Quantitative Aptitude PYQs

Shortcut Trick

Effective contribution of A = Rs. 1,50,000 (Capital) + Rs. 50,000 (Labor) = Rs. 2,00,000

Effective contribution of B = Rs. 2,00,000 (Capital)

Profit sharing ratio = A : B = 2,00,000 : 2,00,000 = 1 : 1

A's share = (1 ÷ 2) × Total Profit = (1 ÷ 2) × Rs. 1,20,000 = Rs. 60,000

∴ The correct answer is Rs. 60,000.

Alternate Method

Given:

Capital brought by A = Rs. 1,50,000

Equivalent labor value contributed by A = Rs. 50,000

Capital brought by B = Rs. 2,00,000

Total profit earned = Rs. 1,20,000

Formula Used:

Total Effective Contribution = Capital Contribution + Equivalent Labor Value

Profit Share Ratio = Ratio of Total Effective Contributions

Capital: 1.5L Labor: 0.5L Partner A (2.0L) Capital: 2.0L Partner B (2.0L) Effective Contribution Ratio = 1 : 1 Total Profit = Rs. 1,20,000

Calculations:

⇒ Total effective contribution of A = Rs. 1,50,000 + Rs. 50,000 = Rs. 2,00,000

⇒ Total effective contribution of B = Rs. 2,00,000

⇒ Ratio of effective contributions (A : B) = 2,00,000 : 2,00,000 = 1 : 1

⇒ Sum of the ratio parts = 1 + 1 = 2

⇒ A's share in the profit = (1 ÷ 2) × Rs. 1,20,000

⇒ A's share = Rs. 60,000

∴ The correct answer is Rs. 60,000.

Additional Information

Simple Partnership

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

Compound Partnership

When investments are for different durations, the profit ratio is calculated as: Profit Ratio = (C₁ × T₁) : (C₂ × T₂), where C is capital and T is time.

Working/Active Partner

A partner who acts as the manager of a business usually receives a certain percentage of the profit as salary for their labor, after which the remaining profit is distributed in proportion to capital. In this question, the labor value was directly capitalized.

Shortcut Trick

Average of parallel bases = (5 + 3) ÷ 2 = 4 m

Area = Average base × height = 4 × 4 = 16 m²

Total cost = Area × Rate = 16 × 250 = ₹4,000

∴ The correct answer is ₹4,000.

Alternate Method

Given:

Parallel bases of the trapezium, a = 5 m and b = 3 m

Height of the trapezium, h = 4 m

Printing cost rate = ₹250/m²

Formula Used:

Area of Trapezium = 1/2 × (Sum of parallel bases) × height

Total Cost = Area × Rate per m²

3 m 5 m 4 m Area = 16 m²

Calculations:

⇒ Area = 1/2 × (5 + 3) × 4

⇒ Area = 1/2 × 8 × 4

⇒ Area = 16 m²

⇒ Total Cost = 16 × 250

⇒ Total Cost = ₹4,000

∴ The correct answer is ₹4,000.

Additional Information

Area of a Parallelogram

Area = Base × Height. A trapezium becomes a parallelogram if both pairs of opposite sides are equal and parallel.

Area of a Rhombus

Area = 1/2 × d₁ × d₂, where d₁ and d₂ are the lengths of the two diagonals.

Total Cost Calculation in Mensuration

Total Cost = Area (or Perimeter / Volume) × Unit Rate. This standard formula is widely used to evaluate the expense of painting, flooring, printing, or fencing a given space.

Shortcut Trick

Total Area = Area(T1) + Area(T2)

⇒ [0.5 × (70 + 50) × 20] + [0.5 × (50 + 30) × 15]

⇒ (0.5 × 120 × 20) + (0.5 × 80 × 15)

⇒ 1200 + 600 = 1800 m²

Total Cost = 1800 × 60 = ₹1,08,000

∴ The correct answer is ₹1,08,000.

Alternate Method

Given:

Trapezoid 1 (T1) bases = 70 m and 50 m, height = 20 m

Trapezoid 2 (T2) bases = 50 m and 30 m, height = 15 m

Land cost = ₹60 / m²

Formula Used:

Area of Trapezoid = 0.5 × (Sum of parallel sides) × Height

  15 m 20 m 30 m 50 m 70 m T2 T1

Calculations:

⇒ Area of T1 = 0.5 × (70 + 50) × 20

⇒ Area of T1 = 0.5 × 120 × 20 = 1200 m²

⇒ Area of T2 = 0.5 × (50 + 30) × 15

⇒ Area of T2 = 0.5 × 80 × 15 = 600 m²

⇒ Total Area of land = 1200 + 600 = 1800 m²

⇒ Total Cost = Total Area × Cost per m²

⇒ Total Cost = 1800 × 60 = 108000

∴ The correct answer is ₹1,08,000.

Additional Information

Properties of a Trapezoid

A trapezoid is a quadrilateral with at least one pair of parallel sides. The perpendicular distance between the parallel sides is called its height. Area = 0.5 × (Sum of parallel sides) × Height.

Median of a Trapezoid

The segment connecting the midpoints of the non-parallel sides is called the median. Its length is equal to the average of the two parallel bases. Area = Median × Height.

Isosceles Trapezoid

When the non-parallel sides of a trapezoid are equal in length, it is an isosceles trapezoid. In this case, the base angles are equal, and the diagonals are of equal length.

Shortcut Trick

For any two regular polygons with the same number of sides, they are geometrically similar.

The ratio of their corresponding sides is exactly the square root of the ratio of their areas.

Side ratio = √(Area ratio)

⇒ Side ratio = √(4 / 9) = 2 / 3

∴ The correct answer is 2:3.

Alternate Method

Given:

Ratio of areas of two regular polygons = 4 : 9

Both polygons have an equal number of sides, say 'n'.

Formula Used:

Area of a regular polygon with 'n' sides = (n × a²) ÷ [4 × tan(180°/n)], where 'a' is the side length.

Area = 4 Side 1 Area = 9 Side 2 Ratio

Calculation:

⇒ Let the sides of the two polygons be a₁ and a₂.

⇒ Let their areas be A₁ and A₂ respectively.

⇒ Since the number of sides 'n' is equal, the constant terms n and [4 × tan(180°/n)] will cancel out when calculating the ratio.

⇒ A₁ / A₂ = a₁² / a₂²

⇒ 4 / 9 = (a₁ / a₂)²

⇒ a₁ / a₂ = √(4 / 9)

⇒ a₁ / a₂ = 2 / 3

∴ The correct answer is 2:3.

Additional Information

Similarity of Regular Polygons

All regular polygons with the same number of sides are geometrically similar. The ratio of their areas is strictly the square of the ratio of their corresponding side lengths.

Perimeter Ratio

For any two similar figures, the ratio of their perimeters is precisely equal to the linear scale factor, which is the direct ratio of their corresponding side lengths.

Volume of Similar 3D Solids

If two 3D solids are similar, the ratio of their volumes is identical to the cube of the ratio of their corresponding linear dimensions (V₁ / V₂ = (a₁ / a₂)³).

Shortcut Trick

Since the maximum marks for both subjects are equal, the overall percentage is simply the arithmetic average of the individual percentages.

Overall Percentage = (80% + 60%) ÷ 2

⇒ Overall Percentage = 140% ÷ 2 = 70%

∴ The correct answer is 70%.

Alternate Method

Given:

Percentage scored in Subject 1 = 80%

Percentage scored in Subject 2 = 60%

Maximum marks for both subjects are equal.

Subject 1: 80 / 100 Subject 2: 60 / 100 + Total Marks: 140 / 200 → Overall: 70%

Formula Used:

Overall Percentage = (Total Marks Obtained ÷ Total Maximum Marks) × 100

Calculations:

⇒ Let the maximum marks for each subject be 100.

⇒ Marks obtained in Subject 1 = 80% of 100 = 80

⇒ Marks obtained in Subject 2 = 60% of 100 = 60

⇒ Total marks obtained by the student = 80 + 60 = 140

⇒ Total maximum marks = 100 + 100 = 200

⇒ Overall Percentage = (140 ÷ 200) × 100

⇒ Overall Percentage = 70%

∴ The correct answer is 70%.

Additional Information

Average of Percentages

When the base values (maximum possible marks) are strictly equal, the overall percentage simplifies precisely to the arithmetic mean of the given percentages.

Weighted Average of Percentages

When the base values are unequal (e.g., weights w₁ and w₂), the overall percentage must be found using: Overall % = (P₁ × w₁ + P₂ × w₂) ÷ (w₁ + w₂).

Core Percentage Concept

Percentage is fundamentally a fraction out of 100. It can be interpreted directly as a standard ratio: Percentage = (Part ÷ Whole) × 100.

Shortcut Trick

For a 'Buy X for the price of Y' offer, the number of free items is X − Y.

Loss % = (Free Items ÷ Total Items) × 100

Here, Free items = 3 − 2 = 1

Loss % = (1 ÷ 3) × 100 = 33.33%

∴ The correct answer is 33.33%.

Alternate Method

Given:

Offer: Buy 3 shirts for the price of 2.

Cost Price (CP) of 1 shirt = Rs. 300

Formula Used:

Loss = Total CP − Total SP

Loss % = (Loss ÷ Total CP) × 100

Shirt 1 SP: ₹300 Shirt 2 SP: ₹300 Shirt 3 Free (SP: ₹0) + + Total CP = ₹900   |   Total SP = ₹600 Effective Loss = ₹300

Calculations:

⇒ Total Cost Price (CP) for 3 shirts = 3 × 300 = Rs. 900

⇒ Total Selling Price (SP) for 2 shirts = 2 × 300 = Rs. 600

⇒ Loss incurred by the seller = Total CP − Total SP

⇒ Loss = 900 − 600 = Rs. 300

⇒ Loss % = (Loss ÷ Total CP) × 100

⇒ Loss % = (300 ÷ 900) × 100

⇒ Loss % = (1 ÷ 3) × 100 = 33.33%

∴ The correct answer is 33.33%.

Additional Information

Buy X Get Y Free Offers

Discount or Loss percentage is always calculated on the total number of items leaving the seller. Formula: (Free Items ÷ Total Items) × 100.

Successive Offers

If multiple promotional offers are applied, calculate the successive change using the net percentage formula: Net % = a + b − (a × b ÷ 100).

Marked Price vs Cost Price

If the seller sets a Marked Price (MP) higher than the Cost Price (CP) before offering free items, their actual profit or loss will depend entirely on the markup percentage ratio.

Shortcut Trick

When a value is first decreased by x% and then increased by x%, there is always a net decrease of (x² ÷ 100)%.

Given the overall net decrease is 10%.

⇒ x² ÷ 100 = 10

⇒ x² = 1000

⇒ x = √1000 ≈ 31.62

∴ The correct answer is 31.6.

Alternate Method

Given:

Initial price is decreased by x% and then increased by x%.

Final price is 10% less than the original price.

100 100 − x 90 −x% +x% Net Decrease = 10%

Formula Used:

Effective Percentage Change = a + b + (a × b) ÷ 100

Calculations:

Let the original price be 100.

⇒ Price after x% decrease = 100 − x

⇒ Price after subsequent x% increase = (100 − x) × (100 + x) ÷ 100

⇒ Final price = (10000 − x²) ÷ 100 = 100 − x² ÷ 100

Given that the final price becomes 10% less than the original.

⇒ Final expected price = 100 − 10 = 90

Equating both values:

⇒ 100 − x² ÷ 100 = 90

⇒ x² ÷ 100 = 10

⇒ x² = 1000

⇒ x = √1000 = 10√10

Using √10 ≈ 3.162:

⇒ x ≈ 10 × 3.162 = 31.62

∴ The correct answer is 31.6.

Additional Information

Successive Percentage Change

If a quantity is changed by a% and then by b%, the net percentage change is given by the formula: Net Change = a + b + (a × b) ÷ 100. Use positive sign for increase and negative for decrease.

Equal Percentage Increase and Decrease

When a value is first increased by x% and then decreased by x% (or vice versa), there is always an overall net decrease. The net decrease percentage is calculated directly as (x² ÷ 100)%.

Useful Approximations

Memorising basic square roots is very helpful for estimation questions in competitive exams: √2 ≈ 1.414, √3 ≈ 1.732, √5 ≈ 2.236, and √10 ≈ 3.162.

Shortcut Trick

Using the net deviation method for averages:

The removed smallest number (15) is 9 less than the initial average (24 − 15 = 9).

This deficit of 9 must be compensated by the remaining 4 numbers.

Increase in average for the remaining numbers = 9 ÷ 4 = 2.25

New average = 24 + 2.25 = 26.25

∴ The correct answer is 26.25.

Alternate Method

Given:

Average of 5 numbers = 24

Smallest number = 15

Formula Used:

Sum of observations = Average × Number of observations

15 Remaining 4 numbers Sum = 105 Total Sum = 5 × 24 = 120

Calculations:

⇒ Total sum of 5 numbers = 24 × 5 = 120

⇒ Sum of the remaining 4 numbers = Total sum − Smallest number

⇒ Sum of the remaining 4 numbers = 120 − 15 = 105

⇒ Average of the remaining 4 numbers = 105 ÷ 4

⇒ Average = 26.25

∴ The correct answer is 26.25.

Additional Information

Combined Average

If group 1 has n₁ items with average A₁ and group 2 has n₂ items with average A₂, the combined average of the whole group is given by (n₁ × A₁ + n₂ × A₂) ÷ (n₁ + n₂).

Effect of Operations on Average

If every observation in a given dataset is increased, decreased, multiplied, or divided by a specific constant value 'k', the new overall average will also undergo the exact same mathematical operation with 'k'.

Average of Consecutive Numbers

For an arithmetic progression or any evenly spaced sequence of consecutive numbers, the average is always strictly the middle term, which can be directly calculated as (First Term + Last Term) ÷ 2.

Shortcut Trick

For any right prism, the direct relation is: Lateral Surface Area = Base Perimeter × Height.

The given Base Area (80 cm²) is extraneous data deliberately provided to cause confusion.

Base Perimeter = 384 ÷ 12

Base Perimeter = 32 cm

∴ The correct answer is 32 cm.

Alternate Method

Given:

Base Area = 80 cm²

Height (h) = 12 cm

Lateral Surface Area = 384 cm²

Formula Used:

Lateral Surface Area of a right prism = Perimeter of Base × Height

         h = 12 cm  Base Area = 80 cm²  Lateral Surface Area = 384 cm²

Calculations:

⇒ LSA = Perimeter × Height

⇒ 384 = Perimeter × 12

⇒ Perimeter = 384 ÷ 12

⇒ Perimeter = 32 cm

∴ The correct answer is 32 cm.

Additional Information

Properties of a Right Prism

A prism is a 3D solid with two identical and parallel polygonal bases. In a right prism, the lateral faces are always flat rectangles.

Volume of a Prism

The volume strictly relies on the area of the base. The formula is: Volume = Base Area × Height. For this particular shape, it would be 80 × 12 = 960 cm³.

Total Surface Area (TSA)

TSA = Lateral Surface Area + 2 × Base Area. For the given measurements, TSA = 384 + (2 × 80) = 544 cm².

Extraneous Data in Examinations

Competitive tests often inject extra numerical data to observe if candidates can identify the specific, required formula parameter. The base area was provided purely as a distraction from the straightforward relation.

Shortcut Trick

Net deviation of top 5 students from overall average = 5 × (80 − 68) = +60

Net deviation of lowest 5 students from overall average = 5 × (50 − 68) = −90

Total deviation of these 10 students combined = +60 − 90 = −30

To balance the sum, the remaining 15 students must provide a positive deviation of +30.

Average of remaining 15 students = 68 + (30 ÷ 15) = 68 + 2 = 70

∴ The correct answer is 70 marks.

Alternate Method

Given:

Total number of students = 25

Average score of all 25 students = 68

Average score of top 5 students = 80

Average score of lowest 5 students = 50

Formula Used:

Total Sum = Average × Number of observations

Top 5 Avg: 80

Remaining 15 Avg: x

Low 5 Avg: 50

Total 25 Students → Overall Avg: 68

Calculations:

⇒ Total score of all 25 students = 25 × 68 = 1700

⇒ Total score of the top 5 students = 5 × 80 = 400

⇒ Total score of the lowest 5 students = 5 × 50 = 250

⇒ Total score of the remaining 15 students = Total score − (Top 5 score + Lowest 5 score)

⇒ Score of the remaining 15 students = 1700 − (400 + 250)

⇒ Score of the remaining 15 students = 1700 − 650 = 1050

⇒ Average of the remaining 15 students = 1050 ÷ 15 = 70

∴ The correct answer is 70 marks.

Additional Information

Weighted Average

When combining multiple groups, the overall combined average is calculated as (n₁A₁ + n₂A₂ + ...) ÷ (n₁ + n₂ + ...).

Net Deviation Method

The sum of all deviations from the overall average is always zero. This is a highly effective shortcut to find a missing group's average without calculating large sums.

Combined Average

If group A has an average X and group B has an average Y, the combined average will invariably lie between X and Y based strictly on the ratio of their observation counts.

Shortcut Trick

Successive discount multiplying factor method:

Effective Price = Original Price × (1 − D₁%) × (1 − D₂%)

⇒ Final Price = 20000 × 0.90 × 0.95

⇒ Final Price = 17100

∴ The correct answer is Rs. 17,100.

Alternate Method

Given:

Original order value = Rs. 20,000

Since the order is above Rs. 15,000, the bulk discount of 10% applies.

Additional seasonal discount = 5%

Formula Used:

Selling Price (SP) = Marked Price × (1 − D₁/100) × (1 − D₂/100)

 −10% Discount −5% Discount    20k 18k 17.1k Original Price Intermediate Final Price Rs. 20000 Rs. 18000 Rs. 17100

Calculations:

⇒ Step 1: Apply the 10% bulk discount on Rs. 20,000

⇒ Price after 1^st discount = 20000 − (10 ÷ 100 × 20000)

⇒ Price after 1^st discount = 20000 − 2000 = Rs. 18,000

⇒ Step 2: Apply the 5% additional seasonal discount on the new intermediate price

⇒ Final Price = 18000 − (5 ÷ 100 × 18000)

⇒ Final Price = 18000 − 900 = Rs. 17,100

∴ The correct answer is Rs. 17,100.

Additional Information

Successive Discounts Formula

When two discounts D₁% and D₂% are applied consecutively, the single equivalent effective discount is given by: D = D₁ + D₂ − (D₁ × D₂ ÷ 100)%.

Marked Price vs Selling Price

Discount is always calculated on the Marked Price (MP). The final amount paid by the customer is the Selling Price (SP), where SP = MP − Total Discount.

Shortcut Trick

We know the direct trigonometric relation: tanA = sinA ÷ cosA

Using the fundamental identity: cosA = √(1 − sin²A)

Substitute sinA = x into the relation:

⇒ tanA = x ÷ √(1 − x²)

∴ The correct answer is x / √(1 − x²).

Alternate Method

Given:

sinA = x

Formula Used:

sinθ = Perpendicular ÷ Hypotenuse

tanθ = Perpendicular ÷ Base

Pythagoras Theorem: (Hypotenuse)² = (Perpendicular)² + (Base)²

A x √(1 − x2) 1

Calculations:

⇒ Let Perpendicular = x and Hypotenuse = 1

⇒ Then, sinA = Perpendicular ÷ Hypotenuse = x ÷ 1 = x

⇒ Using Pythagoras Theorem: (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ Base² = 1² − x²

⇒ Base = √(1 − x²)

⇒ tanA = Perpendicular ÷ Base

⇒ tanA = x ÷ √(1 − x²)

∴ The correct answer is x / √(1 − x²).

Additional Information

Trigonometric Identities

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

Trigonometric Ratios

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

Relation Between Ratios

Tangent is the ratio of sine and cosine: tanθ = sinθ ÷ cosθ.

Shortcut Trick

For a right-angled triangle, the Pythagorean triplet (3, 4, 5) gives Perpendicular = 3, Hypotenuse = 5, and Base = 4.

The magnitude of tan A = Perpendicular ÷ Base = 3/4.

Since angle A lies in the 2^nd quadrant, the tangent function must be negative.

⇒ tan A = −3/4

∴ The correct answer is −3/4.

Alternate Method

Given:

sin A = 3/5

Angle A lies in the 2^nd quadrant.

Formula Used:

sin²A + cos²A = 1

tan A = sin A ÷ cos A

−4 3 5 A

X Y

Calculations:

⇒ cos²A = 1 − sin²A

⇒ cos²A = 1 − (3/5)²

⇒ cos²A = 1 − 9/25

⇒ cos²A = 16/25

⇒ cos A = ±4/5

⇒ Since angle A is in the 2^nd quadrant, the cosine function must be negative.

⇒ cos A = −4/5

⇒ tan A = (3/5) ÷ (−4/5)

⇒ tan A = −3/4

∴ The correct answer is −3/4.

Additional Information

ASTC Rule (All Sin Tan Cos)

1^st Quadrant: All ratios are positive. 2^nd Quadrant: Only sin and cosec are positive. 3^rd Quadrant: Only tan and cot are positive. 4^th Quadrant: Only cos and sec are positive.

Pythagorean Identities

The fundamental identities are sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Common Pythagorean Triplets

Standard right-angled triangle sides frequently tested in exams include (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25).

Shortcut Trick

Direct formula for sector area (when angle is in radians):

Area = ½ × r² × θ

⇒ Area = ½ × 8² × (π ÷ 4)

⇒ Area = 8π cm²

∴ The correct answer is 8π cm².

Alternate Method

Given:

Radius of the circle (r) = 8 cm

Angle of the sector (θ) = π/4 radians

     O  π/4 r = 8 cm r = 8 cm

Formula Used:

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

Since π radians = 180°

⇒ θ = π/4 = 180° ÷ 4 = 45°

Calculations:

⇒ Area = (45° ÷ 360°) × π × (8)²

⇒ Area = (1 ÷ 8) × π × 64

⇒ Area = 8π cm²

∴ The correct answer is 8π cm².

Additional Information

Area of a Sector

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

Length of an Arc

The length of an arc that subtends an angle θ (in radians) at the center is given by: L = r × θ. If θ is in degrees, L = (θ ÷ 360) × 2πr.

Area using Arc Length

If the arc length (L) and radius (r) are known, the area of the sector can be calculated directly as: Area = ½ × L × r.

Perimeter of a Sector

The perimeter of a sector includes the arc length and the two straight radii. Perimeter = L + 2r.

Shortcut Trick

For sin A = cos B, the sum of angles A and B is always equal to 90°.

Sum of given angles = x + 2x − 10° = 3x − 10°

Equating the sum to 90°, we directly get 3x = 100°.

x = 100° ÷ 3 = 33.33°

∴ The correct answer is 33.33°.

Alternate Method

Given:

sin x = cos(2x − 10°)

Formula Used:

cos θ = sin(90° − θ)

Calculations:

⇒ sin x = sin(90° − (2x − 10°))

Comparing the angles on both sides:

⇒ x = 90° − 2x + 10°

⇒ x + 2x = 100°

⇒ 3x = 100°

⇒ x = 100° ÷ 3

⇒ x = 33.33°

∴ The correct answer is 33.33°.

Additional Information

Complementary Angles Rule

For any acute angle θ, sin(90° − θ) = cos θ and cos(90° − θ) = sin θ.

Other Complementary Pairs

tan(90° − θ) = cot θ and sec(90° − θ) = cosec θ.

Pythagorean Identities

sin²θ + cos²θ = 1 | sec²θ − tan²θ = 1 | cosec²θ − cot²θ = 1.

Shortcut Trick

Direct Common Tangent (DCT) = √(d² − (r₁ − r₂)²)

Substitute the given values: d = 15, r₁ = 8, and r₂ = 3 directly.

DCT = √(15² − 5²) = √(225 − 25)

DCT = √200 = 10√2 cm.

∴ The correct answer is 10√2 cm.

Alternate Method

Given:

Radius of first circle (r₁) = 8 cm

Radius of second circle (r₂) = 3 cm

Distance between centers (d) = 15 cm

 C1 C2  d = 15 cm    8 cm 3 cm DCT = ?

Formula Used:

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

Calculations:

⇒ DCT = √[15² − (8 − 3)²]

⇒ DCT = √[225 − 5²]

⇒ DCT = √[225 − 25]

⇒ DCT = √200

⇒ DCT = √(100 × 2)

⇒ DCT = 10√2 cm

∴ The correct answer is 10√2 cm.

Additional Information

Transverse Common Tangent (TCT)

The formula for the length of a transverse common tangent is TCT = √(d² − (r₁ + r₂)²).

Number of Common Tangents

If d > r₁ + r₂, the circles do not intersect or touch each other, resulting in a total of 4 common tangents (2 direct, 2 transverse).

When Circles Touch Externally

If d = r₁ + r₂, the circles touch externally at exactly one point. There are 3 common tangents in this case, and the Direct Common Tangent formula directly simplifies to DCT = 2√(r₁ × r₂).

Shortcut Trick

According to the standard circle theorem, a line drawn through the center of a circle to bisect a chord is always perpendicular to that chord.

Therefore, the angle formed between the bisecting line and the chord is exactly a right angle.

∴ The correct answer is 90°.

Alternate Method

       O A B M

Given:

A circle with center O and chord AB.

A line OM drawn from the center bisects the chord AB, meaning AM = MB.

Concept Used:

Congruency of triangles using the SSS (Side-Side-Side) criterion.

Calculations:

⇒ In ΔOMA and ΔOMB:

⇒ OA = OB (Radii of the same circle)

⇒ AM = MB (Given that OM bisects the chord)

⇒ OM = OM (Common side)

⇒ By SSS congruence criterion, ΔOMA ≅ ΔOMB.

⇒ Therefore, their corresponding angles are equal: ∠OMA = ∠OMB.

⇒ Since AMB is a straight line, the sum of angles on the line is 180°.

⇒ ∠OMA + ∠OMB = 180°

⇒ 2 × ∠OMA = 180°

⇒ ∠OMA = 90°

∴ The correct answer is 90°.

Additional Information

Perpendicular from Center Theorem

The converse is also universally true: A perpendicular dropped from the center of a circle to a chord will always bisect the chord into two equal halves.

Equal Chords and Distance

Equal chords of a circle are always equidistant from the center. Conversely, chords that are equidistant from the center are equal in length.

Angle at the Center

The angle subtended by an arc at the center of a circle is double the angle subtended by it at any remaining part of the circle's circumference.

Shortcut Trick

By the central angle theorem, the angle subtended by an arc at the center is exactly double the angle subtended by it at any point on the remaining part of the circle.

The major arc subtends an angle of 110° at the circumference (minor arc).

Central angle = 2 × Angle at circumference

Central angle = 2 × 110° = 220°

∴ The correct answer is 220°.

Alternate Method

              220° 110° O P A B Major Arc AB Minor Arc AB

Given:

Angle subtended by the major arc at the circumference = 110°.

Formula Used:

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Calculation:

⇒ Let the circle have center O and chord AB.

⇒ The chord divides the circle into a major arc and a minor arc.

⇒ Let P be a point on the minor arc. The angle subtended by the major arc at point P is ∠APB = 110°.

⇒ The central angle subtended by the major arc is the reflex angle ∠AOB.

⇒ Reflex ∠AOB = 2 × ∠APB

⇒ Reflex ∠AOB = 2 × 110° = 220°

⇒ Thus, the central angle subtended by the major arc is 220°.

∴ The correct answer is 220°.

Additional Information

Central Angle Theorem

The angle subtended by an arc at the center is double the angle subtended by it at any remaining part of the circle.

Cyclic Quadrilateral Property

If we take another point Q on the major arc, quadrilateral PAQB is cyclic. The sum of opposite angles is 180° (∠APB + ∠AQB = 180°).

Major vs Minor Arc

A minor arc subtends an acute or obtuse angle (< 180°) at the center, whereas a major arc subtends a reflex angle (> 180°) at the center.

Shortcut Trick

According to the circle theorem, angles subtended by the same chord in the same segment of a circle are always equal.

Since points C and D lie on the same arc, they form angles in the same segment relative to chord AB.

Therefore, the angle at point D must be exactly equal to the angle at point C.

Given the angle at C is 50°, the angle at D is also 50°.

∴ The correct answer is 50°.

Alternate Method

Given:

Angle subtended by chord AB at point C, ∠ACB = 50°

Point D is located on the same arc as point C.

Concept Used:

Angles subtended by the same arc (or chord) at any point on the remaining part of the circle (i.e., in the same segment) are equal.

A B C D O

50° 50°

Angles in the same segment are equal

Calculations:

⇒ Both points C and D lie in the major segment of the circle formed by the chord AB.

⇒ By the fundamental property of circles, any angle subtended by a chord in the same segment will measure exactly the same.

⇒ ∠ADB = ∠ACB

⇒ ∠ADB = 50°

∴ The correct answer is 50°.

Additional Information

Angle at Center Theorem

The angle subtended by an arc at the center is exactly double the angle subtended by it at any point on the remaining part of the circle. (Center Angle = 2 × Inscribed Angle)

Cyclic Quadrilateral Property

If all four vertices of a quadrilateral lie on the circumference of a circle, it forms a cyclic quadrilateral. The sum of opposite angles in a cyclic quadrilateral is always supplementary (180°).

Angle in a Semicircle

The angle subtended by the diameter of a circle at any point on the circumference is always a right angle (90°).

Shortcut Trick

For a triangle formed by two tangents from an external point and a third intersecting tangent, a direct property applies:

Perimeter of ▵PRS = 2 × Length of the tangent from the external point

Perimeter = 2 × PA

Perimeter = 2 × 10 = 20 cm.

∴ The correct answer is 20 cm.

Alternate Method

Given:

PA = 10 cm

PA and PB are tangents to the circle from point P.

RS is a tangent at point Q intersecting PA at R and PB at S.

           P A B R S Q O PA = 10 cm PB = 10 cm

Formula Used:

Tangents drawn from an external point to a circle are equal in length.

Calculations:

⇒ By tangent properties, PA = PB = 10 cm.

⇒ Similarly, from point R, tangents RA and RQ are equal in length (RA = RQ).

⇒ From point S, tangents SB and SQ are equal in length (SB = SQ).

⇒ Perimeter of ▵PRS = PR + RS + SP

⇒ Since RS = RQ + QS, we can substitute it into the perimeter equation:

⇒ Perimeter of ▵PRS = PR + (RQ + QS) + SP

⇒ Substitute RQ with RA and QS with SB:

⇒ Perimeter of ▵PRS = PR + RA + SB + SP

⇒ Grouping the terms gives:

⇒ Perimeter of ▵PRS = (PR + RA) + (SP + SB)

⇒ Since PR + RA = PA and SP + SB = PB:

⇒ Perimeter of ▵PRS = PA + PB

⇒ Perimeter of ▵PRS = 10 + 10 = 20 cm

∴ The correct answer is 20 cm.

Additional Information

Length of Tangents from an External Point

Tangents drawn from a point outside a circle are always equal in length and they subtend equal angles at the center of the circle.

Perimeter of a Tangent Triangle

When a tangent intersects two other tangents drawn from a single external point, the perimeter of the resulting triangle formed is exactly twice the length of the original tangent segments.

Direct Common Tangent (DCT)

The length of a DCT between two circles is given by √(d² − (r₁ − r₂)²), where d is the distance between their centers and r₁, r₂ are their radii.