Quantitative Aptitude PYQs

Shortcut Trick

Use the column multiplication method for continuous ratios:

P = 6 × 5 × 9 = 270

Q = 11 × 5 × 9 = 495

R = 11 × 4 × 9 = 396

S = 11 × 4 × 7 = 308

∴ The correct answer is 270 : 495 : 396 : 308.

Alternate Method

Given:

⇒ P : Q = 6 : 11

⇒ Q : R = 5 : 4

⇒ R : S = 9 : 7

Visual Method: Fill empty spaces with adjacent numbers

P Q R S

6

11

11

11

5

5

4

4

9

9

9

7

× Multiply columns vertically

270

495

396

308

Ratio → 270 : 495 : 396 : 308

Calculations:

⇒ Equalize Q in the first two ratios (LCM of 11 and 5 is 55).

⇒ P : Q = (6 × 5) : (11 × 5) = 30 : 55

⇒ Q : R = (5 × 11) : (4 × 11) = 55 : 44

⇒ Combining them, P : Q : R = 30 : 55 : 44

⇒ Now, equalize R in P : Q : R and R : S (LCM of 44 and 9 is 396).

⇒ P : Q : R = (30 × 9) : (55 × 9) : (44 × 9) = 270 : 495 : 396

⇒ R : S = (9 × 44) : (7 × 44) = 396 : 308

⇒ Combine all to get the final ratio:

⇒ P : Q : R : S = 270 : 495 : 396 : 308

∴ The correct answer is 270 : 495 : 396 : 308.

Additional Information

Compound Ratio

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

Duplicate & Sub-duplicate Ratio

The duplicate ratio of a : b is a² : b². The sub-duplicate ratio is √a : √b.

Componendo and Dividendo

If a/b = c/d, then (a + b) / (a − b) = (c + d) / (c − d). This rule is heavily used to simplify complex algebraic fraction statements.

Shortcut Trick

Total value ratio = (10 × 1) + (5 × 2) + (2 × 3) = 10 + 10 + 6 = 26 units.

Total given money = ₹390.

Value of 1 ratio unit = 390 ÷ 26 = 15.

Multiply the initial coin ratio (1 : 2 : 3) by 15.

Number of coins = (1 × 15), (2 × 15), and (3 × 15) = 15, 30, and 45.

∴ The correct answer is 15 coins of ₹10, 30 of ₹5, and 45 of ₹2.

Alternate Method

Given:

Denominations of coins = ₹10, ₹5, and ₹2

Ratio of number of coins = 1 : 2 : 3

Total amount in the bag = ₹390

₹10

₹5

₹2

Ratio: x Ratio: 2x Ratio: 3x

Value: 10x

Value: 10x

Value: 6x

Total Value = 10x + 10x + 6x = 26x = ₹390

Calculation:

⇒ Let the number of ₹10, ₹5, and ₹2 coins be x, 2x, and 3x respectively.

⇒ The values of these coins will be ₹(10 × x), ₹(5 × 2x), and ₹(2 × 3x) respectively.

⇒ Total value in terms of x = 10x + 10x + 6x = 26x

⇒ We are given that the total amount is ₹390.

⇒ 26x = 390

⇒ x = 390 ÷ 26 = 15

⇒ Number of ₹10 coins = x = 15

⇒ Number of ₹5 coins = 2x = 2 × 15 = 30

⇒ Number of ₹2 coins = 3x = 3 × 15 = 45

∴ The correct answer is 15 coins of ₹10, 30 of ₹5, and 45 of ₹2.

Additional Information

Total Value of Coins

The total value of any specific coin denomination is found by multiplying the face value of the coin by the number of coins of that denomination.

Coin Ratio vs Value Ratio

If the ratio of the number of coins is given, multiply it by their respective denominations to get the ratio of their values. Conversely, if the ratio of values is given, divide by their denominations to get the ratio of the number of coins.

Proportion Rule

When values are given in a ratio a : b : c, they can be represented as ax, bx, and cx where x is a common multiplier.

Shortcut Trick

Convert the fractions into decimals to make calculation faster.

Total capacity = 5¾ = 5.75 L

Total drained = 2.375 × 2 = 4.75 L

Remaining water = 5.75 − 4.75 = 1 L

∴ The correct answer is 1 liter.

Alternate Method

Given:

Total water = 5¾ L

Water drained each time = 2.375 L

Number of times drained = 2

Drained: 4.75 LLeft: 1 LTotal Water: 5.75 L

Calculations:

⇒ Total water = 23/4 L

⇒ Total water drained = 2 × (19/8) = 19/4 L

⇒ Remaining water = Total water − Total water drained

⇒ Remaining water = 23/4 − 19/4 = 4/4 = 1 L

∴ The correct answer is 1 liter.

Additional Information

Converting Mixed Fractions

To convert a mixed fraction a(b/c) to an improper fraction, multiply the whole number by the denominator and add the numerator: (a × c + b) / c.

Fraction to Decimal Conversion

Common standard conversions: 1/2 = 0.5, 1/4 = 0.25, 3/4 = 0.75, 1/8 = 0.125, 3/8 = 0.375. Memorizing these saves time in exams.

Operations on Fractions

When adding or subtracting fractions, always take the Least Common Multiple (LCM) of the denominators to equate them before computing the numerators.

Shortcut Trick

Solve Numerator: 2/3 + (4/5 × 7/2) = 2/3 + 14/5 = 52/15

Solve Denominator: 3/2 − (15/24) = 3/2 − 5/8 = 7/8

Divide them: 52/15 ÷ 7/8 = 52/15 × 8/7 = 416/105

∴ The correct answer is 416/105.

Alternate Method

Given:

Expression: [(2/3) + (4/5 ÷ 2/7)] ÷ [(3/2) − (5/6 × 3/4)]

Formula Used:

BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Calculations:

⇒ Numerator = (2/3) + (4/5 ÷ 2/7)

⇒ Numerator = (2/3) + (4/5 × 7/2)

⇒ Numerator = (2/3) + (14/5)

⇒ Numerator = (10/15) + (42/15) = 52/15

⇒ Denominator = (3/2) − (5/6 × 3/4)

⇒ Denominator = (3/2) − (15/24)

⇒ Denominator = (3/2) − (5/8)

⇒ Denominator = (12/8) − (5/8) = 7/8

⇒ Result = (52/15) ÷ (7/8)

⇒ Result = (52/15) × (8/7)

⇒ Result = (52 × 8) / (15 × 7)

⇒ Result = 416/105

∴ The correct answer is 416/105.

Additional Information

BODMAS Rule

Dictates the correct sequence of operations in mathematical expressions: Brackets first, then Orders (roots/powers), Division, Multiplication, Addition, and finally Subtraction.

Fraction Division

To divide by a fraction, multiply by its reciprocal: (a/b) ÷ (c/d) = (a/b) × (d/c).

Fraction Addition and Subtraction

Find the Least Common Multiple (LCM) of the denominators to ensure fractions are equivalent before proceeding to add or subtract their respective numerators.

Shortcut Trick

Convert decimals to fractions for faster calculation.

⇒ (5/2 + 18/5) ÷ 3/4 + (5/4 × 4/5)

⇒ (61/10) × 4/3 + 1

⇒ 122/15 + 1 = 137/15

∴ The correct answer is 137/15.

Alternate Method

Given:

(2 1/2 + 3.6) ÷ 0.75 + (1.25 × 4/5)

Method Used:

BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Calculations:

⇒ Step 1: Solve the brackets first.

⇒ First bracket: 2 1/2 + 3.6 = 2.5 + 3.6 = 6.1

⇒ Second bracket: 1.25 × 4/5 = 1.25 × 0.8 = 1

⇒ Step 2: Substitute values back into the expression.

⇒ 6.1 ÷ 0.75 + 1

⇒ Step 3: Perform division.

⇒ 6.1 ÷ (3/4) = 6.1 × 4/3

⇒ 24.4 / 3 = 244 / 30 = 122 / 15

⇒ Step 4: Perform addition.

⇒ 122 / 15 + 1 = (122 + 15) / 15

⇒ 137 / 15

∴ The correct answer is 137/15.

Additional Information

BODMAS Rule

Always strictly follow the sequence of operations: Brackets → Orders → Division → Multiplication → Addition → Subtraction to ensure the final result is accurate.

Fraction to Decimal Conversion

Converting common decimals into fractions (e.g., 0.75 to 3/4 and 1.25 to 5/4) allows division to become simple multiplication by the reciprocal.

Adding Fractions

To add a whole number to a fraction, express the whole number as a fraction with the same denominator: a/b + c = (a + b × c) / b.

Shortcut Trick

Time Ratio = Profit Ratio ÷ Investment Ratio

Time Ratio of A : B = (10 ÷ 5) : (9 ÷ 6) = 2 : 1.5 = 4 : 3

Here, B invested for 12 months, which corresponds to 3 parts in the ratio.

So, A's investment time (4 parts) = (12 ÷ 3) × 4 = 16 months.

∴ The correct answer is 16 months.

Alternate Method

Given:

Ratio of investments of A and B = 5 : 6

Ratio of profits of A and B = 10 : 9

Time period of investment for B = 12 months

Formula Used:

Profit = Investment × Time

⇒ Profit Ratio = (Capital of A × Time of A) : (Capital of B × Time of B)

Partnership: Capital × Time = Profit

Partner A

Capital = 5 units Time = t months Profit Ratio = 10

Partner B

Capital = 6 units Time = 12 months Profit Ratio = 9 VS

Calculations:

Let the time period for A's investment be T_A months.

⇒ P_A / P_B = (I_A × T_A) / (I_B × T_B)

⇒ 10 / 9 = (5 × T_A) / (6 × 12)

⇒ 10 / 9 = 5T_A / 72

⇒ 10 × 72 = 9 × 5T_A

⇒ 720 = 45T_A

⇒ T_A = 720 ÷ 45 = 16

∴ The correct answer is 16 months.

Additional Information

Simple Partnership

When investments of all the partners are for the same time period, the profit or loss is distributed purely in the ratio of their original capital investments.

Compound Partnership

When investments of the partners are for different time periods, the profit or loss is distributed in the ratio of the product of their capital and respective time periods (Profit Ratio = Capital × Time).

Working and Sleeping Partner

A working partner actively manages the business and usually receives a fixed salary or a specific percentage of the profit before the final division. A sleeping partner simply invests money and only receives a share of the remaining profit based on their capital and time invested.

Shortcut Trick

Removed Number = Old Average + Remaining Numbers × Decrease in Average

⇒ Removed Number = 56 + 8 × (56 − 54)

⇒ Removed Number = 56 + 8 × 2

⇒ Removed Number = 56 + 16

⇒ Removed Number = 72

∴ The correct answer is 72.

Alternate Method

Given:

Average of 9 numbers = 56

Average of remaining 8 numbers (after one is removed) = 54

Formula Used:

Total Sum = Average × Number of items

Total Sum of 9 numbers = 9 × 56 = 504

Sum of 8 numbers = 8 × 54 = 432

72

Removed

Calculation:

⇒ Initial total sum of 9 numbers = 9 × 56 = 504

⇒ New total sum of remaining 8 numbers = 8 × 54 = 432

⇒ The removed number = Initial total sum − New total sum

⇒ The removed number = 504 − 432 = 72

∴ The correct answer is 72.

Additional Information

Averages

The average is the central value of a set of numbers, calculated as the sum of all values divided by the number of values: Average = Sum of observations ÷ Total number of observations.

Deviation Method

The net deviation from the average across all terms is always zero. This property helps in quickly finding missing numbers, replacing numbers, or discovering new averages without multiplying large values.

Weighted Average

When combining groups, the combined average is calculated as (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n represents the number of items and A is the average of each respective group.

Shortcut Trick

Average of Truck 1 = (18 + 20 + 22 + 21 + 19) ÷ 5 = 100 ÷ 5 = 20 litres

Average of Truck 2 = (25 + 24 + 26 + 23 + 24) ÷ 5 = 122 ÷ 5 = 24.4 litres

Overall Average = (20 + 24.4) ÷ 2 = 44.4 ÷ 2 = 22.2 litres

∴ The correct answer is 22.2 litres.

Alternate Method

Given:

Fuel usage of Truck 1 for 5 days: 18, 20, 22, 21, 19

Fuel usage of Truck 2 for 5 days: 25, 24, 26, 23, 24

Formula Used:

Average = Sum of all observations ÷ Total number of observations

Calculations:

⇒ Total fuel used by Truck 1 = 18 + 20 + 22 + 21 + 19 = 100 litres

⇒ Total fuel used by Truck 2 = 25 + 24 + 26 + 23 + 24 = 122 litres

⇒ Total fuel used by both trucks = 100 + 122 = 222 litres

⇒ Total number of observations = 5 days × 2 trucks = 10

⇒ Overall average = 222 ÷ 10 = 22.2 litres

∴ The correct answer is 22.2 litres.

Additional Information

Combined Average

If two groups have sizes n₁ and n₂ with averages A₁ and A₂, the combined average is calculated as (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Data Interpretation (Tabulation)

In tabular data interpretation, it is often faster to find sums of individual rows or columns first before computing total averages, avoiding multiple complex calculations.

Percentage Change

Frequently tested in DI questions. Percentage change is mathematically given as ((New Value − Old Value) ÷ Old Value) × 100.

Shortcut Trick

Rate = 10% = 1 / 10. Multiplying factor = 11 / 10.

Ratio of Principal to Amount for 2 years = (10)² : (11)² = 100 : 121.

Here, 121 units correspond to ₹12100.

1 unit = ₹12100 ÷ 121 = ₹100.

Principal = 100 units = 100 × ₹100 = ₹10000.

∴ The correct answer is ₹10000.

Alternate Method

Given:

Amount (A) = ₹12100

Time (T) = 2 years

Rate (R) = 10% per annum

Formula Used:

A = P × (1 + R / 100)^T

Where A = Amount, P = Principal, R = Rate, and T = Time.

+10% CI +10% CI

Year 0 Year 1 Year 2

P = ? A = ₹12100

Calculations:

⇒ 12100 = P × (1 + 10 / 100)²

⇒ 12100 = P × (1 + 1 / 10)²

⇒ 12100 = P × (11 / 10)²

⇒ 12100 = P × (121 / 100)

⇒ P = (12100 × 100) ÷ 121

⇒ P = 100 × 100 = 10000

∴ The correct answer is ₹10000.

Additional Information

Effective Compound Interest Rate

For 2 years, the effective CI rate can be found using the successive percentage formula: R + R + (R × R) / 100. For 10%, the effective rate is 10 + 10 + (10 × 10) / 100 = 21%.

Difference Between SI and CI (2 Years)

The difference between Compound Interest and Simple Interest for exactly 2 years can be directly calculated using the formula: Difference = P × (R / 100)².

Half-Yearly Compounding

When interest is compounded half-yearly, the rate of interest is halved (R / 2) and the given time period is doubled (2T) because interest is computed every 6 months.

Shortcut Trick

Find the profit or loss per kg for each variety relative to the selling price.

Profit on 50 kg rice = (48 − 40) × 50 = +8 × 50 = +₹400

Loss on 20 kg rice = (48 − 55) × 20 = −7 × 20 = −₹140

Net Profit = 400 − 140 = ₹260

∴ The correct answer is Profit of ₹260.

Alternate Method

Given:

Quantity 1 = 50 kg, Rate 1 = ₹40/kg

Quantity 2 = 20 kg, Rate 2 = ₹55/kg

Selling rate of mixture = ₹48/kg

Formula Used:

Total CP = (Q₁ × R₁) + (Q₂ × R₂)

Profit = Total SP − Total CP

50 kg @ ₹40 CP: ₹2000

20 kg @ ₹55 CP: ₹1100

Total CP ₹3100

70 kg @ ₹48 SP: ₹3360

Profit = SP − CP = 3360 − 3100 = ₹260

Calculations:

⇒ Cost Price of 50 kg of rice = 50 × 40 = ₹2000

⇒ Cost Price of 20 kg of rice = 20 × 55 = ₹1100

⇒ Total Cost Price (CP) = 2000 + 1100 = ₹3100

⇒ Total quantity of mixture = 50 + 20 = 70 kg

⇒ Total Selling Price (SP) of the mixture = 70 × 48 = ₹3360

⇒ Since SP > CP, there is a profit.

⇒ Profit = Total SP − Total CP = 3360 − 3100 = ₹260

∴ The correct answer is Profit of ₹260.

Additional Information

Profit and Loss Basic Formula

Profit occurs when Selling Price > Cost Price. Profit = SP − CP.

Profit and Loss Percentage

Profit % = (Profit ÷ CP) × 100. Loss % = (Loss ÷ CP) × 100.

Weighted Average Concept

Average Cost Price per unit of a mixture is calculated as (Q₁ × C₁ + Q₂ × C₂) ÷ (Q₁ + Q₂).

Shortcut Trick

Intermediate Price after 1^st discount =  ₹20,000 × 85/100 = ₹17,000

Value of 2^nd discount = ₹17,000 − ₹15,300 = ₹1,700

Percentage of 2^nd discount = (1700 ÷ 17000) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Marked Price (MP) = ₹20,000

First discount = 15%

Final Selling Price (SP) = ₹15,300

Formula Used:

Price after discount = MP × (100 − D%)/100

Discount % = (Discount Amount ÷ Base Price) × 100

₹20,000 Marked Price

− 15%

₹17,000

2^nd Discount (−₹1,700)

₹15,300 Selling Price

Calculations:

⇒ Price after 1^st discount = 20000 × (100 − 15)/100

⇒ Price after 1^st discount = 20000 × 85/100 = ₹17,000

⇒ 2^nd discount amount = Price after 1^st discount − Final SP

⇒ 2^nd discount amount = 17000 − 15300 = ₹1,700

⇒ 2^nd discount percentage = (1700 ÷ 17000) × 100

⇒ 2^nd discount percentage = 1/10 × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Equivalent Successive Discount

For two successive discounts of a% and b%, the single equivalent discount formula is (a + b − ab/100)%.

Marked Price and Cost Price Relation

MP/CP = (100 + Profit%) ÷ (100 − Discount%). This directly gives the ratio of Marked Price to Cost Price without knowing the Selling Price.

Multiplier Method for Successive Discounts

Final Selling Price = Marked Price × (1 − d₁/100) × (1 − d₂/100) × ...

Shortcut Trick

Let the Cost Price (CP) be 100.

Since the product is marked 30% above CP, Marked Price (MP) = 130.

With a profit of 17%, Selling Price (SP) = 117.

Discount = MP − SP = 130 − 117 = 13.

Discount % = (13 ÷ 130) × 100 = 10%.

∴ The correct answer is 10%.

Alternate Method

Given:

Mark-up percentage = 30%

Profit percentage = 17%

Formula Used:

MP = CP × (1 + Mark-up% ÷ 100)

SP = CP × (1 + Profit% ÷ 100)

Discount % = [(MP − SP) ÷ MP] × 100

CP: 100

SP: 117

MP: 130

Mark Up: +30%

Profit: +17%

Discount: ?

Calculations:

Let the Cost Price (CP) of the product be 100.

⇒ Marked Price (MP) = 100 × (1 + 30 ÷ 100) = 130

⇒ Selling Price (SP) = 100 × (1 + 17 ÷ 100) = 117

⇒ Discount given = MP − SP = 130 − 117 = 13

⇒ Rate of Discount = (Discount ÷ MP) × 100

⇒ Rate of Discount = (13 ÷ 130) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Direct Relation Between CP and MP

There is a direct ratio relationship connecting Cost Price, Marked Price, Profit% and Discount%: CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%).

Successive Discounts

If two successive discounts of a% and b% are offered, the single equivalent discount is calculated as: (a + b − a × b ÷ 100)%.

Dishonest Dealer Concept

When a dealer cheats by using false weights while selling at CP, the profit percentage is calculated as: (Error ÷ True Value) × 100.

Shortcut Trick

Alcohol from Container A = (4 ÷ 5) × 10 = 8 L and Water from A = 10 − 8 = 2 L

Alcohol from Container B = (5 ÷ 7) × 14 = 10 L and Water from B = 14 − 10 = 4 L

Total Alcohol in new mixture = 8 + 10 = 18 L

Total Water in new mixture = 2 + 4 = 6 L

Required Ratio of Alcohol to Water = 18 : 6 = 3 : 1

∴ The correct answer is 3:1.

Alternate Method

Given:

Ratio of alcohol to water in Container A = 4:1

Ratio of alcohol to water in Container B = 5:2

Mixture drawn from Container A = 10 liters

Mixture drawn from Container B = 14 liters

Formula Used:

Quantity of component = (Ratio of component ÷ Total ratio parts) × Total quantity

Container A 10 L (Ratio 4:1) Alc: 8L | Wat: 2L

Container B 14 L (Ratio 5:2) Alc: 10L | Wat: 4L

New Mixture Alc: 18L | Wat: 6L Ratio → 3:1

Calculations:

⇒ In Container A, the sum of ratio parts = 4 + 1 = 5

⇒ Quantity of alcohol drawn from A = (4 ÷ 5) × 10 = 8 liters

⇒ Quantity of water drawn from A = (1 ÷ 5) × 10 = 2 liters

⇒ In Container B, the sum of ratio parts = 5 + 2 = 7

⇒ Quantity of alcohol drawn from B = (5 ÷ 7) × 14 = 10 liters

⇒ Quantity of water drawn from B = (2 ÷ 7) × 14 = 4 liters

⇒ Total alcohol in the third container = 8 + 10 = 18 liters

⇒ Total water in the third container = 2 + 4 = 6 liters

⇒ Ratio of alcohol to water in the new mixture = 18 : 6

⇒ Simplifying the ratio by dividing both sides by 6, we get 3 : 1

∴ The correct answer is 3:1.

Additional Information

Multiple Mixtures Concept

When components from multiple different mixtures are mixed together, individually calculate the absolute quantities of each component from every source. Sum these respective quantities up to find the final total amount before constructing the new ratio.

Alligation Method

Used to find the ratio in which two mixtures must be mixed to obtain a desired concentration. The fundamental rule relates the quantities: (Quantity of Cheaper) ÷ (Quantity of Dearer) = (Price of Dearer − Mean Price) ÷ (Mean Price − Price of Cheaper).

Replacement Method

If a certain volume 'y' is replaced 'n' times from a total pure volume 'x', the final quantity of the original pure substance left is given by the formula: Final Quantity = Initial Quantity × (1 − y/x)ⁿ.

Shortcut Trick

For any 3D shape, if its linear dimensions (like radius) are multiplied by a factor 'n', its surface area becomes n² times and volume becomes n³ times.

Here, the radius is doubled, meaning the scaling factor n = 2.

Ratio of New Surface Area to Original = 2² : 1 = 4 : 1

Ratio of New Volume to Original = 2³ : 1 = 8 : 1

∴ The correct answer is 4:1 and 8:1.

Alternate Method

r Original Sphere SA = 4×π×r²

2r New Sphere SA = 16×π×r²

Given:

Let the original radius of the sphere be r.

The new radius of the sphere is doubled, so new radius = 2r.

Formula Used:

Surface Area of a Sphere = 4 × π × radius²

Volume of a Sphere = 4/3 × π × radius³

Calculations:

⇒ Original Surface Area = 4 × π × r²

⇒ New Surface Area = 4 × π × (2r)² = 16 × π × r²

⇒ Ratio of New Surface Area to Original = (16 × π × r²) : (4 × π × r²) = 4 : 1

⇒ Original Volume = 4/3 × π × r³

⇒ New Volume = 4/3 × π × (2r)³ = 4/3 × π × 8r³ = 8 × (4/3 × π × r³)

⇒ Ratio of New Volume to Original = 8 × (4/3 × π × r³) : (4/3 × π × r³) = 8 : 1

∴ The correct answer is 4:1 and 8:1.

Additional Information

Scaling Principle

For any 3D geometric solid, if all its linear dimensions are scaled by a factor 'k', its total surface area directly increases by a factor of k², and its volume increases by a factor of k³.

Hemisphere Volume

When a sphere is cut exactly in half, it forms a hemisphere. The volume of a solid hemisphere is given by 2/3 × π × r³.

Hemisphere Surface Area

The Curved Surface Area (CSA) of a hemisphere is 2 × π × r², whereas the Total Surface Area (TSA), which includes the flat circular base, is 3 × π × r².

Shortcut Trick

Since the hemisphere and the cone have equal volumes and share the same base radius R, we directly equate their volume formulas.

⇒ (2/3) × πR³ = (1/3) × πR²h

⇒ 2R = h

∴ The correct answer is 2R.

Alternate Method

Given:

Volume of Hemisphere = Volume of Cone

Radius of Hemisphere = Radius of Cone = R

R Hemisphere V = (2/3)×π×R³

R

h Cone V = (1/3)×π×R²h

=

Formula Used:

Volume of a Hemisphere = (2/3) × πr³

Volume of a Cone = (1/3) × πr²h

Calculation:

Equating the volume of the hemisphere and the cone:

⇒ (2/3) × πR³ = (1/3) × πR²h

Multiplying both sides by 3 to remove the fraction:

⇒ 2 × πR³ = πR²h

Dividing both sides by πR²:

⇒ 2R = h

⇒ h = 2R

∴ The correct answer is 2R.

Additional Information

Cylinder vs Cone Relationship

A cylinder with base radius r and height h has a volume of V = πr²h. Its volume is exactly 3 times that of a cone with the identical base and height.

Sphere vs Hemisphere Volume

A complete sphere has a volume of V = (4/3)πr³, while a hemisphere (which is half a sphere) has a volume of V = (2/3)πr³.

Surface Area of a Cone

The total surface area of a cone is calculated as TSA = πr(r + l), where l is the slant height given by l = √(r² + h²).

Shortcut Trick

When a hemisphere is melted into 2 equal cones, their total volumes remain equal.

Volume of Hemisphere = Volume of 2 Cones

2/3 × π × R³ = 2 × (1/3 × π × r² × h)

⇒ R³ = r² × h

⇒ 10³ = 5² × h ⇒ 1000 = 25 × h ⇒ h = 40

∴ The correct answer is 40 cm.

Alternate Method

Given:

Radius of the solid hemisphere (R) = 10 cm

Radius of each newly formed cone (r) = 5 cm

Number of cones formed = 2

Formula Used:

Volume of a hemisphere = 2/3 × π × R³

Volume of a cone = 1/3 × π × r² × h

R = 10 Hemisphere

Melted

r = 5 h = ?

r = 5 h = ?

Two Equal Cones

Calculations:

⇒ According to the problem, Total Volume of 2 Cones = Volume of the Hemisphere

⇒ 2 × (1/3 × π × r² × h) = 2/3 × π × R³

⇒ 2/3 × π × (5)² × h = 2/3 × π × (10)³

⇒ Canceling (2/3 × π) from both sides:

⇒ 25 × h = 1000

⇒ h = 1000 ÷ 25

⇒ h = 40 cm

∴ The correct answer is 40 cm.

Additional Information

Principle of Recasting

When a solid object is melted and recast into new shapes, the total volume always remains constant regardless of the new shape forms.

Hemisphere Formulas

Volume = 2/3 × π × r³ | Curved Surface Area = 2 × π × r² | Total Surface Area = 3 × π × r²

Cone Formulas

Volume = 1/3 × π × r² × h | Slant height (l) = √(r² + h²) | Curved Surface Area = π × r × l

Shortcut Trick

For any regular pyramid, the lateral surface area can be calculated directly using its base perimeter and slant height.

Lateral Surface Area = ½ × Base Perimeter × Slant Height

⇒ LSA = ½ × 40 × 10

⇒ LSA = 200 cm²

∴ The correct answer is 200 cm².

Alternate Method

Given:

Base perimeter = 40 cm

Slant height (l) = 10 cm

l = 10 cm a = 10 cm Base Perimeter = 40 cm

Formula Used:

Area of one lateral triangle = ½ × Base × Slant height

Side of square base (a) = Perimeter ÷ 4

Calculation:

⇒ Side of the base (a) = 40 ÷ 4 = 10 cm

⇒ A square pyramid has 4 identical lateral triangular faces.

⇒ Area of one triangular face = ½ × 10 × 10

⇒ Area of one triangular face = 50 cm²

⇒ Total Lateral Surface Area = 4 × Area of one face

⇒ Total LSA = 4 × 50 = 200 cm²

∴ The correct answer is 200 cm².

Additional Information

Total Surface Area (TSA)

The TSA of a regular pyramid is the sum of its Lateral Surface Area and Base Area. For a square pyramid of base side 'a': TSA = LSA + a².

Volume of a Pyramid

The volume of any pyramid is one-third of the product of its base area and true vertical height. V = ⅓ × Base Area × h.

Relation between Heights

For a regular square pyramid, the true vertical height (h), slant height (l), and base side (a) form a right-angled triangle. By Pythagoras theorem: l² = h² + (a/2)².

Shortcut Trick

Ratio of areas = (Ratio of radii)²

Area ratio = 2² : 3² = 4 : 9

Given 4 units = 100π cm² ⇒ 1 unit = 25π cm²

Area of larger pizza = 9 units = 9 × 25π = 225π cm²

∴ The correct answer is 225π cm².

Alternate Method

Given:

Ratio of radii of two circular pizzas, r₁ : r₂ = 2 : 3

Area of smaller pizza = 100π cm²

Formula Used:

Area of a circle = πr²

r₁ = 2x A₁ = 100π

r₂ = 3x A₂ = ?

Calculations:

Let the radii of the smaller and larger pizzas be 2x and 3x respectively.

⇒ Area of smaller pizza = π(2x)²

⇒ 4πx² = 100π

⇒ 4x² = 100

⇒ x² = 25

⇒ Area of larger pizza = π(3x)²

⇒ 9πx²

Substituting the value of x²:

⇒ Area of larger pizza = 9π × 25

⇒ Area of larger pizza = 225π cm²

∴ The correct answer is 225π cm².

Additional Information

Area of a Sector

The area of a sector of a circle with radius r and central angle θ is given by A = (θ ÷ 360) × πr².

Circumference of a Circle

The total boundary length of a circle is its circumference, calculated as C = 2πr.

Similar Figures Property

For any two similar geometric figures, the ratio of their areas is strictly equal to the square of the ratio of their corresponding linear dimensions (e.g., radii, diameters, or sides).

Shortcut Trick

The intersection of medians of a triangle is called its centroid.

The centroid always divides a median in the ratio 2 : 1 (from vertex to base).

Length of AG = (2/3) × Total length of median AD

AG = (2/3) × 12 cm = 8 cm

∴ The correct answer is 8 cm.

Alternate Method

Given:

Length of median AD = 12 cm

Intersection of medians AD and BE is G.

Concept Used:

The point where the medians of a triangle intersect is called the centroid.

The centroid divides each median in the ratio 2 : 1, measuring from the vertex to the midpoint of the opposite side.

 A B C D E G 2 Units 1 Unit

Calculations:

⇒ AG : GD = 2 : 1

⇒ Total parts on median AD = 2 + 1 = 3 parts

⇒ 3 parts = 12 cm

⇒ 1 part = 12 ÷ 3 = 4 cm

⇒ Length of AG = 2 parts = 2 × 4 cm

⇒ AG = 8 cm

∴ The correct answer is 8 cm.

Additional Information

Centroid

The point of intersection of the three medians of a triangle. It divides each median in the ratio 2 : 1. The centroid always lies strictly inside the triangle.

Incenter

The point of intersection of the internal angle bisectors of the triangle. It is equidistant from all three sides and forms the center of the inscribed circle (incircle).

Circumcenter

The point of intersection of the perpendicular bisectors of the sides. It is equidistant from all three vertices and forms the center of the circumscribed circle (circumcircle).

Orthocenter

The point of intersection of the three altitudes (perpendiculars dropped from vertices to opposite sides) of the triangle.

Shortcut Trick

In the given right-angled triangles ABC and DEF, we are given:

∠B = ∠E = 90° (Right angles)

AC = DF (Hypotenuses are equal)

AB = DE (One pair of corresponding sides are equal)

Since we have matching right angles, equal hypotenuses, and one pair of equal sides, the RHS (Right angle − Hypotenuse − Side) congruence criterion is perfectly satisfied.

∴ The correct answer is Yes, by RHS.

Alternate Method

Given:

⇒ Two right-angled triangles, ΔABC and ΔDEF.

⇒ ∠B = ∠E = 90°

⇒ Hypotenuse AC = Hypotenuse DF

⇒ Side AB = Side DE

A B C

D E F

Proof:

⇒ In right-angled ΔABC and right-angled ΔDEF:

⇒ ∠B = ∠E = 90° (Given right angles)

⇒ AC = DF (Given equal hypotenuses)

⇒ AB = DE (Given equal corresponding sides)

⇒ According to the RHS (Right angle − Hypotenuse − Side) congruence rule, if the hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of another right-angled triangle, then the two triangles are congruent.

⇒ ∴ ΔABC ≅ ΔDEF directly by the RHS congruence criterion.

∴ The correct answer is Yes, by RHS.

Additional Information

SSS (Side-Side-Side) Congruence

If all three sides of one triangle are exactly equal to the corresponding three sides of another triangle, the two triangles are congruent.

SAS (Side-Angle-Side) Congruence

If two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, they are congruent.

ASA (Angle-Side-Angle) Congruence

If two angles and the included side of one triangle are equal to the corresponding two angles and the included side of another triangle, they are congruent.

AAS (Angle-Angle-Side) Congruence

If two angles and a non-included side of one triangle are equal to the corresponding parts of another triangle, the two triangles are congruent.