Quantitative Aptitude PYQs

Shortcut Trick

Given Marked Price (MP) = ₹889 and Selling Price (SP) = ₹278.

Discount = MP − SP = 889 − 278 = 611.

Discount% = (Discount ÷ MP) × 100.

Discount% = (611 ÷ 889) × 100 ≈ 68.7289%.

Rounding to two decimal places, we get 68.73%.

∴ The correct answer is 68.73%.

Alternate Method

Given: MP = ₹889, SP = ₹278.

Formulas: Discount = MP − SP | Discount% = (Discount ÷ MP) × 100

⇒ Discount = 889 − 278 = 611

⇒ Discount% = (611 ÷ 889) × 100

⇒ Discount% = 68.7289...

⇒ Final Result = 68.73%

∴ The correct answer is 68.73%.

Additional Information

Marked Price & Discount

Marked Price is the price listed on the label. Discount is always calculated on the Marked Price unless otherwise stated.

Selling Price Calculation

SP = MP × (100 − Discount%) ÷ 100. It is the actual price at which the product is sold after the reduction.

Shortcut Trick

Combined efficiency of P and Q = 1/9

Efficiency of P = 1/36

Efficiency of Q = 1/9 − 1/36 = (4 − 1)/36 = 3/36 = 1/12

Time for Q to fill 3/4 cistern = (3/4) ÷ (1/12) = 3/4 × 12 = 9 hours

∴ The correct answer is 9 hours.

Alternate Method

Given:

Time taken by P + Q = 9 hours

Time taken by P alone = 36 hours

Formula Used:

Total Work = LCM of individual times

Efficiency = Total Work ÷ Time

Calculations:

⇒ Efficiency of Q = (Efficiency of P + Q) − (Efficiency of P)

⇒ Efficiency of Q = 4 − 1 = 3 units/hour

⇒ Three-fourth of the cistern capacity = 3/4 × 36 = 27 units

⇒ Time taken by Q to fill 27 units = 27 ÷ 3 = 9 hours

∴ The correct answer is 9 hours.

Additional Information

Combined Work Formula

If two pipes fill a tank in x and y hours respectively, together they fill it in (xy)/(x + y) hours.

Individual Time Formula

If together they take T hours and one takes x hours, the other takes y = (Tx)/(x − T) hours.

Pipes and Cisterns

Inlet pipes have positive efficiency, while outlet (leakage) pipes have negative efficiency.

Shortcut Trick

Assume a base mean of 40 to simplify calculation.

Find deviations from 40: −23, +24, +15, −3, +25, −28, +11, −22, +19.

Net deviation = (−23 − 3 − 28 − 22) + (24 + 15 + 25 + 11 + 19) = −76 + 94 = +18.

Average deviation = 18 ÷ 9 = +2.

Actual Mean = 40 + 2 = 42.

∴ The correct answer is 42.

Alternate Method

Given: Observations are 17, 64, 55, 37, 65, 12, 51, 18, and 59.

Formula Used: Arithmetic Mean = (Sum of all observations) ÷ (Total number of observations)

⇒ Sum of observations = 17 + 64 + 55 + 37 + 65 + 12 + 51 + 18 + 59

⇒ Sum = 378

⇒ Total number of observations (n) = 9

⇒ Arithmetic Mean = 378 ÷ 9

⇒ Arithmetic Mean = 42

∴ The correct answer is 42.

Additional Information

Properties of Arithmetic Mean

If each observation is increased, decreased, multiplied, or divided by a constant k, the mean also changes by that same constant k.

Empirical Relationship

In a moderately skewed distribution, the relationship between Mean, Median, and Mode is: Mode = 3 Median − 2 Mean.

Sum of Deviations

The sum of the deviations of all observations from their Arithmetic Mean is always zero (Σ(x_i − Mean) = 0).

Shortcut Trick

Let Y share be y.

Total sum = (y + 4000) + y + (y + 8000) = 3y + 12000 = 33600.

3y = 21600 → y = 7200.

X = 7200 + 4000 = 11200; Y = 7200; Z = 7200 + 8000 = 15200.

Ratio X : Y : Z = 11200 : 7200 : 15200 = 112 : 72 : 152 = 14 : 9 : 19.

∴ The correct answer is 14 : 9 : 19.

Alternate Method

Given: Total sum = ₹33,600. X = Y + ₹4,000. Z = Y + ₹8,000.

Formula Used: Total Sum = X + Y + Z

⇒ (Y + 4000) + Y + (Y + 8000) = 33600

⇒ 3Y + 12000 = 33600

⇒ 3Y = 33600 − 12000 = 21600

⇒ Y = 21600 ÷ 3 = ₹7,200

⇒ X = 7200 + 4000 = ₹11,200

⇒ Z = 7200 + 8000 = ₹15,200

⇒ Ratio X : Y : Z = 11200 : 7200 : 15200

⇒ Dividing by 800: 11200/800 : 7200/800 : 15200/800 = 14 : 9 : 19

∴ The correct answer is 14 : 9 : 19.

Additional Information

Ratio and Proportion

Ratio is the comparison of two quantities by division. If a sum S is divided in ratio a:b:c, parts are S×a/(a+b+c), S×b/(a+b+c), and S×c/(a+b+c).

Compound Ratio

The compound ratio of (a : b) and (c : d) is (ac : bd), obtained by multiplying corresponding terms.

Inverse Ratio

The inverse ratio of a : b is 1/a : 1/b, which simplifies to b : a.

Shortcut Trick

The angle formed by the bisectors of two adjacent angles in a quadrilateral is equal to half the sum of the other two angles:

∠STR = (∠P + ∠Q) ÷ 2

⇒ 27.5° = (∠P + ∠Q) ÷ 2

⇒ ∠P + ∠Q = 55°

Since the ratio of ∠P : ∠Q is 5 : 6, the sum of ratio parts is 5 + 6 = 11 units.

⇒ 11 units = 55° ⇒ 1 unit = 5°

The difference between ∠Q and ∠P is 6 − 5 = 1 unit = 5°.

∴ The correct answer is 5°.

Alternate Method

Given:

⇒ In quadrilateral PQRS, the bisectors of ∠R and ∠S meet at point T.

⇒ ∠STR = 27.5°

⇒ Ratio of ∠P : ∠Q = 5 : 6

Formula Used:

⇒ Sum of angles in a triangle = 180°

⇒ Sum of angles in a quadrilateral = 360°

Calculations:

In ▵STR:

⇒ ∠STR + ∠TSR + ∠TRS = 180°

Since ST and RT are angle bisectors of ∠S and ∠R respectively:

⇒ ∠TSR = ∠S ÷ 2 and ∠TRS = ∠R ÷ 2

⇒ 27.5° + (∠S ÷ 2) + (∠R ÷ 2) = 180°

⇒ (∠S + ∠R) ÷ 2 = 180° − 27.5° = 152.5°

⇒ ∠S + ∠R = 305°

In quadrilateral PQRS, the sum of all interior angles is 360°:

⇒ ∠P + ∠Q + ∠R + ∠S = 360°

⇒ ∠P + ∠Q + 305° = 360°

⇒ ∠P + ∠Q = 55°

Let ∠P = 5x and ∠Q = 6x:

⇒ 5x + 6x = 55°

⇒ 11x = 55°

⇒ x = 5°

The difference between the measures of ∠Q and ∠P is:

⇒ ∠Q − ∠P = 6x − 5x = x = 5°

∴ The correct answer is 5°.

Additional Information

Angle Bisector Theorem in Quadrilaterals

The angle formed by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the other two angles: ∠STR = (∠P + ∠Q) ÷ 2.

Sum of Interior Angles of a Polygon

The sum of all interior angles of any n-sided polygon is given by (n − 2) × 180°. For a quadrilateral (n = 4), the sum is always 360°.

Angle Bisectors of Opposite Angles

If the bisectors of opposite angles of a quadrilateral meet, the angle between them is equal to half the difference of the other two angles.

Shortcut Trick

If every observation in a data set is increased by a constant k, then the average of the set also increases by k.

Current Average of 7 numbers = 23

Increase in each number = 4

New Average = 23 + 4 = 27

∴ The correct answer is 27.

Alternate Method

Given: Average of 7 numbers = 23. Each number is increased by 4.

Formula Used: Average = Sum of observations ÷ Number of observations

⇒ Total sum of 7 numbers = 23 × 7 = 161

⇒ Since each of the 7 numbers is increased by 4, the total sum increase = 7 × 4 = 28

⇒ New total sum = 161 + 28 = 189

⇒ New Average = 189 ÷ 7

⇒ New Average = 27

∴ The correct answer is 27.

Additional Information

Property of Multiplication

If each number is multiplied by a constant k, the new average becomes (Original Average × k).

Property of Subtraction

If each number is decreased by a constant k, the new average becomes (Original Average − k).

Property of Division

If each number is divided by a non-zero constant k, the new average becomes (Original Average ÷ k).

Shortcut Trick

Total profit required = ₹195.

Profit per apple = CP × Profit% = 15 × 13/100 = ₹1.95.

Number of apples = Total Profit ÷ Profit per apple.

Number of apples = 195 ÷ 1.95 = 100.

∴ The correct answer is 100.

Alternate Method

Given: Cost Price (CP) of one apple = ₹15, Profit percentage = 13%, Total Profit required = ₹195.

Formulas: Profit per unit = CP × (Profit% ÷ 100) | Number of units = Total Profit ÷ Profit per unit.

⇒ Profit on 1 apple = 15 × (13 ÷ 100) = 15 × 0.13 = ₹1.95.

⇒ Let the number of apples be n.

⇒ Total Profit = n × Profit on 1 apple.

⇒ 195 = n × 1.95.

⇒ n = 195 ÷ 1.95 = 100.

∴ The correct answer is 100.

Additional Information

Total Cost Price and Total Sales

Total CP = Number of items × CP per item. Similarly, Total SP = Number of items × SP per item.

Profit Calculation

Profit can be calculated as (Total SP − Total CP) or (Number of items × Profit per item).

Shortcut Trick

Combine all fractions: (1/5) × (1/2) × (5/9) × Number = 364

⇒ (1/18) × Number = 364

⇒ Number = 364 × 18

⇒ Number = 6552

∴ The correct answer is 6552.

Alternate Method

Given: Result = 364

Formula: Value = Result ÷ (Product of Fraction equivalents)

⇒ Let the number be x.

⇒ 20% × 50% × (5/9) × x = 364

⇒ (20/100) × (50/100) × (5/9) × x = 364

⇒ (1/5) × (1/2) × (5/9) × x = 364

⇒ (1/18) × x = 364

⇒ x = 364 × 18

⇒ x = 6552

∴ The correct answer is 6552.

Additional Information

Percentage to Fraction Conversion

To convert percentage to fraction, divide by 100. For example, 25% = 25/100 = 1/4.

Successive Percentage Application

When taking percentages of percentages, multiply the corresponding fractions or decimals directly: a% of b% of x = (a/100) × (b/100) × x.

Reverse Calculation Trick

To find the original number when a fractional part is given, multiply the result by the reciprocal of the total fraction.

Shortcut Trick

Total distance = 150 + 350 = 500 m

Time = 30 seconds

Speed (m/s) = 500 ÷ 30 = 50/3 m/s

Speed (km/hr) = (50/3) × (18/5) = 10 × 6 = 60 km/hr

∴ The correct answer is 60 km/hr.

Alternate Method

Given:

Length of Train (L₁) = 150 m

Length of Platform (L₂) = 350 m

Time taken (T) = 30 seconds

Formula Used:

Speed = Total Distance ÷ Total Time

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

Calculations:

⇒ Total Distance = L₁ + L₂ = 150 + 350 = 500 m

⇒ Speed in m/s = 500 ÷ 30 = 50/3 m/s

⇒ Speed in km/hr = (50/3) × (18/5)

⇒ Speed = 10 × 6 = 60 km/hr

∴ The correct answer is 60 km/hr.

Additional Information

Unit Conversion

1 m/s = 18/5 km/hr and 1 km/hr = 5/18 m/s.

Crossing Stationary Objects

When a train crosses a stationary object of length 'L', the total distance covered is (Length of train + L).

Relative Speed

If two objects move in the same direction, relative speed = S₁ − S₂. If opposite directions, relative speed = S₁ + S₂.

Shortcut Trick

Length = Area ÷ Breadth = 100.64 ÷ 7.4 = 13.6 cm

Perimeter = 2 × (Length + Breadth)

Perimeter = 2 × (13.6 + 7.4) = 2 × 21 = 42 cm

∴ The correct answer is 42 cm.

Alternate Method

Given:

Area of the rectangle = 100.64 cm²

Breadth (b) of the rectangle = 7.4 cm

Formula Used:

Area = Length (l) × Breadth (b)

Perimeter = 2 × (l + b)

Calculations:

⇒ 100.64 = Length × 7.4

⇒ Length = 100.64 ÷ 7.4

⇒ Length = 13.6 cm

⇒ Perimeter = 2 × (13.6 + 7.4)

⇒ Perimeter = 2 × 21

⇒ Perimeter = 42 cm

∴ The correct answer is 42 cm.

Additional Information

Diagonal of a Rectangle

The diagonal 'd' of a rectangle can be calculated using the Pythagorean theorem: d = √(l² + b²).

Properties of a Rectangle

A rectangle is a quadrilateral where all four angles are right angles (90°) and opposite sides are equal and parallel.

Square as a Special Rectangle

A square is a special type of rectangle where the length and breadth are equal (l = b = a), resulting in Area = a² and Perimeter = 4a.

Shortcut Trick

Net difference in Annual Interest Rate = 7% − 3.5% = 3.5%.

Total difference in Simple Interest for 2 years = 3.5% × 2 = 7%.

Interest Difference = 7% of Principal (₹5,00,000).

⇒ Difference = (7 ÷ 100) × 5,00,000 = ₹35,000.

∴ The correct answer is ₹35,000.

Alternate Method

Given:

Principal (P) = ₹5,00,000 for each bank.

Rate for Bank A (R₁) = 3.5% per annum.

Rate for Bank B (R₂) = 7% per annum.

Time (T) = 2 years.

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ SI paid to Bank A = (5,00,000 × 3.5 × 2) ÷ 100

⇒ SI_A = 5,000 × 7 = ₹35,000

⇒ SI paid to Bank B = (5,00,000 × 7 × 2) ÷ 100

⇒ SI_B = 5,000 × 14 = ₹70,000

⇒ Positive Difference = SI_B − SI_A

⇒ Difference = 70,000 − 35,000 = ₹35,000

∴ The correct answer is ₹35,000.

Additional Information

Simple Interest Formula

The basic formula for SI is (P × R × T) ÷ 100, where interest is calculated only on the initial principal amount throughout the duration.

Direct Difference Calculation

When principal and time are constant, the difference in interest is directly proportional to the difference in rates: ΔSI = P × (R₂ − R₁) × T ÷ 100.

Total Amount

The total amount (A) payable after the loan period is A = P + SI = P × [1 + (R × T ÷ 100)].

Shortcut Trick

Equation: 4x − 8 = 3x + 6

Direct difference method:

Difference in multiples = 4 − 3 = 1 part

Difference in constants = 6 − (−8) = 14

1 part = 14

∴ The correct answer is 14.

Alternate Method

Given:

Let the unknown number be x.

Conditions:

⇒ 4 times the number = 4x

⇒ 8 subtracted from 4 times the number = 4x − 8

⇒ 6 more than thrice the number = 3x + 6

Calculations:

⇒ 4x − 8 = 3x + 6

⇒ 4x − 3x = 6 + 8

⇒ x = 14

∴ The correct answer is 14.

Additional Information

Linear Equation in One Variable

An equation of the form ax + b = c, where a, b, and c are constants and x is the variable.

Transposition Method

Moving a term from one side of the equation to another changes its sign (+ becomes − and vice versa).

Verification Rule

Substitute the obtained value back into the original equation to ensure LHS = RHS.

Shortcut Trick

3125 = 5⁵ ⇒ x + y = 5

2187 = 3⁷ ⇒ 3x − y = 7

Add both ⇒ 4x = 12 ⇒ x = 3

y = 5 − 3 = 2

∴ 41x − 32y = 123 − 64 = 59

Alternate Method

Given:

5^x+y = 3125

3^3x−y = 2187

Formula used:

If a^m = aⁿ then m = n

Calculation:

⇒ 3125 = 5⁵ ⇒ x + y = 5 ...(1)

⇒ 2187 = 3⁷ ⇒ 3x − y = 7 ...(2)

⇒ Add (1) and (2): 4x = 12 ⇒ x = 3

⇒ Substitute in (1): 3 + y = 5 ⇒ y = 2

⇒ Required value = 41x − 32y

⇒ = 41 × 3 − 32 × 2

⇒ = 123 − 64 = 59

∴ The correct answer is 59.

Additional Information

Exponential Rule

If bases are same, equate powers: a^m = aⁿ ⇒ m = n

Standard Powers

3125 = 5⁵, 2187 = 3⁷

Linear Equations

Solve simultaneous equations using substitution or elimination

Shortcut Tip

Convert numbers into powers first to simplify quickly

Shortcut Trick

Total distance = Sum of lengths = 346 + 319 = 665 m.

Relative speed (Opposite direction) = 67 + 59 = 126 km/h.

Relative speed in m/s = 126 × (5/18) = 35 m/s.

Time = 665 ÷ 35 = 19 seconds.

∴ The correct answer is 19 seconds.

Alternate Method

Given:

Length of Train 1 (L₁) = 346 m, Length of Train 2 (L₂) = 319 m

Speed of Train 1 (S₁) = 67 km/h, Speed of Train 2 (S₂) = 59 km/h

Formula Used:

Time = (Total Length) ÷ (Relative Speed)

Relative Speed (Opposite) = S₁ + S₂

Conversion: 1 km/h = 5/18 m/s

Calculations:

⇒ Total distance = 346 + 319 = 665 m

⇒ Relative Speed = 67 + 59 = 126 km/h

⇒ Relative Speed in m/s = 126 × (5/18)

⇒ Relative Speed in m/s = 7 × 5 = 35 m/s

⇒ Time required = 665 ÷ 35

⇒ Time required = 19 seconds

∴ The correct answer is 19 seconds.

Additional Information

Relative Speed Concepts

When moving in opposite directions, speeds are added: S_rel = S₁ + S₂. When moving in the same direction, speeds are subtracted: S_rel = |S₁ − S₂|.

Unit Conversion

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

Crossing Objects

When a train crosses another train or a bridge, the total distance is the sum of the lengths of both entities.

Shortcut Trick

Put θ = 45° to simplify the expression quickly.

Numerator: sin²45° × (1 + cot²45°) = (1/2) × (1 + 1) = 1.

Denominator: sec²45° − 1 = (√2)² − 1 = 2 − 1 = 1.

Result: 1 ÷ 1 = 1. Checking options: cot²45° = 1² = 1.

∴ The correct answer is cot²θ.

Alternate Method

Given: [sin²θ(1 + cot²θ)] ÷ (sec²θ − 1)

Formula Used:

1 + cot²θ = cosec²θ

sec²θ − 1 = tan²θ

cosecθ = 1/sinθ

Calculations:

⇒ Numerator = sin²θ × cosec²θ

⇒ Numerator = sin²θ × (1/sin²θ) = 1

⇒ Denominator = tan²θ

⇒ Expression = 1/tan²θ

⇒ Expression = cot²θ

∴ The correct answer is cot²θ.

Additional Information

Fundamental Identities

sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.

Reciprocal Relations

tanθ = sinθ/cosθ and cotθ = cosθ/sinθ = 1/tanθ.

Shortcut Trick

Identify the numbers as multiples of 13: 13×1, 13×5, and 13×7.

The LCM must contain the common factor 13 and all unique prime factors 5 and 7.

LCM = 13 × 5 × 7 = 65 × 7 = 455.

∴ The correct answer is 455.

Alternate Method

Given: Numbers are 13, 65, and 91.

Formula Used: Prime Factorization Method.

⇒ Prime factorization of 13 = 13¹

⇒ Prime factorization of 65 = 5¹ × 13¹

⇒ Prime factorization of 91 = 7¹ × 13¹

⇒ LCM = Product of highest powers of all prime factors = 13¹ × 5¹ × 7¹

⇒ LCM = 13 × 35 = 455

∴ The correct answer is 455.

Additional Information

LCM and HCF Relationship

For two numbers a and b, the product of the numbers is equal to the product of their LCM and HCF: a × b = LCM(a, b) × HCF(a, b).

HCF of 13, 65, 91

The Highest Common Factor is the largest number that divides all given numbers. Here, HCF = 13.

Co-prime Numbers

If the HCF of two numbers is 1, they are called co-prime, and their LCM is simply their product.

Shortcut Trick

Convert quickly using decimal form

60% of 45 = 0.6 × 45 = 27

1/5 of 55 = 11

Difference = 27 − 11

∴ Answer = 16

Alternate Method

Given:

60% of 45 and 1/5 of 55

Formula used:

x% of y = (x/100) × y

Calculation:

⇒ 60% of 45 = (60/100) × 45 = 27

⇒ 1/5 of 55 = 55 ÷ 5 = 11

⇒ Difference = 27 − 11 = 16

∴ The correct answer is 16.

Additional Information

Percentage Formula

x% of y = (x/100) × y

Fraction to Percentage

1/5 = 20%, so 1/5 of any number = 20% of it

Quick Multiplication Trick

0.6 × 45 = (6 × 45)/10 = 27

Difference Concept

If A is greater than B, then Difference = A − B

Shortcut Trick

Let the total valid votes be 100%.

Votes for Candidate A = 40%.

Votes for Candidate B = 40% − 12% = 28%.

Votes for Candidate C = 100% − (40% + 28%) = 32%.

Wait, if Candidate B received 12% fewer votes than A (as a direct percentage of valid votes):

Votes for C = 100% − 40% − 35.2% = 24.8% (since 12% of 40 is 4.8%).

∴ The correct answer is 24.8%.

Alternate Method

Given: Total voters = 75,000. Valid votes = 95% of 75,000.

Formula: Total Valid Votes = 100% | Candidate C% = 100% − (%A + %B).

⇒ Let total valid votes cast = 100%.

⇒ Percentage of votes for Candidate A = 40%.

⇒ Candidate B received 12% fewer votes than A ⇒ 40% − (12% of 40%).

⇒ Candidate B = 40% − 4.8% = 35.2%.

⇒ Percentage of votes for Candidate C = 100% − (40% + 35.2%).

⇒ Candidate C = 100% − 75.2% = 24.8%.

∴ The correct answer is 24.8%.

Additional Information

Percentage Comparison

When a value is x% less than y, the value is calculated as y × (100 − x)/100. In this case, B = 40 × (88/100).

Valid vs Invalid Votes

Total Votes Cast = Valid Votes + Invalid Votes. Winners and losers are always calculated based on Valid Votes unless specified otherwise.

Voter Turnout

Voter turnout is the percentage of eligible voters who cast a ballot in an election. Here, 95% of 75,000 represents the actual turnout.

Shortcut Trick

In a proportion a : b :: c : d, the product of extremes equals the product of means.

P × 6 = 17 × 42

P = 17 × (42 ÷ 6) = 17 × 7 = 119

∴ The correct answer is 119.

Alternate Method

Given: P : 17 :: 42 : 6

Formula Used: If a : b :: c : d, then a/b = c/d

⇒ P/17 = 42/6

⇒ P/17 = 7

⇒ P = 7 × 17

⇒ P = 119

∴ The correct answer is 119.

Additional Information

Fourth Proportional

If four quantities a, b, c, and d are in proportion, then d is called the fourth proportional to a, b, and c, where d = (b × c) ÷ a.

Mean Proportional

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

Continued Proportion

Three numbers a, b, and c are said to be in continued proportion if a : b :: b : c, which implies b² = a × c.

Shortcut Trick

When a large sphere is melted to form small spheres, the number of laddoos is the cube of the ratio of their radii.

Ratio of radii (k) = 32 cm ÷ 2 cm = 16

Number of small laddoos = k³ = 16³

Number = 16 × 16 × 16 = 4096

∴ The correct answer is 4096 laddoos.

Alternate Method

Given:

Radius of the large laddoo (R) = 32 cm

Radius of the small laddoo (r) = 2 cm

Formula Used:

Volume of a sphere = (4/3) × π × r³

Number of laddoos = Volume of large laddoo ÷ Volume of small laddoo

Calculations:

⇒ Number = [(4/3) × π × (32)³] ÷ [(4/3) × π × (2)³]

⇒ Number = (32 × 32 × 32) ÷ (2 × 2 × 2)

⇒ Number = (32 ÷ 2) × (32 ÷ 2) × (32 ÷ 2)

⇒ Number = 16 × 16 × 16

⇒ Number = 256 × 16

⇒ Number = 4096

∴ The correct answer is 4096 laddoos.

Additional Information

Volume of a Sphere

The total space occupied by a 3D sphere is given by V = (4/3)πr³.

Surface Area of a Sphere

The total external area of a sphere is given by A = 4πr².

Volume of a Hemisphere

Half of a sphere has a volume given by V = (2/3)πr³.

Conservation of Volume

When one solid object is recast into multiple smaller objects of the same material, the total volume remains constant.