SSC CGL Tier 2 Quant Previous Year Paper (Held On: 18 November 2020)

Shortcut Trick

Total percentage of students = 100%

Failure percentage = 100% − 92% = 8%

Given: 8% = 480 students

100% = (480 ÷ 8) × 100 = 60 × 100 = 6000

∴ The correct answer is 6000 students.

Alternate Method

Given:

Pass percentage = 92%

Number of failed students = 480

Formula Used:

Total Students = (Number of Failed Students ÷ Fail Percentage) × 100

Calculations:

Let the total number of students be x.

⇒ Fail Percentage = 100% − 92%

⇒ Fail Percentage = 8%

⇒ 8% of x = 480

⇒ (8 ÷ 100) × x = 480

⇒ x = (480 × 100) ÷ 8

⇒ x = 60 × 100

⇒ x = 6000

∴ The correct answer is 6000 students.

Additional Information

Percentage to Fraction Conversion

8% is equivalent to 8/100 or 2/25. This means if 2 units represent the failures (480), then 1 unit = 240, and the total students (25 units) = 25 × 240 = 6000.

Concept of Base Change

When solving percentage problems, always identify the base (100%). In exams, the total number of students or the maximum marks usually represents the 100% base.

Total Students Rule

Total Students = Number of Passed Students + Number of Failed Students. Since 92% passed, the number of passed students = 6000 − 480 = 5520.

Shortcut Trick

Marked Price = ₹7,600

Price after 1st discount (10%) = 7,600 − 760 = ₹6,840

Remaining amount to be discounted = 6,840 − 5,814 = ₹1,026

Additional discount % = (1,026 ÷ 6,840) × 100 = 15%

∴ The correct answer is 15%.

Alternate Method

Given:

Listed Price (MP) = ₹7,600

First Discount (D₁) = 10%

Net Selling Price (SP) = ₹5,814

Formula Used:

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

Calculations:

⇒ Price after 1st discount = 7,600 × (90/100) = ₹6,840

⇒ Let the second discount be x%.

⇒ 5,814 = 6,840 × (100 − x)/100

⇒ 5,814 ÷ 6,840 = (100 − x)/100

⇒ 0.85 = (100 − x)/100

⇒ 85 = 100 − x

⇒ x = 100 − 85 = 15%

∴ The correct answer is 15%.

Additional Information

Successive Discount Formula

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

Marked Price and Selling Price Relation

Selling Price = Marked Price − Discount. Also, SP = CP × (100 + Profit%)/100.

Discount Percentage

Discount% is always calculated on the Marked Price (MP): Discount% = (Discount ÷ MP) × 100.

Shortcut Trick

Efficiency Ratio (A : B) = 3 : 1

Combined Efficiency (A + B) = 3 + 1 = 4 units/day

Total Work = Combined Efficiency × Combined Time = 4 × 18 = 72 units

Days taken by B alone = Total Work ÷ Efficiency of B = 72 ÷ 1 = 72 days

∴ The correct answer is 72 days.

Alternate Method

Given: Time taken by A and B together = 18 days. A is three times as efficient as B.

Formula Used: Total Work = Efficiency × Time

⇒ Let the efficiency of B be 1 unit/day.

⇒ Then, efficiency of A = 3 × 1 = 3 units/day.

⇒ Total efficiency of A and B = 3 + 1 = 4 units/day.

⇒ Total work = Combined efficiency × Number of days taken together = 4 × 18 = 72 units.

⇒ Time taken by B alone = Total work ÷ B’s efficiency = 72 ÷ 1 = 72 days.

∴ The correct answer is 72 days.

Additional Information

Efficiency-Time Relationship

Efficiency is inversely proportional to the time taken. If Efficiency ratio is a : b, then Time ratio is b : a.

Combined Work Formula

If A does a work in ’x’ days and B does it in ’y’ days, then together they take (x × y) ÷ (x + y) days.

Man-Days Concept

The amount of work is proportional to the product of efficiency and duration, expressed as M₁D₁ = M₂D₂.

Shortcut Trick

Number of identical smaller spheres obtained by melting a larger sphere is the cube of the ratio of their dimensions.

Ratio of Diameters = Large Diameter ÷ Small Diameter = 42 ÷ 3 = 14.

Number of balls = (Ratio)³ = 14³.

Number of balls = 14 × 14 × 14 = 2744.

∴ The correct answer is 2744 balls.

Alternate Method

Given:

Diameter of the solid lead sphere (D) = 42 cm

Diameter of each small lead ball (d) = 3 cm

Formula Used:

Volume of a sphere = (4/3) × π × r³

Number of balls = Volume of large sphere ÷ Volume of one small ball

Calculations:

⇒ Radius of large sphere (R) = 42 ÷ 2 = 21 cm

⇒ Radius of small ball (r) = 3 ÷ 2 = 1.5 cm

⇒ Number of balls (N) = [(4/3) × π × R³] ÷ [(4/3) × π × r³]

⇒ N = R³ ÷ r³ = (R ÷ r)³

⇒ N = (21 ÷ 1.5)³

⇒ N = (14)³

⇒ N = 14 × 14 × 14 = 2744

∴ The correct answer is 2744 balls.

Additional Information

Recasting Rule

When one solid object is melted and recast into multiple smaller objects of the same or different shapes, the total volume remains constant (Volume_Initial = Volume_Final).

Volume of Hemisphere

The volume of a hemisphere is exactly half of the volume of a sphere: Volume = (2/3) × π × r³.

Total Surface Area of Sphere

The total surface area of a sphere with radius r is given by the formula: Area = 4 × π × r².

Shortcut Trick

Direct formula for Total Surface Area (TSA) when Sum (S) and Diagonal (D) are given:

TSA = (Sum of dimensions)² − (Length of diagonal)²

⇒ TSA = (20)² − (12)²

⇒ TSA = 400 − 144 = 256 cm²

∴ The correct answer is 256 cm².

Alternate Method

Given:

Sum of length, breadth, and height (l + b + h) = 20 cm

Length of the diagonal (d) = 12 cm

Formula Used:

1. Diagonal (d) = √(l² + b² + h²)
2. Total Surface Area (TSA) = 2(lb + bh + hl)
3. Algebraic Identity: (l + b + h)² = l² + b² + h² + 2(lb + bh + hl)

Calculations:

⇒ From the diagonal formula: d² = l² + b² + h²

⇒ 12² = l² + b² + h² ⇒ 144 = l² + b² + h²

⇒ Using the identity: (20)² = (l² + b² + h²) + TSA

⇒ 400 = 144 + TSA

⇒ TSA = 400 − 144

⇒ TSA = 256 cm²

∴ The correct answer is 256 cm².

Additional Information

Volume of a Cuboid

The volume of a cuboid is the product of its dimensions: V = l × b × h.

Lateral Surface Area (LSA)

The area of the four side walls of the cuboid is given by: LSA = 2h(l + b).

Properties of a Cube

A cube is a special cuboid where l = b = h = a. Here, TSA = 6a² and Diagonal = a√3.

Diagonal and Sum relation

For any rectangular parallelepiped, the sum of dimensions, the diagonal, and the surface area are related by (l+b+h)² = d² + TSA.

Shortcut Trick

Ratio of CP of A to CP of B by Alligation = |0 − (−10)| : |15 − 0| = 10 : 15 = 2 : 3

Total units = 2 + 3 = 5 units → ₹10,000

CP of article A = 2 units = (10,000 ÷ 5) × 2 = ₹4,000

SP of article A = 115% of CP = 1.15 × 4,000 = ₹4,600

∴ The correct answer is ₹4,600.

Alternate Method

Given:

Total Cost Price (CP) of A and B = ₹10,000

Profit on article A = 15%

Loss on article B = 10%

Net Profit/Loss = 0%

Formula Used:

Selling Price (SP) = CP × (100 + Profit%)/100

Using Alligation to find the ratio of CP of two articles:

Calculations:

⇒ Let CP of A be 2x and CP of B be 3x.

⇒ 2x + 3x = 10,000

⇒ 5x = 10,000

⇒ x = 2,000

⇒ Cost Price of A (CP_A) = 2 × 2,000 = ₹4,000

⇒ Selling Price of A (SP_A) = CP_A + 15% of CP_A

⇒ SP_A = 4,000 + (15/100) × 4,000

⇒ SP_A = 4,000 + 600 = ₹4,600

∴ The correct answer is ₹4,600.

Additional Information

Alligation Rule in Profit and Loss

When two articles are sold, the ratio of their Cost Prices is given by the ratio of the differences between the individual profit/loss percentages and the overall profit/loss percentage.

No Profit No Loss Condition

In a "no profit no loss" scenario, the absolute value of the profit earned on one article must be exactly equal to the absolute value of the loss incurred on the other article: Profit on A = Loss on B.

Selling Price Formula

SP = CP × (100 ± P/L)% where '+' is used for profit and '−' is used for loss.

Shortcut Trick

Ratio of Boys : Girls = 27 : 23

Difference in ratio units = 27 − 23 = 4 units

Given: 4 units = 200 ⇒ 1 unit = 50

Number of boys = 27 units = 27 × 50 = 1350

∴ The correct answer is 1350.

Alternate Method

Given:

Ratio of boys to girls = 27 : 23

Difference between boys and girls = 200

Let the number of boys be 27x and the number of girls be 23x.

According to the question:

⇒ 27x − 23x = 200

⇒ 4x = 200

⇒ x = 200 ÷ 4

⇒ x = 50

Now, find the number of boys:

⇒ Number of boys = 27x

⇒ Number of boys = 27 × 50

⇒ Number of boys = 1350

∴ The correct answer is 1350.

Additional Information

Ratio and Proportion Basics

A ratio a:b can be represented as a fraction a/b. If a:b = c:d, then ad = bc (Product of extremes = Product of means).

Total Value from Ratio

To find the total quantity when parts are given in ratio a:b and difference is D, Total = [(a + b) × D] ÷ |a − b|.

Duplicate and Triplicate Ratios

Duplicate ratio of a:b is a²:b² and triplicate ratio is a³:b³.

Shortcut Trick

Interior Angle (I) + Exterior Angle (E) = 180°

Given: I − E = 90°

Adding the two: 2I = 270° ⇒ I = 135°

Subtracting the two: 2E = 90° ⇒ E = 45°

Number of sides (n) = 360° ÷ Exterior Angle = 360° ÷ 45° = 8

∴ The correct answer is 8.

Alternate Method

Given:

Interior angle (I) − Exterior angle (E) = 90°

Formula Used:

1. Interior angle + Exterior angle at any vertex = 180°
2. Number of sides of a regular polygon (n) = 360° ÷ Exterior angle

Calculations:

⇒ Let the interior angle be I and the exterior angle be E.

⇒ I − E = 90° --- (Equation 1)

⇒ I + E = 180° --- (Equation 2)

⇒ Subtracting (1) from (2):

⇒ (I + E) − (I − E) = 180° − 90°

⇒ 2E = 90°

⇒ E = 45°

⇒ Number of sides (n) = 360° ÷ E

⇒ n = 360° ÷ 45°

⇒ n = 8

∴ The correct answer is 8.

Additional Information

Sum of Interior Angles

The sum of all interior angles of a polygon with n sides is given by (n − 2) × 180°.

Each Interior Angle of a Regular Polygon

For a regular polygon, each interior angle = [(n − 2) × 180°] ÷ n.

Sum of Exterior Angles

The sum of all exterior angles of any convex polygon is always equal to 360°.

Relation between Side and Exterior Angle

For a regular polygon, each exterior angle = 360° ÷ n, where n is the number of sides.

Shortcut Trick

Volume of a cylinder (V) ∝ r²h

Ratio of volumes V₁ : V₂ = (r₁/r₂)² × (h₁/h₂)

V₁ : V₂ = (3/4)² × (8/5) = (9/16) × (8/5)

V₁ : V₂ = (9 × 1) : (2 × 5) = 9 : 10

∴ The correct answer is 9 : 10.

Alternate Method

Given:

Ratio of radii (r₁ : r₂) = 3 : 4

Ratio of heights (h₁ : h₂) = 8 : 5

Formula Used:

Volume of a cylinder (V) = πr²h

Calculations:

Let the radii be 3x and 4x, and the heights be 8y and 5y.

⇒ Volume of first cylinder (V₁) = π × (3x)² × 8y

⇒ V₁ = π × 9x² × 8y = 72πx²y

⇒ Volume of second cylinder (V₂) = π × (4x)² × 5y

⇒ V₂ = π × 16x² × 5y = 80πx²y

⇒ Ratio V₁ : V₂ = 72πx²y : 80πx²y

⇒ Ratio V₁ : V₂ = 72 : 80

⇒ Ratio V₁ : V₂ = 9 : 10

∴ The correct answer is 9 : 10.

Additional Information

Curved Surface Area (CSA) of a Cylinder

The lateral surface area of a cylinder is given by the formula CSA = 2πrh, where r is the radius and h is the height.

Total Surface Area (TSA) of a Cylinder

The total surface area includes the curved surface and the two circular bases: TSA = 2πrh + 2πr² = 2πr(h + r).

Ratio of Volumes of Cones

Similar to cylinders, the ratio of the volumes of two cones with the same radii and height ratios is also (r₁/r₂)² × (h₁/h₂), as V = 1/3 πr²h.

Shortcut Trick

Identify the number that is exactly divisible by 7 from the given options.

Option 1: 366 ÷ 7 → Remainder = 2

Option 2: 634 ÷ 7 → Remainder = 4

Option 3: 384 ÷ 7 → Remainder = 6

Option 4: 364 ÷ 7 = 52 → Remainder = 0

Since 364 is the only option divisible by 7, it is the required number.

∴ The correct answer is 364.

Alternate Method

Given: Divisors = 12, 18, 24, 30; Common Remainder = 4; Special Divisor = 7.

Formula Used: Number = LCM(Divisors) × k + Remainder

⇒ Prime factors of 12 = 2² × 3

⇒ Prime factors of 18 = 2 × 3²

⇒ Prime factors of 24 = 2³ × 3

⇒ Prime factors of 30 = 2 × 3 × 5

⇒ LCM (12, 18, 24, 30) = 2³ × 3² × 5 = 8 × 9 × 5 = 360

⇒ General form of the number = 360k + 4

⇒ To satisfy divisibility by 7: (360k + 4) ÷ 7 must have remainder 0

⇒ (360k + 4) = (7 × 51k) + (3k + 4)

⇒ For the number to be divisible by 7, (3k + 4) must be divisible by 7.

⇒ If k = 1: 3(1) + 4 = 7 (Divisible by 7)

⇒ Least number = 360(1) + 4 = 364

∴ The correct answer is 364.

Additional Information

LCM and Remainder Concept

When a number is divided by x, y, z and leaves remainder r in each case, the general form is LCM(x, y, z) × k + r.

Divisibility Rule of 7

Double the last digit and subtract it from the remaining number. If the result is divisible by 7, the original number is also divisible by 7.

Modular Arithmetic Shortcut

For (Ak + B) mod n = 0, reduce A and B by dividing with n to find the smallest value of k quickly.

Shortcut Trick

Interest = 6525 − 4500 = ₹2025

Total Interest Percentage = (Interest ÷ Principal) × 100

⇒ Total Interest % = (2025 ÷ 4500) × 100 = 45%

Rate per annum = Total Interest % ÷ Time

⇒ Rate = 45% ÷ 5 = 9%

∴ The correct answer is 9%.

Alternate Method

Given:

Principal (P) = ₹4500

Amount (A) = ₹6525

Time (T) = 5 years

Formula Used:

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

Amount (A) = Principal (P) + Simple Interest (SI)

Calculations:

⇒ SI = Amount − Principal

⇒ SI = 6525 − 4500 = ₹2025

⇒ 2025 = (4500 × R × 5) ÷ 100

⇒ 2025 = 45 × R × 5

⇒ 2025 = 225 × R

⇒ R = 2025 ÷ 225

⇒ R = 9

∴ The correct answer is 9%.

Additional Information

Simple Interest Formula

Simple Interest remains constant every year and is calculated only on the initial principal: SI = (P × R × T) ÷ 100.

Relationship between A, P, and R

Amount is the sum of Principal and Interest: A = P × (1 + (R × T) ÷ 100). Rate can be derived as R = (SI × 100) ÷ (P × T).

Units of Time

Always ensure time (T) is in years. If given in months, divide by 12; if given in days, divide by 365.

Shortcut Trick

Speed = Distance ÷ Time

Speed = 800 ÷ 96 m/s

To directly convert m/s to km/h, multiply by 18/5.

Speed = (800 ÷ 96) × (18 ÷ 5) = (25 ÷ 3) × (18 ÷ 5) = 30 km/h

∴ The correct answer is 30 km/h.

Alternate Method

Given:

Distance (D) = 800 m

Time (T) = 96 seconds

Formula Used:

Speed (S) = Distance ÷ Time

Conversion: 1 m/s = 18/5 km/h

Calculations:

⇒ Speed in m/s = 800 ÷ 96

⇒ Speed = 100 ÷ 12 = 25 ÷ 3 m/s

⇒ Converting the speed to km/h:

⇒ Speed in km/h = (25 ÷ 3) × (18 ÷ 5)

⇒ Speed = 5 × 6

⇒ Speed = 30 km/h

∴ The correct answer is 30 km/h.

Additional Information

Conversion of Units

To convert km/h to m/s, multiply the speed by 5/18. To convert m/s to km/h, multiply the speed by 18/5.

Average Speed

Average Speed = Total Distance ÷ Total Time taken.

Relative Speed

If two bodies move in the same direction, Relative Speed = Difference of their speeds. If they move in opposite directions, Relative Speed = Sum of their speeds.

Shortcut Trick

Discount amount = Marked Price − Selling Price = 150 − 138 = 12

Discount % = (Discount ÷ Marked Price) × 100

Discount % = (12 ÷ 150) × 100 = 8%

∴ The correct answer is 8%.

Alternate Method

Given:

Marked Price (MP) = ₹150

Selling Price (SP) = ₹138

Formula Used:

Discount Percentage = [(Marked Price − Selling Price) ÷ Marked Price] × 100

Calculations:

⇒ Discount = 150 − 138

⇒ Discount = ₹12

⇒ Discount % = (12 ÷ 150) × 100

⇒ Discount % = (4 ÷ 50) × 100

⇒ Discount % = 4 × 2

⇒ Discount % = 8%

∴ The correct answer is 8%.

Additional Information

Marked Price (MP)

The price printed on the label of an article is known as the Marked Price or List Price. Discount is always calculated on this price.

Selling Price (SP) formula

SP = MP − Discount. If a discount % is given, then SP = MP × (100 − Discount %) ÷ 100.

Successive Discounts

If two discounts d₁% and d₂% are given, the single equivalent discount is (d₁ + d₂ − d₁d₂/100)%.

Shortcut Trick

Volume of Hemisphere = (2/3) × π × r³

Convert mixed fraction: 2425 1/2 = 4851/2 cm³

4851/2 = (2/3) × (22/7) × r³

r³ = (4851 × 21) / (2 × 44) = (441 × 21) / 8 = 21³ / 2³

r = 21/2 = 10.5 cm

∴ The correct answer is 10.5 cm.

Alternate Method

Given:

Volume of hemisphere (V) = 2425 1/2 cm³ = 4851/2 cm³

Value of π = 22/7

Formula used:

Volume of a hemisphere = (2/3) × π × r³

Calculations:

⇒ 4851/2 = (2/3) × (22/7) × r³

⇒ 4851/2 = (44/21) × r³

⇒ r³ = (4851 × 21) / (2 × 44)

⇒ r³ = (441 × 11 × 21) / (2 × 4 × 11)

⇒ r³ = (21² × 21) / 8

⇒ r³ = 9261 / 8

⇒ r = ³√(9261 / 8)

⇒ r = 21 / 2

⇒ r = 10.5 cm

∴ The correct answer is 10.5 cm.

Additional Information

Curved Surface Area of Hemisphere

The area of the curved part of a hemisphere is 2πr².

Total Surface Area of Hemisphere

The total surface area (including the flat circular base) is 3πr².

Volume of a Sphere

A hemisphere is exactly half of a sphere. Volume of a full sphere is (4/3)πr³.

Surface Area of a Sphere

The total surface area of a full sphere is 4πr².

Shortcut Trick Loss = CP − SP ⇒ CP of 6 balls = CP of 38 balls − SP of 38 balls. Rearranging: SP of 38 balls = CP of (38 − 6) balls. ⇒ SP of 38 balls = CP of 32 balls. ⇒ ₹ 2,240 = 32 × CP of 1 ball. ⇒ CP of 1 ball = 2240 ÷ 32 = 70. ∴ The correct answer is ₹ 70.

Alternate Method Given: Total balls sold = 38 Total Selling Price (SP) = ₹ 2,240 Loss = Cost Price (CP) of 6 balls

Calculations: Let the CP of one ball be ₹ x. ⇒ CP of 38 balls = 38x ⇒ Loss = CP of 6 balls = 6x We know that: SP = CP − Loss ⇒ 2240 = 38x − 6x ⇒ 2240 = 32x ⇒ x = 2240 ÷ 32 ⇒ x = 70 ∴ The correct answer is ₹ 70.

Additional Information Basic Profit and Loss Formulas Profit = SP − CP (when SP > CP) and Loss = CP − SP (when CP > SP). Percentage Profit/Loss Profit% = (Profit ÷ CP) × 100 and Loss% = (Loss ÷ CP) × 100. Both are always calculated on the CP unless stated otherwise. Selling Price in terms of CP SP = CP × (100 + Profit%) ÷ 100 or SP = CP × (100 − Loss%) ÷ 100.

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Shortcut Trick

 

 

Supplement of x = 104° is 180° − 104° = 76°

Supplement of y = 121° is 180° − 121° = 59°

Sum of angles of triangle = 180° ⇒ z = 180° − 76° − 59° = 45°

Largest angle among 76°, 59°, 45° is 76°

∴ The correct answer is 76°.

Alternate Method

Given:

First exterior angle = 121°

Second exterior angle = 104°

Concept Used:

Sum of all exterior angles of a triangle = 360°

Interior angle + Exterior angle = 180°

Calculation:

⇒ Let the third exterior angle = x

⇒ 121° + 104° + x = 360°

⇒ x = 360° - 225°

⇒ x = 135°

⇒ First interior angle = 180° - 121° = 59°

⇒ Second interior angle = 180° - 104° = 76°

⇒ Third interior angle = 180° - 135° = 45°

⇒ Largest angle = 76°

∴ The measure of the largest angle of the triangle is 76°.

Additional Information

Sum of Exterior Angles

The sum of the exterior angles of any polygon (one at each vertex) is always 360°.

Exterior Angle Theorem

An exterior angle of a triangle is equal to the sum of the two opposite interior angles (e.g., Ext ∠ABD = ∠BAC + ∠ACB).

Types of Triangles by Angles

Acute triangle (all angles < 90°), Right triangle (one angle = 90°), and Obtuse triangle (one angle > 90°).

Shortcut Trick

In ΔABC, AB = BC ⇒ ∠BCA = ∠BAC = 35°.

Angle in same segment: ∠BDC = ∠BAC = 35°.

In ΔBCD, ∠BCD = 180° − (∠DBC + ∠BDC) = 180° − (65° + 35°) = 80°.

∠ECD (same as ∠ACD) = ∠BCD − ∠BCA = 80° − 35° = 45°.

∴ The correct answer is 45°.

Alternate Method

Given: ∠DBC = 65°, ∠BAC = 35°, AB = BC. Points A, B, C, D are on the circumference of a circle.

Formula Used:

1. Angles subtended by the same chord at the circumference are equal.
2. Angles opposite to equal sides in a triangle are equal.
3. Sum of angles in a triangle = 180°.

Calculations:

⇒ In ΔABC, given AB = BC.

⇒ ∴ ∠BCA = ∠BAC = 35° (Angles opposite to equal sides).

⇒ ∠BDC = ∠BAC = 35° (Angles in the same segment subtended by chord BC).

⇒ In ΔBCD, ∠DBC + ∠BDC + ∠BCD = 180°.

⇒ 65° + 35° + ∠BCD = 180°.

⇒ 100° + ∠BCD = 180°.

⇒ ∠BCD = 180° − 100° = 80°.

⇒ From the figure, ∠BCD = ∠BCA + ∠ACD.

⇒ 80° = 35° + ∠ACD.

⇒ ∠ACD = 80° − 35° = 45°.

⇒ Point E lies on the diagonal BD, so ∠ECD is the same as ∠ACD.

⇒ ∠ECD = 45°.

∴ The correct answer is 45°.

Additional Information

Angles in Same Segment

Angles subtended by a chord at any point on the same arc of the circle are equal.

Isosceles Triangle Property

In a triangle, if two sides are equal, the angles opposite to those sides are also equal.

Cyclic Quadrilateral

In a cyclic quadrilateral, the sum of opposite angles is 180° (∠A + ∠C = 180° and ∠B + ∠D = 180°).

Shortcut Trick

Profit Ratio = (Capital × Time)_Ramesh : (Capital × Time)_Kevin

Ramesh : Kevin = (40,000 × 12) : (20,000 × 6)

Ratio = 480,000 : 120,000 = 4 : 1

Kevin’s share = 1 / (4 + 1) × 10,000 = 1/5 × 10,000 = Rs. 2,000

∴ The correct answer is Rs. 2,000.

Alternate Method

Given:

Ramesh’s Investment = Rs. 40,000

Kevin’s Investment = Rs. 20,000

Total Profit = Rs. 10,000

Time period = 1 year (12 months)

Formula: Profit Share ∝ Capital × Time

⇒ Ramesh’s total investment units = 40,000 × 12 = 4,80,000

⇒ Kevin’s total investment units = 20,000 × 6 = 1,20,000

⇒ Ratio of their profits = 4,80,000 : 1,20,000

⇒ Ratio = 4 : 1

⇒ Sum of ratio parts = 4 + 1 = 5

⇒ Kevin’s share = (1 / 5) × 10,000

⇒ Kevin’s share = Rs. 2,000

∴ The correct answer is Rs. 2,000.

Additional Information

Simple Partnership

If the investment period is same for all partners, the profit is divided in the ratio of their capitals.

Compound Partnership

When investments are for different time periods, Profit Ratio = (C₁ × T₁) : (C₂ × T₂).

Ratio of Capitals

Capital Ratio = (Profit₁ / Time₁) : (Profit₂ / Time₂).

Shortcut Trick

Ratio of new fares = (Initial Fare₁ × Increase Multiplier₁) : (Initial Fare₂ × Increase Multiplier₂)

⇒ New Ratio = (2500 × 120%) : (2000 × 110%)

⇒ New Ratio = (25 × 120) : (20 × 110)

⇒ New Ratio = (5 × 12) : (4 × 11) = 60 : 44 = 15 : 11

∴ The correct answer is 15 : 11.

Alternate Method

Given:

Initial Fare of 2^nd AC = ₹2500

Initial Fare of 3^rd AC = ₹2000

Increase in 2^nd AC fare = 20%

Increase in 3^rd AC fare = 10%

Formula Used:

New Fare = Original Fare × (100 + Increase)%

Calculations:

⇒ New Fare of 2^nd AC = 2500 × (120/100) = 25 × 120 = ₹3000

⇒ New Fare of 3^rd AC = 2000 × (110/100) = 20 × 110 = ₹2200

⇒ New Ratio = 3000 : 2200

⇒ New Ratio = 300 / 220 = 15 / 11

⇒ New Ratio = 15 : 11

∴ The correct answer is 15 : 11.

Additional Information

Percentage Increase Formula

New Value = Original Value × (1 + Rate/100). If the rate is 20%, the multiplying factor is 1.20.

Ratio Simplification

To simplify a ratio A : B, divide both terms by their Highest Common Factor (HCF). Here, HCF of 3000 and 2200 is 200.

Direct Ratio of Change

New Ratio = (A_old × %Change_A) : (B_old × %Change_B). This bypasses calculating absolute values.

Given:

$\frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \frac{1}{99} + \frac{1}{143}$ = ?

Formula used:

Sum = $\frac{1}{d} (\frac{1}{a} - \frac{1}{l})$

Where d = difference between factors, a = first factor of first term, l = last factor of last term

Calculation:

⇒ $\frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \frac{1}{9 \times 11} + \frac{1}{11 \times 13}$

⇒ $\frac{1}{2} \times (\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \frac{1}{9} - \frac{1}{11} + \frac{1}{11} - \frac{1}{13})$

⇒ $\frac{1}{2} \times (\frac{1}{3} - \frac{1}{13})$

⇒ $\frac{1}{2} \times (\frac{13 - 3}{39})$

⇒ $\frac{1}{2} \times \frac{10}{39}$

⇒ $\frac{5}{39}$

∴ The correct answer is option (2).

Shortcut Trick

For secants intersecting externally at point P, the property is: PA × PB = PC × PD.

Given: AB = 4 cm, CD = 11 cm, PD = 15 cm.

PC = PD − CD = 15 − 11 = 4 cm.

Product PC × PD = 4 × 15 = 60.

We need factors of 60 (PA and PB) such that their difference |PB − PA| = chord AB = 4 cm.

Factors of 60 with a difference of 4 are 10 and 6 (since 10 × 6 = 60 and 10 − 6 = 4).

∴ The correct answer is 10 cm.

Alternate Method

Given:

Length of chord AB = 4 cm

Length of chord CD = 11 cm

Length of external segment PD = 15 cm

Formula Used:

The Secant-Secant Power Theorem states that if two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.

PA × PB = PC × PD

Calculations:

⇒ PC = PD − CD

⇒ PC = 15 − 11 = 4 cm

⇒ Let the length of segment PB be x cm.

⇒ Then, PA = PB − AB = x − 4 cm

⇒ Applying the theorem: PA × PB = PC × PD

⇒ (x − 4) × x = 4 × 15

⇒ x² − 4x = 60

⇒ x² − 4x − 60 = 0

⇒ x² − 10x + 6x − 60 = 0

⇒ x(x − 10) + 6(x − 10) = 0

⇒ (x − 10)(x + 6) = 0

⇒ x = 10 or x = −6 (Length cannot be negative)

⇒ x = 10 cm

∴ The correct answer is 10 cm.

Additional Information

Intersecting Chords Theorem (Internal)

When two chords intersect inside a circle, the product of the segments of one chord is equal to the product of the segments of the other: PA × PB = PC × PD.

Tangent-Secant Theorem

If a tangent and a secant are drawn to a circle from an exterior point, the square of the tangent segment equals the product of the whole secant and its external part: PT² = PA × PB.

Secant Segment Relation

The length of the entire secant segment is the sum of the external segment and the chord length: Whole Secant = External Segment + Chord.

Shortcut Trick

Combine ratios: A:B = 2:3 and B:C = 4:5.

To equalize B, multiply first ratio by 4 and second by 3.

Combined Ratio A : B : C = (2×4) : (3×4) : (3×5) = 8 : 12 : 15.

Sum of ratio parts = 8 + 12 + 15 = 35 units.

35 units = 280 → 1 unit = 8. ∴ 2nd number = 12 × 8 = 96.

∴ The correct answer is 96.

Alternate Method

Given:

Sum of three numbers (A + B + C) = 280

Ratio A : B = 2 : 3

Ratio B : C = 4 : 5

Formula Used:

Combined Ratio A : B : C = (a × c) : (b × c) : (b × d) where A:B = a:b and B:C = c:d.

Calculations:

⇒ First : Second = 2 : 3

⇒ Second : Third = 4 : 5

⇒ Multiplying the ratios to make the Second number common:

⇒ A : B = (2 × 4) : (3 × 4) = 8 : 12

⇒ B : C = (4 × 3) : (5 × 3) = 12 : 15

⇒ Combined Ratio A : B : C = 8 : 12 : 15

⇒ Let the numbers be 8x, 12x, and 15x.

⇒ Sum = 8x + 12x + 15x = 35x

⇒ 35x = 280

⇒ x = 280 ÷ 35 = 8

⇒ Second number (B) = 12x = 12 × 8 = 96

∴ The correct answer is 96.

Additional Information

Ratio Balancing

To combine two ratios A:B and B:C, find the LCM of the common term B to make its value equal in both ratios.

Direct Combination Formula

If A:B = a:b and B:C = c:d, then the continuous ratio A:B:C is calculated as (a×c) : (b×c) : (b×d).

Proportional Parts

Value of any part = (Ratio of that part ÷ Sum of all ratios) × Total Value.

Shortcut Trick

Rationalize the denominator by multiplying the numerator and denominator by the conjugate (5 − 3√3).

Numerator: (8 + 2√3) × (5 − 3√3) = 40 − 24√3 + 10√3 − 18 = 22 − 14√3.

Denominator: (5 + 3√3) × (5 − 3√3) = 5² − (3√3)² = 25 − 27 = −2.

Simplified Result: (22 − 14√3) ÷ (−2) = −11 + 7√3 = 7√3 − 11.

Comparing with a√3 − b, we get a = 7 and b = 11.

Calculation: a + b = 7 + 11 = 18.

∴ The correct answer is 18.

Alternate Method

Given: (8 + 2√3) ÷ (3√3 + 5) = a√3 − b

Formula Used: Rationalization using (x + y)(x − y) = x² − y²

⇒ Expression = (8 + 2√3) / (5 + 3√3)

⇒ Multiply by conjugate (5 − 3√3): [(8 + 2√3) × (5 − 3√3)] / [(5 + 3√3) × (5 − 3√3)]

⇒ Numerator = 8(5) − 8(3√3) + 2√3(5) − 2√3(3√3)

⇒ Numerator = 40 − 24√3 + 10√3 − 6(3) = 40 − 14√3 − 18 = 22 − 14√3

⇒ Denominator = 5² − (3√3)² = 25 − 27 = −2

⇒ Value = (22 − 14√3) / (−2) = −11 + 7√3

⇒ Compare 7√3 − 11 with a√3 − b

⇒ a = 7, b = 11

⇒ a + b = 7 + 11 = 18

∴ The correct answer is 18.

Additional Information

Rationalization of Surds

To rationalize a denominator of the form (x + √y), multiply both numerator and denominator by its conjugate (x − √y) to eliminate the radical.

Conjugate Rule

The conjugate of (a + √b) is (a − √b). Their product always results in a rational number: (a + √b)(a − √b) = a² − b.

Equality of Irregular Numbers

If x + y√n = a + b√n (where x, y, a, b are rational and √n is irrational), then x = a and y = b.

Shortcut Trick

Convert mixed fraction: 2 51/79 = (2 × 79 + 51) ÷ 79 = 209/79.

Apply Componendo and Dividendo: (sec θ + tan θ + sec θ − tan θ) ÷ (sec θ + tan θ − (sec θ − tan θ)) = (209 + 79) ÷ (209 − 79).

2 sec θ ÷ 2 tan θ = 288 ÷ 130 ⇒ sec θ ÷ tan θ = 144 ÷ 65.

Since sec θ ÷ tan θ = (1/cos θ) ÷ (sin θ/cos θ) = 1/sin θ.

1/sin θ = 144/65 ⇒ sin θ = 65/144.

∴ The correct answer is 65/144.

Alternate Method

Given:

(sec θ + tan θ) ÷ (sec θ − tan θ) = 2 51/79

Formula Used:

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

Trigonometric relations: sec θ = 1/cos θ and tan θ = sin θ/cos θ.

Calculations:

⇒ (sec θ + tan θ) ÷ (sec θ − tan θ) = (2 × 79 + 51) ÷ 79

⇒ (sec θ + tan θ) ÷ (sec θ − tan θ) = 209/79

⇒ Applying Componendo and Dividendo:

⇒ [(sec θ + tan θ) + (sec θ − tan θ)] ÷ [(sec θ + tan θ) − (sec θ − tan θ)] = (209 + 79) ÷ (209 − 79)

⇒ (2 sec θ) ÷ (2 tan θ) = 288 ÷ 130

⇒ sec θ ÷ tan θ = 144 ÷ 65

⇒ (1/cos θ) ÷ (sin θ/cos θ) = 144/65

⇒ 1/sin θ = 144/65

⇒ sin θ = 65/144

∴ The correct answer is 65/144.

Additional Information

Componendo and Dividendo Rule

It is a method used to simplify equations involving ratios. If a/b = c/d, then (a+b)/(a−b) = (c+d)/(c−d).

Pythagorean Identity

A fundamental identity in trigonometry is sec²θ − tan²θ = 1, which implies (sec θ + tan θ)(sec θ − tan θ) = 1.

Reciprocal Ratios

sec θ = 1/cos θ, cosec θ = 1/sin θ, and cot θ = 1/tan θ.

Shortcut Trick

Since the point (a, 4) lies on both lines, substitute y = 4 in the first equation.

4x − 2(4) = 10 ⇒ 4x − 8 = 10 ⇒ 4x = 18.

In the second equation 4x + ky = 2, substitute 4x = 18 and y = 4 directly.

18 + k × 4 = 2 ⇒ 4k = 2 − 18 ⇒ 4k = −16.

k = −16 ÷ 4 = −4.

∴ The correct answer is −4.

Alternate Method

Given:

Equation 1: 4x − 2y = 10

Equation 2: 4x + ky = 2

Intersection point: (a, 4)

Formula Used:

For a point (x₁, y₁) to be the intersection of two lines, it must satisfy both equations simultaneously.

Step 1: Substitute the coordinates of the intersection point (a, 4) into Equation 1.

⇒ 4(a) − 2(4) = 10

⇒ 4a − 8 = 10

⇒ 4a = 18

⇒ a = 18 ÷ 4 = 4.5

Step 2: Substitute x = 4.5 and y = 4 into Equation 2.

⇒ 4(4.5) + k(4) = 2

⇒ 18 + 4k = 2

⇒ 4k = 2 − 18

⇒ 4k = −16

⇒ k = −16 ÷ 4

⇒ k = −4

∴ The correct answer is −4.

Additional Information

Condition for Unique Solution

For two linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to have a unique intersection point, a₁/a₂ ≠ b₁/b₂.

Parallel Lines (No Solution)

Lines are parallel and never intersect if a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Coincident Lines (Infinite Solutions)

Lines overlap completely if a₁/a₂ = b₁/b₂ = c₁/c₂.

Shortcut Trick

Identify the series as an Arithmetic Progression (AP) where the first term is 6 and the last term is 40.

Average of an AP = (First term + Last term) ÷ 2 = (6 + 40) ÷ 2 = 23.

Number of terms (n) = [(Last − First) ÷ Common Difference] + 1 = [(40 − 6) ÷ 2] + 1 = 18.

Sum of the series = Average × n = 23 × 18 = 414.

∴ The correct answer is 414.

Alternate Method

Given series: 6, 8, 10, 12, 14, ............., 40

First term (a) = 6

Last term (l) = 40

Common difference (d) = 8 − 6 = 2

Formula for the n^th term of an AP: l = a + (n − 1)d

⇒ 40 = 6 + (n − 1) × 2

⇒ 40 − 6 = (n − 1) × 2

⇒ 34 = (n − 1) × 2

⇒ n − 1 = 17

⇒ n = 18

Formula for the Sum (S_n) of an AP: S_n = (n/2) × (a + l)

⇒ S₁₈ = (18 ÷ 2) × (6 + 40)

⇒ S₁₈ = 9 × 46

⇒ S₁₈ = 414

∴ The correct answer is 414.

Additional Information

Arithmetic Progression (AP)

A sequence where the difference between consecutive terms is constant. n^th term T_n = a + (n − 1)d.

Sum of First n Natural Numbers

The sum of the first n positive integers is calculated as [n × (n + 1)] ÷ 2.

Sum of First n Even Numbers

The sum of the first n even natural numbers starting from 2 is given by n × (n + 1).

General Sum Formula

For any AP, the sum is also S_n = (n/2) × [2a + (n − 1)d].

Shortcut Trick

In right-angled ΔADP, AP = √(AD² + DP²) = √(24² + 10²) = 26 cm.

By similar triangles, ΔADP ∼ ΔQCP (since AD || QC and A, P, Q are collinear).

The side ratio is DP : CP = 10 : (15 − 10) = 10 : 5 = 2 : 1.

Therefore, AP : PQ = 2 : 1 ⇒ PQ = 26 ÷ 2 = 13 cm.

Total length AQ = AP + PQ = 26 + 13 = 39 cm.

∴ The correct answer is 39 cm.

Alternate Method

Given: ABCD is a rectangle, BC = 24 cm, DP = 10 cm, CD = 15 cm.

Formula: AP = √(AD² + DP²) and Similarity Ratio: AP/PQ = DP/PC.

⇒ In rectangle ABCD, AD = BC = 24 cm and CD = AB = 15 cm.

⇒ Point P is on CD such that DP = 10 cm, so PC = CD − DP = 15 − 10 = 5 cm.

⇒ In right-angled ΔADP, using Pythagoras Theorem:

⇒ AP² = AD² + DP² = 24² + 10² = 576 + 100 = 676

⇒ AP = √676 = 26 cm.

⇒ In ΔADP and ΔQCP, ∠ADP = ∠QCP = 90° and ∠APD = ∠QPC (vertically opposite).

⇒ ∴ ΔADP ∼ ΔQCP (by AA similarity).

⇒ AP / PQ = DP / CP ⇒ 26 / PQ = 10 / 5

⇒ 26 / PQ = 2 ⇒ PQ = 13 cm.

⇒ AQ = AP + PQ = 26 + 13 = 39 cm.

∴ The correct answer is 39 cm.

Additional Information

AA Similarity Criterion

If two angles of one triangle are equal to two angles of another triangle, then the two triangles are similar, and their corresponding sides are in the same ratio.

Pythagoras Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: a² + b² = c².

Properties of a Rectangle

Opposite sides are equal and parallel, and each interior angle is 90°. In this problem, AD || BC implies AD || CQ.

Shortcut Trick

Total time spent = 11:25 a.m. − 10:00 a.m. = 1 hour 25 minutes = 85 minutes.

Actual travelling time = 85 minutes − 15 minutes (wait) = 70 minutes = 7/6 hours.

Average speed for the entire journey = (2 × S₁ × S₂) / (S₁ + S₂) = (2 × 32 × 24) / 56 = 192/7 km/h.

Total distance = Average speed × Total travelling time = (192/7) × (7/6) = 32 km.

∴ The correct answer is 32 km.

Alternate Method

Given:

Departure time = 10:00 a.m. | Arrival time = 11:25 a.m. | Wait time = 15 minutes.

Speed to delivery point (S₁) = 32 km/h | Speed back to office (S₂) = 24 km/h.

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Total duration = 1 hour 25 minutes = 85 minutes.

⇒ Total travel time = 85 − 15 = 70 minutes = 70/60 hours = 7/6 hours.

⇒ Let the distance to the delivery point be d km.

⇒ (d / 32) + (d / 24) = 7 / 6

⇒ LCM of 32 and 24 is 96.

⇒ (3d + 4d) / 96 = 7 / 6

⇒ 7d / 96 = 7 / 6

⇒ d / 96 = 1 / 6

⇒ d = 96 / 6 = 16 km.

⇒ Total distance travelled = 2 × d = 2 × 16 = 32 km.

∴ The correct answer is 32 km.

Additional Information

Average Speed for Equal Distances

When a person travels the same distance at two different speeds x and y, the average speed for the whole journey is given by (2xy) / (x + y).

Speed-Time Relation

For a constant distance, speed is inversely proportional to time (S ∝ 1/T). The ratio of speeds is the inverse of the ratio of times taken.

Relative Speed in Same Direction

When two objects move in the same direction, their relative speed is the difference of their individual speeds |S₁ − S₂|.

Shortcut Trick

If a sin x + b cos x = c, then b sin x − a cos x = √(a² + b² − c²).

Here, a = 3, b = 4, and c = 2.

Value = √(3² + 4² − 2²)

Value = √(9 + 16 − 4) = √21

∴ The correct answer is √21.

Alternate Method

Given: 3 sin x + 4 cos x = 2

To find: 3 cos x − 4 sin x

Formula: (a sin x + b cos x)² + (a cos x − b sin x)² = a² + b²

⇒ Let 3 cos x − 4 sin x = y

⇒ Squaring and adding both equations:

⇒ (3 sin x + 4 cos x)² + (3 cos x − 4 sin x)² = 2² + y²

⇒ (9 sin²x + 16 cos²x + 24 sin x cos x) + (9 cos²x + 16 sin²x − 24 sin x cos x) = 4 + y²

⇒ 9(sin²x + cos²x) + 16(cos²x + sin²x) = 4 + y²

⇒ 9(1) + 16(1) = 4 + y²

⇒ 25 = 4 + y²

⇒ y² = 21

⇒ y = √21

∴ The correct answer is √21.

Additional Information

Trigonometric Coefficient Identity

For any angle x and constants a and b, the sum of the squares of (a sin x + b cos x) and (a cos x − b sin x) is always independent of x and equal to a² + b².

Maximum and Minimum Values

The expression a sin x + b cos x has a maximum value of √(a² + b²) and a minimum value of −√(a² + b²).

Secant and Tangent Relation

If a sec x + b tan x = c, then a related identity for tangent and secant is (a tan x + b sec x)² = c² − (a² − b²).

Given:

Total boys and girls in school B is 2600

Difference of boys and girls in school B is 520

Formula used:

% Boys in B = Total boys in B/Total students in school B  × 100

Calculation:

Let the boys and girls in School B be x, y

⇒ x + y = 2600, x – y = 520

Adding those two equations we get 2x = 3120

⇒ x = 1560

∴ Percentage of boys in School B = 1560 × 100/2600 = 60%

Shortcut Trick

Total sum of 5 numbers = 56 × 5 = 280.

Ratio of First Number : Sum of last 4 numbers = 3 : 4.

Total parts = 3 + 4 = 7 parts → 280.

1 part = 280 ÷ 7 = 40.

Sum of last 4 numbers = 4 parts = 4 × 40 = 160.

Average of last 4 numbers = 160 ÷ 4 = 40.

∴ The correct answer is 40.

Alternate Method

Given:

Average of 5 numbers = 56

First number = 3/4 × (Sum of last 4 numbers)

Formula Used:

Sum of observations = Average × Number of observations

Average = Sum of observations ÷ Number of observations

Calculations:

⇒ Total sum of 5 numbers = 56 × 5 = 280

⇒ Let the sum of the last four numbers be S.

⇒ First number = 3/4 × S = 3S/4

⇒ Sum of all 5 numbers = First number + Sum of last 4 numbers

⇒ 280 = 3S/4 + S

⇒ 280 = (3S + 4S) ÷ 4

⇒ 280 = 7S/4

⇒ S = (280 × 4) ÷ 7

⇒ S = 40 × 4 = 160

⇒ Average of last 4 numbers = S ÷ 4

⇒ Average of last 4 numbers = 160 ÷ 4 = 40

∴ The correct answer is 40.

Additional Information

Basic Average Formula

The average of n numbers is calculated as Sum of all numbers ÷ n. To find the sum, use Average × n.

Weighted Average

When two groups with averages x₁ and x₂ and sizes n₁ and n₂ are combined, the new average is (n₁x₁ + n₂x₂) ÷ (n₁ + n₂).

Average of Consecutive Numbers

For any arithmetic progression, the average is exactly the middle term or (First term + Last term) ÷ 2.

Shortcut Trick

Volume of Cylinder ∝ r²h

Successive change in Radius (r × r): −20% and −20% ⇒ −20 − 20 + (−20 × −20)/100 = −36%

Net change with Height (+20%): −36 + 20 + (−36 × 20)/100

Net change = −16 − 7.2 = −23.2%

∴ The correct answer is 23.2%.

Alternate Method

Given: Radius (r) decreased by 20%, Height (h) increased by 20%.

Formula: Volume (V) = πr²h

⇒ Let original radius r₁ = 5 and original height h₁ = 5.

⇒ Original Volume V₁ = π × (5)² × 5 = 125π.

⇒ New radius r₂ = 5 − (20% of 5) = 4.

⇒ New height h₂ = 5 + (20% of 5) = 6.

⇒ New Volume V₂ = π × (4)² × 6 = 16 × 6 × π = 96π.

⇒ Decrease in volume = 125π − 96π = 29π.

⇒ Percentage decrease = (29π / 125π) × 100

⇒ Percentage decrease = (29 / 125) × 100 = 23.2%.

∴ The correct answer is 23.2%.

Additional Information

Successive Percentage Change

For two percentage changes a% and b%, the net change is given by [a + b + (ab/100)]%. For three changes, apply this formula twice.

Cylinder Formulas

Volume = πr²h | Curved Surface Area (CSA) = 2πrh | Total Surface Area (TSA) = 2πr(h + r).

Proportionality in Geometry

When dimensions of a 3D shape change, the volume changes according to the product of the change factors of its dimensions (e.g., V ∝ r²h for cylinders, V ∝ a³ for cubes).

Shortcut Trick

Direct formula for height (h) when shadow increases by distance (d) as angle changes from θ₁ to θ₂:

h = d ÷ (cot θ₂ − cot θ₁)

Here, d = 10 m, θ₂ = 30°, and θ₁ = 45°

h = 10 ÷ (cot 30° − cot 45°) = 10 ÷ (√3 − 1)

By rationalizing: h = [10 × (√3 + 1)] ÷ [(√3 − 1)(√3 + 1)] = 10(√3 + 1) ÷ 2 = 5(√3 + 1)

∴ The correct answer is 5(√3 + 1) m.

Alternate Method

Given:

Increase in shadow length (CD) = 10 m

Angle of elevation at initial position (C) = 45°

Angle of elevation at final position (D) = 30°

Calculations:

⇒ Let the height of the tower AB be h.

⇒ In right-angled ▵ABC, tan 45° = AB ÷ BC

⇒ 1 = h ÷ BC → BC = h

⇒ In right-angled ▵ABD, tan 30° = AB ÷ BD

⇒ 1 ÷ √3 = h ÷ BD → BD = h√3

⇒ Given CD = BD − BC = 10 m

⇒ h√3 − h = 10

⇒ h(√3 − 1) = 10

⇒ h = 10 ÷ (√3 − 1)

⇒ h = [10 × (√3 + 1)] ÷ [(√3 − 1)(√3 + 1)]

⇒ h = 10(√3 + 1) ÷ (3 − 1) = 10(√3 + 1) ÷ 2

⇒ h = 5(√3 + 1) m

∴ The correct answer is 5(√3 + 1) m.

Additional Information

Trigonometric Ratios for Standard Angles

tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. Cot is the reciprocal: cot 30° = √3, cot 45° = 1, cot 60° = 1/√3.

Shadow and Angle Relation

As the altitude of the sun decreases, the length of the shadow of a vertical object increases.

Rationalization Rule

To rationalize a denominator of the form (a − √b), multiply both numerator and denominator by its conjugate (a + √b).

Shortcut Trick

Ratio of sides AP : AB = 1 : (1 + 3) = 1 : 4

Ratio of sides AQ : AC = 1.5 : (1.5 + 4.5) = 1.5 : 6 = 1 : 4

Ratio of Area(ΔAPQ) : Area(ΔABC) = (Side Ratio)² = (1 : 4)² = 1 : 16

Area(ΔABC) = 12 × 16 = 192 cm²

Area(BPQC) = Area(ΔABC) − Area(ΔAPQ) = 192 − 12 = 180 cm²

∴ The correct answer is 180 cm².

Alternate Method

Given: AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, CQ = 4.5 cm, Area(ΔAPQ) = 12 cm².

Formula Used: Area of Similar Triangles Ratio = (Corresponding Sides Ratio)²

⇒ Total length AB = AP + PB = 1 + 3 = 4 cm

⇒ Total length AC = AQ + CQ = 1.5 + 4.5 = 6 cm

⇒ In ΔAPQ and ΔABC:

⇒ AP/AB = 1/4 and AQ/AC = 1.5/6 = 1/4

⇒ ∠A is common to both triangles.

⇒ By SAS similarity criterion, ΔAPQ ~ ΔABC.

⇒ Area(ΔAPQ) / Area(ΔABC) = (AP / AB)² = (1/4)²

⇒ 12 / Area(ΔABC) = 1/16

⇒ Area(ΔABC) = 12 × 16 = 192 cm²

⇒ Area of quadrilateral BPQC = Area(ΔABC) − Area(ΔAPQ)

⇒ Area(BPQC) = 192 − 12 = 180 cm²

∴ The correct answer is 180 cm².

Additional Information

SAS Similarity Theorem

If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.

Area Ratio Property

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides: Area₁/Area₂ = (s₁/s₂)².

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Calculation:

Total Students who got marks between 30 – 40 is 20

Total Students = 12 + 16 + 20 + 28 + 8 + 12 + 4 = 100

∴ Ratio is 20 ∶ 100 = 1 ∶ 5

Shortcut Trick

For an isosceles right triangle with hypotenuse h, the Perimeter is given by h × (√2 + 1).

Given Perimeter = 8 × (√2 + 1) cm.

By comparing the two expressions: h(√2 + 1) = 8(√2 + 1).

⇒ h = 8 cm.

∴ The correct answer is 8 cm.

Alternate Method

Given: Perimeter of isosceles right triangle = 8(√2 + 1) cm.

Formula: Perimeter = a + a + h, where a is the leg and h is the hypotenuse (h = a√2).

⇒ Perimeter = 2a + h

⇒ Perimeter = 2(h/√2) + h = h√2 + h

⇒ Perimeter = h(√2 + 1)

⇒ 8(√2 + 1) = h(√2 + 1)

⇒ h = 8 cm

∴ The correct answer is 8 cm.

Additional Information

Area of Isosceles Right Triangle

Area can be calculated using legs as a²/2 or using hypotenuse as h²/4.

Side Ratio Property

In a 45°−45°−90° triangle, the sides are always in the ratio 1 : 1 : √2.

Inradius and Circumradius

Circumradius (R) = h/2 and Inradius (r) = (2a − h)/2 for an isosceles right triangle.

Shortcut Trick

By Alligation: Initial mixture (10% acid) vs Pure acid (100% acid) to get 25% acid.

Ratio of Initial mixture to Added acid = (100 − 25) : (25 − 10) = 75 : 15 = 5 : 1.

Initial volume is 20 L, which represents 5 units.

5 units = 20 L ⇒ 1 unit = 4 L.

∴ The correct answer is 4 L.

Alternate Method

Initial Mixture = 20 L

Initial Acid percentage = 10%

Required Acid percentage = 25%

⇒ Quantity of acid in initial mixture = 10% of 20 = 2 L

⇒ Quantity of water in initial mixture = 20 − 2 = 18 L

⇒ Since only acid is added, the quantity of water remains constant at 18 L.

⇒ In the final solution, acid is 25%, so water must be 100% − 25% = 75%.

⇒ 75% of New Total Mixture = 18 L

⇒ New Total Mixture = 18 ÷ (75/100) = 18 × (4/3) = 24 L

⇒ Quantity of acid added = New Total − Initial Total = 24 − 20 = 4 L

∴ The correct answer is 4 L.

Additional Information

Alligation Rule

It is used to find the ratio in which two ingredients at given prices or concentrations must be mixed to produce a mixture of a desired price or concentration: (Quantity of Cheaper / Quantity of Dearer) = (CP of Dearer − Mean Price) / (Mean Price − CP of Cheaper).

Constant Product / Constant Quantity Rule

When a substance is added to or removed from a mixture, the quantity of the other component (which is neither added nor removed) remains constant. Using this constant value to equate percentages is often the fastest method.

Successive Dilution

If x units are removed from a container of C units of pure liquid and replaced by water, after n operations, the quantity of pure liquid remaining = C × [1 − (x/C)]ⁿ.

Shortcut Trick

If x² + 1/x² = a, then x + 1/x = √(a + 2).

Given a = 7 ⇒ x + 1/x = √(7 + 2) = √9 = 3.

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

Here k = 3, so x³ + 1/x³ = 3³ − 3(3) = 27 − 9 = 18.

∴ The correct answer is 18.

Alternate Method

Given: x² + 1/x² = 7 and x > 0

Formula: (a + b)² = a² + b² + 2ab and (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ x² + 1/x² + 2 = 7 + 2

⇒ (x + 1/x)² = 9

⇒ x + 1/x = 3 (since x > 0)

⇒ (x + 1/x)³ = 3³

⇒ x³ + 1/x³ + 3(x)(1/x)(x + 1/x) = 27

⇒ x³ + 1/x³ + 3(3) = 27

⇒ x³ + 1/x³ = 27 − 9

⇒ x³ + 1/x³ = 18

∴ The correct answer is 18.

Additional Information

Algebraic Power Reduction

To go from power 2 to power 1: x + 1/x = √(x² + 1/x² + 2) and x − 1/x = √(x² + 1/x² − 2).

Difference of Cubes

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

Higher Power Relation

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

Shortcut Trick

Effective interest rate for 2 years at 10% = 10 + 10 + (10 × 10)/100 = 21%

Effective interest rate for 2 years at 12% = 12 + 12 + (12 × 12)/100 = 25.44%

Given: 21% of Sum = ₹1,050

Interest for Scheme B = (1050 ÷ 21) × 25.44 = 50 × 25.44 = ₹1,272

∴ The correct answer is Rs. 1,272.

Alternate Method

Given:

Sum invested (P) = Equal for both schemes

Rate of interest for Scheme A (r₁) = 10% p.a.

Rate of interest for Scheme B (r₂) = 12% p.a.

Time (n) = 2 years

Compound Interest Scheme A (CI_A) = ₹1,050

Formula Used:

CI = P[(1 + r/100)ⁿ − 1]

Calculations:

For Scheme A:

⇒ 1050 = P[(1 + 10/100)² − 1]

⇒ 1050 = P[(11/10)² − 1]

⇒ 1050 = P[121/100 − 1]

⇒ 1050 = P[21/100]

⇒ P = (1050 × 100) ÷ 21

⇒ P = 50 × 100 = ₹5,000

For Scheme B (using same P = 5000):

⇒ CI_B = 5000[(1 + 12/100)² − 1]

⇒ CI_B = 5000[(1.12)² − 1]

⇒ CI_B = 5000[1.2544 − 1]

⇒ CI_B = 5000 × 0.2544

⇒ CI_B = ₹1,272

∴ The correct answer is Rs. 1,272.

Additional Information

Successive Percentage Change

Compound interest for 2 years can be calculated using the formula: Effective Rate = (x + y + xy/100)% where x and y are annual rates.

Difference between SI and CI

For 2 years, the difference between CI and SI on the same principal is given by: D = P(r/100)².

Amount in Compound Interest

The total amount A after n years is A = P(1 + r/100)ⁿ. Then Interest CI = A − P.

Shortcut Trick

Total work = LCM(18, 30) = 90 units.

Efficiency of (A + B) = 90 ÷ 18 = 5 units and (B + C) = 90 ÷ 30 = 3 units.

Difference in efficiency: (A + B) − (B + C) = A − C = 5 − 3 = 2 units.

Given A = 2C, so 2C − C = 2 ⇒ C's efficiency = 2 units.

Efficiency of B = Eff(B + C) − Eff(C) = 3 − 2 = 1 unit.

Time taken by B alone = 90 ÷ 1 = 90 days.

∴ The correct answer is 90 days.

Alternate Method

Given:

Time taken by (A + B) = 18 days

Time taken by (B + C) = 30 days

Efficiency of A = 2 × Efficiency of C

Formula Used:

Efficiency = Total Work ÷ Time taken

⇒ Efficiency of (A + B) = 5

⇒ Efficiency of (B + C) = 3

⇒ (A + B) − (B + C) = 5 − 3

⇒ A − C = 2

⇒ Since A = 2C, substitute A: 2C − C = 2 ⇒ C = 2 units/day

⇒ Efficiency of B = (B + C) − C = 3 − 2 = 1 unit/day

⇒ Days taken by B alone = Total Work ÷ Efficiency of B

⇒ Days taken by B = 90 ÷ 1 = 90 days

∴ The correct answer is 90 days.

Additional Information

Total Work Concept

Total Work = Efficiency × Time. When work is constant, Efficiency is inversely proportional to Time.

Efficiency Comparison

If A is n times as good as B, then the ratio of their efficiencies is n : 1 and the ratio of time taken to complete the same work is 1 : n.

Combined Work Rule

If person X takes x days and Y takes y days, together they take (x × y) ÷ (x + y) days to complete the work.

Shortcut Trick

Check the relationship between the squares of the sides: AB = 6√3, AC = 12, BC = 6.

Calculate squares: AB² = 108, AC² = 144, BC² = 36.

Observe that 108 + 36 = 144, which means AB² + BC² = AC².

According to the Converse of Pythagoras Theorem, the angle opposite to the longest side (AC) must be 90°.

∴ The correct answer is 90°.

Alternate Method

Given: In ΔABC, AB = 6√3 cm, AC = 12 cm, BC = 6 cm.

Formula Used: Cosine Rule: cos B = (a² + c² − b²) ÷ (2ac)

Calculations:

⇒ Let a = BC = 6, b = AC = 12, c = AB = 6√3

⇒ cos B = (6² + (6√3)² − 12²) ÷ (2 × 6 × 6√3)

⇒ cos B = (36 + 108 − 144) ÷ (72√3)

⇒ cos B = (144 − 144) ÷ (72√3)

⇒ cos B = 0 ÷ 72√3

⇒ cos B = 0

⇒ Since cos 90° = 0, ∴ ∠B = 90°

∴ The correct answer is 90°.

Additional Information

Converse of Pythagoras Theorem

If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle (90°).

General Cosine Rule

For any triangle with sides a, b, c and angles A, B, C, the rule is b² = a² + c² − 2ac cos B.

30-60-90 Triangle Property

In a right triangle where sides are in ratio 1 : √3 : 2, the angles are always 30°, 60°, and 90° respectively.

Shortcut Trick

In a rhombus, adjacent angles are supplementary: ∠ABC + ∠BCD = 180°.

Given ∠ABC = 52°, then ∠BCD = 180° − 52° = 128°.

Diagonals of a rhombus bisect the vertex angles, so diagonal AC bisects ∠BCD.

∠ACD = ∠BCD ÷ 2 = 128° ÷ 2 = 64°.

∴ The correct answer is 64°.

Alternate Method

Given: ABCD is a rhombus and ∠ABC = 52°.

In a rhombus, opposite sides are parallel (AB || DC and AD || BC).

⇒ ∠ABC + ∠BCD = 180° (Sum of interior angles on the same side of a transversal)

⇒ 52° + ∠BCD = 180°

⇒ ∠BCD = 180° − 52° = 128°

⇒ In a rhombus, the diagonals bisect the interior angles of the vertices.

⇒ ∠ACD = ∠ACB = ∠BCD ÷ 2

⇒ ∠ACD = 128° ÷ 2

⇒ ∠ACD = 64°

∴ The correct answer is 64°.

Additional Information

Diagonals Property

The diagonals of a rhombus bisect each other at right angles (90°).

Area of a Rhombus

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

Relation between Side and Diagonals

4 × (side)² = d₁² + d₂².

Angle Bisector Property

Each diagonal of a rhombus bisects the internal angles of the vertices it joins.

Shortcut Trick

Step 1: Divide the Sum by the HCF → 240 ÷ 15 = 16.

Step 2: Find pairs of co-prime numbers (a, b) such that a + b = 16.

Step 3: Co-prime pairs are (1, 15), (3, 13), (5, 11), and (7, 9).

Total number of such valid pairs = 4.

∴ The correct answer is 4 pairs.

Alternate Method

Given:

Sum of two numbers = 240

HCF of the numbers = 15

Formula Used:

If the HCF of two numbers is H, then the numbers can be represented as H × a and H × b, where a and b are co-prime integers (&gcd;(a, b) = 1).

Calculations:

⇒ Let the two positive numbers be 15a and 15b.

⇒ According to the problem: 15a + 15b = 240

⇒ 15(a + b) = 240

⇒ a + b = 240 ÷ 15

⇒ a + b = 16

⇒ Since a and b must be co-prime (having no common factor other than 1), we check possible pairs:

⇒ (1, 15) → &gcd;(1, 15) = 1 (Valid)

⇒ (2, 14) → &gcd;(2, 14) = 2 (Invalid)

⇒ (3, 13) → &gcd;(3, 13) = 1 (Valid)

⇒ (4, 12) → &gcd;(4, 12) = 4 (Invalid)

⇒ (5, 11) → &gcd;(5, 11) = 1 (Valid)

⇒ (6, 10) → &gcd;(6, 10) = 2 (Invalid)

⇒ (7, 9) → &gcd;(7, 9) = 1 (Valid)

⇒ (8, 8) → &gcd;(8, 8) = 8 (Invalid)

⇒ Total valid pairs = 4.

∴ The correct answer is 4 pairs.

Additional Information

HCF and Sum Property

The HCF of two numbers always divides their sum, their difference, and their LCM. If H is the HCF, then Sum = H × (a + b).

Co-prime Numbers

Two numbers are co-prime if their highest common factor is 1. When finding pairs using HCF, the remaining factors must be co-prime to maintain the given HCF.

Euler's Totient Function φ(n)

For finding the number of co-prime integers less than n, we use φ(n). The number of pairs (a, b) such that a + b = n and &gcd;(a, b) = 1 is φ(n) ÷ 2.

Shortcut Trick

SP₁ = SP₂ ⇒ 85% of CP₁ = 75% of CP₂

Ratio CP₁ : CP₂ = 75 : 85 = 15 : 17

Total units = 15 + 17 = 32 units = ₹640 ⇒ 1 unit = ₹20

CP₁ = 15 × 20 = ₹300

SP = 85% of 300 = ₹255

∴ The correct answer is ₹255.

Alternate Method

Given:

Sum of Cost Prices (CP₁ + CP₂) = ₹640

Discount on CP₁ = 15%

Discount on CP₂ = 25%

Selling Price SP₁ = SP₂

Calculations:

⇒ Let the cost price of the first article be CP₁ and the second be CP₂.

⇒ SP₁ = CP₁ − 15% of CP₁ = 85% of CP₁

⇒ SP₂ = CP₂ − 25% of CP₂ = 75% of CP₂

⇒ Since SP₁ = SP₂:

⇒ 0.85 × CP₁ = 0.75 × CP₂

⇒ CP₁ / CP₂ = 0.75 / 0.85 = 75 / 85 = 15 / 17

⇒ Let CP₁ = 15k and CP₂ = 17k

⇒ 15k + 17k = 640

⇒ 32k = 640

⇒ k = 640 ÷ 32 = 20

⇒ CP₁ = 15 × 20 = ₹300

⇒ CP₂ = 17 × 20 = ₹340

⇒ SP = 85% of 300 = (85/100) × 300 = 85 × 3 = ₹255

⇒ Verification: SP = 75% of 340 = (75/100) × 340 = 0.75 × 340 = ₹255

∴ The correct answer is ₹255.

Additional Information

Selling Price and Discount

Generally, discount is offered on Marked Price (MP), so SP = MP × (100 − D%)/100. If specifically stated on CP, then SP = CP × (100 − D%)/100.

Ratio of Cost Prices

When selling prices are equal and losses/discounts are given on CP, the ratio of cost prices is inversely proportional to the remaining value: CP₁ : CP₂ = (100 − L₂%) : (100 − L₁%).

Successive Discounts

The single equivalent discount for two successive discounts a% and b% is given by (a + b − ab/100)%.

Shortcut Trick

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

Given k = 3 ⇒ x + 1/x = 3² − 2 = 7.

If x + 1/x = m, then x³ + 1/x³ = m³ − 3m.

Using m = 7 ⇒ x³ + 1/x³ = 7³ − 3(7) = 343 − 21 = 322.

∴ The correct answer is 322.

Alternate Method

Given:

√x + 1/√x = 3

Formula Used:

(a + b)² = a² + b² + 2ab

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculations:

Squaring the given expression on both sides:

⇒ (√x + 1/√x)² = 3²

⇒ (√x)² + (1/√x)² + 2 × (√x) × (1/√x) = 9

⇒ x + 1/x + 2 = 9

⇒ x + 1/x = 7

Now, cubing both sides of the derived equation:

⇒ (x + 1/x)³ = 7³

⇒ x³ + (1/x)³ + 3 × (x) × (1/x) × (x + 1/x) = 343

⇒ x³ + 1/x³ + 3 × (x + 1/x) = 343

⇒ x³ + 1/x³ + 3(7) = 343

⇒ x³ + 1/x³ + 21 = 343

⇒ x³ + 1/x³ = 343 − 21

⇒ x³ + 1/x³ = 322

∴ The correct answer is 322.

Additional Information

Algebraic Identities for Squares

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

Algebraic Identities for Cubes

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

Higher Powers

To find x⁴ + 1/x⁴, square the expression for x² + 1/x² and subtract 2. For x⁵, use (x² + 1/x²)(x³ + 1/x³) − (x + 1/x).

Shortcut Trick Test the given options in the equation: cos2 θ / (cot2 θ − cos2 θ) = 3. Put θ = 60°, then cos 60° = 1/2 and cot 60° = 1/√3. LHS = (1/2)2 / ((1/√3)2 − (1/2)2) LHS = (1/4) / (1/3 − 1/4) = (1/4) / (1/12) = 12/4 = 3. Since LHS = RHS = 3, the value of θ is 60°. ∴ The correct answer is 60°.Alternate Method Given: cos2 θ / (cot2 θ − cos2 θ) = 3 Where 0° < θ < 90°. Formula Used: cot θ = cos θ / sin θ sin 60° = √3 / 2

Calculations: ⇒ cos2 θ / (cot2 θ − cos2 θ) = 3 ⇒ cos2 θ = 3 × (cot2 θ − cos2 θ) ⇒ cos2 θ = 3cot2 θ − 3cos2 θ ⇒ 4cos2 θ = 3 × (cos2 θ / sin2 θ) ⇒ 4 = 3 / sin2 θ (Since θ ≠ 90°, cos θ ≠ 0) ⇒ sin2 θ = 3/4 ⇒ sin θ = √3 / 2 ⇒ θ = 60° ∴ The correct answer is 60°.Additional Information Trigonometric Identities Fundamental identities include sin2 θ + cos2 θ = 1, 1 + tan2 θ = sec2 θ, and 1 + cot2 θ = cosec2 θ. Standard Trigonometric Values For θ = 60°: sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3, and cot 60° = 1/√3. Ratio Relations tan θ = sin θ / cos θ and cot θ = cos θ / sin θ = 1 / tan θ.

Shortcut Trick

Use the direct formula for finding the time of the second person when combined time and one person's time are given:

Days taken by B = (X × Y) ÷ (Y − X)

Where X = combined days of A and B = 12, and Y = days of A alone = 18.

Days taken by B = (18 × 12) ÷ (18 − 12) = (18 × 12) ÷ 6 = 3 × 12 = 36 days.

∴ The correct answer is 36 days.

Alternate Method

Given:

Time taken by (A + B) = 12 days

Time taken by A alone = 18 days

Calculations:

⇒ Total work = LCM of 12 and 18 = 36 units.

⇒ Efficiency of (A + B) = Total Work ÷ Days = 36 ÷ 12 = 3 units/day.

⇒ Efficiency of A = Total Work ÷ Days = 36 ÷ 18 = 2 units/day.

⇒ Efficiency of B = Efficiency of (A + B) − Efficiency of A = 3 − 2 = 1 unit/day.

⇒ Time taken by B alone = Total Work ÷ Efficiency of B

⇒ Time taken by B alone = 36 ÷ 1 = 36 days.

∴ The correct answer is 36 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a work. If the ratio of efficiency is x:y, the ratio of time taken is y:x.

Work and Wages

Wages are always distributed in the ratio of the work done or in the ratio of efficiencies when the time period of work for all workers is the same.

Pipes and Cisterns

The same LCM logic applies where filling pipes have positive efficiency and leakage or outlet pipes have negative efficiency.

Shortcut Trick

Perimeter of square base = 4 × 15 = 60 cm

Lateral Surface Area (LSA) = 60 × 8 = 480 cm²

Area of two square bases = 2 × (15 × 15) = 450 cm²

Total Surface Area (TSA) = 480 + 450 = 930 cm²

∴ The correct answer is 930 cm².

Alternate Method

Given:

Side of the square base (s) = 15 cm

Height of the prism (h) = 8 cm

Formula Used:

Total Surface Area (TSA) of Prism = Lateral Surface Area + 2 × Base Area

Lateral Surface Area (LSA) = Perimeter of Base × Height

Calculations:

⇒ Area of the square base = s² = 15² = 225 cm²

⇒ Perimeter of the square base = 4 × s = 4 × 15 = 60 cm

⇒ Lateral Surface Area = Perimeter × height = 60 × 8 = 480 cm²

⇒ Total Surface Area = LSA + 2 × Base Area

⇒ Total Surface Area = 480 + 2 × 225

⇒ Total Surface Area = 480 + 450 = 930 cm²

∴ The correct answer is 930 cm².

Additional Information

Volume of a Prism

The volume is calculated by multiplying the area of the base by the vertical height: Volume = Base Area × Height.

Right Prism Properties

In a right prism, the joining edges and faces are perpendicular to the base faces. This means the height is equal to the length of the lateral edges.

Diagonal of Square Base Prism

The longest diagonal of a prism with a square base can be found using the formula √(2s² + h²).

Shortcut Trick

Surface Area of Sphere = 4πr² = 1386 cm².

r² = (1386 × 7) ÷ (4 × 22) ⇒ r² = (63 × 7) ÷ 4 = 441 ÷ 4 ⇒ r = 21 ÷ 2 = 10.5 cm.

Volume must be a multiple of 11 since π = 22/7 is in the formula.

Check options for 11 divisibility: For 4851, (4+5) − (8+1) = 9 − 9 = 0. Hence, 4851 is divisible by 11.

∴ The correct answer is 4851 cm³.

Alternate Method

Given:

Surface area of sphere = 1386 cm²

π = 22/7

Formula Used:

Surface area of sphere = 4πr²

Volume of sphere = (4/3)πr³

Calculations:

⇒ 4 × (22/7) × r² = 1386

⇒ r² = (1386 × 7) ÷ (4 × 22)

⇒ r² = (63 × 7) ÷ 4 = 441 ÷ 4

⇒ r = √(441 ÷ 4) = 21 ÷ 2 = 10.5 cm

⇒ Volume = (4/3) × (22/7) × (21/2)³

⇒ Volume = (4/3) × (22/7) × (21/2) × (21/2) × (21/2)

⇒ Volume = 11 × 21 × 21

⇒ Volume = 11 × 441 = 4851 cm³

∴ The correct answer is 4851 cm³.

Additional Information

Hemisphere Formulas

Curved Surface Area = 2πr² | Total Surface Area = 3πr² | Volume = (2/3)πr³.

Sphere vs. Hemisphere

A sphere does not have a base, so its CSA and TSA are identical (4πr²), whereas a hemisphere has a flat circular base.

Ratio of Volume to Surface Area

For a sphere, the ratio of Volume to Surface Area is [(4/3)πr³] ÷ [4πr²] = r ÷ 3.

Given:

Numerator = Denominator - 6

New Numerator = Numerator - 1

New Denominator = Denominator + 5

New Denominator = 4 × New Numerator

Formula used:

Fraction = $\frac{Numerator}{Denominator}$

Calculation:

Let Denominator = x

Numerator = x - 6

⇒ x + 5 = 4 × (x - 6 - 1)

⇒ x + 5 = 4 × (x - 7)

⇒ x + 5 = 4x - 28

⇒ 33 = 3x

⇒ x = 11

⇒ Numerator = 11 - 6 = 5

⇒ Fraction = $\frac{5}{11}$

∴ The correct answer is option (3).

Given:

Total number of students in school E = 2800

Difference between the number of boys and girls in school E = 700

Formula used:

Number of boys in a school = (Total number of students in the school + Difference between the number of boys and girls in school E)/2

Number of girls in a school = Total number of students in the school - Number of boys in the school.

Calculations:

Number of boys in a school = (Total number of students in the school + Difference between the number of boys and girls in school E)/2

⇒ (2800 + 700)/2

⇒ 1750

Number of girls in a school = Total number of students in the school - Number of boys in the school.

⇒ 2800 - 1750

⇒ 1050

The ratio of number of boys to the number of girls in school E = 1750 ∶ 1050

⇒ 5 ∶ 3

∴ The ratio of number of boys to the number of girls in school E is 5 ∶ 3.

Shortcut Trick

Given b/a = 0.7 = 7/10. Let a = 10 and b = 7.

Value of (a − b) / (a + b) = (10 − 7) / (10 + 7) = 3/17.

Expression = 3/17 + 11/34 = 6/34 + 11/34.

Expression = 17/34 = 0.5.

∴ The correct answer is 0.5.

Alternate Method

Given: b/a = 0.7

Formula: Ratio substitution and addition of fractions.

⇒ b/a = 0.7 = 7/10

⇒ We can write (a − b) / (a + b) by dividing both numerator and denominator by a:

⇒ (1 − b/a) / (1 + b/a)

⇒ (1 − 0.7) / (1 + 0.7) = 0.3 / 1.7

⇒ 0.3 / 1.7 = 3/17

⇒ Total Value = 3/17 + 11/34

⇒ Total Value = (3 × 2) / 34 + 11/34

⇒ Total Value = 6/34 + 11/34 = 17/34

⇒ Total Value = 1/2 = 0.5

∴ The correct answer is 0.5.

Additional Information

Componendo and Dividendo Rule

If x/y = p/q, then (x + y)/(x − y) = (p + q)/(p − q). This is useful for solving symmetric fractional expressions quickly.

Fraction Addition

To add fractions like a/b + c/d, find the Least Common Multiple (LCM) of the denominators to create a common base.

Decimal to Ratio Conversion

Converting decimals to simplest fractions (e.g., 0.7 = 7/10) allows for easier substitution in algebraic expressions involving ratios.

Shortcut Trick

Divide the numerator and denominator by x to simplify the expression.

Numerator becomes x³ − 27/x³ and Denominator becomes x − 3 − 3/x.

Given x − 3/x = 6, the denominator is 6 − 3 = 3.

Using the cube shortcut: x³ − 27/x³ = 6³ + 3 × 3 × 6 = 216 + 54 = 270.

The final value is 270 ÷ 3 = 90.

∴ The correct answer is 90.

Alternate Method

Given: x − 3/x = 6, x ≠ 0

Formula Used: (a − b)³ = a³ − b³ − 3ab(a − b)

Expression = (x⁴ − 27/x²) / (x² − 3x − 3)

⇒ Dividing numerator and denominator by x:

⇒ [ (x⁴ − 27/x²) ÷ x ] / [ (x² − 3x − 3) ÷ x ]

⇒ (x³ − 27/x³) / (x − 3 − 3/x)

⇒ Denominator = (x − 3/x) − 3 = 6 − 3 = 3

⇒ Cubing the given equation: (x − 3/x)³ = 6³

⇒ x³ − (3/x)³ − 3 × x × (3/x) × (x − 3/x) = 216

⇒ x³ − 27/x³ − 9 × (6) = 216

⇒ x³ − 27/x³ = 216 + 54 = 270

⇒ Expression = 270 ÷ 3 = 90

∴ The correct answer is 90.

Additional Information

Algebraic Cubes with Reciprocals

If x − k/x = n, then x³ − k³/x³ = n³ + 3kn. This is derived from (a−b)³ = a³−b³−3ab(a−b).

Algebraic Squares with Reciprocals

If x − k/x = n, then x² + k²/x² = n² + 2k. Note that the sign changes to positive for the square term.

Expression Simplification Trick

In complex algebraic fractions involving powers of x, dividing both numerator and denominator by xⁿ often reduces the problem to basic reciprocal identities.

Shortcut Trick Given: Remainder = 40 Divisor = 3 × Remainder = 3 × 40 = 120 Quotient = Divisor ÷ 15 = 120 ÷ 15 = 8 Dividend = (Divisor × Quotient) + Remainder Dividend = (120 × 8) + 40 = 960 + 40 = 1000 ∴ The correct answer is 1000. Alternate Method Given: Remainder (R) = 40 Divisor (D) = 3 × R = 15 × Q (where Q is Quotient) Formula Used: Dividend = (Divisor × Quotient) + Remainder

⇒ Divisor (D) = 3 × 40 = 120 ⇒ 15 × Quotient (Q) = 120 ⇒ Q = 120 ÷ 15 = 8 ⇒ Dividend = (120 × 8) + 40 ⇒ Dividend = 960 + 40 ⇒ Dividend = 1000 ∴ The correct answer is 1000. Additional Information Division Algorithm For any integer dividend a and divisor b, there exist unique integers q and r such that a = b × q + r, where 0 ≤ r < |b|. Remainder Property The remainder is always smaller than the divisor. If the remainder is 0, the dividend is exactly divisible by the divisor. Finding Divisor or Quotient Divisor = (Dividend − Remainder) ÷ Quotient | Quotient = (Dividend − Remainder) ÷ Divisor.

GIVEN:

Sum of the boys and girls(B + G) of School A = 1800 and Difference between Boys and Girls (B - G )of School A = 350

Sum of the boys and girls(B + G) of School C = 2000 and Difference between Boys and Girls (B - G )of School C = 500

CALCULATION:

For School A,

B + G = 1800      ----(I)

B - G = 350      ----(II)

Adding I and II we get

So 2G = 1450      ----(III)

⇒ Girls in school A = 725

For School C,

B + G = 2000      ----(IV)

B - G = 500      ----(V)

Adding I and II we get

So 2G = 1500

⇒ G = 750      ----(VI)

∴ Difference = 750 - 725 = 25

Shortcut Trick

Assume Total Work = LCM(18, 36, 54) = 108 units.

Efficiencies: A = 6, B = 3, C = 2. Combined efficiency = 11.

Shortcut for 'leaving before completion': Add the work those people would have done to the total work.

New Total Work = 108 + (B's 5 days work) + (C's 10 days work) = 108 + (5 × 3) + (10 × 2) = 143 units.

Total time taken = New Total Work ÷ Total Efficiency = 143 ÷ 11 = 13 days.

∴ The correct answer is 13 days.

Alternate Method

Given:

Time taken by A = 18 days

Time taken by B = 36 days

Time taken by C = 54 days

Let the total time to complete the work be 'x' days.

⇒ A worked for all x days.

⇒ B left 5 days before, so B worked for (x − 5) days.

⇒ C left 10 days before, so C worked for (x − 10) days.

⇒ Total work = (Eff_A × x) + (Eff_B × (x − 5)) + (Eff_C × (x − 10))

⇒ 108 = (6 × x) + (3 × (x − 5)) + (2 × (x − 10))

⇒ 108 = 6x + 3x − 15 + 2x − 20

⇒ 108 = 11x − 35

⇒ 11x = 108 + 35

⇒ 11x = 143

⇒ x = 143 ÷ 11 = 13 days

∴ The correct answer is 13 days.

Additional Information

Work and Time Relationship

Total Work = Efficiency × Time. For a constant work, Efficiency is inversely proportional to Time.

Leaving After Starting

If a person leaves after 'n' days of starting, subtract their 'n' days of work from the total work and then solve for remaining workers.

Work and Wages

Wages are always distributed in the ratio of the total units of work actually completed by each individual.

GIVEN: 

Total number of students admitted to a university in various fields = 5000

CALCULATION:

 

⇒ Total number of students admitted to a university in various fields = 5000

⇒ 100% = 5000

⇒ 1% = 50

⇒  the average number of boys in CS, ECE and EEE fields = (330 + 576 + 612)/3

⇒ 1518/3

⇒ 506

∴  the average number of boys in CS, ECE and EEE fields is 506

Shortcut Trick

In trigonometry, if sin A = cos B, then the sum of the angles is 90° (A + B = 90°).

Here, (x + y) + (x − y) = 90°.

2x = 90° ⇒ x = 45°.

cos² 45° = (1 / √2)² = 1 / 2.

∴ The correct answer is 1 / 2.

Alternate Method

Given: sin (x + y) = cos (x − y)

Formula Used: cos θ = sin (90° − θ)

⇒ sin (x + y) = sin [90° − (x − y)]

⇒ x + y = 90° − x + y

⇒ x + x = 90° + y − y

⇒ 2x = 90°

⇒ x = 45°

⇒ cos x = cos 45° = 1 / √2

⇒ cos² x = (1 / √2)²

⇒ cos² x = 1 / 2

∴ The correct answer is 1 / 2.

Additional Information

Complementary Ratios

Trigonometric ratios of complementary angles satisfy: sin θ = cos (90° − θ), tan θ = cot (90° − θ), and sec θ = cosec (90° − θ).

Standard Values

Standard values for cos θ are: cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, and cos 90° = 0.

Pythagorean Identity

For any angle θ, the identity sin² θ + cos² θ = 1 always holds true.

Given:

$5 - \frac{8 + 2\sqrt{15}}{4} - \frac{1}{8 + 2\sqrt{15}}$ = ?

Formula used:

Rationalization: (a + b)(a - b) = a² - b²

Calculation:

$5 - \frac{8 + 2\sqrt{15}}{4} - \frac{1}{8 + 2\sqrt{15}}$

⇒ $\frac{1}{8 + 2\sqrt{15}} = \frac{1 \times (8 - 2\sqrt{15})}{(8 + 2\sqrt{15}) \times (8 - 2\sqrt{15})}$

⇒ $\frac{1}{8 + 2\sqrt{15}} = \frac{8 - 2\sqrt{15}}{8^2 - (2\sqrt{15})^2}$

⇒ $\frac{1}{8 + 2\sqrt{15}} = \frac{8 - 2\sqrt{15}}{64 - 60} = \frac{8 - 2\sqrt{15}}{4}$

Substitute back into the expression:

⇒ $5 - \frac{8 + 2\sqrt{15}}{4} - \frac{8 - 2\sqrt{15}}{4}$

⇒ $5 - (\frac{8 + 2\sqrt{15} + 8 - 2\sqrt{15}}{4})$

⇒ $5 - \frac{16}{4}$

⇒ 5 - 4

⇒ 1

∴ The correct answer is option (1).

Shortcut Trick

Convert all angles to 39° using 51° + 39° = 90°.

1/cosec² 51° = sin² 51° = cos² 39° and tan² 51° = cot² 39°.

The term 1/(sin² 51° sec² 39°) = 1/(cos² 39° sec² 39°) = 1/1 = 1.

Expression = (cos² 39° + sin² 39°) + cot² 39° − 1.

Expression = 1 + cot² 39° − 1 = cot² 39° = cosec² 39° − 1.

Given cosec 39° = x, the value is x² − 1.

∴ The correct answer is x² − 1.

Alternate Method

Given: cosec 39° = x

Formula Used: sin(90°−θ) = cosθ, tan(90°−θ) = cotθ, sin²θ + cos²θ = 1, cot²θ = cosec²θ−1

Let the expression be E:

⇒ E = 1/cosec² 51° + sin² 39° + tan² 51° − 1/(sin² 51° sec² 39°)

⇒ E = sin² 51° + sin² 39° + cot² 39° − 1/(cos² 39° sec² 39°) [Since 51° = 90° − 39°]

⇒ E = (cos² 39° + sin² 39°) + cot² 39° − 1/1 [Since cos θ × sec θ = 1]

⇒ E = 1 + cot² 39° − 1

⇒ E = cot² 39°

⇒ E = cosec² 39° − 1 [Using identity 1 + cot²θ = cosec²θ]

⇒ E = x² − 1

∴ The correct answer is x² − 1.

Additional Information

Complementary Angle Identities

For any angle θ, sin(90°−θ) = cosθ, tan(90°−θ) = cotθ, and sec(90°−θ) = cosecθ.

Pythagorean Identities

Standard identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Relations

Primary reciprocal pairs are sinθ = 1/cosecθ, cosθ = 1/secθ, and tanθ = 1/cotθ.

Shortcut Trick

If sec θ + tan θ = x, then sec θ = (x² + 1) / 2x.

Given the value of sec θ + tan θ = 3, we take x = 3.

Substituting x = 3 into the shortcut formula:

sec θ = (3² + 1) / (2 × 3) = (9 + 1) / 6 = 10 / 6

sec θ = 5 / 3

∴ The correct answer is 5 / 3.

Alternate Method

Given: sec θ + tan θ = 3

Formula Used: sec² θ − tan² θ = 1

⇒ We know that sec² θ − tan² θ = 1

⇒ (sec θ + tan θ)(sec θ − tan θ) = 1

⇒ 3 × (sec θ − tan θ) = 1

⇒ sec θ − tan θ = 1 / 3 ... (Eqn 1)

⇒ Given: sec θ + tan θ = 3 ... (Eqn 2)

⇒ Adding (Eqn 1) and (Eqn 2):

⇒ (sec θ − tan θ) + (sec θ + tan θ) = (1 / 3) + 3

⇒ 2 sec θ = (1 + 9) / 3

⇒ 2 sec θ = 10 / 3

⇒ sec θ = 10 / (3 × 2)

⇒ sec θ = 10 / 6 = 5 / 3

∴ The correct answer is 5 / 3.

Additional Information

Trigonometric Identity

The fundamental identity connecting sec and tan is 1 + tan² θ = sec² θ, which implies sec² θ − tan² θ = 1.

Conjugate Property

If (sec θ + tan θ) = x, then (sec θ − tan θ) = 1/x because their product must equal 1.

Component Formulas

For a given sum x = sec θ + tan θ, sec θ = (x² + 1) / 2x and tan θ = (x² − 1) / 2x.

Pythagorean Triplets

In this case, the ratios correspond to a 3-4-5 triangle where Base = 3, Perpendicular = 4, and Hypotenuse = 5.

Given:

Number of students in Economics = 12% of total students

Numbers of boys in Economics = 56% of total students in economics

Calculations:

⇒ Total number of student Economics = 5000 × 12% = 600

⇒ Number of boys students in Economics = 56% × 600 = 336

⇒ Boys in economics / total number of student in economics = 336/600 

⇒ Boys in economics/total number of student = 14 : 25

∴ The ratio of Boys in economics and the total number of student is 14 : 25

Shortcut Trick

Walking time = (Total distance ÷ Speed) × 60 = (6 ÷ 8) × 60 = 45 minutes.

Number of rest stops = (Total distance ÷ Interval) − 1 = (6 ÷ 1) − 1 = 5 stops.

Total rest time = 5 × 4 = 20 minutes.

Total time = Walking time + Rest time = 45 + 20 = 65 minutes.

∴ The correct answer is 65 minutes.

Alternate Method

Given:

Speed = 8 km/h

Total Distance = 6 km

Rest Interval = After every 1 km

Rest Duration = 4 minutes

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Walking time in hours = Distance ÷ Speed = 6 ÷ 8 hours

⇒ Walking time in minutes = (6 ÷ 8) × 60 = 45 minutes

⇒ Number of times the man rests = (6 ÷ 1) − 1 = 5 times

⇒ (Note: The rest after the last kilometer is not counted as he reaches the destination)

⇒ Total rest time = 5 × 4 = 20 minutes

⇒ Total time taken = Walking time + Rest time

⇒ Total time taken = 45 + 20 = 65 minutes

∴ The correct answer is 65 minutes.

Additional Information

Rest Stops in Distance Problems

If a person rests after every x km for a total distance D, the number of rests is (D/x) − 1 if D is perfectly divisible by x. If not divisible, use the floor value of (D/x).

Speed Conversion

To convert Speed from km/h to m/s, multiply by 5/18. To convert from m/s to km/h, multiply by 18/5.

Average Speed with Rests

Average Speed = Total Distance ÷ (Walking Time + Stoppage Time). Stoppages always reduce the average speed of the journey.

Shortcut Trick

Given: cos θ = 5/13. In a right-angled triangle, (Base, Hypotenuse) = (5, 13).

Using Pythagorean triplet (5, 12, 13), Perpendicular = 12.

tan θ = Perpendicular / Base = 12/5 and sec θ = Hypotenuse / Base = 13/5.

Required value = (12/5)² + (13/5)² = 144/25 + 169/25 = 313/25.

∴ The correct answer is 313/25.

Alternate Method

Given:

cos θ = 5/13

Formula Used:

1. sec θ = 1/cos θ
2. tan² θ = sec² θ − 1
3. In a right triangle, P² + B² = H²

Calculations:

⇒ cos θ = Base / Hypotenuse = 5 / 13

⇒ Base (B) = 5, Hypotenuse (H) = 13

⇒ Perpendicular (P) = √(H² − B²) = √(13² − 5²)

⇒ P = √(169 − 25) = √144 = 12

⇒ tan θ = P / B = 12 / 5

⇒ sec θ = H / B = 13 / 5

⇒ Expression = tan² θ + sec² θ

⇒ Expression = (12/5)² + (13/5)²

⇒ Expression = 144/25 + 169/25

⇒ Expression = (144 + 169) / 25

⇒ Expression = 313/25

∴ The correct answer is 313/25.

Additional Information

Trigonometric Identities

Standard identities include: sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = cosec² θ.

Pythagorean Triplets

Sets of three integers (a, b, c) that satisfy a² + b² = c², such as (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25).

Reciprocal Ratios

Secant (sec θ) is the reciprocal of Cosine (cos θ), and Tangent (tan θ) is the ratio of Sine to Cosine (sin θ / cos θ).

Shortcut Trick

Ratio of sides AB : AC : BC = 1 : 1 : √2 (since AB = AC and BC = √2AB).

Sum of ratios for perimeter = 1 + 1 + √2 = 2 + √2.

Comparing with given perimeter 8(2 + √2), the multiplier is 8.

Sides are AB = 8 cm and AC = 8 cm.

Since 8² + 8² = (8√2)², it is a right-angled triangle at A.

Area = (1/2) × 8 × 8 = 32 cm².

∴ The correct answer is 32 cm².

Alternate Method

Given:

AB = AC

BC = √2 × AB

Perimeter of ΔABC = 8(2 + √2) cm

Formula Used:

Perimeter = Sum of all sides

Area of right triangle = (1/2) × base × height

Calculations:

Let the length of AB be x cm. Then AC = x cm and BC = √2x cm.

⇒ x + x + √2x = 8(2 + √2)

⇒ 2x + √2x = 8(2 + √2)

⇒ x(2 + √2) = 8(2 + √2)

⇒ x = 8 cm

Now, checking for right-angled triangle property:

⇒ AB² + AC² = 8² + 8² = 64 + 64 = 128

⇒ BC² = (8√2)² = 64 × 2 = 128

⇒ Since AB² + AC² = BC², the triangle is right-angled at A.

⇒ Area = (1/2) × AB × AC

⇒ Area = (1/2) × 8 × 8

⇒ Area = 32 cm²

∴ The correct answer is 32 cm².

Additional Information

Right Isosceles Triangle

A triangle where two sides are equal and the angle between them is 90°. The side ratios are always 1 : 1 : √2.

Heron's Formula

For any triangle with sides a, b, and c, Area = √[s(s−a)(s−b)(s−c)], where s is the semi-perimeter (a+b+c)/2.

Pythagoras Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: a² + b² = c².