒ Value = 1.4 − 1.2

⇒ Value = 0.2

∴ The correct answer is 0.2.

Additional Information

Sum of Cubes Identity

The corresponding identity for addition is a³ + b³ = (a + b)(a² − ab + b²), where the middle term in the second bracket is negative.

Algebraic Simplification

Always check for standard patterns like (a + b)², (a − b)², and a² − b² to avoid lengthy decimal calculations.

Cubing Decimals

Remember that (0.x)³ will have three decimal places. For example, 1.4³ = 2.744 and 1.2³ = 1.728.

Shortcut Trick

Efficiency of Pipe X = +4 and Pipe Y = −3 (Total work = LCM of 15, 20 = 60).

Combined efficiency for first 10 hours = 4 − 3 = 1 unit/hr.

Work completed in 10 hours = 1 × 10 = 10 units.

Remaining work to be filled by Pipe X alone = 60 − 10 = 50 units.

Time taken by Pipe X for remaining part = 50 ÷ 4 = 12.5 hr.

Total time = 10 + 12.5 = 22.5 hr.

∴ The correct answer is 22.5 hr.

Alternate Method

Given: Filling time of Pipe X = 15 hr, Emptying time of Pipe Y = 20 hr, Opening duration of Pipe Y = 10 hr.

Formula: Total Work = Efficiency × Time; Net Efficiency = Efficiency_Fill − Efficiency_Empty

⇒ Let total capacity of the reservoir = LCM(15, 20) = 60 units.

⇒ Efficiency of Pipe X = 60 ÷ 15 = 4 units/hr.

⇒ Efficiency of Pipe Y = 60 ÷ 20 = 3 units/hr (Negative as it empties).

⇒ Work done in 10 hours with both pipes open = (4 − 3) × 10 = 10 units.

⇒ Remaining capacity to be filled = 60 − 10 = 50 units.

⇒ Time taken by Pipe X to fill the remaining 50 units = 50 ÷ 4 = 12.5 hr.

⇒ Total time required = Initial 10 hr + Subsequent 12.5 hr = 22.5 hr.

∴ The correct answer is 22.5 hr.

Additional Information

Pipes and Cisterns Basics

If a pipe fills a tank in 'a' hours and another empties it in 'b' hours, the net work done in 1 hour is (1/a − 1/b).

Negative Efficiency

An outlet pipe or leak is always treated as having negative efficiency in work calculations.

Time and Efficiency Relation

Efficiency is inversely proportional to the time taken to complete a task (Efficiency = Total Work ÷ Time).

Additional Information

` And the three concepts:

Product of Two Numbers Formula

For any two numbers A and B, the product of the numbers is equal to the product of their HCF and LCM: A × B = HCF × LCM.

LCM of Ratio Numbers

If two numbers are in the ratio a : b (where a and b are co-prime), then their HCF = x, the numbers are ax and bx

Shortcut Trick

Mean proportional of two numbers a and b = √(a × b)

Given: a = 0.16 and b = 0.49

Mean proportional = √(0.16 × 0.49) = 0.4 × 0.7 = 0.28

∴ The correct answer is 0.28.

Alternate Method

Given:

First number (a) = 0.16

Second number (b) = 0.49

Formula Used:

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

This implies: x² = a × b or x = √(a × b)

Calculations:

⇒ Let the mean proportional be x.

⇒ 0.16/x = x/0.49

⇒ x² = 0.16 × 0.49

⇒ x = √(0.16 × 0.49)

⇒ x = √((16/100) × (49/100))

⇒ x = (4/10) × (7/10)

⇒ x = 28/100

⇒ x = 0.28

∴ The correct answer is 0.28.

Additional Information

Third Proportional

For two numbers a and b, the third proportional c is such that a:b = b:c, which gives c = b²/a.

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d is such that a:b = c:d, which gives d = (b × c)/a.

Continued Proportion

Three quantities a, b, and c are said to be in continued proportion if a:b = b:c.

Shortcut Trick

Since the entire selling price of the first article is reinvested to buy the second article, the overall percentage change is completely independent of the initial cost price of ₹347.

We can directly apply the successive percentage change formula:

Net Percentage Change = a + b + (a × b)/100

Here, a = −6% (loss) and b = +65% (gain).

Overall Gain = −6 + 65 + (−6 × 65)/100 = 59 − 3.9 = 55.1%.

∴ The correct answer is 55.1%.

Alternate Method

Given:

Initial Cost Price (C₁) = ₹347

Loss percentage on the first article = 6%

Gain percentage on the second article = 65%

Formula Used:

Selling Price (SP) = Cost Price (CP) × (1 − Loss%/100) or

Shortcut Trick

Net efficiency = Efficiency of Tap A + Efficiency of Tap B − Efficiency of Tap C.

Since both Tap B and Tap C take 15 hr, their filling and emptying rates are equal and cancel each other out.

Net efficiency = Efficiency of Tap A = 1/4 of the cistern per hour.

Time required to fill 1/4 of the empty cistern = (1/4) ÷ (1/4) = 1 hour.

∴ The correct answer is 1 hr.

Alternate Method

Given: Tap A fills in 4 hr, Tap B fills in 15 hr, and Tap C empties in 15 hr.

Let the Total Capacity (Total Work) of the cistern be the LCM of 4, 15, and 15.

⇒ Total Capacity = 60 units.

⇒ Net Efficiency of all taps together = Efficiency(A) + Efficiency(B) − Efficiency(C)

⇒ Net Efficiency = 15 + 4 − 4 = 15 units/hr.

⇒ Capacity to be filled = One-fourth of the cistern = 1/4 × 60 = 15 units.

⇒ Time taken = Capacity to be filled ÷ Net Efficiency

⇒ Time = 15 ÷ 15 = 1 hour.

∴ The correct answer is 1 hr.

Additional Information

Pipes and Cisterns Concept

Filling pipes are considered as positive work (+), while emptying or outlet pipes (leaks) are considered as negative work (−).

Net Rate Formula

If Pipe A fills in 'x' hours and Pipe B empties in 'y' hours, the net work done in one hour is (1/x − 1/y).

Cistern Capacity

The capacity of the tank is often assumed as the LCM of the individual times taken by the pipes to simplify calculations.

Shortcut Trick

Net multiplier for successive discounts of 20% and 30% = 0.80 × 0.70 = 0.56

Final Selling Price = Marked Price × 0.56

SP = 5000 × 0.56 = 2800

∴ The correct answer is ₹2,800.

Alternate Method

Given:

Marked Price (MP) of the jacket = ₹5,000

First discount (d₁) = 20%

Second discount (d₂) = 30%

Formula Used:

Final SP = MP × (100 − d₁)/100 × (100 − d₂)/100

Calculations:

⇒ Price after 1^st discount = 5000 − (20% of 5000)

⇒ Price after 1^st discount = 5000 − 1000 = 4000

⇒ Price after 2^nd discount = 4000 − (30% of 4000)

⇒ Price after 2^nd discount = 4000 − 1200 = 2800

⇒ Final Selling Price = ₹2,800

∴ The correct answer is ₹2,800.

Additional Information

Single Equivalent Discount

For two successive discounts x% and y%, the single equivalent discount is given by (x + y − xy/100)%.

Relationship between MP and SP

The Selling Price is always calculated on the Marked Price after deducting the discount: SP = MP × (100 − Discount%)/100.

Order of Discounts

The final Selling Price remains the same regardless of the order in which the successive discounts are applied (e.g., 20% then 30% is the same as 30% then 20%).

Shortcut Trick

A diagonal divides a parallelogram into two congruent triangles.

The sides of the triangle are 9 cm, 12 cm, and 15 cm.

Recognize the Pythagorean triplet: 9² + 12² = 81 + 144 = 225 = 15².

Since it is a right-angled triangle, the area of the parallelogram is simply the product of its adjacent sides.

Area = 12 cm × 9 cm = 108 cm².

∴ The correct answer is 108 cm².

Alternate Method

Given: Adjacent sides of the parallelogram are a = 12 cm and b = 9 cm. Length of the diagonal is d = 15 cm.

Formula Used: Area of parallelogram = 2 × Area of the triangle formed by its sides and diagonal.

⇒ Let the triangle be ΔABC with sides 9 cm, 12 cm, and 15 cm.

⇒ Checking for a right-angled triangle: 9² + 12² = 81 + 144 = 225.

⇒ Square of the longest side (diagonal) = 15² = 225.

⇒ Since 9² + 12² = 15², the triangle is right-angled at the vertex between sides 9 cm and 12 cm.

⇒ Area of ΔABC = 1 ÷ 2 × base × height = 1 ÷ 2 × 12 × 9 = 54 cm².

⇒ Area of parallelogram = 2 × Area of ΔABC = 2 × 54 = 108 cm².

∴ The correct answer is 108 cm².

Additional Information

Properties of Parallelogram Diagonals

Each diagonal of a parallelogram divides it into two triangles of equal area.

Heron's Formula

For any triangle with sides a, b, and c, Area = √[s × (s − a) × (s − b) × (s − c)], where s = (a + b + c) ÷ 2.

Parallelogram Law

The sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of its two diagonals: 2(a² + b²) = d₁² + d₂².