Search Previous Year Questions

Shortcut Trick

Substitute the given point coordinates directly into the line equation.

⇒ y = mx + 2

⇒ 8 = m × 3 + 2

⇒ 3m = 8 − 2 = 6

⇒ m = 2

∴ The correct answer is 2.

Alternate Method

0 X Y

y-intercept (0, 2) Point (3, 8) Run = 3 Rise = 6

y = 2x + 2

Given:

Equation of the line: y = mx + 2

Point on the line = (3, 8)

Method Used:

If a point (x₁, y₁) lies on a straight line, substituting its coordinates must perfectly satisfy the algebraic equation of the line.

Calculations:

⇒ Substitute the coordinates x = 3 and y = 8 into the given equation.

⇒ 8 = m × 3 + 2

⇒ 8 − 2 = 3m

⇒ 6 = 3m

⇒ m = 6 ÷ 3

⇒ m = 2

∴ The correct answer is 2.

Additional Information

Slope-Intercept Form

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

Slope Formula (Two Points)

If a line passes through two specified points (x₁, y₁) and (x₂, y₂), the slope is calculated as m = (y₂ − y₁) ÷ (x₂ − x₁).

Point-Slope Form

When the slope m and a single point (x₁, y₁) on the line are known, the equation can be formed using y − y₁ = m(x − x₁).

Shortcut Trick

For identical circles, the area of a sector is directly proportional to its central angle.

Convert the second sector's angle to degrees: 2π/3 radians = (2 × 180°) ÷ 3 = 120°.

Since both sectors have identical central angles (120°) and belong to the same circle (same radius), their areas are equal.

Therefore, the ratio of their areas is exactly 1:1.

∴ The correct answer is 1:1.

Alternate Method

120° 5 cm

First Sector

2π/3 rad 5 cm

Second Sector

Given:

Radius of the circle (r) = 5 cm

Central angle of 1^st sector (θ₁) = 120°

Central angle of 2^nd sector (θ₂) = 2π/3 radians

Formula Used:

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

Area of a sector (in radians) = (1/2)r²θ

Calculations:

⇒ Area of 1^st sector = (120°/360°) × π × (5)²

⇒ Area of 1^st sector = (1/3) × 25π = 25π/3 cm²

⇒ Area of 2^nd sector = (1/2) × (5)² × (2π/3)

⇒ Area of 2^nd sector = (25/2) × (2π/3) = 25π/3 cm²

⇒ Ratio of their areas = (25π/3) ÷ (25π/3) = 1/1

∴ The correct answer is 1:1.

Additional Information

Area of Sector

A sector is a portion of a circle enclosed by two radii and an arc. If the central angle θ is in degrees, Area = (θ/360°) × πr². If θ is in radians, Area = (1/2)r²θ.

Length of an Arc

The arc length is the distance along the curved line making up the sector. For an angle in degrees, Length = (θ/360°) × 2πr. For radians, Length = rθ.

Radian to Degree Conversion

To convert an angle from radians to degrees, multiply by 180°/π. To convert from degrees to radians, multiply by π/180°.

Shortcut Trick

For any regular polygon, the sum of an interior angle and an exterior angle is 180°.

Exterior angle of a regular polygon = 360° ÷ n

⇒ Exterior angle = 360° ÷ 8 = 45°

⇒ Interior angle = 180° − 45° = 135°

∴ The correct answer is 135°.

Alternate Method

Given:

Number of sides in the regular polygon (octagon), n = 8

Formula Used:

Measure of each interior angle = (n − 2) × 180° ÷ n

135°

45° Regular Octagon n = 8

Calculations:

⇒ Sum of all interior angles = (8 − 2) × 180°

⇒ Sum of all interior angles = 6 × 180° = 1080°

⇒ Measure of each interior angle = 1080° ÷ 8

⇒ Measure of each interior angle = 135°

∴ The correct answer is 135°.

Additional Information

Sum of Interior Angles

For any convex polygon with n sides, the sum of its interior angles is (n − 2) × 180°.

Number of Diagonals

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

Exterior Angle Property

The sum of the exterior angles (taking one at each vertex) of any convex polygon is always exactly 360°, regardless of the number of sides.

Shortcut Trick

Sum of parallel sides = (2 × Area) ÷ Height = (2 × 240) ÷ 10 = 48 cm.

Ratio of parallel sides = 3 : 5 (Total 8 parts).

Value of 1 part = 48 ÷ 8 = 6 cm.

Lengths are 3 × 6 = 18 cm and 5 × 6 = 30 cm.

∴ The correct answer is 18 cm and 30 cm.

Alternate Method

Given:

Ratio of parallel sides = 3 : 5

Height (h) = 10 cm

Total Area = 240 cm²

Formula Used:

Area of Trapezoid = 1/2 × (Sum of parallel sides) × Height

3x 5x h = 10 cm Area = 240 cm²

⇒ Let the parallel sides be 3x and 5x.

⇒ Area = 1/2 × (3x + 5x) × 10 = 240

⇒ 1/2 × 8x × 10 = 240

⇒ 40x = 240

⇒ x = 240 ÷ 40 = 6

⇒ Length of first parallel side = 3 × 6 = 18 cm

⇒ Length of second parallel side = 5 × 6 = 30 cm

∴ The correct answer is 18 cm and 30 cm.

Additional Information

Isosceles Trapezoid

A trapezoid in which the non-parallel sides are equal in length. The base angles are equal and the diagonals are of equal length.

Midsegment of a Trapezoid

The line connecting the midpoints of the non-parallel sides. Its length is the average of the two parallel sides: Midsegment = (a + b) ÷ 2.

Area of Parallelogram

Area = Base × Height. Opposite sides are equal and parallel, meaning a parallelogram is considered a special type of trapezoid.

Shortcut Trick

If two triangles are congruent, their corresponding sides and perimeters are identical.

Perimeter of ΔABC = AB + BC + AC = 5 + 6 + 7 = 18 cm.

Since ΔXYZ is exactly the same shape and size as ΔABC, its perimeter remains exactly the same.

∴ The correct answer is 18 cm.

Alternate Method

Given:

Sides of ΔABC: AB = 5 cm, BC = 6 cm, AC = 7 cm.

ΔXYZ is congruent to ΔABC (ΔXYZ ≅ ΔABC).

Formula Used:

Perimeter of a triangle = Sum of all three sides.

For congruent triangles, corresponding sides are equal (CPCT).

A B C 5 cm 6 cm 7 cm

X Y Z 5 cm 6 cm 7 cm

≅ Perimeter = 5 + 6 + 7 = 18 cm

Calculations:

⇒ Since ΔXYZ ≅ ΔABC, their corresponding side lengths strictly match.

⇒ XY = AB = 5 cm

⇒ YZ = BC = 6 cm

⇒ XZ = AC = 7 cm

⇒ Perimeter of ΔXYZ = XY + YZ + XZ

⇒ Perimeter of ΔXYZ = 5 + 6 + 7 = 18 cm

∴ The correct answer is 18 cm.

Additional Information

Congruence of Triangles

Two triangles are congruent if all three corresponding sides and all three corresponding angles are equal in measure, meaning they superimpose exactly when placed over each other.

Conditions for Congruence

Triangles can be formally proven congruent using established criteria such as SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), AAS, and RHS (Right Angle-Hypotenuse-Side).

Properties of Congruent Triangles (CPCT)

Because congruent triangles possess the exact same overall dimensions, their calculated perimeters, internal areas, corresponding altitudes, and circumradii are mathematically identical.

Shortcut Trick

If a + 1 ÷ a = k, then a² + 1 ÷ a² = k² − 2.

Let a = √x, then a² = x. We are given k = 4.

⇒ x + 1 ÷ x = 4² − 2

⇒ x + 1 ÷ x = 16 − 2 = 14

∴ The correct answer is 14.

Alternate Method

Given:

√x + 1 ÷ √x = 4

Formula Used:

(a + b)² = a² + b² + 2ab

Calculations:

⇒ Squaring both sides of the given equation:

⇒ (√x + 1 ÷ √x)² = 4²

⇒ (√x)² + (1 ÷ √x)² + 2 × √x × (1 ÷ √x) = 16

⇒ x + 1 ÷ x + 2 = 16

⇒ x + 1 ÷ x = 16 − 2

⇒ x + 1 ÷ x = 14

∴ The correct answer is 14.

Additional Information

Minus Identity Variant

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

Cube Identity (Positive)

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

Cube Identity (Negative)

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

Power of 4 Variant

If x + 1 ÷ x = k, then x⁴ + 1 ÷ x⁴ = (k² − 2)² − 2.

Shortcut Trick

Given, cotA = cosA ÷ sinA = 2 ÷ 1

This gives a direct ratio of cosA to sinA as 2 : 1. Let Base = 2 and Perpendicular = 1.

Using Pythagorean triplet logic, Hypotenuse = √(2² + 1²) = √5

⇒ sinA = 1/√5 and cosA = 2/√5

⇒ sinA − cosA = 1/√5 − 2/√5 = −1/√5

∴ The correct answer is −1/√5.

Alternate Method

Given:

cotA = 2 = 2/1

Formula Used:

cotA = Base ÷ Perpendicular

sinA = Perpendicular ÷ Hypotenuse

cosA = Base ÷ Hypotenuse

ABase = 2Perp = 1Hyp = √5

Calculations:

⇒ Base (B) = 2, Perpendicular (P) = 1

⇒ Hypotenuse (H) = √(B² + P²) = √(2² + 1²) = √5

⇒ sinA = 1/√5

⇒ cosA = 2/√5

⇒ sinA − cosA = 1/√5 − 2/√5 = −1/√5

∴ The correct answer is −1/√5.

Additional Information

Trigonometric Identities

The primary Pythagorean identities are: sin²A + cos²A = 1, sec²A − tan²A = 1, and cosec²A − cot²A = 1.

Basic Trigonometric Ratios

For any right-angled triangle: sinA = Perpendicular/Hypotenuse, cosA = Base/Hypotenuse, and tanA = Perpendicular/Base.

Reciprocal Relations

The reciprocal ratios are: cosecA = 1/sinA, secA = 1/cosA, and cotA = 1/tanA = cosA/sinA.

Shortcut Trick

The radius, the tangent, and the line connecting the center to the external point always form a right-angled triangle.

The given lengths for the legs are 7 cm and 24 cm.

Using the standard Pythagorean triplet (7, 24, 25), the hypotenuse directly equals 25.

Thus, the distance from the point to the center is 25 cm.

∴ The correct answer is 25 cm.

Alternate Method

Given:

Radius of the circle (r) = 7 cm

Length of the tangent from external point P (t) = 24 cm

O

T

P

7 cm

24 cm

d = ?

Formula Used:

A tangent at any point of a circle is perpendicular to the radius through the point of contact.

By Pythagoras theorem: Hypotenuse² = Perpendicular² + Base²

⇒ OP² = OT² + PT²

Calculation:

⇒ OP² = 7² + 24²

⇒ OP² = 49 + 576

⇒ OP² = 625

⇒ OP = √625

⇒ OP = 25 cm

∴ The correct answer is 25 cm.

Additional Information

Tangent Perpendicularity Theorem

A tangent to a circle is always perpendicular to the radius drawn to the point of tangency. This forms a 90° angle, enabling the direct use of the Pythagorean theorem.

Number of Tangents from a Point

From any external point, exactly two tangents can be drawn to a circle. Both tangents will always be equal in length.

Common Pythagorean Triplets

Triplets are integer combinations that satisfy the a² + b² = c² rule. Common examples include (3, 4, 5), (5, 12, 13), (7, 24, 25), and (8, 15, 17).

Shortcut Trick

According to the fundamental properties of a circle, the lengths of tangents drawn from a single external point to a circle are always equal.

Since the length of one tangent from the external point is given as 12 cm, the other tangent must also be exactly equal in length.

∴ The correct answer is 12 cm.

Alternate Method

O P A B

12 cm ? cm

Given:

⇒ Let the circle have center O and the external point be P.

⇒ Two tangents PA and PB are drawn to the circle, where PA = 12 cm.

Calculations:

⇒ We know that the radius drawn to the point of tangency is perpendicular to the tangent line.

⇒ ∠OAP = ∠OBP = 90°

⇒ In right-angled triangles ΔOAP and ΔOBP:

⇒ OA = OB (Radii of the same circle)

⇒ OP = OP (Common hypotenuse)

⇒ By RHS (Right-angle Hypotenuse Side) congruence criterion, ΔOAP ≅ ΔOBP.

⇒ Therefore, PA = PB (Corresponding parts of congruent triangles).

⇒ PB = 12 cm.

∴ The correct answer is 12 cm.

Additional Information

Properties of Tangents

A tangent to a circle is perpendicular to the radius at the point of tangency. Lengths of the two tangents drawn from a common external point to a circle are always equal.

Direct Common Tangent (DCT)

Length of DCT = √[d² − (R − r)²], where d is the distance between centers and R, r are the radii of the two circles.

Transverse Common Tangent (TCT)

Length of TCT = √[d² − (R + r)²].

Shortcut Trick

Observe the multiplicative relation between the numerator and denominator: 0.04 = 0.2 × 0.2 and 0.02 = 0.2 × 0.1

⇒ Numerator = (0.2 × 0.2)³ + (0.2 × 0.1)³ = 0.2³ × (0.2³ + 0.1³)

The term (0.2³ + 0.1³) is common in both the numerator and denominator and cancels out.

⇒ Result = 0.2³ = 0.008

∴ The correct answer is 0.008.

Alternate Method

Given:

(0.04³ + 0.02³) ÷ (0.2³ + 0.1³)

Calculations:

Evaluate the cubes of the given decimals directly:

⇒ 0.04³ = 0.04 × 0.04 × 0.04 = 0.000064

⇒ 0.02³ = 0.02 × 0.02 × 0.02 = 0.000008

⇒ Numerator sum = 0.000064 + 0.000008 = 0.000072

Now evaluate the denominator:

⇒ 0.2³ = 0.2 × 0.2 × 0.2 = 0.008

⇒ 0.1³ = 0.1 × 0.1 × 0.1 = 0.001

⇒ Denominator sum = 0.008 + 0.001 = 0.009

Divide the numerator sum by the denominator sum:

⇒ 0.000072 ÷ 0.009

Multiply both numerator and denominator by 1000000 to remove decimals:

⇒ 72 ÷ 9000

⇒ 8 ÷ 1000 = 0.008

∴ The correct answer is 0.008.

Additional Information

Laws of Exponents

When factors are multiplied and raised to a power, the power distributes to each factor: (a × b)ⁿ = aⁿ × bⁿ.

Sum of Cubes Formula

The sum of two cubes can be factorised using the algebraic identity: a³ + b³ = (a + b)(a² − ab + b²).

Decimal Multiplication Rules

When multiplying decimals, the result has a number of decimal places equal to the sum of the decimal places in the individual numbers being multiplied.