Quantitative Aptitude PYQs

Shortcut Trick

Price after 20% discount = 1250 − 250 = ₹1,000

Final Selling Price = ₹900

Amount of second discount = 1000 − 900 = ₹100

Second Discount percentage = (100 ÷ 1000) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given: Marked Price (MP) = ₹1,250, First Discount (D₁) = 20%, Final Selling Price (SP) = ₹900

Formula: Selling Price = MP × (100 − D₁)% × (100 − D₂)%

⇒ Price after first discount = 1250 × (100 − 20) ÷ 100

⇒ Price after first discount = 1250 × 0.8 = ₹1,000

⇒ Now, the second discount is applied on ₹1,000 to get ₹900

⇒ Second Discount amount = 1000 − 900 = ₹100

⇒ Second Discount % = (Discount Amount ÷ Previous Price) × 100

⇒ Second Discount % = (100 ÷ 1000) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Successive Discounts Formula

When two discounts 'a%' and 'b%' are given successively, the equivalent single discount is given by: [a + b − (a × b ÷ 100)]%.

Marked Price (MP) and Discount

Discount is always calculated on the Marked Price. The price obtained after subtracting the discount from the MP is the Selling Price (SP).

Relationship between SP and MP

SP = MP × [(100 − D₁) ÷ 100] × [(100 − D₂) ÷ 100] × ... × [(100 − D_n) ÷ 100].

Shortcut Trick

Simplify the second term: 10 − √75 = 10 − 5√3 = 5(2 − √3).

The numbers are (2 + √3) and 5(2 − √3).

Product = 5 × (2 + √3)(2 − √3) = 5 × (4 − 3) = 5.

Mean proportional = √Product = √5.

∴ The correct answer is √5.

Alternate Method

Given:

First number (a) = 2 + √3

Second number (b) = 10 − √75

Formula Used:

The mean proportional between two numbers a and b is given by x = √(a × b).

⇒ First, simplify b = 10 − √(25 × 3) = 10 − 5√3

⇒ Factor out 5: b = 5(2 − √3)

⇒ Now, x = √[(2 + √3) × 5(2 − √3)]

⇒ x = √[5 × (2 + √3)(2 − √3)]

⇒ Using algebraic identity (a + b)(a − b) = a² − b²:

⇒ x = √[5 × (2² − (√3)²)]

⇒ x = √[5 × (4 − 3)]

⇒ x = √[5 × 1] = √5

∴ The correct answer is √5.

Additional Information

Mean Proportional

The mean proportional of two numbers a and b is x if a:x = x:b, which results in the formula x = √(ab).

Third Proportional

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

Conjugate Surds

Surds of the form (p + √q) and (p − √q) are conjugates; their product is always a rational number (p² − q).

Shortcut Trick

For any closed cylinder, the ratio of Curved Surface Area (CSA) to Total Surface Area (TSA) is always h : (r + h).

Given Height (h) = 10 m and Radius (r) = 12 ÷ 2 = 6 m.

Required Ratio = 10 : (6 + 10) = 10 : 16 = 5 : 8.

∴ The correct answer is 5:8.

Alternate Method

Given values for the cylindrical tank:

⇒ Diameter = 12 m

⇒ Height (h) = 10 m

⇒ Radius (r) = Diameter ÷ 2 = 12 ÷ 2 = 6 m

⇒ Curved Surface Area (CSA) = 2 × π × r × h

⇒ Total Surface Area (TSA) = 2 × π × r × (r + h)

⇒ Ratio = [2 × π × r × h] ÷ [2 × π × r × (r + h)]

⇒ Ratio = h ÷ (r + h)

⇒ Ratio = 10 ÷ (6 + 10) = 10 ÷ 16

⇒ Ratio = 5 : 8

∴ The correct answer is 5:8.

Additional Information

Volume of a Cylinder

The total space occupied by a cylinder is given by V = πr²h.

Base Area of a Cylinder

The area of each circular end of the cylinder is πr².

Surface Area for Open Tanks

If a cylindrical tank is open at the top, its total surface area is 2πrh + πr².

Shortcut Trick

Net efficiency of all taps = 1/4 + 1/22 − 1/22 = 1/4 cistern/hr.

The question asks for the time to fill 1/4^th of the empty cistern.

Time required = (Work to be done) ÷ (Net efficiency)

Time = (1/4) ÷ (1/4) = 1 hr.

∴ The correct answer is 1 hr.

Alternate Method

Given:

Time taken by Tap 1 (filling) = 4 hr

Time taken by Tap 2 (filling) = 22 hr

Time taken by Tap 3 (emptying) = 22 hr

Formula Used:

Total Work = LCM of individual times

Efficiency = Total Work ÷ Individual Time

Calculations:

⇒ Total capacity of the cistern (LCM of 4 and 22) = 44 units.

⇒ Net efficiency when all taps are open = 11 + 2 − 2 = 11 units/hr.

⇒ Work to be completed (1/4^th of the cistern) = 44 × (1/4) = 11 units.

⇒ Time taken to fill 11 units = 11 ÷ 11 = 1 hr.

∴ The time required to fill one-fourth of the cistern is 1 hr.

∴ The correct answer is 1 hr.

Additional Information

Pipes and Cisterns Basics

If a pipe can fill a tank in x hours, then the part filled in 1 hour is 1/x. This is known as the efficiency of the pipe.

Inlet and Outlet Taps

An inlet tap performs positive work (filling), while an outlet or emptying tap performs negative work. The net rate is the algebraic sum of their individual efficiencies.

LCM Method

Using the Least Common Multiple (LCM) of the given time periods as the total capacity helps avoid working with complex fractions, making calculations faster.

Shortcut Trick

Assume Total Capacity of the tank = LCM(10, 15, 20) = 60 units.

Net efficiency of (A + B + C) = 6 + 4 − 3 = 7 units/hr.

Work done in 4 hours = 7 × 4 = 28 units.

Remaining work = 60 − 28 = 32 units.

Time for A and B to fill remaining = 32 ÷ (6 + 4) = 3.2 hr.

Total Time = 4 + 3.2 = 7.2 hr.

∴ The correct answer is 7.2 hr.

Alternate Method

Given: Pipe A fills in 10 hr, Pipe B fills in 15 hr, and Pipe C empties in 20 hr.

Formula Used: Total Work = Net Efficiency × Time

⇒ Net efficiency of all three pipes (A + B − C) = 6 + 4 − 3 = 7 units/hr.

⇒ Work completed in the first 4 hours = 7 × 4 = 28 units.

⇒ Remaining work to be completed after closing Pipe C = 60 − 28 = 32 units.

⇒ Efficiency of remaining pipes (A + B) = 6 + 4 = 10 units/hr.

⇒ Time taken to fill the remaining part = 32 ÷ 10 = 3.2 hr.

⇒ Total time required = Initial time + Remaining time = 4 hr + 3.2 hr = 7.2 hr.

∴ The correct answer is 7.2 hr.

Additional Information

Inlet and Outlet Pipes

An Inlet pipe fills the tank (positive work), while an Outlet pipe empties the tank (negative work).

Net Efficiency

If multiple pipes operate simultaneously, the Net Efficiency is the algebraic sum of their individual efficiencies: Net Eff = Eff_Inlet1 + Eff_Inlet2 − Eff_Outlet.

Time and Work Relation

Efficiency is inversely proportional to the Time taken to complete the work (Efficiency = 1 ÷ Time), assuming the work is constant.

Shortcut Trick

Length of train = (Difference in speeds × 5/18) × (t₁ × t₂) ÷ (t₂ − t₁)

Difference in speeds = 7 − 5 = 2 km/h

Length = (2 × 5/18) × (18 × 20) ÷ (20 − 18)

Length = (5/9) × 360 ÷ 2 = 5 × 20 = 100 m

∴ The correct answer is 100 m.

Alternate Method

Given: Speed of runner₁ = 5 km/h, Speed of runner₂ = 7 km/h, Time taken t₁ = 18 s, t₂ = 20 s.

Formula: Relative Speed (same direction) = Speed of train − Speed of object.

Let Speed of the train be V km/h.

⇒ Relative Speed with first runner = (V − 5) × 5/18 m/s

⇒ Relative Speed with second runner = (V − 7) × 5/18 m/s

⇒ Length L = (V − 5) × (5/18) × 18 = (V − 7) × (5/18) × 20

⇒ (V − 5) × 1 = (V − 7) × (20/18)

⇒ 18(V − 5) = 20(V − 7)

⇒ 18V − 90 = 20V − 140

⇒ 2V = 50 ⇒ V = 25 km/h

⇒ L = (25 − 5) × (5/18) × 18 = 20 × 5 = 100 m

∴ The correct answer is 100 m.

Additional Information

Relative Speed in Same Direction

When two objects move in the same direction, the relative speed is the difference between their speeds: (u − v).

Relative Speed in Opposite Direction

When two objects move towards each other, the relative speed is the sum of their speeds: (u + v).

Unit Conversion

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

Length of Train

When a train passes a point object (man, pole, runner), the distance covered equals the length of the train.

Shortcut Trick

Assume all 120 pens are of Type B (₹8 each): Total sale = 120 × 8 = ₹960.

Actual total sale is ₹1120, so the difference is 1120 − 960 = ₹160.

Difference in price per pen between Type A and Type B = 12 − 8 = ₹4.

Number of Type A pens = 160 ÷ 4 = 40.

∴ The correct answer is 40 pens.

Alternate Method

Given:

Total pens = 120 | Price of Type A = ₹12 | Price of Type B = ₹8 | Total Amount = ₹1120

Formula Used:

Weighted Average Total = (n₁ × Price₁) + (n₂ × Price₂)

Let the number of Type A pens be x and Type B pens be (120 − x).

⇒ 12 × x + 8 × (120 − x) = 1120

⇒ 12x + 960 − 8x = 1120

⇒ 4x + 960 = 1120

⇒ 4x = 1120 − 960

⇒ 4x = 160

⇒ x = 40

∴ The correct answer is 40 pens.

Additional Information

Rule of Alligation

It is a rule that enables us to find the ratio in which two or more ingredients at their respective prices must be mixed to produce a mixture of a specified price.

Linear Equation Method

Standard method where unknown quantities are represented as variables (x, y) and solved using algebraic manipulation.

Weighted Average

The average of a data set that contains values with different levels of importance or frequency: Average = (Sum of all values) ÷ (Total number of items).

Given:

(340 / 17) × (156 / 13) + (1484 / 28) − (14 / 10)

Formula Used:

The BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction) is applied to simplify the expression.

Calculations:

(340 ÷ 17) × (156 ÷ 13) + (1484 ÷ 28) − (14 ÷ 10)

⇒ 20 × 12 + 53 − 1.4

⇒ 240 + 53 − 1.4

⇒ 293 − 1.4

⇒ 291.6

∴ The correct answer is option (1).

Shortcut Trick

Check the sum of the base numbers: 22 + (−18) + (−4) = 0.

When the sum of three numbers is zero, the sum of their cubes is 3 × Product of the numbers.

Required value = 3 × 22 × (−18) × (−4) = 66 × 72 = 4752.

∴ The correct answer is 4752.

Alternate Method

Given:

The expression to evaluate is 22³ + (−18)³ + (−4)³.

Let a = 22, b = −18, and c = −4.

Formula Used:

We use the algebraic identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Calculations:

⇒ First, calculate the sum of the variables: a + b + c = 22 + (−18) + (−4)

⇒ a + b + c = 22 − 22 = 0

⇒ Since a + b + c = 0, the identity simplifies to: a³ + b³ + c³ = 3abc

⇒ Substituting the values: 22³ + (−18)³ + (−4)³ = 3 × 22 × (−18) × (−4)

⇒ Value = 66 × 72

⇒ Value = 4752

Therefore, the correct option is 2.

∴ The correct answer is 4752.

Additional Information

Cubic Identity Properties

The expression a³ + b³ + c³ is equal to 3abc only if a + b + c = 0 or if a = b = c. This is a fundamental conditional identity in algebra.

General Expansion

For any three numbers, a³ + b³ + c³ − 3abc = ½ (a + b + c)[(a − b)² + (b − c)² + (c − a)²].

Applications in Exams

This identity is frequently used to simplify large cubic calculations by checking if the sum of the bases equals zero before performing traditional multiplication.

Shortcut Trick

Mean proportional of 9 and 16 = √(9 × 16) = 12.

Given mean proportional of 12 and the number is twice this value.

Target MP = 2 × 12 = 24.

Using the property (MP)² = Product of numbers ⇒ 24² = 12 × Number.

Number = 576 ÷ 12 = 48.

∴ The correct answer is 48.

Alternate Method

Given:

First set of numbers = 9 and 16

Second set of numbers = 12 and x (let the number be x)

Formula Used:

Mean proportional (MP) between two numbers a and b is √(a × b).

Calculations:

⇒ Mean proportional between 9 and 16 = √(9 × 16)

⇒ Mean proportional_(9,16) = √144 = 12

⇒ According to the question, MP_(12,x) = 2 × 12 = 24

⇒ √(12 × x) = 24

⇒ Squaring both sides: 12 × x = 24²

⇒ 12 × x = 576

⇒ x = 576 ÷ 12

⇒ x = 48

∴ The correct answer is 48.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is √(a × b). If x is the mean proportional, then a : x :: x : b.

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, giving d = (b × c) ÷ a.

Shortcut Trick

Step 1: Calculate the LCM of 10, 11, and 8 ⇒ 440.

Step 2: Find the largest 4-digit multiple of 440 ⇒ 440 × 22 = 9680.

Step 3: Add the required remainder (7) to this multiple ⇒ 9680 + 7 = 9687.

∴ The correct answer is 9687.

Alternate Method

Given: Divisors are 10, 11, and 8; Remainder in each case is 7.

Formula Used: Required Number = (LCM of divisors × k) + Remainder

⇒ LCM(10, 11, 8) = 2³ × 5 × 11 = 8 × 5 × 11 = 440

⇒ The largest four-digit number is 9999.

⇒ Dividing 9999 by 440: 9999 ÷ 440 = 22 with a remainder of 319.

⇒ The largest multiple of 440 below 10000 = 9999 − 319 = 9680.

⇒ The required number = 9680 + 7 = 9687.

∴ The correct answer is 9687.

Additional Information

LCM (Least Common Multiple)

It is the smallest positive integer that is evenly divisible by all numbers in a given set.

The Remainder Property

If a number N leaves a constant remainder R when divided by a, b, and c, then N = LCM(a, b, c) × k + R.

Divisibility Rule of 11

A number is divisible by 11 if the difference between the sum of digits at odd places and even places is 0 or a multiple of 11.

Shortcut Trick

Total deviation from old average 38 = (50 − 38) + (62 − 38) = 12 + 24 = 36.

Total numbers now = 12 + 2 = 14.

Average increase = Total deviation ÷ Total count = 36 ÷ 14 ≈ 2.57.

New Average = 38 + 2.57 = 40.57.

∴ The correct answer is 40.57.

Alternate Method

Given: Average of 12 numbers = 38, Included numbers = 50 and 62.

Formula: New Average = (Old Sum + New Values) ÷ New Count

⇒ Sum of 12 numbers = 12 × 38 = 456

⇒ Sum of 14 numbers = 456 + 50 + 62 = 568

⇒ Total count = 12 + 2 = 14

⇒ New Average = 568 ÷ 14 = 40.5714...

⇒ Rounding to two decimal places, New Average = 40.57

∴ The correct answer is 40.57.

Additional Information

Standard Average Formula

Average = (Sum of all observations) ÷ (Total number of observations).

Weighted Average

When two groups are combined, New Average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Inclusion/Exclusion Rule

If the newly added value is greater than the current average, the new average will increase.

Shortcut Trick

Ratio of Profit = (Capital × Time)_X : (Capital × Time)_Y : (Capital × Time)_Z

Ratio = (60 × 12) : (80 × 6) : (120 × 4) = 720 : 480 : 480

Simplified Ratio = 3 : 2 : 2

Total units = 3 + 2 + 2 = 7 units → ₹77,000

Z's share (2 units) = (77,000 ÷ 7) × 2 = ₹22,000

∴ The correct answer is ₹22,000.

Alternate Method

Given:

Investment of X = ₹60,000, Time = 12 months

Investment of Y = ₹80,000, Time = 6 months

Investment of Z = ₹1,20,000, Time = 4 months

Total Profit = ₹77,000

Formula Used:

Profit Share = Capital × Time Period

Calculations:

⇒ Profit share of X = 60,000 × 12 = 7,20,000

⇒ Profit share of Y = 80,000 × 6 = 4,80,000

⇒ Profit share of Z = 1,20,000 × 4 = 4,80,000

⇒ Ratio of profits X : Y : Z = 7,20,000 : 4,80,000 : 4,80,000

⇒ Ratio = 72 : 48 : 48 = 3 : 2 : 2

⇒ Sum of the ratio parts = 3 + 2 + 2 = 7

⇒ Share of Z = (2 ÷ 7) × 77,000

⇒ Share of Z = 2 × 11,000 = ₹22,000

∴ The correct answer is ₹22,000.

Additional Information

Compound Partnership

A partnership where the partners invest their capitals for different periods of time. The profit is divided based on the product of capital and time.

Types of Partners

An Active Partner manages the business and may receive an extra share or salary, while a Sleeping Partner only invests money without participating in daily operations.

Ratio of Capitals

If the profit and time periods are given, the ratio of capitals C₁ : C₂ is calculated as (P₁/T₁) : (P₂/T₂).

Shortcut Trick

Net percentage change formula: Net % = a + b + (ab/100)

Here, a = −30 (decrease) and b = +15 (increase).

Net change = −30 + 15 + [(−30 × 15) / 100] = −15 − 4.5 = −19.5%.

New Price = 800 − (19.5% of 800) = 800 − 156 = 644.

∴ The correct answer is ₹644.

Alternate Method

Given: Original Price of the shirt = ₹800, First change = 30% decrease, Second change = 15% increase.

Method: Step-by-step price calculation using basic percentage.

⇒ Price after 30% decrease = Original Price × [ (100 − 30) / 100 ]

⇒ Intermediate Price = 800 × (70 / 100) = ₹560

⇒ Final price after 15% increase = Intermediate Price × [ (100 + 15) / 100 ]

⇒ New Price = 560 × (115 / 100) = 5.6 × 115 = ₹644

Thus, the final price of the shirt is ₹644, which corresponds to Option (4).

∴ The correct answer is ₹644.

Additional Information

Successive Percentage Rule

When a value changes by x% and then y%, the effective change is [x + y + (xy/100)]%. Remember to use a negative sign for decrease/discount.

Multiplying Factor Method

For a decrease of d%, multiply the value by (1 − d/100). For an increase of i%, multiply the value by (1 + i/100).

Fractional Equivalents

Calculations are often faster using fractions: 30% = 3/10 (multiplier 7/10) and 15% = 3/20 (multiplier 23/20).

Shortcut Trick

Ratio of ages of Shyam and Rajat 4 years ago = 7 : 1 (Difference = 6 units)

Ratio of ages 8 years later = 3 : 1 (Multiply by 3 to balance difference) → 9 : 3 (Difference = 6 units)

Increase in ratio units = 9 − 7 = 2 units

Actual time interval = 4 + 8 = 12 years

2 units = 12 years ⇒ 1 unit = 6 years

Shyam's age 4 years ago = 7 × 6 = 42 years ⇒ Present age = 42 + 4 = 46 years

∴ The correct answer is 46 years.

Alternate Method

Given:

Age of Shyam was 7 times Rajat 4 years ago.

Age of Shyam will be 3 times Rajat 8 years later.

Calculations:

Let the age of Rajat 4 years ago be x years.

⇒ Age of Shyam 4 years ago = 7x years.

⇒ Present age of Rajat = x + 4

⇒ Present age of Shyam = 7x + 4

⇒ Age after 8 years: Rajat = (x + 4) + 8 = x + 12; Shyam = (7x + 4) + 8 = 7x + 12

⇒ According to the condition: 7x + 12 = 3 × (x + 12)

⇒ 7x + 12 = 3x + 36

⇒ 7x − 3x = 36 − 12

⇒ 4x = 24

⇒ x = 6

⇒ Present age of Shyam = 7(6) + 4 = 42 + 4 = 46 years.

⇒ Therefore, the age of Shyam is 46 years, which corresponds to Option (4).

∴ The correct answer is 46 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals remains the same throughout their lives, regardless of how many years pass.

Ratio Balancing Method

To solve age problems using ratios, ensure the difference between ratio parts is equal for both time periods by multiplying with the cross-difference.

Average Age Shift

If the average age of n people is A years, then after k years, their average age will be A + k years.

Shortcut Trick

Gain of 50% = 1/2 ⇒ Ratio of CP : SP for Article 1 = 2 : 3.

Since Selling Price is the same for both, SP of Article 2 = 3 units.

No profit, no loss condition implies: Profit on Article 1 = Loss on Article 2.

Profit on Article 1 = 3 − 2 = 1 unit. So, Loss on Article 2 = 1 unit.

CP of Article 2 = SP + Loss = 3 + 1 = 4 units.

Loss % on Article 2 = (1 ÷ 4) × 100 = 25%.

∴ The correct answer is 25%.

Alternate Method

Given:

Selling Price (SP) of Article 1 = ₹2,973

Selling Price (SP) of Article 2 = ₹2,973

Gain % on Article 1 = 50%

Formula Used:

Cost Price (CP) = (SP × 100) ÷ (100 + Gain%)

Loss % = (Loss ÷ CP) × 100

Calculations:

⇒ CP of Article 1 = (2973 × 100) ÷ (100 + 50)

⇒ CP₁ = (2973 × 100) ÷ 150 = ₹1,982

⇒ Profit on Article 1 = 2973 − 1982 = ₹991

⇒ Since overall there is no profit, no loss:

⇒ Loss on Article 2 = Profit on Article 1 = ₹991

⇒ CP of Article 2 = SP of Article 2 + Loss

⇒ CP₂ = 2973 + 991 = ₹3,964

⇒ Loss % on Article 2 = (991 ÷ 3964) × 100

⇒ Loss % = 1/4 × 100 = 25%

∴ The correct answer is 25%.

Additional Information

No Profit, No Loss Condition

This occurs when the total Cost Price of all articles equals the total Selling Price. Mathematically, Total CP = Total SP, or Profit from one item = Loss from another.

Cost Price with Gain/Loss

When gain is given, CP = SP × 100 / (100 + Gain%). When loss is given, CP = SP × 100 / (100 − Loss%).

Successive Changes

If an article's price is changed successively by x% and y%, the net change is given by the formula: [x + y + (xy ÷ 100)]%.

Shortcut Trick

For a regular polygon, the sum of an interior angle and an exterior angle at any vertex is always 180°.

⇒ Exterior angle = 180° − Interior angle = 180° − 162° = 18°.

The number of sides (n) of a regular polygon is given by 360° ÷ exterior angle.

⇒ n = 360° ÷ 18° = 20.

∴ The correct answer is 20 sides.

Alternate Method

Given:

Measure of each interior angle = 162°

Formula Used:

Each interior angle of a regular polygon with n sides = [(n − 2) × 180°] ÷ n

Calculations:

⇒ 162° = [(n − 2) × 180°] ÷ n

⇒ 162n = (n − 2) × 180

⇒ 162n = 180n − 360

⇒ 180n − 162n = 360

⇒ 18n = 360

⇒ n = 360 ÷ 18

⇒ n = 20

∴ The correct answer is 20 sides.

Additional Information

Sum of Interior Angles

The total sum of all interior angles of a polygon with n sides is always (n − 2) × 180°.

Sum of Exterior Angles

For any convex polygon, the sum of all exterior angles (one at each vertex) is exactly 360°.

Regular Polygon Properties

In a regular polygon, all sides are equal and all interior angles are equal. Each exterior angle = 360° ÷ n.

Shortcut Trick

The distance d between two objects in the same direction from a cliff of height h is given by: d = h × (cot θ₁ − cot θ₂).

Given: h = 80 m, θ₁ = 30°, and θ₂ = 60°.

d = 80 × (cot 30° − cot 60°) = 80 × (√3 − 1/√3) = 80 × (2/√3).

d = 160 / 1.732 ≈ 92.38 m.

∴ The correct answer is 92.38 m.

Alternate Method

Given: Height of the cliff (AB) = 80 m. Angles of depression of ships C and D are 60° and 30° respectively.

Formula Used: tan θ = Perpendicular / Base

In right-angled triangle ABC:

⇒ tan 60° = AB / BC

⇒ √3 = 80 / BC

⇒ BC = 80 / √3 m

In right-angled triangle ABD:

⇒ tan 30° = AB / BD

⇒ 1 / √3 = 80 / BD

⇒ BD = 80√3 m

Distance between ships (CD) = BD − BC:

⇒ CD = 80√3 − 80 / √3

⇒ CD = (80 × 3 − 80) / √3 = 160 / √3

⇒ CD = 160 / 1.73205 ≈ 92.376 m

Rounding off to two decimal places, we get 92.38 m.

∴ The correct answer is 92.38 m.

Additional Information

Angle of Depression

The angle formed between the horizontal line and the line of sight when an observer looks downward at an object.

Distance for Opposite Sides

If ships are on opposite sides of the cliff, the distance between them is d = h(cot α + cot β).

Trigonometric Identities

For calculations, remember tan 30° = 1/√3, tan 60° = √3, and √3 ≈ 1.732.

Given:

$\dfrac{3^{-5} \times 8^5 \times 125^3 \times 36^3}{100^4 \times 256 \times 216 \times 2^2}$

Formula used:

Exponent Laws:

(a^m)ⁿ = a^{m × n}

a^m × aⁿ = a^{m + n}

a^m ÷ aⁿ = a^{m − n}

a⁻ⁿ = 1 / aⁿ

Calculations:

Express each term in its prime factor form (2, 3, and 5):

⇒ $\dfrac{3^{-5} \times (2^3)^5 \times (5^3)^3 \times (2^2 \times 3^2)^3}{(2^2 \times 5^2)^4 \times 2^8 \times (2^3 \times 3^3) \times 2^2}$

⇒ $\dfrac{3^{-5} \times 2^{15} \times 5^9 \times 2^6 \times 3^6}{2^8 \times 5^8 \times 2^8 \times 2^3 \times 3^3 \times 2^2}$

Group the bases in the numerator and denominator:

⇒ $\dfrac{2^{15 + 6} \times 3^{6 - 5} \times 5^9}{2^{8 + 8 + 3 + 2} \times 3^3 \times 5^8}$

⇒ $\dfrac{2^{21} \times 3^1 \times 5^9}{2^{21} \times 3^3 \times 5^8}$

Divide by subtracting exponents of the same base:

⇒ 2^{21 − 21} × 3^{1 − 3} × 5^{9 − 8}

⇒ 2⁰ × 3⁻² × 5¹

⇒ 1 × (1 / 3²) × 5

⇒ 5 / 9

∴ The correct answer is option (2).

Shortcut Trick

Since the distance is the same for both ways (48 km), we use the Harmonic Mean formula for average speed.

Average Speed = (2 × s₁ × s₂) / (s₁ + s₂)

Average Speed = (2 × 32 × 24) / (32 + 24) = (2 × 32 × 24) / 56

Average Speed = (32 × 24) / 28 = (8 × 24) / 7 = 192 / 7 ≈ 27.43 km/hr.

∴ The correct answer is 27.43 km/hr.

Alternate Method

Given:

Distance from Warehouse A to Store B = 48 km

Speed during onward journey (s₁) = 32 km/hr

Speed during return journey (s₂) = 24 km/hr

Formula used:

Average Speed = Total Distance ÷ Total Time

⇒ Total Distance = 48 + 48 = 96 km

⇒ Time taken for onward journey (t₁) = Distance ÷ s₁ = 48 ÷ 32 = 1.5 hours

⇒ Time taken for return journey (t₂) = Distance ÷ s₂ = 48 ÷ 24 = 2 hours

⇒ Total Time = t₁ + t₂ = 1.5 + 2 = 3.5 hours

⇒ Average Speed = 96 ÷ 3.5

⇒ Average Speed = 960 ÷ 35 = 192 ÷ 7 ≈ 27.4285...

∴ The correct answer is 27.43 km/hr.

Additional Information

Average Speed (Fixed Distance)

When an object covers equal distances at speeds x and y, the average speed is 2xy ÷ (x + y).

Average Speed (Fixed Time)

When an object travels for the same duration at speeds x and y, the average speed is the arithmetic mean: (x + y) ÷ 2.

General Average Speed

For any journey, Average Speed = (d₁ + d₂ + ... + d_n) ÷ (t₁ + t₂ + ... + t_n).

Unit Conversion

To convert km/hr to m/s, multiply the speed by 5/18. To convert m/s to km/hr, multiply by 18/5.