Search Previous Year Questions

Shortcut Trick

There is a direct ratio relation between Cost Price and Marked Price:

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%)

⇒ CP ÷ 800 = (100 − 12) ÷ (100 + 8)

⇒ CP = 800 × (88 ÷ 108)

⇒ CP = 17600 ÷ 27 ≈ 651.85

∴ The correct answer is ₹652.

Alternate Method

CP = ?

SP = 704

MP = 800

• 8%

− 12%

Finding CP from SP and Profit %

Given:

⇒ Marked Price (MP) = ₹800

⇒ Discount = 12%

⇒ Profit = 8%

Calculations:

⇒ Selling Price (SP) = MP × (100 − Discount%) ÷ 100

⇒ SP = 800 × (100 − 12) ÷ 100

⇒ SP = 800 × (88 ÷ 100) = ₹704

⇒ SP = CP × (100 + Profit%) ÷ 100

⇒ 704 = CP × (100 + 8) ÷ 100

⇒ 704 = CP × (108 ÷ 100)

⇒ CP = (704 × 100) ÷ 108

⇒ CP = 70400 ÷ 108 ≈ 651.85

∴ The correct answer is ₹652.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price. Formula: SP = MP − Discount.

Direct CP & MP Relation

A very useful formula directly connects Cost Price and Marked Price: CP/MP = (100 − D%) ÷ (100 + P%).

Successive Discount

For two successive discounts of a% and b%, the effective single discount is calculated as: a + b − (a × b) ÷ 100 %.

Shortcut Trick

Cost Price (CP) for the showroom = Profit ÷ Profit% = 4500 ÷ 30% = ₹15,000

Net multiplying factor from MP to CP = (1 − 25%) × (1 + 10%) = 0.75 × 1.10 = 0.825

Marked Price (MP) = CP ÷ Net Factor

MP = 15,000 ÷ 0.825 = 2,00,000 ÷ 11 = 18,181.818... ≈ ₹18,181.82

∴ The correct answer is ₹18,181.82.

Alternate Method

Given:

Discount offered by manufacturer = 25%

Handling fee added on discounted price = 10%

Profit earned by showroom = 30%

Profit amount = ₹4500

Formula Used:

Cost Price (CP) = Profit ÷ (Profit% ÷ 100)

Final Cost Price = MP × (1 − Discount%) × (1 + Fee%)

MP−25% Disc.+10% Fee(× 0.825)CPProfit = 30%Profit = ₹4500CP × 30% = 4500SPMP × 0.75 × 1.10 = CP

Calculations:

⇒ Showroom CP × 0.30 = 4500

⇒ Showroom CP = 4500 ÷ 0.30 = ₹15,000

Let the Marked Price be MP.

⇒ Discounted Price = MP × 0.75

⇒ Price after handling fee (Showroom CP) = MP × 0.75 × 1.10

⇒ 0.825 MP = 15,000

⇒ MP = 15,000 ÷ 0.825

⇒ MP = 1,50,00,000 ÷ 825 = 2,00,000 ÷ 11

⇒ MP = 18,181.818... ≈ ₹18,181.82

∴ The correct answer is ₹18,181.82.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP). Selling Price (SP) = MP − Discount = MP × (100 − D%) ÷ 100.

Successive Percentage Changes

When multiple changes occur on a value, the final value is computed as Initial Value × (1 ± R₁/100) × (1 ± R₂/100). The overall percentage change is not simply the sum of individual percentages.

Profit & Loss Basics

Profit or Loss is always calculated on the Cost Price (CP) unless stated otherwise. Profit = SP − CP, and Profit% = (Profit ÷ CP) × 100.

Shortcut Trick

Use the net deviation method from the average mixture price of ₹55/kg.

Deviation of 1^st type = (45 − 55) × 1 part = −10

Deviation of 2^nd type = (60 − 55) × 2 parts = +10

Net deviation of the first two types = −10 + 10 = 0

Since the net deviation is 0, the 3^rd type must have 0 deviation to maintain the average, meaning its price is exactly ₹55.

∴ The correct answer is ₹55.00.

Alternate Method

Given:

⇒ Price of 1^st type of sugar = ₹45/kg

⇒ Price of 2^nd type of sugar = ₹60/kg

⇒ Ratio of mixtures = 1 : 2 : 3

⇒ Average price of the final mixture = ₹55/kg

⇒ Let the price of the 3^rd type of sugar be ₹x/kg.

Formula Used:

⇒ Total Value = Sum of (Quantity × Price)

Calculation:

1 part ₹45

2 parts ₹60

3 parts ₹x

=

6 parts (Mixture) ₹55

Total Cost: (1 × 45) + (2 × 60) + (3 × x) = (6 × 55) 165 + 3x = 330 ⇒ x = 55

⇒ Total quantity of the mixture = 1 + 2 + 3 = 6 parts

⇒ Total cost of the mixture = 6 × 55 = ₹330

⇒ Total cost of individual parts = (1 × 45) + (2 × 60) + (3 × x)

⇒ 45 + 120 + 3x = 330

⇒ 165 + 3x = 330

⇒ 3x = 330 − 165

⇒ 3x = 165

⇒ x = 165 ÷ 3 = 55

∴ The correct answer is ₹55.00.

Additional Information

Weighted Average Formula

Used to find the combined average when multiple groups are mixed: Average = (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n).

Rule of Alligation

Used when two ingredients are mixed: Quantity of Cheaper / Quantity of Dearer = (Dearer Price − Mean Price) / (Mean Price − Cheaper Price).

Repeated Replacement

When 'y' units of a pure liquid are replaced by water 'n' times from a total 'x' units: Final amount of pure liquid = x × (1 − y/x)ⁿ.

Shortcut Trick

If the entire sum of ₹18,000 was invested at the lower rate of 10% p.a., the total interest generated would be ₹1,800.

The extra interest earned = ₹2,160 − ₹1,800 = ₹360.

This extra ₹360 is solely due to the additional 5% interest rate on Scheme X (since 15% − 10% = 5%).

Therefore, 5% of Scheme X investment = ₹360.

100% of Scheme X investment = ₹360 × (100 ÷ 5) = ₹7,200.

∴ The correct answer is ₹7,200.

Alternate Method

Given:

Total Principal = ₹18,000

Rate for Scheme X = 15% p.a. | Rate for Scheme Y = 10% p.a.

Total Simple Interest for 1 year = ₹2,160

Method Used:

We can find the overall combined interest rate and apply the Alligation method to determine the ratio of investments.

⇒ Overall interest rate = (Total Interest ÷ Total Principal) × 100

⇒ Overall rate = (2160 ÷ 18000) × 100 = 12%

Scheme X (15%)

Scheme Y (10%)

12%

12 − 10 = 2

15 − 12 = 3 Ratio (X : Y) = 2 : 3

Calculations:

⇒ Total ratio parts = 2 + 3 = 5 parts

⇒ 5 parts = ₹18,000

⇒ 1 part = 18000 ÷ 5 = ₹3,600

⇒ Investment in Scheme X (2 parts) = 2 × 3600 = ₹7,200

∴ The correct answer is ₹7,200.

Additional Information

Alligation Method

When an amount is divided into two components and invested at different rates, the ratio of the invested parts can be quickly found using the diagonal cross-difference between the individual component rates and the overall average rate.

Shift Method (Shortcut Trick)

This assumes the entire sum is invested at the lower percentage. The remaining unassigned extra interest corresponds directly to the percentage difference between the two rates applied only on the higher-rate amount.

Simple Interest (SI)

SI = (P × R × T) ÷ 100. Under simple interest, the principal never changes, meaning the interest accrued remains strictly constant for every equivalent time period.

Shortcut Trick

For a cone and a hemisphere with equal volumes and the same base radius (r):

⇒ (1/3) × π × r² × h = (2/3) × π × r³

⇒ h = 2r

⇒ h : r = 2 : 1

∴ The correct answer is 2 : 1.

Alternate Method

Given:

Base radius of the cone = Base radius of the hemisphere = r

Let the vertical height of the cone be h.

Volume of the cone = Volume of the hemisphere

h r Cone

r r Hemisphere

Formula Used:

⇒ Volume of a cone = (1/3) × π × r² × h

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

Calculations:

According to the given condition in the question:

⇒ (1/3) × π × r² × h = (2/3) × π × r³

Canceling out the common terms ((1/3) × π × r²) from both sides:

⇒ h = 2 × r

⇒ h / r = 2 / 1

⇒ h : r = 2 : 1

∴ The correct answer is 2 : 1.

Additional Information

Volume of a Cylinder

The volume of a cylinder with base radius r and height h is three times the volume of a cone with the identical base and height. Formula: V = π × r² × h

Total Surface Area of a Hemisphere

The total surface area encompasses both the curved surface area and the flat circular base. Formula: TSA = 3 × π × r²

Curved Surface Area of a Cone

The curved surface area relies on its slant height (l), which can be found using the Pythagorean theorem where l = √(r² + h²). Formula: CSA = π × r × l

Shortcut Trick

Ratio of areas = (Ratio of circumferences)²

Area ratio = 3² : 5² = 9 : 25

9 parts = 150 m²

25 parts = (150 ÷ 9) × 25 = 416.67 m²

Approximate area ≈ 417 m²

∴ The correct answer is 417 m².

Alternate Method

Given:

Ratio of circumferences of two circular ponds = 3 : 5

Area of the smaller pond = 150 m²

Formula Used:

Circumference of a circle = 2 × π × r

Area of a circle = π × r²

3x Smaller Pond Area = 150 m²

5x Larger Pond Area = ?

Calculations:

⇒ Ratio of circumferences = (2 × π × r₁) : (2 × π × r₂) = 3 : 5

⇒ Ratio of radii (r₁ : r₂) = 3 : 5

⇒ Ratio of their areas = (π × r₁²) : (π × r₂²) = r₁² : r₂²

⇒ Ratio of areas = 3² : 5² = 9 : 25

⇒ Let the area of the larger pond be A.

⇒ 9 / 25 = 150 / A

⇒ 9 × A = 150 × 25

⇒ 9 × A = 3750

⇒ A = 3750 ÷ 9

⇒ A = 416.66... m²

⇒ Approximating to the nearest whole number, A ≈ 417 m²

∴ The correct answer is 417 m².

Additional Information

Scaling Principle for 2D Shapes

For any two similar 2D figures, the ratio of their areas is strictly equal to the square of the ratio of their corresponding linear dimensions (e.g., radius, diameter, circumference, perimeter).

Ratio of Circumference vs Radius

Since the circumference of a circle is directly proportional to its radius (C = 2πr), the ratio of the circumferences of any two circles is exactly the same as the ratio of their radii or diameters.

Ratio of Volumes for 3D Shapes

If the objects were 3D spheres instead of 2D circles, the ratio of their volumes would be equal to the cube of the ratio of their linear dimensions (Volume Ratio = a³ : b³).

Shortcut Trick

Volume of a regular hexagonal prism = (3√3 ÷ 2) × a² × h

Given base side (a) = 6 cm and height (h) = 10 cm

Volume = (3√3 ÷ 2) × 6² × 10

Volume = (3√3 ÷ 2) × 36 × 10 = 54√3 × 10 = 540√3 cm³

∴ The correct answer is 540√3 cm³.

Alternate Method

Given:

Side of the regular hexagonal base (a) = 6 cm

Height of the prism (h) = 10 cm

Formula Used:

Area of a regular hexagon = 6 × (√3 ÷ 4) × a² = (3√3 ÷ 2) × a²

Volume of a prism = Base Area × Height

6 cm

10 cm

Regular Hexagonal Prism

Calculations:

⇒ Base Area = (3√3 ÷ 2) × (6)²

⇒ Base Area = (3√3 ÷ 2) × 36

⇒ Base Area = 54√3 cm²

⇒ Volume of the prism = 54√3 × 10

⇒ Volume = 540√3 cm³

∴ The correct answer is 540√3 cm³.

Additional Information

Volume of a Prism

The volume of any right prism is calculated by multiplying the area of its base by its height. Formula: Volume = Base Area × Height.

Area of a Regular Hexagon

A regular hexagon consists of six equilateral triangles. The area is calculated as Area = 6 × (√3 ÷ 4) × a² = (3√3 ÷ 2) × a², where 'a' is the side length.

Total Surface Area of Hexagonal Prism

The total surface area includes the lateral surface area and the areas of the two bases. Formula: TSA = (6 × a × h) + 2 × [(3√3 ÷ 2) × a²].

Shortcut Trick

Since the Circumference (C) is directly proportional to the diameter (D) as C = π × D, the circumferences will also be in Arithmetic Progression (AP).

The middle term of an AP is the exact average of the first and last terms.

C_middle = (C_smallest + C_largest) ÷ 2

C_middle = (6π + 14π) ÷ 2 = 10π cm

∴ The correct answer is 10π cm.

Alternate Method

Given:

Circumference of the smallest circle (C₁) = 6π cm

Circumference of the largest circle (C₃) = 14π cm

The diameters D₁, D₂, and D₃ are in Arithmetic Progression (AP).

Outer C = 14π cm Middle C = ? Inner C = 6π cm

Diameters in AP

Formula Used:

Circumference of a circle, C = π × D

If three values a, b, c are in AP, then 2 × b = a + c

Calculations:

⇒ D₁ = C₁ ÷ π = 6π ÷ π = 6 cm

⇒ D₃ = C₃ ÷ π = 14π ÷ π = 14 cm

⇒ Since D₁, D₂, D₃ are in AP:

⇒ 2 × D₂ = D₁ + D₃

⇒ 2 × D₂ = 6 + 14

⇒ 2 × D₂ = 20

⇒ D₂ = 10 cm

⇒ Circumference of the middle circle (C₂) = π × D₂

⇒ C₂ = π × 10 = 10π cm

∴ The correct answer is 10π cm.

Additional Information

Concentric Circles

Circles having a common center but different radii are called concentric circles. The gap region bounded by any two concentric circles is known as an annulus.

Arithmetic Progression (AP)

A sequence of numbers in which the difference between consecutive terms is constant. For terms a, b, c in AP, the middle term is the arithmetic mean: b = (a + c) ÷ 2.

Area vs. Circumference

While circumferences of circles are directly proportional to their radii (C = 2 × π × r), areas are proportional to the squares of their radii (A = π × r²). Thus, if radii are in AP, areas will not be in AP.

Shortcut Trick

Since the height of the cylinder and the holes is the same, the ratio of volumes is equal to the ratio of their base areas.

Base area of the original solid cylinder = π × 5² = 25π.

Total base area of the 4 removed holes = 4 × (π × 1²) = 4π.

Percentage of volume removed = (4π ÷ 25π) × 100.

⇒ (4 ÷ 25) × 100 = 16%.

∴ The correct answer is 16%.

Alternate Method

Given:

Radius of the solid cylinder (R) = 5 cm

Radius of each drilled hole (r) = 1 cm

Number of drilled holes = 4

Let the full height of the cylinder be h.

Formula Used:

Volume of a cylinder = πr²h

Percentage removed = (Volume removed ÷ Total original volume) × 100

R = 5 cm

r = 1 cm

Top View of the Cylinder

Calculations:

⇒ Original volume of the solid cylinder = π × R² × h = π × 5² × h = 25πh

⇒ Volume of one cylindrical hole = π × r² × h = π × 1² × h = 1πh

⇒ Total volume removed by the 4 holes = 4 × 1πh = 4πh

⇒ Percentage of volume removed = (4πh ÷ 25πh) × 100

⇒ Percentage removed = (4 ÷ 25) × 100

⇒ Percentage removed = 16%

∴ The correct answer is 16%.

Additional Information

Cylinder Volume vs Surface Area

While drilling holes decreases the overall volume of the solid cylinder, it actually increases the total surface area by exposing new inner cylindrical surfaces.

Volume of a Cylinder

The total space occupied by a right circular cylinder is universally given by V = πr²h, where r is the base radius and h is the height.

Volume of a Cone

A cone with the exact same base radius and height as a cylinder contains exactly one-third of the cylinder's volume, evaluated as V = (1 ÷ 3)πr²h.

Shortcut Trick

For sectors of the same circle, the area is directly proportional to the central angle.

Central angle of the first sector = 135°

Central angle of the second sector = (3π ÷ 4) radians = (3 ÷ 4) × 180° = 135°

Since both central angles are equal, the areas of both sectors are identical.

Ratio of their areas = 1 : 1

∴ The correct answer is 1:1.

Alternate Method

Given:

Radius of the circle (r) = 8 cm

Central angle of the first sector (θ₁) = 135°

Central angle of the second sector (θ₂) = 3π ÷ 4 radians

Formula Used:

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

Area of a sector (in radians) = (1 ÷ 2) × θ × r²

r = 8 cm 135°

Sector 1

r = 8 cm 3π/4 rad

Sector 2

Calculations:

⇒ Area of the first sector (A₁) = (135° ÷ 360°) × π × (8)²

⇒ A₁ = (3 ÷ 8) × 64π = 24π cm²

⇒ Area of the second sector (A₂) = (1 ÷ 2) × (3π ÷ 4) × (8)²

⇒ A₂ = (3π ÷ 8) × 64 = 24π cm²

⇒ Ratio of their areas = A₁ : A₂

⇒ Ratio = 24π : 24π = 1 : 1

∴ The correct answer is 1:1.

Additional Information

Area of a Sector

When the central angle θ is in degrees: Area = (θ ÷ 360°) × πr². When θ is in radians: Area = (1 ÷ 2) × θr².

Arc Length of a Sector

When θ is in degrees: Arc length = (θ ÷ 360°) × 2πr. When θ is in radians: Arc length = r × θ.

Angle Conversions

Degrees to Radians: Multiply the degree value by (π ÷ 180°). Radians to Degrees: Multiply the radian value by (180° ÷ π).