Quantitative Aptitude PYQs

Shortcut Trick

When an algebraic equation is given as x + 1/x = −2, the only possible real value for x is −1.

For any even power, x^even = (−1)^even = 1.

Substitute x = −1 directly into the required expression:

⇒ (1 + 1) − 3 × (1 + 1) + 4 × (1 + 1)

⇒ 2 − 6 + 8 = 4

∴ The correct answer is 4.

Alternate Method

Given:

x + 1/x = −2

Formula Used:

(a + b)² = a² + 2ab + b²

Calculations:

⇒ (x² + 1) / x = −2

⇒ x² + 1 = −2x

⇒ x² + 2x + 1 = 0

⇒ (x + 1)² = 0

⇒ x = −1

Now, substitute x = −1 into the given expression:

Value = x⁶ + 1/x⁶ − 3(x⁴ + 1/x⁴) + 4(x² + 1/x²)

⇒ (−1)⁶ + 1/(−1)⁶ − 3 × [(−1)⁴ + 1/(−1)⁴] + 4 × [(−1)² + 1/(−1)²]

⇒ (1 + 1/1) − 3 × (1 + 1/1) + 4 × (1 + 1/1)

⇒ (1 + 1) − 3 × (1 + 1) + 4 × (1 + 1)

⇒ 2 − 3 × 2 + 4 × 2

⇒ 2 − 6 + 8

⇒ 4

∴ The correct answer is 4.

Additional Information

Standard Algebraic Case (x = 1)

If x + 1/x = 2, then the only real solution is x = 1. Consequently, xⁿ + 1/xⁿ = 2 for any real number n.

Standard Algebraic Case (x = −1)

If x + 1/x = −2, then the only real solution is x = −1. The value of xⁿ + 1/xⁿ will be 2 if n is even, and −2 if n is odd.

Algebraic Identities involving Squares and Cubes

x² + 1/x² = (x + 1/x)² − 2

x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)

Shortcut Trick

For simple interest, the time taken for a sum to become n times is given by the ratio formula:

⇒ T₁ ÷ (n₁ − 1) = T₂ ÷ (n₂ − 1)

⇒ 5 ÷ (2.5 − 1) = T₂ ÷ (5 − 1)

⇒ 5 ÷ 1.5 = T₂ ÷ 4

⇒ T₂ = (5 × 4) ÷ 1.5 = 20 ÷ 1.5 = 40 ÷ 3 = 13¹/₃ years.

∴ The correct answer is 13¹/₃ years.

Alternate Method

Given:

A sum becomes 2.5 times in 5 years.

We need to find the time it takes to become 5 times.

P 0 Years

2.5P 5 Years

5P T Years

+1.5P (SI)

+4P (SI)

Calculations:

Let the initial principal be P.

⇒ Amount after 5 years (A₁) = 2.5P

⇒ Simple Interest generated in 5 years (SI₁) = 2.5P − P = 1.5P

⇒ We need the target Amount (A₂) to be 5P.

⇒ Required Simple Interest (SI₂) = 5P − P = 4P

⇒ At a constant rate, Simple Interest is directly proportional to time.

⇒ Required Time = (SI₂ ÷ SI₁) × T₁

⇒ T₂ = (4P ÷ 1.5P) × 5

⇒ T₂ = (4 ÷ 1.5) × 5 = 20 ÷ 1.5

⇒ T₂ = 40 ÷ 3 = 13¹/₃ years.

∴ The correct answer is 13¹/₃ years.

Additional Information

Simple Interest Formula

SI = (P × R × T) ÷ 100, where P is the principal, R is the rate of interest, and T is the time period.

Amount in Simple Interest

The total amount A is the sum of the principal and interest: A = P + SI = P × (1 + (R × T) ÷ 100).

Proportionality of SI

For a given principal and fixed rate, the simple interest earned is directly proportional to the time period (SI ∝ T). This allows for direct ratio comparisons.

Shortcut Trick

When the simple interest obtained from different schemes is equal, the ratio of principal investments is inversely proportional to the rates: P₁ : P₂ : P₃ = 1/R₁ : 1/R₂ : 1/R₃.

Ratio of investments = 1/2 : 1/4 : 1/5 = 10 : 5 : 4

Total parts = 10 + 5 + 4 = 19 parts

19 parts = ₹1900 ⇒ 1 part = ₹100

Amount invested at 4% corresponds to 5 parts = 5 × 100 = ₹500

∴ The correct answer is ₹500.

Alternate Method

Given:

Total sum invested = ₹1900

Interest rates = 2%, 4%, and 5%

Time period (T) = 1 year

Simple Interest from all three schemes is equal.

Formula Used:

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

Total Investment = ₹1900 (19 parts)

10 parts (2%) ₹1000

5 parts (4%) ₹500

4 parts (5%) ₹400

Interest is same → Ratio = 1/R1 : 1/R2 : 1/R3 1/2 : 1/4 : 1/5 → 10 : 5 : 4

Calculations:

⇒ Let the amounts invested be P₁, P₂, and P₃.

⇒ Interest from scheme 1 = (P₁ × 2 × 1) ÷ 100

⇒ Interest from scheme 2 = (P₂ × 4 × 1) ÷ 100

⇒ Interest from scheme 3 = (P₃ × 5 × 1) ÷ 100

⇒ Since interests are equal: 2P₁ = 4P₂ = 5P₃

⇒ To find the ratio, take the LCM of the coefficients 2, 4, and 5, which is 20.

⇒ Divide the entire equation by 20:

⇒ 2P₁ ÷ 20 = 4P₂ ÷ 20 = 5P₃ ÷ 20

⇒ P₁/10 = P₂/5 = P₃/4

⇒ The ratio of investments is P₁ : P₂ : P₃ = 10 : 5 : 4

⇒ Sum of the ratio parts = 10 + 5 + 4 = 19 parts

⇒ The amount invested at 4% corresponds to P₂, which has 5 parts.

⇒ P₂ = (5 ÷ 19) × 1900

⇒ P₂ = 5 × 100 = 500

∴ The correct answer is ₹500.

Additional Information

Equal Simple Interest Rule

When a sum is divided into multiple parts and invested such that the simple interest earned on each part is the same, the ratio of the invested parts is given by P₁ : P₂ : P₃ = 1/(R₁T₁) : 1/(R₂T₂) : 1/(R₃T₃).

Equal Amount Rule

If the final amounts (Principal + Simple Interest) are equal, then the ratio of the invested parts is given by P₁ : P₂ : P₃ = 1/(100 + R₁T₁) : 1/(100 + R₂T₂) : 1/(100 + R₃T₃).

Difference between SI and CI

For a duration of exactly 1 year (when compounded annually), the Simple Interest and Compound Interest are identical. The difference arises from the second year onwards because CI earns interest on the accumulated interest as well.

Shortcut Trick

In a 45°-45°-90° right-angled triangle, the ratio of the sides (Perpendicular : Base : Hypotenuse) is always 1 : 1 : √2.

Here, the perpendicular (height of the wall) corresponds to 1 unit = 8 m.

Therefore, the hypotenuse (length of the ladder) corresponds to √2 units = 8√2 m.

Length ≈ 8 × 1.414 = 11.312 m.

Rounding to the nearest whole number gives 11 m.

∴ The correct answer is 11 m.

Alternate Method

Given:

Height of the window on the wall (Perpendicular) = 8 m

Angle of elevation of the ladder from the ground (θ) = 45°

Formula Used:

sin θ = Perpendicular ÷ Hypotenuse

45°

8 m Ladder (L) Window Wall Base

Calculations:

⇒ Let the length of the ladder (Hypotenuse) be L.

⇒ sin 45° = 8 ÷ L

⇒ 1 / √2 = 8 ÷ L

⇒ L = 8 × √2

⇒ L = 8 × 1.414

⇒ L = 11.312 m

⇒ Rounding 11.312 m to the nearest whole number gives 11 m.

∴ The correct answer is 11 m.

Additional Information

Trigonometric Ratios of Standard Angles

For an angle of 45°, the common values are sin 45° = 1/√2, cos 45° = 1/√2, and tan 45° = 1. These ratios are essential for solving basic heights and distances problems quickly.

Angle of Elevation

The angle of elevation is the angle formed between the horizontal line of sight and the upward line of sight when looking at an object above the observer.

Pythagoras Theorem

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: H² = P² + B².

Shortcut Trick

Volume of a hollow cylinder = Base Area × Height

Volume = π × (R² − r²) × H

Volume = 3.14 × (5² − 3²) × 10

Volume = 3.14 × 16 × 10 = 502.4 cm³

∴ The correct answer is 502 cm³.

Alternate Method

Given:

Outer radius (R) = 5 cm

Inner radius (r) = 3 cm

Height (H) = 10 cm

Formula Used:

Volume of a hollow cylinder = πH(R² − r²)

r=3 R=5

H = 10 cm

Remaining Volume = πH(R² − r²)

Calculations:

⇒ Volume of the original solid cylinder = πR²H = π × 5² × 10 = 250π

⇒ Volume of the cylindrical hole = πr²H = π × 3² × 10 = 90π

⇒ Remaining Volume = 250π − 90π

⇒ Remaining Volume = 160π

⇒ Using π ≈ 3.14

⇒ Remaining Volume = 160 × 3.14 = 502.4 cm³

⇒ Approximating to the nearest integer, we get 502 cm³.

∴ The correct answer is 502 cm³.

Additional Information

Total Surface Area of Hollow Cylinder

The TSA includes the outer curved surface, inner curved surface, and the two circular rings at the top and bottom. Formula: 2πR H + 2πr H + 2π(R² − r²).

Curved Surface Area of Hollow Cylinder

The CSA only considers the lateral vertical surfaces (both inside and outside). Formula: CSA = 2πH(R + r).

Solid Cylinder Formulas

For a completely solid right circular cylinder with radius R and height H, Volume = πR²H and Total Surface Area = 2πR(R + H).

Shortcut Trick

Volume of a rectangular parallelepiped (cuboid) = Length × Breadth × Height

Volume = 50 × 40 × 3 = 6000 m³

Since 1 m³ of volume holds exactly 1000 litres (which is 1 kilolitre), the capacity in m³ is numerically equal to the capacity in kilolitres.

∴ The correct answer is 6,000 kilolitres.

Alternate Method

Given:

Length of the tank (l) = 50 m

Breadth of the tank (b) = 40 m

Height of the tank (h) = 3 m

50 m 40 m 3 m Rectangular Parallelepiped Tank

Formula Used:

Volume of a cuboid = l × b × h

Relation between Volume and Capacity: 1 m³ = 1000 litres = 1 kilolitre

Calculations:

⇒ Volume of the tank = 50 × 40 × 3

⇒ Volume of the tank = 6000 m³

⇒ Capacity in kilolitres = Volume in m³ × 1

⇒ Capacity = 6000 kilolitres

∴ The correct answer is 6,000 kilolitres.

Additional Information

Unit Conversions for Volume & Capacity

1 m³ = 1000 Litres = 1 Kilolitre

1 Litre = 1000 cm³

Volume of a Cylinder

Volume = πr²h, where r is the base radius and h is the height.

Volume of a Cone

Volume = (1/3)πr²h, where r is the base radius and h is the vertical height.

Shortcut Trick

The parallel planes are at heights 10 cm and 20 cm from the base, meaning heights from the top are 10 cm and 20 cm.

Heights of the three similar cones from the top vertex are 10 cm, 20 cm, and 30 cm.

Ratio of their heights = 10 : 20 : 30 = 1 : 2 : 3

Since volumes of similar 3D shapes are proportional to the cube of their heights, the ratio of their volumes = 1³ : 2³ : 3³ = 1 : 8 : 27

Ratio of volumes of the individual three parts (top to bottom) = 1 : (8 − 1) : (27 − 8) = 1 : 7 : 19

∴ The correct answer is 1 : 7 : 19.

Alternate Method

Given:

Total height of the original cone (H) = 30 cm

Cutting planes are parallel to the base at heights 10 cm and 20 cm from the base.

10 cm

10 cm

10 cm

Part 1 Part 2 Part 3

Formula Used:

For a right circular cone cut by planes parallel to the base, the smaller cones formed are similar to the original cone.

If two cones are similar with heights h₁ and h₂, the ratio of their volumes is:

⇒ V₁ / V₂ = (h₁ / h₂)³

Calculations:

⇒ Total height of the cone (H) = 30 cm

⇒ Height of the first cut from base = 20 cm → Height from the top vertex (h₁) = 30 − 20 = 10 cm

⇒ Height of the second cut from base = 10 cm → Height from the top vertex (h₂) = 30 − 10 = 20 cm

⇒ Height of the entire cone (h₃) = 30 cm

Let V₁, V₂, and V₃ be the volumes of the three similar cones starting from the top with heights 10 cm, 20 cm, and 30 cm respectively.

⇒ V₁ : V₂ : V₃ = (10)³ : (20)³ : (30)³

⇒ V₁ : V₂ : V₃ = 1000 : 8000 : 27000

⇒ V₁ : V₂ : V₃ = 1 : 8 : 27

The volumes of the individual parts from top to bottom are:

⇒ Volume of Part 1 (Top cone) = V₁ = 1 part

⇒ Volume of Part 2 (Middle frustum) = V₂ − V₁ = 8 − 1 = 7 parts

⇒ Volume of Part 3 (Bottom frustum) = V₃ − V₂ = 27 − 8 = 19 parts

⇒ Ratio of volumes of the three parts (Top to Bottom) = 1 : 7 : 19

∴ The correct answer is 1 : 7 : 19.

Additional Information

Similar Solids Property

When a 3D solid is cut parallel to its base to form a smaller similar solid, the ratio of their volumes equals the cube of the ratio of their corresponding linear dimensions (height, radius, or slant height).

Volume of a Cone

The volume of a right circular cone is given by V = (1/3)πr²h, where r is the base radius and h is the perpendicular height.

Volume of a Frustum of a Cone

The volume of a frustum of a cone (the shape remaining after cutting the top off parallel to the base) is V = (1/3)πh(R² + r² + Rr), where h is the frustum's height, and R and r are the radii of its circular bases.

Shortcut Trick

When a 3D object is melted and recast into smaller objects of similar shape, the number of objects is simply the ratio of their volumes.

Since the volume of a hemisphere is strictly proportional to the cube of its radius, we can directly find the count.

Number of objects (n) = Volume of large hemisphere ÷ Volume of small hemisphere

n = R³ ÷ r³ = (R/r)³

∴ The correct answer is (R/r)³.

Alternate Method

R Solid Hemisphere

Melted & Recast

r

r

r

n Smaller Hemispheres

Given:

Radius of large solid hemisphere = R

Radius of each small hemisphere = r

Number of small hemispheres produced = n

Formula Used:

Volume of a hemisphere = (2/3) × π × (Radius)³

When an object is melted and recast, the total volume remains constant.

Calculations:

⇒ Volume of large hemisphere = n × Volume of one small hemisphere

⇒ (2/3) × π × R³ = n × (2/3) × π × r³

⇒ By dividing both sides by (2/3) × π, we get:

⇒ R³ = n × r³

⇒ n = R³ / r³

⇒ n = (R/r)³

∴ The correct answer is (R/r)³.

Additional Information

Volume of a Sphere

The volume of a full sphere of radius r is calculated using the formula: V = (4/3) × π × r³.

Surface Area of a Hemisphere

The Curved Surface Area (CSA) is 2 × π × r². The Total Surface Area (TSA) of a solid hemisphere is 3 × π × r² because it includes the circular base.

Melting and Recasting Principle

When a 3D geometric solid is melted down to form new objects, the total volume always remains conserved, even though the total combined surface area may change.

Shortcut Trick

Given cotA = √3 ⇒ A = 30° (since the standard value of cot30° is √3).

Substitute A = 30° in the expression (1 + sinA)(1 + cosA):

⇒ (1 + sin30°) × (1 + cos30°)

⇒ (1 + 1/2) × (1 + √3/2)

⇒ (3/2) × ((2 + √3) / 2) = (6 + 3√3) / 4

∴ The correct answer is (6+3√3)/4.

Alternate Method

Given:

cotA = √3

Formula Used:

In a right-angled triangle, cotA = Base / Perpendicular

Pythagoras Theorem: Hypotenuse = √(Base² + Perpendicular²)

sinA = Perpendicular / Hypotenuse

cosA = Base / Hypotenuse

A

A B C

Base = √3 Perpendicular = 1 Hypotenuse = 2

Calculations:

⇒ cotA = Base / Perpendicular = √3 / 1

⇒ Let Base = √3 and Perpendicular = 1

⇒ Hypotenuse = √((√3)² + 1²)

⇒ Hypotenuse = √(3 + 1) = √4 = 2

⇒ sinA = Perpendicular / Hypotenuse = 1 / 2

⇒ cosA = Base / Hypotenuse = √3 / 2

Now, substitute these values in the given expression:

⇒ (1 + sinA) × (1 + cosA)

⇒ (1 + 1/2) × (1 + √3/2)

⇒ (3/2) × ((2 + √3) / 2)

⇒ (6 + 3√3) / 4

∴ The correct answer is (6+3√3)/4.

Additional Information

Trigonometric Ratios

In a right-angled triangle, sinθ = Perpendicular/Hypotenuse, cosθ = Base/Hypotenuse, and tanθ = Perpendicular/Base.

Values of Standard Angles

For 30° angle: sin30° = 1/2, cos30° = √3/2, and tan30° = 1/√3.

Pythagoras Theorem

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: Hypotenuse² = Base² + Perpendicular².

Fundamental Trigonometric Identities

The three basic identities are: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Shortcut Trick

We know that the maximum value of a × sin x + b × cos x is √(a² + b²).

For sin x + cos x, maximum value = √(1² + 1²) = √2.

This maximum value is obtained only when x = 45°.

Substitute x = 45° in the target expression: sin x − cos x

⇒ sin 45° − cos 45° = 1/√2 − 1/√2 = 0

∴ The correct answer is 0.

Alternate Method

Given:

sin x + cos x = √2

Formula Used:

(a − b)² = a² + b² − 2 × a × b

sin² x + cos² x = 1

Calculations:

Squaring both sides of the given equation:

⇒ (sin x + cos x)² = (√2)²

⇒ sin² x + cos² x + 2 × sin x × cos x = 2

Substitute sin² x + cos² x = 1:

⇒ 1 + 2 × sin x × cos x = 2

⇒ 2 × sin x × cos x = 1

Now, we need the value of sin x − cos x. Consider its square:

⇒ (sin x − cos x)² = sin² x + cos² x − 2 × sin x × cos x

Substitute the known values back into the expression:

⇒ (sin x − cos x)² = 1 − 1

⇒ (sin x − cos x)² = 0

⇒ sin x − cos x = 0

∴ The correct answer is 0.

Additional Information

Trigonometric Identities

sin² x + cos² x = 1 | sec² x − tan² x = 1 | cosec² x − cot² x = 1

Maximum and Minimum Values

For the expression a × sin x + b × cos x, the maximum value is √(a² + b²) and the minimum value is −√(a² + b²).

Complementary Angles

sin(90° − x) = cos x | cos(90° − x) = sin x | tan(90° − x) = cot x.

Shortcut Trick

For any point to lie on a given line, its (x, y) coordinates must satisfy the line's equation.

Substitute the given points into y = −3x + 7 directly:

(0, 7) ⇒ 7 = −3(0) + 7 ⇒ 7 = 7 (True)

(2, 1) ⇒ 1 = −3(2) + 7 ⇒ 1 = 1 (True)

(3, −2) ⇒ −2 = −3(3) + 7 ⇒ −2 = −2 (True)

∴ The correct answer is (0,7), (2,1) and (3,-2) all.

Alternate Method

Given:

Equation of the straight line: y = −3x + 7

Points to verify: (0, 7), (2, 1), and (3, −2)

Formula Used:

A point (x₁, y₁) lies on the line y = mx + c if substituting x = x₁ gives y = y₁.

(0, 7) (2, 1) (3, -2)

y = -3x + 7

X Y O

Calculations:

Let's verify each given point individually by putting the values in the equation y = −3x + 7.

1. Checking Point (0, 7):

⇒ Put x = 0 and y = 7 in the equation.

⇒ LHS = 7

⇒ RHS = −3(0) + 7 = 7

⇒ LHS = RHS, so the point (0, 7) lies on the line.

2. Checking Point (2, 1):

⇒ Put x = 2 and y = 1 in the equation.

⇒ LHS = 1

⇒ RHS = −3(2) + 7 = −6 + 7 = 1

⇒ LHS = RHS, so the point (2, 1) lies on the line.

3. Checking Point (3, −2):

⇒ Put x = 3 and y = −2 in the equation.

⇒ LHS = −2

⇒ RHS = −3(3) + 7 = −9 + 7 = −2

⇒ LHS = RHS, so the point (3, −2) lies on the line.

⇒ Since all three points successfully satisfy the linear equation, they all lie on the same straight line.

∴ The correct answer is (0,7), (2,1) and (3,-2) all.

Additional Information

Equation of a Straight Line (Slope-Intercept Form)

The equation y = mx + c represents a straight line, where 'm' is the slope (gradient) of the line and 'c' is the y-intercept (the point where the line crosses the Y-axis).

Slope of a Line

The slope 'm' of a line passing through any two points (x₁, y₁) and (x₂, y₂) is calculated using the formula: m = (y₂ − y₁) ÷ (x₂ − x₁).

Collinear Points

Three or more points are said to be collinear if they all lie on the same straight line. The area of a triangle formed by collinear points is always equal to 0.

Shortcut Trick

Convert the central angle of the second sector from radians to degrees: π/2 radians × (180° / π) = 90°.

Both sectors have the exact same central angle (90°) and belong to circles of the same radius (6 cm).

Since both the radius and the central angle are identical, the areas of the two sectors must be equal.

Ratio of their areas = 1:1.

∴ The correct answer is 1:1.

Alternate Method

Given:

Radius of 1^st sector (r₁) = 6 cm

Central angle of 1^st sector (θ₁) = 90°

Central angle of 2^nd sector (θ₂) = π/2 radians

Radius of 2^nd sector (r₂) = 6 cm (assumed identical as no other variation is mentioned)

Formulas Used:

Area of a sector (when angle is in degrees) = (θ / 360°) × πr²

Area of a sector (when angle is in radians) = (1/2) × r² × θ

90° r = 6 cm Sector 1

π/2 rad r = 6 cm Sector 2

Calculations:

⇒ Area of 1^st sector = (90° / 360°) × π × (6)²

⇒ Area of 1^st sector = (1 / 4) × 36π = 9π cm²

⇒ Area of 2^nd sector = (1 / 2) × (6)² × (π / 2)

⇒ Area of 2^nd sector = (1 / 2) × 36 × (π / 2) = 9π cm²

⇒ Ratio of areas = Area of 1^st sector : Area of 2^nd sector

⇒ Ratio = 9π : 9π = 1 : 1

∴ The correct answer is 1:1.

Additional Information

Degrees to Radians Conversion

Angle in radians = Angle in degrees × (π / 180°). Conversely, to convert from radians to degrees, multiply by (180° / π).

Area of a Sector

The area represents the region bounded by two radii and an arc. Formula: A = (θ / 360°) × πr² if θ is in degrees, and A = (1/2)r²θ if θ is in radians.

Length of an Arc

The distance along the curved line making up the arc. Formula: l = (θ / 360°) × 2πr (degrees) or l = rθ (radians).

Shortcut Trick

The sum of an interior angle and its corresponding exterior angle is 180°.

Exterior angle = 180° − 150° = 30°

For any regular polygon, the number of sides = 360° ÷ Exterior angle.

Number of sides = 360° ÷ 30° = 12

∴ The correct answer is 12 sides.

Alternate Method

Given:

Each interior angle of the regular polygon = 150°

Formula Used:

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

150°

30°

Polygon Vertex Exterior Angle = 180° − 150° = 30°

Calculations:

⇒ 150° = [(n − 2) × 180°] ÷ n

⇒ 150n = 180n − 360

⇒ 180n − 150n = 360

⇒ 30n = 360

⇒ n = 360 ÷ 30 = 12

∴ The correct answer is 12 sides.

Additional Information

Sum of Interior Angles

The sum of all interior angles of a polygon with n sides is evaluated using the formula (n − 2) × 180°.

Sum of Exterior Angles

For any convex polygon, regardless of its number of sides, the sum of its exterior angles is consistently 360°.

Number of Diagonals

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

Shortcut Trick

The radius, tangent, and distance to the center form a right-angled triangle.

Radius r = √(Distance² − Tangent²)

⇒ r = √(13² − 12²) = √(169 − 144) = √25 = 5 cm.

Area of circle = πr² = π × 5² = 25π cm².

∴ The correct answer is 25π cm².

Alternate Method

Given:

Distance of point P from the center of the circle O (OP) = 13 cm

Length of the tangent drawn from P (PT) = 12 cm

O P T 13 cm 12 cm r = ? Radius r = √(13² − 12²) = 5 cm

Formula Used:

The tangent to a circle is always perpendicular to the radius at the point of contact. Therefore, ΔOTP is a right-angled triangle at T.

Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular²

Area of a circle = πr²

Calculations:

⇒ In right-angled ΔOTP, OP² = OT² + PT²

⇒ 13² = r² + 12²

⇒ 169 = r² + 144

⇒ r² = 169 − 144

⇒ r² = 25

⇒ r = √25 = 5 cm

⇒ Area of the circle = π × (5)²

⇒ Area = 25π cm²

∴ The correct answer is 25π cm².

Additional Information

Tangent Properties

A tangent touches a circle at exactly one point. The radius drawn from the center to the point of tangency is always strictly perpendicular to the tangent line.

Two Tangents from an External Point

If two tangents are drawn from a single external point to a circle, their lengths are always equal, and they subtend equal angles at the center.

Direct Common Tangent (DCT)

Length of DCT = √[d² − (r₁ − r₂)²], where d is the straight-line distance between the centers and r₁, r₂ are the radii of the two circles.

Transverse Common Tangent (TCT)

Length of TCT = √[d² − (r₁ + r₂)²]. The TCT crosses the connecting line segment joining the centers of the two circles.

Shortcut Trick

To prove congruence, check the matching parts between the two triangles.

⇒ Side: AB = AC (Given)

⇒ Angle: ∠BAD = ∠CAD (AD is the angle bisector)

⇒ Side: AD = AD (Common base)

Since two sides and the included angle are equal, the triangles are congruent by the SAS rule.

∴ The correct answer is Yes, by SAS.

Alternate Method

Given:

⇒ In ΔABC, AD bisects ∠BAC.

⇒ AB = AC.

A B C D Side Side

Proof:

⇒ In ΔABD and ΔACD:

⇒ AB = AC (Given data)

⇒ ∠BAD = ∠CAD (Since AD is the angle bisector of ∠BAC)

⇒ AD = AD (Common side for both triangles)

⇒ Here, two sides and the included angle of ΔABD are exactly equal to the corresponding two sides and the included angle of ΔACD.

⇒ According to the Side-Angle-Side (SAS) congruence criterion, ΔABD ≅ ΔACD.

∴ The correct answer is Yes, by SAS.

Additional Information

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, the triangles are congruent.

SSS (Side-Side-Side) Congruence

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

Properties of Isosceles Triangles

In an isosceles triangle, the angle bisector drawn from the vertex angle to the base is also the median and the altitude. Therefore, BD = DC and ∠ADB = ∠ADC = 90°.

Shortcut Trick

For any expression of the form 1/(√x − √y) + 1/(√x + √y), the direct shortcut formula is 2√x ÷ (x − y).

Here, x = 7 and y = 3.

⇒ Result = 2√7 ÷ (7 − 3)

⇒ Result = 2√7 ÷ 4 = √7 / 2

∴ The correct answer is √7/2.

Alternate Method

Given:

Expression = 1/(√7 − √3) + 1/(√7 + √3)

Formula Used:

(a − b)(a + b) = a² − b²

Calculations:

⇒ To simplify, take the LCM of the denominators:

⇒ Expression = [(√7 + √3) + (√7 − √3)] ÷ [(√7 − √3) × (√7 + √3)]

⇒ Numerator = √7 + √3 + √7 − √3 = 2√7

⇒ Denominator = (√7)² − (√3)² = 7 − 3 = 4

⇒ Expression = 2√7 ÷ 4

⇒ Expression = √7 / 2

∴ The correct answer is √7/2.

Additional Information

Rationalization

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

Algebraic Identities

The identity (a + b)(a − b) = a² − b² is a crucial rule for clearing out radical denominators.

Conjugate Pairs

The conjugate of the binomial surd √x + √y is √x − √y. Their product directly gives a rational number (x − y).

Shortcut Trick

From fundamental trigonometric identities, we know that sin²A + cos²A = 1.

⇒ cos²A = 1 − sin²A

By substituting the given value sinA = x into the equation:

⇒ cos²A = 1 − (x)²

∴ The correct answer is 1 − x².

Alternate Method

A x 1 √(1 - x²)

Given:

sinA = x

Formula Used:

sinA = Perpendicular ÷ Hypotenuse

cosA = Base ÷ Hypotenuse

Pythagoras Theorem: Hypotenuse² = Perpendicular² + Base²

Calculations:

⇒ sinA = x ÷ 1 = Perpendicular ÷ Hypotenuse

⇒ By comparing, we get Perpendicular = x and Hypotenuse = 1

⇒ Base² = Hypotenuse² − Perpendicular²

⇒ Base² = 1² − x² = 1 − x²

⇒ Base = √(1 − x²)

⇒ cosA = Base ÷ Hypotenuse = √(1 − x²) ÷ 1 = √(1 − x²)

⇒ cos²A = (√(1 − x²))² = 1 − x²

∴ The correct answer is 1 − x².

Additional Information

Fundamental Trigonometric Identities

1. sin²A + cos²A = 1

2. sec²A − tan²A = 1

3. cosec²A − cot²A = 1

Basic Trigonometric Ratios

• sinA = Perpendicular ÷ Hypotenuse

• cosA = Base ÷ Hypotenuse

• tanA = Perpendicular ÷ Base

Pythagoras Theorem

For any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the perpendicular and the base: Hypotenuse² = Perpendicular² + Base².

Shortcut Trick

The radius drawn to a tangent at the point of contact is perpendicular to it, forming a 90° angle.

This forms a quadrilateral where two opposite angles (at the points of contact) are 90° each.

The sum of the remaining two opposite angles (the angle at the center and the angle between the tangents) is exactly 180°.

Angle at the center = 180° − Angle between tangents = 180° − 60° = 120°.

∴ The correct answer is 120°.

Alternate Method

Given:

Angle between the two tangents = 60°

Formula Used:

The angle between a tangent to a circle and the radius drawn to the point of contact is 90°.

The sum of all interior angles of a quadrilateral is 360°.

60°120°OPAB

Calculations:

⇒ Let PA and PB be the two tangents drawn from an external point P to a circle with center O.

⇒ Let A and B be the respective points of contact on the circle.

⇒ ∠APB = 60° (Given)

⇒ Since a radius is always perpendicular to a tangent at the point of contact:

⇒ ∠OAP = 90° and ∠OBP = 90°

⇒ In the quadrilateral OAPB, the sum of all interior angles is 360°.

⇒ ∠AOB + ∠OAP + ∠OBP + ∠APB = 360°

⇒ ∠AOB + 90° + 90° + 60° = 360°

⇒ ∠AOB + 240° = 360°

⇒ ∠AOB = 360° − 240° = 120°

∴ The correct answer is 120°.

Additional Information

Properties of Tangents

Tangents drawn from an external point to a circle are exactly equal in length, and they subtend equal angles at the center of the circle.

Perpendicular Radius

The tangent at any point of a circle is strictly perpendicular to the radius passing through the exact point of contact.

Cyclic Quadrilateral Formation

The quadrilateral formed by the two tangents originating from an external point and their corresponding radii has opposite angles that continuously sum to 180°, classifying it perfectly as a cyclic quadrilateral.

Shortcut Trick

By the Tangent-Secant theorem, PA² = PB × PC

⇒ 6² = PB × (PB + 5)

⇒ 36 = PB × (PB + 5)

Checking the given options, if PB = 4 cm, then 4 × (4 + 5) = 36. This matches the equation perfectly.

∴ The correct answer is 4 cm.

Alternate Method

Given:

Length of the tangent segment PA = 6 cm

Length of the internal secant segment BC = 5 cm

Let the length of the external segment PB be x cm.

The total length of the secant segment PC = PB + BC = (x + 5) cm.

A P B C O 6 cm PB = x 5 cm

Formula Used:

By the Tangent-Secant Theorem, when a tangent and a secant are drawn to a circle from an external point, the square of the tangent segment is equal to the product of the external secant segment and the total secant length.

⇒ PA² = PB × PC

Calculation:

Substitute the given values into the geometric equation:

⇒ 6² = x × (x + 5)

⇒ 36 = x² + 5x

⇒ x² + 5x − 36 = 0

⇒ x² + 9x − 4x − 36 = 0

⇒ x(x + 9) − 4(x + 9) = 0

⇒ (x − 4)(x + 9) = 0

Since the length of a geometric segment cannot be negative, we discard x = −9.

⇒ x = 4 cm

∴ The correct answer is 4 cm.

Additional Information

Secant-Secant Theorem

If two secant segments are drawn to a circle from an exterior point, the product of the external segment and the total segment of one secant equals the product of the external and total segments of the other. Formula for secants PAB and PCD: PA × PB = PC × PD.

Intersecting Chords Theorem

If two chords intersect inside a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. Formula for chords AB and CD intersecting at P: PA × PB = PC × PD.

Length of Direct Common Tangent (DCT)

For two circles of radii R and r with their centers at a distance d apart, the length of DCT = √(d² − (R − r)²).

Length of Transverse Common Tangent (TCT)

For two circles of radii R and r with their centers at a distance d apart, the length of TCT = √(d² − (R + r)²).

Shortcut Trick

Observe the ratio between the corresponding terms in the numerator and denominator.

Each term in the denominator (0.4 and 0.04) is exactly 2 times the respective term in the numerator (0.2 and 0.02).

The expression follows the pattern: (a³ + b³) ÷ ((2a)³ + (2b)³)

This simplifies to: 1 ÷ 2³ = 1 ÷ 8

1 ÷ 8 = 0.125

∴ The correct answer is 0.125.

Alternate Method

Given:

Expression: (0.2³ + 0.02³) ÷ (0.4³ + 0.04³)

Formula Used:

(a × b)ⁿ = aⁿ × bⁿ

Calculation:

⇒ Numerator = (0.2)³ + (0.02)³

⇒ Denominator = (0.4)³ + (0.04)³

⇒ We can rewrite 0.4 as (2 × 0.2) and 0.04 as (2 × 0.02).

⇒ Denominator = (2 × 0.2)³ + (2 × 0.02)³

⇒ Using the exponent formula, expand the terms:

⇒ Denominator = (2³ × 0.2³) + (2³ × 0.02³)

⇒ Factor out the common term 2³ (which equals 8):

⇒ Denominator = 8 × [(0.2)³ + (0.02)³]

⇒ Now, substitute this back into the original expression:

⇒ Expression = [(0.2)³ + (0.02)³] ÷ {8 × [(0.2)³ + (0.02)³]}

⇒ The common bracket [(0.2)³ + (0.02)³] cancels out from both the numerator and denominator:

⇒ Expression = 1 ÷ 8

⇒ Expression = 0.125

∴ The correct answer is 0.125.

Additional Information

Laws of Exponents

When multiplying two numbers within a common power, you can distribute the power: (a × b)ⁿ = aⁿ × bⁿ. This rule is highly useful for factoring out common numeric multiples from complex equations.

BODMAS Rule

Always follow the sequence of Brackets, Orders (powers/roots), Division/Multiplication, and Addition/Subtraction to ensure standard and accurate calculations.

Common Decimal Fractions

Memorizing basic fraction-to-decimal conversions saves significant time in arithmetic exams. Common equivalencies include: 1 ÷ 2 = 0.5, 1 ÷ 4 = 0.25, and 1 ÷ 8 = 0.125.