Shortcut Trick

Apply the BODMAS rule and use fraction equivalents for faster decimal multiplication.

⇒ (2.4 + 3.125) × 0.25 − (0.75 ÷ 0.3)

⇒ 5.525 × (1/4) − 2.5

⇒ 1.38125 − 2.5 = −1.11875

∴ The correct answer is -1.11875.

Alternate Method

Given:

Expression: (2.4 + 3.125) × 0.25 − (0.75 ÷ 0.3)

Formula Used:

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

Calculation:

⇒ Step 1: Solve the terms inside the parentheses first.

⇒ First bracket: (2.4 + 3.125) = 5.525

⇒ Second bracket: (0.75 ÷ 0.3) = 7.5 ÷ 3 = 2.5

⇒ Step 2: Substitute the solved bracket values back into the main expression.

⇒ 5.525 × 0.25 − 2.5

⇒ Step 3: Perform the multiplication operation.

⇒ 5.525 × 0.25 = 1.38125

⇒ Step 4: Perform the final subtraction operation.

⇒ 1.38125 − 2.5 = −1.11875

∴ The correct answer is -1.11875.

Additional Information

BODMAS Rule

It defines the sequence of operations for mathematical expressions: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction. Expressions inside brackets must always be evaluated first.

Decimal Division Technique

When dividing by a decimal number, shift the decimal point to the right in both the dividend and divisor until the divisor becomes a whole number. For instance, 0.75 ÷ 0.3 simplifies to 7.5 ÷ 3.

Fraction Equivalents of Decimals

Converting decimals to fractions often speeds up calculations. For example, multiplying any value by 0.25 is mathematically equivalent to dividing it by 4, and multiplying by 0.5 is equivalent to dividing by 2.

Shortcut Trick

Given 20% of (P + Q) = 50% of (P − Q).

Write as a ratio: (P + Q) ÷ (P − Q) = 50 ÷ 20 = 5 ÷ 2.

Apply Componendo and Dividendo directly:

P ÷ Q = (5 + 2) ÷ (5 − 2) = 7 ÷ 3.

∴ The correct answer is 7 : 3.

Alternate Method

Given:

20% of (P + Q) = 50% of (P − Q)

20(P + Q)

50(P − Q)

P : Q = 7 : 3

Calculations:

⇒ (20 ÷ 100) × (P + Q) = (50 ÷ 100) × (P − Q)

⇒ 20(P + Q) = 50(P − Q)

⇒ 2(P + Q) = 5(P − Q)

⇒ 2P + 2Q = 5P − 5Q

⇒ 2Q + 5Q = 5P − 2P

⇒ 7Q = 3P

⇒ P ÷ Q = 7 ÷ 3

⇒ P : Q = 7 : 3

∴ The correct answer is 7 : 3.

Additional Information

Componendo and Dividendo Rule

If a ÷ b = c ÷ d, then (a + b) ÷ (a − b) = (c + d) ÷ (c − d). This rule is very useful to directly extract the individual ratio values from the sum and difference equations without lengthy cross-multiplication.

Percentage Equality

When x% of A = y% of B, the relationship can be instantly simplified to A ÷ B = y ÷ x by cancelling out the common percentage base (100) on both sides.

Compound Ratio

When multiple ratios are combined by multiplication, the ratios a:b and c:d become ac:bd. This is often used in proportional reasoning.

Shortcut Trick

Assume the quantities of the two varieties are 3 kg and 1 kg.

Total Cost Price = (80 × 3) + (100 × 1) = ₹340 for 4 kg.

Mean Cost Price = 340 ÷ 4 = ₹85 per kg.

Given Selling Price (SP) = ₹91.80 per kg.

Gain % = [(91.80 − 85) ÷ 85] × 100 = (6.80 ÷ 85) × 100 = 8%.

∴ The correct answer is 8% Gain.

Alternate Method

Given:

Cost of 1^st variety = ₹80/kg

Cost of 2^nd variety = ₹100/kg

Ratio of mixing = 3 : 1

Selling Price (SP) of mixture = ₹91.80/kg

₹80 (CP1) ₹100 (CP2)

₹85 Mean CP

3 1 Ratio

(100 − 85 = 15) (85 − 80 = 5)

Formula Used:

Average CP = (CP₁ × Q₁ + CP₂ × Q₂) ÷ (Q₁ + Q₂)

Gain % = [(SP − CP) ÷ CP] × 100

Calculations:

⇒ Average CP = (80 × 3 + 100 × 1) ÷ (3 + 1)

⇒ Average CP = (240 + 100) ÷ 4

⇒ Average CP = 340 ÷ 4 = ₹85 per kg

⇒ Since SP (₹91.80) is greater than CP (₹85), there is a gain.

⇒ Gain = SP − CP = 91.80 − 85.00 = ₹6.80

⇒ Gain % = (6.80 ÷ 85) × 100

⇒ Gain % = 680 ÷ 85 = 8%

∴ The correct answer is 8% Gain.

Additional Information

Rule of Alligation

When two items of different prices are mixed, the ratio of their quantities is inversely proportional to the differences of their prices from the mean price. Ratio = (Price of dearer − Mean Price) : (Mean Price − Price of cheaper).

Weighted Average

Instead of the Alligation cross method, you can directly use the weighted average formula: A_w = (w₁x₁ + w₂x₂) ÷ (w₁ + w₂), where w represents ratio parts and x represents the price values.

Profit and Loss Basics

Profit occurs when the Selling Price (SP) is strictly greater than the Cost Price (CP). Profit percentage is always calculated on the base of Cost Price unless explicitly stated otherwise. Formula: Profit % = (Profit ÷ CP) × 100.

Shortcut Trick

Profit ratio = (Investment of A × Time of A) : (Investment of B × Time of B)

Profit ratio = (45000 × 8) : (60000 × 6)

Profit ratio = 360000 : 360000 = 1 : 1

Total profit = ₹36,000, so A's share = 1 / 2 × 36000

A's share = ₹18,000

∴ The correct answer is ₹18,000.

Alternate Method

Given:

Investment of A = ₹45,000

Time period of A = 8 months

Investment of B = ₹60,000

Time period of B = 6 months

Total Profit = ₹36,000

Formula Used:

Profit Ratio = (C₁ × T₁) : (C₂ × T₂)

Partner A ₹45,000 × 8 mo Profit: 360000

Partner B ₹60,000 × 6 mo Profit: 360000

Ratio = 1 : 1

Calculations:

⇒ Profit share of A : Profit share of B = (45000 × 8) : (60000 × 6)

⇒ Ratio = 360000 : 360000

⇒ Simplified Ratio = 1 : 1

⇒ Total profit ratio parts = 1 + 1 = 2 parts

⇒ Given, 2 parts = ₹36,000

⇒ 1 part = 36000 ÷ 2 = ₹18,000

⇒ A's share = 1 part = ₹18,000

∴ The correct answer is ₹18,000.

Additional Information

Working vs Sleeping Partner

A working partner receives a fixed salary or a specific percentage of the profit for managing the business. The remaining profit is distributed among all partners based on their capital and time invested. A sleeping partner only receives their share based on their capital investment.

Joining Mid-way in Business

If a partner joins the business x months after it starts and the total accounting period is 1 year (12 months), their effective investment duration for calculating the profit ratio is taken as (12 − x) months.

Multiple Partners

For a business with multiple partners, the profit ratio is always the compound ratio of their investments and respective time periods. Profit Ratio = (C₁ × T₁) : (C₂ × T₂) : (C₃ × T₃).

Shortcut Trick

Weight of removed student = Old Average + (Remaining Students × Drop in Average)

⇒ Weight = 60 + (29 × 0.5)

⇒ Weight = 60 + 14.5 = 74.5

∴ The correct answer is 74.5 kg.

Alternate Method

Given:

Average weight of 30 students = 60 kg

Average weight of remaining 29 students = 59.5 kg

Total Weight of 30 Students = 1800 kg

29 Students = 1725.5 kg

1 Left = ? kg

Formula Used:

Total Weight = Average × Number of Students

Weight of removed student = Total initial weight − Total new weight

Calculation:

⇒ Total weight of 30 students = 30 × 60 = 1800 kg

⇒ Total weight of 29 students = 29 × 59.5 = 1725.5 kg

⇒ Weight of removed student = 1800 − 1725.5

⇒ Weight of removed student = 74.5 kg

∴ The correct answer is 74.5 kg.

Additional Information

Average Change on Removal

If removing an entity decreases the group's average, the value of the removed entity must be greater than the initial average. The direct logic formula is: Removed Value = Old Average + (Remaining Items × Decrease in Average).

Average Change on Addition

If a new entity joins and increases the average, its value is greater than the initial average. The direct logic formula is: Added Value = Old Average + (New Total Items × Increase in Average).

Combined Average

When combining two groups of sizes n₁ and n₂ having averages A₁ and A₂ respectively, the total combined average is calculated as (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Shortcut Trick

Net change in total sum = Correct value − Incorrect value = 156 − 165 = −9 cm.

Change in average = Net change ÷ Total boys = −9 ÷ 40 = −0.225 cm.

Correct average = Initial average + Change in average = 160 − 0.225 = 159.775 cm.

∴ The correct answer is 159.775 cm.

Alternate Method

159.775 cm Correct Avg

160 cm Initial Avg

Drop: -0.225 cm

Total sum drop = 9 cm for 40 boys

Given:

Total number of boys = 40

Initial average height = 160 cm

Incorrectly recorded height = 165 cm

Correct height = 156 cm

Formula Used:

Total Sum = Average × Number of Items

Correct Sum = Initial Sum − Incorrect Value + Correct Value

Calculations:

⇒ Initial sum of heights = 40 × 160 = 6400 cm

⇒ Correct sum of heights = 6400 − 165 + 156

⇒ Correct sum of heights = 6400 − 9 = 6391 cm

⇒ Accurate average height = 6391 ÷ 40

⇒ Accurate average height = 159.775 cm

∴ The correct answer is 159.775 cm.

Additional Information

Deviation Method in Averages

When an existing value is replaced, the net change in average is simply the difference between the new and old values divided by the total number of items.

Combined Average

If group A has n₁ items with average A₁ and group B has n₂ items with average A₂, the combined average is calculated as (n₁ × A₁ + n₂ × A₂) ÷ (n₁ + n₂).

Adding or Removing an Item

If a new entity joins a group and the overall average increases, the value of the new entity is strictly greater than the original average. Conversely, if the average decreases, its value is less than the original average.

Shortcut Trick

Ratio of Amount to Principal = 6050 / 5000 = 121 / 100

For 2 years compound interest, the multiplying factor is the square root of the ratio.

⇒ Multiplying factor per year = √(121 / 100) = 11 / 10

⇒ Annual Interest Rate = (11 − 10) / 10 × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

0 Years 1 Year 2 Years P = ₹5000 A = ₹6050

Compound Interest (2 Years) Rate = R% p.a.

Given:

Principal (P) = ₹5000

Amount (A) = ₹6050

Time (T) = 2 years

Formula Used:

A = P × (1 + R / 100)^T

Calculation:

⇒ 6050 = 5000 × (1 + R / 100)²

⇒ 6050 / 5000 = (1 + R / 100)²

⇒ 121 / 100 = (1 + R / 100)²

⇒ (11 / 10)² = (1 + R / 100)²

Taking the square root on both sides:

⇒ 11 / 10 = 1 + R / 100

⇒ R / 100 = 11 / 10 − 1

⇒ R / 100 = 1 / 10

⇒ R = 10%

∴ The correct answer is 10%.

Additional Information

Compound Interest Formula

The amount after T years is calculated using A = P(1 + R/100)^T, where P is the principal and R is the rate of interest per annum.

Half-Yearly Compounding

When interest is compounded half-yearly, the rate is halved (R/2) and the time duration doubles (2T). The formula becomes A = P(1 + R/200)^2T.

SI vs CI Difference

For a 2-year period at rate R%, the difference between Compound Interest and Simple Interest on the same principal is D = P(R/100)².

Effective Rate of Interest

For 2 years of compounding at an annual rate of R%, the total effective compound interest rate is (2R + R²/100)%.

Shortcut Trick

When profit is given in terms of Cost Price (CP), the total transaction amount (₹750) is not needed.

Base (Total Cost Price) = CP of 15 kg

Profit = CP of 5 kg

Profit % = (Profit ÷ Total CP) × 100

Profit % = (5 ÷ 15) × 100 = 33.33%

∴ The correct answer is 33.33%.

Alternate Method

Given:

Quantity sold = 15 kg

Selling Price (SP) of 15 kg = ₹750

Profit = Cost Price (CP) of 5 kg

Formula Used:

Profit = Total SP − Total CP

Profit % = (Profit ÷ Total CP) × 100

Total CP ₹562.5

Total SP ₹750

• Profit (₹187.5) (CP of 5 kg)

Calculations:

⇒ Let the CP of 1 kg of apples be x.

⇒ Total CP for 15 kg = 15 × x = 15x

⇒ Profit = CP of 5 kg = 5 × x = 5x

⇒ Total SP for 15 kg = Total CP + Profit = 15x + 5x = 20x

⇒ We are given that Total SP = ₹750

⇒ 20x = 750

⇒ x = 750 ÷ 20 = ₹37.5 (CP of 1 kg)

⇒ Total CP = 15 × 37.5 = ₹562.5

⇒ Profit = 5 × 37.5 = ₹187.5

⇒ Profit % = (187.5 ÷ 562.5) × 100

⇒ Profit % = (1 ÷ 3) × 100 = 33.33%

∴ The correct answer is 33.33%.

Additional Information

Profit and Loss Base Rule

Profit or loss is always calculated on the Cost Price (CP) unless explicitly stated otherwise. Profit % = (Profit ÷ CP) × 100.

Dishonest Dealer Concept

When a dealer claims to sell at cost price but uses a false weight, the profit percentage is given by (Error ÷ (True Weight − Error)) × 100.

Same Selling Price with Same Profit/Loss %

If two articles are sold at the same price, one at a profit of x% and the other at a loss of x%, there is always an overall loss of (x² ÷ 100)%.

Shortcut Trick

Intermediate SP (before 20% second discount) = 8400 × (100 ÷ 80) = ₹10,500

First discount amount = Marked Price − Intermediate SP = 12000 − 10500 = ₹1,500

First discount % = (1500 ÷ 12000) × 100

First discount % = (1 ÷ 8) × 100 = 12.5%

∴ The correct answer is 12.5%.

Alternate Method

Given:

Marked Price (MP) = ₹12,000

Final Selling Price (SP) = ₹8,400

Second Discount (D₂) = 20%

Formula Used:

Final SP = MP × (1 − D₁ / 100) × (1 − D₂ / 100)

MP ₹12000

− D₁%

Intermediate ₹10500

− 20%

Final SP ₹8400

Successive Discount Process

Calculations:

⇒ Let the first discount be D₁%.

⇒ 8400 = 12000 × (1 − D₁ / 100) × (1 − 20 / 100)

⇒ 8400 = 12000 × (1 − D₁ / 100) × (80 / 100)

⇒ 8400 = 12000 × (1 − D₁ / 100) × 0.8

⇒ 8400 = 9600 × (1 − D₁ / 100)

⇒ 8400 ÷ 9600 = 1 − D₁ / 100

⇒ 7 / 8 = 1 − D₁ / 100

⇒ D₁ / 100 = 1 − 7 / 8

⇒ D₁ / 100 = 1 / 8

⇒ D₁ = 100 ÷ 8 = 12.5%

∴ The correct answer is 12.5%.

Additional Information

Successive Discount Formula

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

Marked Price (MP) Base

Any discount provided by a shopkeeper is fundamentally calculated on the Marked Price of the article, not the Cost Price. Selling Price (SP) = MP − Discount amount.

Multiplying Factor Method

For a standard discount of D%, the resultant value is found directly by multiplying the base with a factor of (1 − D ÷ 100). For continuous successive discounts, these multiplying factors are applied sequentially.

Shortcut Trick

Initial Ratio → Alcohol : Water = 7 : 4

Final Ratio → Alcohol : Water = 7 : 6

Since the alcohol quantity remains unchanged, its ratio parts (7) are identical.

Increase in water parts = 6 − 4 = 2 parts, which corresponds to the 12 litres added.

2 parts = 12 litres ⇒ 1 part = 6 litres.

Initial total quantity = (7 + 4) parts = 11 × 6 = 66 litres.

∴ The correct answer is 66 litres.

Alternate Method

Given:

Initial ratio of alcohol to water = 7 : 4

Amount of water added = 12 litres

Final ratio of alcohol to water = 7 : 6

Calculations:

Initial

Alcohol (7 parts)

Water (4)

Final

Alcohol (7 parts)

Water (4)

• 2

• 12 Litres

2 Parts = 12 L → 1 Part = 6 L Initial Total = 11 Parts = 66 Litres

⇒ Let the initial quantity of alcohol and water be 7x and 4x respectively.

⇒ Total initial mixture = 7x + 4x = 11x

⇒ After adding 12 litres of water, quantity of water = 4x + 12

⇒ According to the new ratio: 7x / (4x + 12) = 7 / 6

⇒ 7x × 6 = 7 × (4x + 12)

⇒ 42x = 28x + 84

⇒ 14x = 84

⇒ x = 6

⇒ Initial total quantity = 11x = 11 × 6 = 66 litres

∴ The correct answer is 66 litres.

Additional Information

Constant Quantity Principle

In mixture problems, when only one component is added or removed, the absolute amount of the other component stays constant. Matching the ratio values for this constant part is the fastest method to solve the equation.

Rule of Alligation

Used to find the ratio in which two ingredients at given concentrations must be mixed to produce a mixture at a desired concentration. Formula: (Quantity 1 / Quantity 2) = (Price 2 − Mean Price) / (Mean Price − Price 1).

Replacement Method

When a specific volume of a mixture is drained and replaced by one of its constituent liquids, the ratio of the remaining mixture right before the replacement occurs is identical to the original mixture.

Shortcut Trick

Let Cost Price (CP) be 100%.

Marked Price (MP) = 180% (since markup is 80% above CP).

Selling Price (SP) after 30% discount = 180% × 0.7 = 126%.

With an additional profit of 12%, the actual SP = CP + 12% profit = 112%.

The difference in SP represents the loyalty discount: 126% − 112% = 14%.

CP (100%) = (₹250 ÷ 14) × 100 ≈ ₹1,785.71.

∴ The correct answer is ₹1,785.71.

Alternate Method

CP

SP

MP

Markup: +80%

Discount: -30% & -₹250

Profit: +12%

Given:

Markup = 80%

Discounts = 30% and an additional loyalty discount = ₹250

Profit = 12%

Formula Used:

MP = CP × (1 + Markup% ÷ 100)

SP = MP × (1 − Discount% ÷ 100) − Loyalty Discount

SP = CP × (1 + Profit% ÷ 100)

Calculations:

⇒ Let the Cost Price (CP) be 100x.

⇒ MP = 100x × 1.80 = 180x

⇒ SP after standard 30% discount = 180x × (1 − 0.30) = 180x × 0.70 = 126x

⇒ Final SP = 126x − 250

⇒ According to the profit margin, SP = 100x × 1.12 = 112x

⇒ Equating the selling prices: 126x − 250 = 112x

⇒ 126x − 112x = 250

⇒ 14x = 250

⇒ x = 250 ÷ 14 = 125 ÷ 7

⇒ CP = 100x = 100 × (125 ÷ 7) = 12500 ÷ 7

⇒ CP ≈ ₹1,785.71

∴ The correct answer is ₹1,785.71.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP) unless specified otherwise. Net Selling Price = MP − Discount.

Successive Discounts

If two successive discounts of a% and b% are given, the equivalent single discount is [a + b − (a × b)÷100]%.

Direct CP to MP Ratio

When Profit% (P) and Discount% (D) are given, their relationship is simply derived as CP ÷ MP = (100 − D) ÷ (100 + P).

Shortcut Trick

By Archimedes' principle, the volume of water displaced by a completely submerged object is exactly equal to its own volume.

The cylinder's dimensions are extra information just to confirm the ball fits entirely inside.

Volume of water displaced = Volume of the sphere = 4/3 × π × r³

⇒ Volume = 4/3 × π × 4³ = 256/3 π cm³.

∴ The correct answer is 256/3 π cm³.

Alternate Method

r = 4 cm

R = 5 cm

h = 30 cm

Displaced Volume = Volume of Submerged Sphere

Given:

Radius of the cylindrical container (R) = 5 cm

Height of the cylindrical container (h) = 30 cm

Radius of the spherical ball (r) = 4 cm

Formula Used:

Volume of a sphere = 4/3 × π × r³

Calculations:

⇒ When an object is completely immersed in water, the volume of water displaced is equal to its own volume.

⇒ The spherical ball has a diameter of 8 cm, which is less than the cylinder's 10 cm diameter. It fits completely inside the container.

⇒ Volume of water displaced = Volume of the spherical ball

⇒ Volume = 4/3 × π × (4)³

⇒ Volume = 4/3 × π × 64

⇒ Volume = 256/3 π cm³

∴ The correct answer is 256/3 π cm³.

Additional Information

Archimedes' Principle

When an object is fully immersed in a liquid, it displaces a volume of liquid exactly equal to its own total volume, regardless of the container's shape or size.

Identifying Redundant Data

In competitive exam mensuration problems, dimensions of the outer container are often provided merely to act as distractors or verify physical constraints. Only extract the required dimensions for your calculation.

Volume of a Cylinder

If required to calculate container capacity, the volume of a cylinder is found using the formula V = π × r² × h.

Shortcut Trick

Radius of the dome (r) = 20 ÷ 2 = 10 m

Curved Surface Area (CSA) of Hemisphere = 2 × π × r²

CSA ≈ 2 × 3.14 × (10)²

CSA ≈ 628 m²

∴ The correct answer is 628 m².

Alternate Method

Given:

Diameter of the circular dome area = 20 m

Radius of the dome (r) = 20 ÷ 2 = 10 m

r = 10 m 10 m

Curved Surface Area = 2 × π × r²

Diameter = 20 m

Formula Used:

Curved Surface Area (CSA) of a Hemisphere = 2 × π × r²

Calculations:

⇒ CSA = 2 × 3.14 × (10)²

⇒ CSA = 2 × 3.14 × 100

⇒ CSA ≈ 628 m²

∴ The correct answer is 628 m².

Additional Information

Total Surface Area of Hemisphere

Includes the curved part and the flat circular base. Formula: TSA = 3 × π × r².

Volume of a Hemisphere

Exactly half the volume of a complete sphere. Formula: Volume = (2 ÷ 3) × π × r³.

Surface Area of a Full Sphere

A hemisphere's curved area is exactly half of a full sphere's area. Formula: SA = 4 × π × r².

Shortcut Trick

The curved surface area of a hemisphere is directly proportional to the square of its radius.

Ratio of outer to inner curved surface area = R² : r²

Given R = 2r, the ratio becomes (2r)² : r²

Ratio = 4r² : r² = 4 : 1

∴ The correct answer is 4 : 1.

Alternate Method

Given:

Inner radius = r

Outer radius = R = 2r

r

R = 2r

Center O Hollow Hemisphere Cross-Section

Formula Used:

Curved Surface Area (CSA) of a hemisphere = 2π × (radius)²

Calculations:

⇒ Outer CSA = 2πR²

⇒ Inner CSA = 2πr²

⇒ Ratio = (Outer CSA) ÷ (Inner CSA)

⇒ Ratio = 2πR² ÷ 2πr²

⇒ Ratio = R² ÷ r²

Substitute R = 2r into the equation:

⇒ Ratio = (2r)² ÷ r²

⇒ Ratio = 4r² ÷ r²

⇒ Ratio = 4 : 1

∴ The correct answer is 4 : 1.

Additional Information

Total Surface Area of a Hollow Hemisphere

TSA = Outer CSA + Inner CSA + Area of base ring = 2πR² + 2πr² + π(R² − r²) = π(3R² + r²).

Volume of a Hollow Hemisphere

Volume = (2 ÷ 3)π(R³ − r³), where R is the outer radius and r is the inner radius.

Solid Hemisphere Properties

For a solid hemisphere of radius r: CSA = 2πr², TSA = 3πr², Volume = (2 ÷ 3)πr³.

Shortcut Trick

For a square pyramid, Lateral Surface Area (LSA) = 2 × base side × slant height = 2al

Base Area = a²

Given LSA = 2 × Base Area

⇒ 2al = 2a² ⇒ l = a

Since the base side (a) is 6 cm, the slant height (l) is also 6 cm.

∴ The correct answer is 6 cm.

Alternate Method

Given:

Base side (a) = 6 cm

Lateral Surface Area = 2 × Base Area

Formula Used:

Area of square base = a²

Lateral Surface Area of square pyramid = 4 × (1/2 × base × slant height) = 2 × a × l

Base side (a) = 6 cm Slant height (l) Height (h) Square Pyramid

Calculations:

⇒ Base Area = 6² = 36 cm²

⇒ Lateral Surface Area = 2 × 36 = 72 cm²

⇒ 2 × a × l = 72

⇒ 2 × 6 × l = 72

⇒ 12l = 72

⇒ l = 72 ÷ 12 = 6 cm

∴ The correct answer is 6 cm.

Additional Information

Total Surface Area (TSA)

The TSA of a square pyramid is the sum of its base area and lateral surface area. Formula: TSA = a² + 2al

Volume of Square Pyramid

The volume is one-third of the product of the base area and the vertical height. Formula: V = (1/3) × a² × h

Relation between Heights

Using the Pythagorean theorem inside the pyramid, the slant height (l), vertical height (h), and base side (a) form a right-angled triangle: l² = h² + (a/2)²

Shortcut Trick

The required fencing length is simply 75% of the total circumference.

Required length = 75% × π × D

⇒ Length = (3/4) × (22/7) × 28

⇒ Length = 3 × 22 = 66 m

∴ The correct answer is 66 m.

Alternate Method

r = 14 m Fenced Path = 75% of Circumference Unfenced (25%)

Given:

Diameter of the circular garden (D) = 28 m

Fraction of the garden to be fenced = 75% = 75/100 = 3/4

Formula Used:

Circumference of a circle (C) = π × D

Calculations:

⇒ Total Circumference (C) = (22/7) × 28

⇒ C = 22 × 4 = 88 m

⇒ The gardener wants to fence 75% of the garden circumference.

⇒ Length of fencing needed = 75% of 88 m

⇒ Length of fencing needed = (3/4) × 88

⇒ Length of fencing needed = 3 × 22 = 66 m

∴ The correct answer is 66 m.

Additional Information

Area of a Circle

The area of a circle is the space enclosed within its boundary, calculated as Area = π × r².

Length of an Arc

An arc is a continuous portion of the circumference. Its length is given by Arc Length = (θ ÷ 360) × 2 × π × r, where θ is the central angle.

Area of a Sector

A sector is a portion of the circle enclosed by two radii and an arc. The formula is Area of Sector = (θ ÷ 360) × π × r².

Shortcut Trick

In an equilateral triangle, the circumcenter, incenter, and centroid all coincide at the exact same point.

The centroid divides the median (which is also the altitude) in the ratio 2:1.

The longer segment from the vertex to the centroid is the circumradius (R).

The shorter segment from the centroid to the base is the inradius (r).

Thus, the ratio of circumradius to inradius (R:r) is directly 2:1 without any calculation.

∴ The correct answer is 2:1.

Alternate Method

Given:

Side length of the equilateral triangle = a

R (Circumradius) r (Inradius) Side = a A B C

Formulas Used:

Circumradius (R) of an equilateral triangle = a ÷ √3

Inradius (r) of an equilateral triangle = a ÷ 2√3

Calculations:

⇒ Ratio = R : r

⇒ R : r = (a ÷ √3) : (a ÷ 2√3)

⇒ R : r = 1 : (1 ÷ 2)

⇒ Multiplying both sides by 2

⇒ R : r = 2 : 1

∴ The correct answer is 2:1.

Additional Information

Area of Equilateral Triangle

The area is given by the formula: (√3 ÷ 4) × a².

Altitude of Equilateral Triangle

The height or altitude (h) is given by: h = (√3 ÷ 2) × a.

Coinciding Centers

In an equilateral triangle, all four key centers — the centroid, orthocenter, circumcenter, and incenter — merge and lie at the exact same point.

Shortcut Trick

By the Tangent-Secant Theorem: AD² = BD × CD

⇒ AD² = (BC + CD) × CD

⇒ 6² = (BC + 4) × 4

⇒ 36 = 4 × BC + 16

⇒ 4 × BC = 20 ⇒ BC = 5

∴ The correct answer is 5 cm.

Alternate Method

Given:

Length of tangent segment AD = 6 cm

Length of outer secant segment CD = 4 cm

Let the length of side BC be x cm.

O A B C D 6 cm 4 cm x

Formula Used:

Tangent-Secant Theorem: If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the length of the tangent segment is equal to the product of the lengths of the external secant segment and the entire secant segment.

⇒ AD² = CD × BD

Calculations:

From the figure, the entire secant segment BD is the sum of the side BC and the external part CD.

⇒ BD = BC + CD

⇒ BD = x + 4

Substitute the given values into the Tangent-Secant theorem equation:

⇒ 6² = 4 × (x + 4)

⇒ 36 = 4 × (x + 4)

⇒ x + 4 = 36 ÷ 4

⇒ x + 4 = 9

⇒ x = 9 − 4

⇒ x = 5 cm

∴ The correct answer is 5 cm.

Additional Information

Intersecting Secants Theorem

When two secants are drawn to a circle from an external point, the product of the entire secant and its external part is equal for both secants: AP × BP = CP × DP.

Intersecting Chords Theorem

When two chords intersect inside a circle, the product of their respective segments is equal: AE × EC = BE × ED.

Alternate Segment Theorem

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment of the circle.

Shortcut Trick

In a trapezoid ABCD with AB || CD, the intersection of diagonals forms two similar triangles along the parallel sides (▵ABE ∼ ▵CDE).

For any two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

⇒ Area(▵ABE) ÷ Area(▵CDE) = (AB ÷ CD)²

∴ The correct answer is The ratio of AB to CD squared.

Alternate Method

ABCDE

Given:

Trapezoid ABCD with side AB parallel to side CD.

Diagonals AC and BD intersect at point E.

Formula Used:

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Calculations:

Consider ▵ABE and ▵CDE:

⇒ ∠EAB = ∠ECD (Alternate interior angles, AB || CD)

⇒ ∠EBA = ∠EDC (Alternate interior angles, AB || CD)

⇒ ∠AEB = ∠CED (Vertically opposite angles)

⇒ Since all three corresponding angles are equal, ▵ABE ∼ ▵CDE by the AAA similarity criterion.

⇒ According to the area theorem for similar triangles:

⇒ Area(▵ABE) ÷ Area(▵CDE) = (AB ÷ CD)²

∴ The correct answer is The ratio of AB to CD squared.

Additional Information

Area of Similar Triangles

When two triangles are similar, the ratio of their areas is strictly equal to the square of the ratio of their corresponding side lengths: Area(▵1) ÷ Area(▵2) = (s₁ ÷ s₂)².

Area of a Trapezoid

The area of a trapezoid is equal to half the product of the sum of parallel sides and the perpendicular height between them: Area = ¹÷₂ × (AB + CD) × h.

Isosceles Trapezoid Properties

In an isosceles trapezoid, the non-parallel sides are equal, both pairs of base angles are equal, and the diagonals are exactly equal in length.

Mid-Segment of a Trapezoid

The segment connecting the midpoints of the non-parallel sides is parallel to the bases, and its length is the arithmetic average of the two parallel bases.

Shortcut Trick

When two circles with radii r₁ and r₂ touch externally, the length of their Direct Common Tangent (DCT) is given directly by the formula 2√(r₁ × r₂).

Given, r₁ = 2x and r₂ = x.

DCT = 2√(2x × x)

DCT = 2√(2x²) = 2x√2

∴ The correct answer is 2x√2.

Alternate Method

C1C2AB2xx3xDCT

Given:

Radius of the first circle, r₁ = 2x

Radius of the second circle, r₂ = x

Since the circles touch each other externally, the distance between their centers (d) is equal to the sum of their radii.

Formula Used:

Distance between centers, d = r₁ + r₂

Length of Direct Common Tangent (DCT) = √[d² − (r₁ − r₂)²]

Calculations:

⇒ d = 2x + x = 3x

⇒ DCT = √[(3x)² − (2x − x)²]

⇒ DCT = √[9x² − x²]

⇒ DCT = √(8x²)

⇒ DCT = 2x√2

∴ The correct answer is 2x√2.

Additional Information

Transverse Common Tangent (TCT)

The length of the Transverse Common Tangent is calculated using the formula: TCT = √[d² − (r₁ + r₂)²].

Number of Common Tangents

If two circles touch each other externally, they have exactly 3 common tangents (2 direct and 1 transverse). If they intersect, they have 2 direct common tangents.

Direct Common Tangent (DCT)

A tangent that touches two separate circles and does not cross the line segment connecting their centers is known as a Direct Common Tangent.