Quantitative Aptitude PYQs

Shortcut Trick

Evaluate the square roots directly: √6400 = 80 and √2401 = 49

Subtract the values inside the bracket: 80 − 49 = 31

Multiply the result by 1/4: 31 × 1/4 = 7.75

∴ The correct answer is 7.75.

Alternate Method

Given:

(√6400 − √2401) × (1/4)

Concept Used:

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

Calculation:

⇒ First, evaluate the square roots present inside the bracket.

⇒ √6400 = 80 (since 80 × 80 = 6400)

⇒ √2401 = 49 (since 50² = 2500, so 49² = 2401)

⇒ Substitute these values back into the expression:

⇒ (80 − 49) × (1/4)

⇒ Solve the subtraction inside the bracket first according to BODMAS:

⇒ 31 × (1/4)

⇒ 31 ÷ 4

⇒ 7.75

∴ The correct answer is 7.75.

Additional Information

BODMAS Rule

Mathematical operations must always be solved in the exact sequence: Brackets, Orders (Powers/Roots), Division, Multiplication, Addition, and Subtraction.

Square Root Shortcut Trick

For finding the square root of a number like 2401, check the nearest base square. 50² = 2500. Since 2401 ends in 1 and is slightly less than 2500, the square root logically must be 49.

Division by 4 Shortcut

To divide any number by 4 quickly without long division, simply halve the number twice. Half of 31 is 15.5, and half of 15.5 is 7.75.

Shortcut Trick

Find square roots quickly using unit digits and base squares.

√5776: 70² = 4900, 80² = 6400. Unit digit 6 → ends in 4 or 6. Since 5776 > 5625 (75²), √5776 = 76.

√4761: 60² = 3600, 70² = 4900. Unit digit 1 → ends in 1 or 9. Since 4761 is close to 4900, √4761 = 69.

Expression = (76 − 69) × (1/2) = 7/2 = 3.5.

∴ The correct answer is 3.5.

Alternate Method

Given:

(√5776 − √4761) × (1/2)

Method Used:

Find the exact square roots of the given perfect squares, then apply the BODMAS rule for simplification.

Calculations:

⇒ √5776 = 76

⇒ √4761 = 69

⇒ Substitute these values back into the main expression:

⇒ (76 − 69) × (1/2)

⇒ 7 × (1/2)

⇒ 3.5

∴ The correct answer is 3.5.

Additional Information

BODMAS Rule

Always follow the order: Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction.

Square Root Shortcut

To find √N, identify the base tens (like 60², 70²) it lies between, check the unit digit, and compare N with the square of the middle number (ends in 5).

Squares Ending in 5

To square a number ending in 5 (like N5), multiply N by (N+1) and append 25. Example: 75² → 7 × 8 = 56 → 5625.

Shortcut Trick

Convert all given values into simple fractions: 40% = 2/5, 0.5 = 1/2, and 1/4 remains 1/4.

The given equation becomes: (2/5)A = (1/2)B = (1/4)C.

To find the ratio A : B : C directly, take the reciprocal of their coefficients.

A : B : C = 5/2 : 2/1 : 4/1.

Multiply the entire ratio by 2 to eliminate the denominator: 5 : 4 : 8.

∴ The correct answer is 5 : 4 : 8.

Alternate Method

Given:

40% of A = 0.5 of B = 1/4 of C

Formula Used:

If p × A = q × B = r × C = k, then A = k/p, B = k/q, and C = k/r

A 5 units

B 4 units

C 8 units

Ratio A : B : C = 5 : 4 : 8

Calculations:

⇒ 40% = 40/100 = 2/5

⇒ 0.5 = 5/10 = 1/2

⇒ (2/5) × A = (1/2) × B = (1/4) × C

⇒ Let (2/5) × A = (1/2) × B = (1/4) × C = k

⇒ A = 5k/2

⇒ B = 2k

⇒ C = 4k

⇒ Ratio A : B : C = 5k/2 : 2k : 4k

⇒ Multiply all terms by 2 to clear fractions:

⇒ A : B : C = 5 : 4 : 8

∴ The correct answer is 5 : 4 : 8.

Additional Information

Equivalent Ratio

Multiplying or dividing each term of a ratio by the same non-zero number does not change the value of the ratio. For example, a : b = 2a : 2b.

Compound Ratio

The ratio compounded of two ratios a : b and c : d is the ratio of the products of the antecedents and consequents, i.e., ac : bd.

Invertendo Property

If two ratios are equal, i.e., a : b = c : d, then their inverse ratios are also equal, making b : a = d : c.

Shortcut Trick

Sum of ratio parts in container 1 = 2 + 3 = 5 units.

Sum of ratio parts in container 2 = 3 + 2 = 5 units.

Since capacities are equal and sum of parts is the same, we can directly add the respective parts.

Total Milk = 2 + 3 = 5 units and Total Water = 3 + 2 = 5 units.

Ratio of Milk to Water in the final mixture = 5 : 5 = 1 : 1.

∴ The correct answer is 1:1.

Alternate Method

Given:

Ratio of milk and water in 1^st container = 2 : 3

Ratio of milk and water in 2^nd container = 3 : 2

Both containers have equal capacities.

 Container 1 Milk = 2x Water = 3x +     Container 2 Milk = 3x Water = 2x =     Final Mixture Milk = 5x Water = 5x  Resulting Ratio = 5x : 5x ⇒ 1 : 1

Calculations:

⇒ Let the uniform capacity of each container be V units.

⇒ Quantity of milk in the 1^st container = 2V ÷ 5

⇒ Quantity of water in the 1^st container = 3V ÷ 5

⇒ Quantity of milk in the 2^nd container = 3V ÷ 5

⇒ Quantity of water in the 2^nd container = 2V ÷ 5

⇒ Total quantity of milk when combined = (2V ÷ 5) + (3V ÷ 5) = 5V ÷ 5 = V

⇒ Total quantity of water when combined = (3V ÷ 5) + (2V ÷ 5) = 5V ÷ 5 = V

⇒ Resulting ratio of milk to water = V : V = 1 : 1

∴ The correct answer is 1:1.

Additional Information

Unequal Capacities

If the containers are of different volumes, multiply the ratios by their respective overall capacities to find the actual proportional quantity of each component before combining them.

Replacement Concept

When a specific volume is removed from a mixture, the removed portion and the remaining mixture both retain the exact same ratio of ingredients as the original mixture.

Rule of Alligation

When two mixtures with varying concentrations are blended to form a third required mixture, the necessary proportion of each initial mixture can be quickly determined using the cross-difference of their concentration fractions.

Shortcut Trick

Selling Price after 1^st discount = 800 × (100 − 10)% = ₹720

Second discount amount = 720 − 612 = ₹108

Second discount % = (108 ÷ 720) × 100 = 15%

∴ The correct answer is 15%.

Alternate Method

Given:

Marked Price (MP) = ₹800

Final Selling Price (SP) = ₹612

First discount (d₁) = 10%

MP = ₹800

−10%

₹720

−d%

SP = ₹612 Discount = ₹720 − ₹612 = ₹108

Formula Used:

SP = MP × (1 − d₁/100) × (1 − d₂/100)

Calculations:

⇒ Price after first discount = 800 − (10% of 800)

⇒ Price after first discount = 800 − 80 = ₹720

⇒ Second discount amount = 720 − 612 = ₹108

⇒ Second discount % = (Discount ÷ Price after 1^st discount) × 100

⇒ d₂ = (108 ÷ 720) × 100

⇒ d₂ = 15%

∴ The correct answer is 15%.

Additional Information

Single Equivalent Discount

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

Marked Price vs Selling Price

Discount is always calculated on the Marked Price. Hence, Selling Price = Marked Price − Discount.

Successive Discounts Formula

Final SP = MP × (1 − d₁/100) × (1 − d₂/100) × ... × (1 − d_n/100).

Shortcut Trick

Let the Cost Price be 100 units.

Marked Price = 160 units (60% markup).

Selling Price = 160 × 3/4 × 9/10 = 108 units.

Profit = 108 − 100 = 8 units.

8 units = ₹160 ⇒ 1 unit = ₹20.

∴ The correct answer is ₹2000.

Alternate Method

CP = 100x

MP = 160x

SP = 108x

+60%

-25%, -10%

Profit = 8x = ₹160

Given:

Markup % = 60%

Successive discounts = 25% and 10%

Profit = ₹160

Formula Used:

Selling Price (SP) = Marked Price × (100 − D₁)/100 × (100 − D₂)/100

Profit = SP − CP

Calculations:

Let the Cost Price (CP) be 100x.

⇒ Marked Price (MP) = 100x + 60% of 100x = 160x

⇒ SP = 160x × (100 − 25)/100 × (100 − 10)/100

⇒ SP = 160x × 75/100 × 90/100

⇒ SP = 160x × 3/4 × 9/10 = 108x

⇒ Profit = SP − CP = 108x − 100x = 8x

Given that the profit earned is ₹160:

⇒ 8x = 160

⇒ x = 20

⇒ Cost Price (CP) = 100x = 100 × 20 = 2000

∴ The correct answer is ₹2000.

Additional Information

Successive Discounts

When two successive discounts of a% and b% are given, the equivalent single discount percentage is calculated as (a + b − a×b/100)%.

Marked Price and Markup Percentage

The Marked Price (MP) is always marked up above the Cost Price (CP). Markup % = (MP − CP) ÷ CP × 100.

Selling Price Relation

Selling Price (SP) acts as the bridge connecting cost and markings. It can be found from CP as SP = CP × (100 + Profit%)/100, and from MP as SP = MP × (100 − Discount%)/100.

Shortcut Trick

If x + a/x = k, then x³ + a³/x³ = k³ − 3ak.

Here, a = 2 and k = 7.

x³ + 8/x³ = 7³ − 3 × 2 × 7

x³ + 8/x³ = 343 − 42 = 301

∴ The correct answer is 301.

Alternate Method

Given:

x + 2/x = 7

Formula Used:

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculations:

⇒ Cube both sides of the given equation:

⇒ (x + 2/x)³ = 7³

⇒ x³ + (2/x)³ + 3 × x × (2/x) × (x + 2/x) = 343

⇒ x³ + 8/x³ + 6(x + 2/x) = 343

⇒ Substitute the value of (x + 2/x) = 7:

⇒ x³ + 8/x³ + 6 × 7 = 343

⇒ x³ + 8/x³ + 42 = 343

⇒ x³ + 8/x³ = 343 − 42

⇒ x³ + 8/x³ = 301

∴ The correct answer is 301.

Additional Information

Cube of sum with reciprocal

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k.

Cube of difference with reciprocal

If x − 1/x = k, then x³ − 1/x³ = k³ + 3k.

Square with reciprocal

If x + 1/x = k, then x² + 1/x² = k² − 2.

Shortcut Trick

Total annual interest = 9% of ₹6,00,000 = ₹54,000.

Sum of the first and third scholarships = ₹18,000 + ₹22,000 = ₹40,000.

Second scholarship = Total interest − Sum of first and third scholarships.

Second scholarship = 54,000 − 40,000 = ₹14,000.

∴ The correct answer is ₹14,000.

Alternate Method

Given:

Educational Fund Principal (P) = ₹6,00,000

Rate of Interest (R) = 9% per annum

Time (T) = 1 year

First scholarship = ₹18,000

Third scholarship = ₹22,000

Formula Used:

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

Fund = ₹6,00,000

9% Interest = ₹54,000

1^st: ₹18k 2^nd: ? 3^rd: ₹22k

Second = 54k - (18k + 22k) = ₹14,000

Calculation:

⇒ Total annual interest = (6,00,000 × 9 × 1) ÷ 100

⇒ Total annual interest = 54,000

⇒ The sum of all three scholarships equals the total annual interest earned.

⇒ First scholarship + Second scholarship + Third scholarship = ₹54,000

⇒ 18,000 + Second scholarship + 22,000 = 54,000

⇒ Second scholarship + 40,000 = 54,000

⇒ Second scholarship = 54,000 − 40,000

⇒ Second scholarship = ₹14,000

∴ The correct answer is ₹14,000.

Additional Information

Simple Interest (SI) Formula

Simple Interest is strictly calculated on the initial principal amount throughout the entire period. Formula: I = (P × R × T) ÷ 100.

Concept of Compound Interest (CI)

Compound Interest is calculated on the principal as well as the accumulated interest from previous periods. Formula for amount: A = P(1 + R/100)^T.

SI vs CI Difference

For the first year, both Simple Interest and Compound Interest remain the same. The difference for 2 years can be directly found using: Difference = P(R ÷ 100)².

Shortcut Trick

Since the rate (R = 5%) and time (T = 2 years) are constant, the ratio of Interest to Principal remains the same in both investments.

Interest ÷ Principal = (R × T) ÷ 100 = (5 × 2) ÷ 100 = 1 ÷ 10

For Case 1: Y ÷ X = 1 ÷ 10

For Case 2: Z ÷ Y = 1 ÷ 10

Equating both ratios: Y ÷ X = Z ÷ Y ⇒ Y² = XZ

∴ The correct answer is Y² = XZ.

Alternate Method

Given:

Principal 1 = ₹X, Interest 1 = ₹Y

Principal 2 = ₹Y, Interest 2 = ₹Z

Rate (R) = 5%, Time (T) = 2 years

Formula Used:

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

₹ X Principal 1

₹ Y Interest 1 / Prin. 2

₹ Z Interest 2

R=5%, T=2

R=5%, T=2 Interest to Principal Ratio is Constant Y ÷ X = Z ÷ Y

Calculations:

⇒ For the first investment, substituting values in the formula:

⇒ Y = (X × 5 × 2) ÷ 100

⇒ Y = 10X ÷ 100 = X ÷ 10

⇒ X = 10Y

⇒ For the second investment:

⇒ Z = (Y × 5 × 2) ÷ 100

⇒ Z = 10Y ÷ 100 = Y ÷ 10

⇒ Y = 10Z

⇒ Let's find the ratio of Interest to Principal in both cases:

⇒ Y ÷ X = 1 ÷ 10

⇒ Z ÷ Y = 1 ÷ 10

⇒ Equating both ratios: Y ÷ X = Z ÷ Y

⇒ Cross multiplying, we get: Y² = XZ

∴ The correct answer is Y² = XZ.

Additional Information

Simple Interest (SI)

It is calculated only on the principal amount, keeping the interest constant every year. Formula: SI = (P × R × T) ÷ 100.

Amount in Simple Interest

The total amount (A) accumulated at the end of the time period is the sum of Principal and Simple Interest. Formula: A = P + [(P × R × T) ÷ 100].

Effective Interest Rate

For simple interest, the total effective percentage of interest accumulated over T years is directly given by (R × T)%. In this given question, it is 5% × 2 = 10%.

Shortcut Trick

For an object at height h, if angles of depression to two points on opposite sides are α and β:

Total distance = h × (cot α + cot β)

Here, h = 300 m, α = 45° and β = 30°

Distance = 300 × (cot 45° + cot 30°)

Distance = 300 × (1 + √3) = 300(√3 + 1) m

∴ The correct answer is 300(√3 + 1) m.

Alternate Method

A B C D

300 m Boat 1 Boat 2

45° 30° 45° 30°

Total Distance = ?

Given:

⇒ Height of airplane (AD) = 300 m

⇒ Angle of depression of Boat 1 (∠B) = 45°

⇒ Angle of depression of Boat 2 (∠C) = 30°

Formula Used:

⇒ tan θ = Perpendicular / Base

Calculations:

⇒ In right-angled ΔADB, tan 45° = AD / BD

⇒ 1 = 300 / BD

⇒ BD = 300 m

⇒ In right-angled ΔADC, tan 30° = AD / DC

⇒ 1 / √3 = 300 / DC

⇒ DC = 300√3 m

⇒ Total distance between the boats (BC) = BD + DC

⇒ BC = 300 + 300√3 = 300(√3 + 1) m

∴ The correct answer is 300(√3 + 1) m.

Additional Information

Same Side vs Opposite Sides

If two objects are located on the same side of the observer, the distance between them is h × |cot α − cot β|.

Angle of Elevation & Depression

The angle of elevation from an object to an observer is always mathematically equal to the angle of depression from the observer to the object due to alternate interior angles.

Shortcut Trick

Height of lower part (h) = Height of upper part (d) ÷ [tan(higher angle) ÷ tan(lower angle) − 1]

Given d = 1.6 m, higher angle = 60°, lower angle = 45°

h = 1.6 ÷ (tan 60° ÷ tan 45° − 1) = 1.6 ÷ (√3 ÷ 1 − 1) = 1.6 ÷ (√3 − 1)

Rationalizing the denominator: h = 1.6 × (√3 + 1) ÷ [(√3 − 1)(√3 + 1)]

h = 1.6(√3 + 1) ÷ (3 − 1) = 0.8(√3 + 1) m

∴ The correct answer is 0.8(√3 + 1) m.

Alternate Method

Given: Height of the statue (AB) = 1.6 m.

Let the height of the pedestal (BC) be h and the horizontal distance to observation point D (CD) be x.

45° 60°

A B C D

x

h

1.6 m

⇒ In right-angled triangle BCD, tan 45° = BC ÷ CD

⇒ 1 = h ÷ x

⇒ x = h

⇒ In right-angled triangle ACD, tan 60° = AC ÷ CD

⇒ √3 = (AB + BC) ÷ x

⇒ √3 = (1.6 + h) ÷ h

⇒ h√3 = 1.6 + h

⇒ h(√3 − 1) = 1.6

⇒ h = 1.6 ÷ (√3 − 1)

⇒ Rationalizing the denominator: h = 1.6(√3 + 1) ÷ [(√3 − 1)(√3 + 1)]

⇒ h = 1.6(√3 + 1) ÷ (3 − 1) = 0.8(√3 + 1) m

∴ The correct answer is 0.8(√3 + 1) m.

Additional Information

Trigonometric Values

Standard trigonometric ratios frequently used in heights and distances: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.

Rationalization of Surds

To simplify an expression of the form a / (√b − c), multiply both the numerator and the denominator by its conjugate (√b + c) to eliminate the square root from the denominator.

Angle of Elevation

It is the angle formed by the line of sight with the horizontal when an object is viewed from below. A higher angle of elevation corresponds to a greater vertical height for objects observed from the same horizontal distance.

Shortcut Trick

Direct formula for the volume of a hollow hemisphere: V = ²/₃ × π × (R³ − r³)

Substitute R = 14 cm, r = 7 cm, and π = 3.14:

V = ²/₃ × 3.14 × (14³ − 7³)

V = ²/₃ × 3.14 × (2744 − 343) = ²/₃ × 7539.14

V ≈ 5026.09 cm³

∴ The correct answer is 5026.09 cm³.

Alternate Method

Given:

Outer radius of the hemispherical shell (R) = 14 cm

Inner radius of the hemispherical shell (r) = 7 cm

Formula Used:

Volume of the metal used = Volume of outer hemisphere − Volume of inner hemisphere

Volume = ²/₃ × π × (R³ − r³)

r = 7 cm

R = 14 cm

Hollow Hemispherical Shell

Calculation:

⇒ Volume = ²/₃ × π × (14³ − 7³)

⇒ Cube values: 14³ = 2744 and 7³ = 343

⇒ Difference of cubes = 2744 − 343 = 2401

⇒ Volume = ²/₃ × 3.14 × 2401   (Taking π ≈ 3.14 for approximation)

⇒ Volume = ²/₃ × 7539.14

⇒ Volume = 5026.0933... cm³

⇒ Volume ≈ 5026.09 cm³

∴ The correct answer is 5026.09 cm³.

Additional Information

Total Surface Area of a Hollow Hemisphere

The total surface area includes the outer curved surface, inner curved surface, and the flat circular ring at the base. Formula: TSA = 2πR² + 2πr² + π(R² − r²) = π(3R² + r²).

Volume of a Solid Hemisphere

For a solid hemisphere of radius R, the volume is exactly half the volume of a full sphere. Formula: V = ²/₃ × πR³.

Approximation of Pi (π)

For standard calculations, π is usually taken as ²²/₇. However, in options involving specific decimal values, taking π ≈ 3.14 or 3.1416 often yields the required exact approximation.

Shortcut Trick

For similar cones, the volume is proportional to the cube of the height: V ∝ H³

Volume of smaller cone = 1 unit ⇒ Volume of larger cone = 1 + 7 = 8 units.

Ratio of their heights (smaller : larger) = ³√1 : ³√8 = 1 : 2

Height of frustum = Height of larger cone − Height of smaller cone = 2 − 1 = 1 unit.

Ratio of heights of frustum and smaller cone = 1 : 1

∴ The correct answer is 1:1.

Alternate Method

Given:

Ratio of the volume of the frustum to the volume of the smaller cone = 7 : 1

Formula Used:

Ratio of volumes of similar geometric solids = (Ratio of their corresponding heights)³

h1

h2

Vol = 1 Vol = 7

Height Ratio = h2 : h1 = 1 : 1

Calculations:

⇒ Let the volume of the smaller cone be V₁ = x

⇒ Volume of the frustum = 7x

⇒ Total volume of the original larger cone V₂ = x + 7x = 8x

⇒ Since the cutting plane is parallel to the base, the smaller cone and the larger cone are similar geometric solids.

⇒ V₁ / V₂ = (h₁ / H)³

⇒ x / 8x = (h₁ / H)³

⇒ 1 / 8 = (h₁ / H)³

⇒ h₁ / H = 1 / 2

⇒ H = 2h₁

⇒ Height of the frustum h₂ = H − h₁

⇒ h₂ = 2h₁ − h₁ = h₁

⇒ Ratio of heights of the frustum and the smaller cone = h₂ : h₁ = h₁ : h₁ = 1 : 1

∴ The correct answer is 1:1.

Additional Information

Similar Solids Property

When a solid is cut by a plane parallel to its base, the smaller shape formed at the top is mathematically similar to the original solid. The ratio of their volumes strictly equals the cube of the ratio of their corresponding linear dimensions (height, radius, or slant height).

Frustum of a Cone

The volume of a frustum of a right circular cone is given by V = πh(r² + R² + rR) ÷ 3, where r and R are the radii of the top and bottom circular bases and h is the vertical height of the frustum.

Curved Surface Area of a Frustum

The lateral or curved surface area is given by CSA = πl(R + r), where l is the slant height of the frustum calculated as l = √(h² + (R − r)²).

Shortcut Trick

For any 2D geometrical figure, the area is directly proportional to the square of its side length.

If the side length is scaled by a factor of n, the area scales by a factor of n².

Given the side is doubled, n = 2.

New Area = 2² = 4 times the original area.

∴ The correct answer is Quadrupled.

Alternate Method

Given:

The side of a regular hexagon is doubled.

Formula Used:

Area of a regular hexagon = (3√3 ÷ 2) × (side)²

a Area = A

2a Area = 4A

Side × 2

Calculations:

⇒ Let the original side of the regular hexagon be a.

⇒ Original Area = (3√3 ÷ 2) × a²

⇒ New side = 2a

⇒ New Area = (3√3 ÷ 2) × (2a)²

⇒ New Area = (3√3 ÷ 2) × 4a²

⇒ New Area = 4 × [(3√3 ÷ 2) × a²]

⇒ New Area = 4 × Original Area

∴ The correct answer is Quadrupled.

Additional Information

Area of a Regular Hexagon

A regular hexagon is made up of 6 equilateral triangles. Hence, Area = 6 × (√3 ÷ 4) × a² = (3√3 ÷ 2) × a².

Scaling of 2D Figures

When the corresponding sides of any 2D shape are scaled by a factor k, the perimeter scales by k, and the area scales by k².

Scaling of 3D Figures

When the corresponding edges of a 3D object are scaled by a factor k, its surface area scales by k², and its volume scales by k³.

Shortcut Trick

Total height of the container = Sum of the arithmetic progression (AP)

⇒ Total height (H) = (n ÷ 2) × (a + l) = (5 ÷ 2) × (2 + 10) = 30 cm

Total Volume = π × r² × H

⇒ Volume = 3.14 × 3² × 30 = 847.8 cm³

∴ The correct answer is 847.8 cm³.

Alternate Method

r = 3 cm

h₅ = 10 cm h₄ = 8 cm h₃ = 6 cm h₂ = 4 cm h₁ = 2 cm

H = 30 cm

Total Volume = π × r² × H

Given:

⇒ Base radius (r) = 3 cm

⇒ Number of sections (n) = 5

⇒ Height of bottom section (a) = 2 cm

⇒ Height of topmost section (l) = 10 cm

Formula Used:

⇒ Sum of an AP = (n ÷ 2) × (a + l)

⇒ Volume of a cylinder = π × r² × H

Calculations:

⇒ Total height (H) of the container is the sum of the heights of 5 sections in an AP.

⇒ H = (5 ÷ 2) × (2 + 10)

⇒ H = 2.5 × 12 = 30 cm

⇒ Total volume = 3.14 × (3)² × 30

⇒ Total volume = 3.14 × 9 × 30

⇒ Total volume = 3.14 × 270 = 847.8 cm³

∴ The correct answer is 847.8 cm³.

Additional Information

Arithmetic Progression (AP)

A sequence of numbers where the difference between consecutive terms is constant. The sum of the first n terms can be calculated directly using S = (n ÷ 2) × (a + l), where a is the first term and l is the last term.

Volume of a Cylinder

The amount of space contained within a cylindrical object. The formula is V = π × r² × h, where r is the base radius and h is the height.

Curved Surface Area (CSA)

The total area of the curved surface excluding the flat circular ends. It is calculated using the formula CSA = 2 × π × r × h.

Shortcut Trick

For any linear equation in the standard form Ax + By + C = 0 or Ax + By = C,

The slope (m) is directly given by the formula: m = −A ÷ B

Here, the equation is 7x + 4y = 8, which gives A = 7 and B = 4.

Slope = −7 ÷ 4 = −7/4

∴ The correct answer is −7/4.

Alternate Method

Given:

The linear equation is 7x + 4y = 8.

Run: +4 Rise: −7

(0, 2)

(4, −5)

(8/7, 0)

Line: 7x + 4y = 8

Formula Used:

The slope-intercept form of a line is y = mx + c, where m represents the slope and c represents the y-intercept.

Calculations:

⇒ The given equation is 7x + 4y = 8

⇒ Rearrange the equation to isolate y on one side:

⇒ 4y = −7x + 8

⇒ Divide the entire equation by 4:

⇒ y = (−7 ÷ 4)x + (8 ÷ 4)

⇒ y = (−7/4)x + 2

⇒ Comparing this directly with y = mx + c, we get:

⇒ Slope (m) = −7/4

∴ The correct answer is −7/4.

Additional Information

Slope of a Line

The slope m defines the steepness and direction of a line. For two points (x₁, y₁) and (x₂, y₂), the formula is m = (y₂ − y₁) ÷ (x₂ − x₁).

Parallel Lines

Two non-vertical lines are parallel if and only if their slopes are exactly equal (m₁ = m₂).

Perpendicular Lines

Two straight lines are perpendicular if the product of their slopes is −1 (m₁ × m₂ = −1).

Intercept Form

The equation x/a + y/b = 1 represents a line with x-intercept a and y-intercept b. Its slope is given by −b/a.

Shortcut Trick

A fundamental property of any convex polygon is that a full trip around its perimeter involves exactly one complete rotation.

Therefore, the total sum of all exterior angles, taking one at each vertex, is equivalent to one full circle.

One complete circle rotation is always equal to 360°.

∴ The correct answer is 360°.

Alternate Method

Given:

Any arbitrary polygon with n sides.

Formula Used:

Sum of interior angles of an n-sided polygon = (n − 2) × 180°

At any vertex, Interior Angle + Exterior Angle = 180° (Linear Pair)

e1

e2

e3

e4

e5

Sum of exterior angles = 360°

Calculations:

⇒ Total sum of all interior and exterior angles for n vertices = n × 180°

⇒ Sum of exterior angles = Total sum − Sum of interior angles

⇒ Sum of exterior angles = n × 180° − (n − 2) × 180°

⇒ Sum of exterior angles = 180° × (n − n + 2)

⇒ Sum of exterior angles = 180° × 2 = 360°

⇒ The sum evaluates to 360° independent of the number of sides.

∴ The correct answer is 360°.

Additional Information

Interior Angles of a Polygon

The sum of interior angles of an n-sided polygon is defined as (n − 2) × 180°. For a regular polygon, each interior angle is equal to ((n − 2) × 180°) ÷ n.

Exterior Angles of a Regular Polygon

Since the total sum of exterior angles is always fixed at 360°, each exterior angle of an n-sided regular polygon is given exactly by 360° ÷ n.

Number of Diagonals

The total number of unique diagonals that can be drawn within an n-sided polygon is calculated using the formula n(n − 3) ÷ 2.

Shortcut Trick

The centroid (G) of any triangle connects to its three vertices to form three smaller triangles of equal area.

These three triangles are ΔGAB, ΔGBC, and ΔGCA.

Area of ΔGBC = 1/3 × Area of ΔABC

Area of ΔGBC = 1/3 × 60 = 20 cm²

∴ The correct answer is 20 cm².

Alternate Method

Given:

⇒ Area of ΔABC = 60 cm²

⇒ AD, BE, and CF are the medians intersecting at the centroid G.

A B C D E F G

10 cm² 10 cm²

Area of ΔGBC = 20 cm²

Property of Medians:

⇒ The three medians of a triangle divide it into 6 smaller triangles of exactly equal area.

⇒ The 6 equal triangles are: ΔAFG, ΔBFG, ΔBDG, ΔCDG, ΔCEG, and ΔAEG.

Calculations:

⇒ Area of each smaller triangle = Area of ΔABC ÷ 6

⇒ Area of each smaller triangle = 60 ÷ 6 = 10 cm²

⇒ ΔGBC is formed by combining two of these smaller triangles: ΔBDG and ΔCDG.

⇒ Area of ΔGBC = Area of ΔBDG + Area of ΔCDG

⇒ Area of ΔGBC = 10 + 10 = 20 cm²

∴ The correct answer is 20 cm².

Additional Information

Centroid Ratio Rule

The centroid divides each median into segments with a 2:1 ratio, with the longer segment being on the side of the vertex. For example, AG : GD = 2 : 1.

Apollonius Theorem

It relates the length of a median to the lengths of its adjacent sides: AB² + AC² = 2 × (AD² + BD²), where AD is the median to side BC.

Incenter vs Centroid

While the centroid is the intersection of medians, the incenter is the intersection of the interior angle bisectors and is exactly equidistant from all three sides of the triangle.

Orthocenter vs Circumcenter

The orthocenter is the intersection of the triangle's altitudes, whereas the circumcenter is the intersection of the perpendicular bisectors of its sides and is equidistant from all vertices.

Shortcut Trick

By Componendo and Dividendo rule: if (a + b) / (a − b) = c / d, then a / b = (c + d) / (c − d).

Given a = √(5+x), b = √(5-x), and c/d = 3/1.

Apply the rule directly: √(5+x) / √(5-x) = (3 + 1) / (3 − 1) = 4 / 2 = 2.

Squaring both sides: (5 + x) / (5 − x) = 4

5 + x = 20 − 4x ⇒ 5x = 15 ⇒ x = 3.

∴ The correct answer is 3.

Alternate Method

Given:

[√(5+x) + √(5-x)] ÷ [√(5+x) − √(5-x)] = 3

Calculation:

⇒ By cross-multiplying, we get:

⇒ √(5+x) + √(5-x) = 3 × [√(5+x) − √(5-x)]

⇒ √(5+x) + √(5-x) = 3√(5+x) − 3√(5-x)

⇒ Grouping similar terms on each side:

⇒ √(5-x) + 3√(5-x) = 3√(5+x) − √(5+x)

⇒ 4√(5-x) = 2√(5+x)

⇒ Dividing both sides by 2:

⇒ 2√(5-x) = √(5+x)

⇒ Squaring both sides to eliminate the square root:

⇒ [2√(5-x)]² = [√(5+x)]²

⇒ 4 × (5 − x) = 5 + x

⇒ 20 − 4x = 5 + x

⇒ 20 − 5 = x + 4x

⇒ 15 = 5x

⇒ x = 3

∴ The correct answer is 3.

Additional Information

Componendo and Dividendo Rule

If a/b = c/d, then according to this rule: (a+b)/(a−b) = (c+d)/(c−d). It helps to quickly simplify fractions involving sum and difference of the same terms.

Rationalization

It is a process to eliminate surds or square roots from the denominator by multiplying the numerator and denominator by the conjugate of the denominator.

Linear Equations

A mathematical equation in which the highest power of the variable is 1. Solving involves isolating the variable on one side by balancing the operations.

Shortcut Trick

By drawing a line parallel to the direct common tangent, a right-angled triangle is formed where hypotenuse = distance between centers, base = DCT, and perpendicular = R − r.

Given the hypotenuse = 10 cm and base = 8 cm.

Using the standard Pythagorean triplet (6, 8, 10), the perpendicular must be 6 cm.

So, R − r = 6 cm. Since R : r = 2 : 1, we know R = 2r.

Thus, 2r − r = 6 cm ⇒ r = 6 cm.

∴ The correct answer is 6 cm.

Alternate Method

d = 10 cm

DCT = 8 cm

R r

C1 C2

Given:

Ratio of radii R : r = 2 : 1

Distance between centers (d) = 10 cm

Direct Common Tangent (DCT) = 8 cm

Formula Used:

DCT = √(d² − (R − r)²)

Calculations:

⇒ Let the radius of the smaller circle be r and the larger circle be 2r.

⇒ 8 = √(10² − (2r − r)²)

⇒ 8 = √(100 − r²)

⇒ Squaring both sides to remove the square root:

⇒ 64 = 100 − r²

⇒ r² = 100 − 64

⇒ r² = 36

⇒ r = 6 cm

∴ The correct answer is 6 cm.

Additional Information

Direct Common Tangent (DCT)

It is a tangent that touches two circles without crossing the line joining their centers. The formula to calculate its length is √(d² − (R − r)²).

Transverse Common Tangent (TCT)

It is a tangent that crosses the line connecting the centers of the two circles. Its length is calculated as √(d² − (R + r)²).

Externally Touching Circles

When two circles touch each other externally, the distance between their centers is exactly R + r. In this case, the length of the Direct Common Tangent simplifies to 2√(Rr).