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Shortcut Trick

For a Rhombus, the area is always half the product of its diagonals.

Area = 1/2 × (d₁ × d₂)

Area = 1/2 × 19 × 29 = 551 ÷ 2 = 275.5

∴ The correct answer is 275.5 cm².

Alternate Method

Given:

Length of first diagonal (d₁) = 19 cm

Length of second diagonal (d₂) = 29 cm

Formula Used:

Area of Rhombus = (d₁ × d₂) ÷ 2

Calculations:

⇒ Area = (19 × 29) ÷ 2

⇒ Area = 551 ÷ 2

⇒ Area = 275.5 cm²

∴ The correct answer is 275.5 cm².

Additional Information

Perpendicular Bisectors

The diagonals of a rhombus always bisect each other at a 90° angle (perpendicularly).

Relation between Side and Diagonals

The side length s is related to the diagonals by: 4s² = d₁² + d₂² or s = 1/2 × √(d₁² + d₂²).

Perimeter of Rhombus

Since all four sides of a rhombus are equal, the Perimeter = 4 × side.

Area with Base and Height

The area can also be calculated as Base × Height, similar to a parallelogram.

Shortcut Trick

For any regular polygon, the sum of an interior angle and its corresponding exterior angle is always 180°.

Exterior Angle = 180° − 150° = 30°

The sum of all exterior angles of any polygon is always 360°.

Number of sides (n) = 360° ÷ Exterior Angle

n = 360° ÷ 30° = 12

∴ The correct answer is 12 sides.

Alternate Method

Given: Each interior angle of a regular polygon = 150°

Formula Used: Interior angle of a regular polygon with n sides = [(n − 2) × 180°] ÷ n

Calculations:

⇒ 150 = [(n − 2) × 180] ÷ n

⇒ 150n = (n − 2) × 180

⇒ 150n = 180n − 360

⇒ 180n − 150n = 360

⇒ 30n = 360

⇒ n = 360 ÷ 30

⇒ n = 12

∴ The correct answer is 12 sides.

Additional Information

Sum of Interior Angles

For any polygon with n sides, the sum of all interior angles is given by (n − 2) × 180°.

Exterior Angle Property

The sum of all exterior angles of any convex polygon is always 360°, regardless of the number of sides.

Regular Polygons

In a regular polygon, all sides are equal and all interior angles are equal. A polygon with 12 sides is specifically called a dodecagon.

Shortcut Trick

Mean Price of the mixture = 1600 ÷ 45 = ₹320/9 per kg.

Apply Alligation between ₹32 and ₹40 with mean price ₹320/9.

Ratio = (40 − 320/9) : (320/9 − 32) = 40/9 : 32/9 = 40 : 32.

Simplifying the ratio: 40 : 32 → 5 : 4.

∴ The correct answer is 5:4.

Alternate Method

Given: Cost of Type 1 rice = ₹32/kg, Cost of Type 2 rice = ₹40/kg, Total Quantity = 45 kg, Total Cost = ₹1600.

Step 1: Calculate Mean Price (M) of the mixture.

⇒ M = Total Cost ÷ Total Quantity

⇒ M = 1600 ÷ 45 = 320/9

Step 2: Let Quantity 1 be Q₁ and Quantity 2 be Q₂.

⇒ Q₁ : Q₂ = (Price₂ − M) : (M − Price₁)

⇒ Q₁ : Q₂ = (40 − 320/9) : (320/9 − 32)

⇒ Q₁ : Q₂ = (360 − 320)/9 : (320 − 288)/9

⇒ Q₁ : Q₂ = 40/9 : 32/9

⇒ Q₁ : Q₂ = 40 : 32 = 5 : 4

∴ The correct answer is 5:4.

Additional Information

Alligation Rule

It states that (Quantity of Cheaper / Quantity of Dearer) = (Rate of Dearer − Mean Rate) / (Mean Rate − Rate of Cheaper).

Mean Price

The Mean Price is the cost price of a unit quantity of the mixture obtained by mixing two or more ingredients.

Mixture Calculations

If two ingredients of concentrations C₁ and C₂ are mixed in ratio n₁:n₂, the resulting concentration C_avg = (n₁C₁ + n₂C₂) ÷ (n₁ + n₂).

Shortcut Trick

Let Time₃ = x, then Time₁ = x + 9 and Time₂ = x + 4.

Apply the rule: If 1/A + 1/B = 1/C, then C = √(a × b) where A = C+a and B = C+b.

Time₃ = √(9 × 4) = 6 hr; therefore Time₂ = 6 + 4 = 10 hr.

Together (2 & 3) = (10 × 6) ÷ (10 + 6) = 60/16 = 3 3/4 hr.

∴ The correct answer is 3 3/4 hr.

Alternate Method

Given:

Let the time taken by Pipe 3 alone be t₃ hr.

⇒ Pipe 2 time (t₂) = t₃ + 4 hr

⇒ Pipe 1 time (t₁) = t₂ + 5 = (t₃ + 4) + 5 = t₃ + 9 hr

Formula Used:

Efficiency(1) + Efficiency(2) = Efficiency(3) ⇒ 1/t₁ + 1/t₂ = 1/t₃

Calculations:

⇒ 1/(t₃ + 9) + 1/(t₃ + 4) = 1/t₃

⇒ (t₃ + 4 + t₃ + 9) / ((t₃ + 9)(t₃ + 4)) = 1/t₃

⇒ (2t₃ + 13) / (t₃² + 13t₃ + 36) = 1/t₃

⇒ 2t₃² + 13t₃ = t₃² + 13t₃ + 36

⇒ t₃² = 36 ⇒ t₃ = 6 hr

⇒ t₂ = 6 + 4 = 10 hr

⇒ Total Efficiency (2 & 3) = 3 + 5 = 8 units/hr

⇒ Time taken together = Total Work ÷ Total Efficiency = 30 ÷ 8

⇒ Time = 15/4 = 3 3/4 hr

∴ The correct answer is 3 3/4 hr.

Additional Information

Root Rule for Combined Work

If A takes 'a' days more and B takes 'b' days more than (A+B) together, then the time taken by (A+B) is √(a × b).

Pipes and Cisterns Basics

If a pipe fills a tank in 'h' hours, the part filled in 1 hour is 1/h. For inlet pipes, efficiency is positive; for outlet pipes, it is negative.

Efficiency and Capacity

Capacity of the tank = Total Time × Combined Efficiency of the filling pipes.

Shortcut Trick

Let the Distance between City A and City B be D.

Forward time = D ÷ 60; Return speed = 60 + 20 = 80 km/h.

Return journey includes 15 km (30 km/h × 0.5 hr) at 30 km/h and (D − 15) km at 80 km/h.

Total time equation: (D/60) + 0.5 + (D − 15)/80 = 7 ⇒ (D/60) + (D − 15)/80 = 6.5.

Multiplying by LCM 240: 4D + 3D − 45 = 1560 ⇒ 7D = 1605 ⇒ D = 229 2/7 km.

∴ The correct answer is 229 2/7 km.

Alternate Method

Given:

Speed (A to B) = 60 km/h

Speed (B to A, normal) = 60 + 20 = 80 km/h

Repair Speed = 30 km/h | Repair Duration = 30 min = 0.5 hr

Total Round Trip Time = 7 hr

Calculations:

⇒ Let Distance between A and B be D km.

⇒ Distance covered during repair = Speed × Time = 30 × 0.5 = 15 km.

⇒ Remaining distance for return journey = D − 15 km.

⇒ Time for forward journey = D/60 hr.

⇒ Time for return journey = 0.5 + (D − 15)/80 hr.

⇒ Total Time = D/60 + 0.5 + (D − 15)/80 = 7

⇒ D/60 + (D − 15)/80 = 6.5

⇒ (4D + 3(D − 15)) ÷ 240 = 6.5

⇒ 4D + 3D − 45 = 6.5 × 240

⇒ 7D − 45 = 1560

⇒ 7D = 1605

⇒ D = 1605 ÷ 7 = 229 2/7 km.

∴ The correct answer is 229 2/7 km.

Additional Information

Average Speed

It is defined as the total distance traveled divided by the total time taken. Average Speed = Total Distance ÷ Total Time.

Speed-Time-Distance Relationship

Distance = Speed × Time. If distance is constant, speed is inversely proportional to time (S ∝ 1/T).

Unit Conversion

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

Shortcut Trick

Scheme A: Successive 50% and 16% ⇒ 50 + 16 − (50 × 16 ÷ 100) = 58%.

Scheme B: Successive 49%, 8%, and 13% ≈ 59.18% (Multiple successive drops).

Scheme C: 8 toys ⇒ (6 × 35% + 2 × 45%) ÷ 8 = (210% + 90%) ÷ 8 = 37.5%.

Scheme D: Buy 8, pay for 4 ⇒ 4 free ÷ 8 total × 100 = 50%.

The least beneficial scheme is the one with the minimum total discount (37.5%).

∴ The correct answer is Scheme C.

Alternate Method

Assume the Marked Price (MP) of each toy is ₹100. For 8 toys, total MP = 8 × 100 = ₹800.

Formula Used: Selling Price (SP) = MP × [ (100 − d₁)/100 ] × [ (100 − d₂)/100 ] ...

Calculation of Total Selling Price (SP) for 8 toys:

⇒ Scheme A: SP = 8 × [ 100 × 0.50 × 0.84 ] = 8 × 42 = ₹336.

⇒ Scheme B: SP = 8 × [ 100 × 0.51 × 0.92 × 0.87 ] ≈ 8 × 40.82 = ₹326.56.

⇒ Scheme C: SP = (6 × 65) + (2 × 55) = 390 + 110 = ₹500.

⇒ Scheme D: SP = Customer billed for 4 toys = 4 × 100 = ₹400.

Comparing the total Selling Prices: ₹326.56 (B) < ₹336 (A) < ₹400 (D) < ₹500 (C).

Since the customer pays the maximum amount in Scheme C, it is the least beneficial.

∴ The correct answer is Scheme C.

Additional Information

Successive Discount Formula

For two successive discounts x% and y%, the single equivalent discount is [x + y − (xy ÷ 100)]%.

Buy X Get Y Free

The effective discount percentage is calculated as: [Free items (Y) ÷ Total items (X + Y)] × 100.

Least vs Most Beneficial

From a customer's perspective, the least beneficial scheme is the one that offers the lowest discount or the highest final price.

Shortcut Trick

Use the Algebraic Identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Here, a = 24, b = −5, and c = −19.

Checking the sum: 24 + (−5) + (−19) = 24 − 24 = 0.

Result = 3 × 24 × (−5) × (−19) = 72 × 95 = 6840.

∴ The correct answer is 6840.

Alternate Method

Given: The expression is 24³ + (−5)³ + (−19)³.

Formula Used: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

⇒ Let a = 24, b = −5, and c = −19.

⇒ Calculate the sum (a + b + c): 24 + (−5) + (−19) = 24 − 5 − 19 = 0.

⇒ Since (a + b + c) = 0, the expression (a + b + c)(a² + b² + c² − ab − bc − ca) becomes 0.

⇒ a³ + b³ + c³ − 3abc = 0

⇒ a³ + b³ + c³ = 3abc

⇒ Substituting the values: 3 × 24 × (−5) × (−19)

⇒ 72 × 95

⇒ 72 × (100 − 5) = 7200 − 360 = 6840.

∴ The correct answer is 6840.

Additional Information

General Cubic Identity

The expression a³ + b³ + c³ − 3abc is always equal to (a + b + c) × 0.5 × [(a−b)² + (b−c)² + (c−a)²].

Conditional Identity

If a³ + b³ + c³ = 3abc, it implies that either a + b + c = 0 or a = b = c.

Square of Trinomial

The square of a sum of three terms is given by (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

Shortcut Trick

To find the ratio between the first term (a) and the last term (c), directly multiply the given ratios.

a/c = (a/b) × (b/c)

a/c = (12/13) × (4/5) = 48/65

∴ The correct answer is 48:65.

Alternate Method

Given: The ratio a:b is 12:13 and b:c is 4:5.

To find a:c, we can express the ratios as fractions:

⇒ a/b = 12/13

⇒ b/c = 4/5

⇒ Multiply the two equations: (a/b) × (b/c) = (12/13) × (4/5)

⇒ The common term 'b' cancels out: a/c = (12 × 4) / (13 × 5)

⇒ a/c = 48/65

This result matches Option 3 provided in the question.

∴ The correct answer is 48:65.

Additional Information

Compound Ratio

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

Continued Proportion

To find the combined ratio a:b:c, make the common term 'b' equal by multiplying the ratios by suitable constants. Here, a:b:c = (12×4) : (13×4) : (13×5) = 48:52:65.

Duplicate Ratio

The duplicate ratio of a:b is simply the ratio of their squares, i.e., a² : b².

Shortcut Trick

Net percentage change formula = x + y + (xy / 100)

Net % change = 20 − 25 − (20 × 25 / 100) = −5 − 5 = −10%

Final Price = 90% of 600 = 0.9 × 600 = 540

∴ The correct answer is ₹540.

Alternate Method

Given: Original price of the jacket = ₹600, First increase = 20%, Second decrease = 25%.

⇒ First price increase = 20% of 600 = (20/100) × 600 = ₹120

⇒ Price after increase = 600 + 120 = ₹720

⇒ Price decrease = 25% of 720 = (25/100) × 720 = ₹180

⇒ Final price of the jacket = 720 − 180 = ₹540

∴ The correct answer is ₹540.

Additional Information

Successive Percentage Change

When a value is changed by x% and then y%, the net change is given by (x + y + xy/100)%. Use positive values for increase and negative for decrease.

Fractional Multiplier Method

For a 20% increase, multiply by 1.2 or 6/5. For a 25% decrease, multiply by 0.75 or 3/4. Final Value = Initial Value × Multiplier₁ × Multiplier₂.

Base Change Property

A change of +20% followed by −20% does not return the value to its original state; the net change would be a 4% decrease (loss).

Shortcut Trick

Net deviation from the overall mean (42) must be zero.

Sum of deviations = (30 − 42) + (38 − 42) + (50 − 42) + 6 × (45 − 42)

⇒ −12 − 4 + 8 + (6 × 3) = −16 + 8 + 18 = +10.

Since the existing 9 numbers are +10 above the mean, the 10^th number must be 10 below the mean.

⇒ Missing observation = 42 − 10 = 32.

∴ The correct answer is 32.

Alternate Method

Given:

Total observations = 10 | Total Arithmetic Mean = 42

First 3 observations = 30, 38, 50 | Average of next 6 = 45

Calculations:

⇒ Total Sum = Mean × Number of observations

⇒ Total Sum = 42 × 10 = 420

⇒ Sum of 3 known observations = 30 + 38 + 50 = 118

⇒ Sum of next 6 observations = 45 × 6 = 270

⇒ Sum of 9 observations = 118 + 270 = 388

⇒ Missing observation = Total Sum − Sum of 9 observations

⇒ Missing observation = 420 − 388

⇒ Missing observation = 32.

∴ The correct answer is 32.

Additional Information

Arithmetic Mean Formula

Mean = (Sum of all observations) ÷ (Number of observations). To find a missing value, subtract the sum of known values from the total sum.

Sum of Deviations

The algebraic sum of the deviations of a set of values from their arithmetic mean is always zero. ∑(x_i − mean) = 0.

Weighted Mean

When different groups have different averages, the combined mean is (n₁ × x₁ + n₂ × x₂ + ...) ÷ (n₁ + n₂ + ...).