Quantitative Aptitude PYQs

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).

Shortcut Trick

Convert the fractions into decimals for faster and easier multiplication.

⇒ 2 ²/₅ = 2.4

⇒ ³/₄ = 0.75

⇒ Product = 1.2 × 2.4 × 0.75

⇒ Rearranging for ease: (1.2 × 0.75) × 2.4 = 0.9 × 2.4 = 2.16

∴ The correct answer is 2.16.

Alternate Method

Given:

Expression: 1.2 × 2 ²/₅ × ³/₄

Calculation:

⇒ Convert 1.2 to a decimal fraction: 1.2 = ¹²/₁₀ = ⁶/₅

⇒ Convert the mixed fraction into an improper fraction: 2 ²/₅ = (2 × 5 + 2) / 5 = ¹²/₅

⇒ Substitute these back into the expression:

⇒ ⁶/₅ × ¹²/₅ × ³/₄

⇒ Divide 12 by 4 to simplify:

⇒ (6 × 3 × 3) / (5 × 5)

⇒ ⁵⁴/₂₅

⇒ Multiply the numerator and denominator by 4 to get a base of 100 for immediate decimal conversion:

⇒ (54 × 4) / (25 × 4) = ²¹⁶/₁₀₀

⇒ 2.16

∴ The correct answer is 2.16.

Additional Information

Fraction to Decimal Conversion

Memorising standard fractions significantly speeds up calculation in exams: ¹/₄ = 0.25, ³/₄ = 0.75, ¹/₅ = 0.2, and ²/₅ = 0.4.

Simplification Trick for Base 100

When dividing any number by 25, simply multiply the numerator by 4 and place the decimal two spots from the right (effectively dividing by 100).

Mixed Fractions Rule

A mixed fraction represented as a(^b/_c) is converted into an improper fraction using the mathematical rule: (a × c + b) ÷ c.

Shortcut Trick

Observe the relationship between the numerator and the denominator.

Each term in the denominator is exactly 3 times the corresponding term in the numerator.

Expression = (0.1³ + 0.03³) ÷ [(3 × 0.1)³ + (3 × 0.03)³]

⇒ 1 ÷ 3³ = 1 ÷ 27

⇒ 1 ÷ 27 ≈ 0.037

∴ The correct answer is 0.037.

Alternate Method

Given:

Expression = (0.1 × 0.1 × 0.1 + 0.03 × 0.03 × 0.03) ÷ (0.3 × 0.3 × 0.3 + 0.09 × 0.09 × 0.09)

Formula Used:

(x × y)ⁿ = xⁿ × yⁿ

Calculations:

⇒ Numerator = 0.1³ + 0.03³

⇒ Denominator = 0.3³ + 0.09³

⇒ Denominator = (3 × 0.1)³ + (3 × 0.03)³

⇒ Denominator = 3³ × 0.1³ + 3³ × 0.03³

⇒ Denominator = 27 × (0.1³ + 0.03³)

⇒ Expression = (0.1³ + 0.03³) ÷ [27 × (0.1³ + 0.03³)]

⇒ Expression = 1 ÷ 27

⇒ 1 ÷ 27 = 0.037037...

∴ The correct answer is 0.037.

Additional Information

Laws of Exponents

When terms are multiplied and raised to a common power, the power distributes directly to both numbers: (xy)ⁿ = xⁿ × yⁿ. This allows easy extraction of common multiples from complex decimals.

Algebraic Simplification

In simplification problems involving decimals that look similar, always check if the denominator is a direct multiple of the numerator before calculating the actual cubes or squares.

Recurring Decimals

Fractions like 1 ÷ 27 result in non-terminating, repeating decimals. The value 1 ÷ 27 equals 0.037037037..., which is generally rounded off to 0.037 as given in the exam options.

Shortcut Trick Heights strictly less than 1.65 m are 1.55 m, 1.60 m, and 1.62 m. Students in Section A = 5 + 9 + 11 = 25 Students in Section B = 4 + 8 + 13 = 25 Required Ratio = 25 : 25 = 1 : 1 ∴ The correct answer is 1:1.

Alternate Method Given: A table containing the heights of students in two sections, A and B. We need the ratio of students with heights less than 1.65 metres in Section A to Section B. Calculations: ⇒ From the table, heights less than 1.65 m correspond to 1.55 m, 1.60 m, and 1.62 m. ⇒ Total number of students in Section A with height < 1.65 m = 5 + 9 + 11 ⇒ Total number of students in Section A = 25 ⇒ Total number of students in Section B with height < 1.65 m = 4 + 8 + 13 ⇒ Total number of students in Section B = 25 ⇒ Required Ratio = (Total students in Section A) : (Total students in Section B) ⇒ Required Ratio = 25 : 25 ⇒ Required Ratio = 1 : 1 ∴ The correct answer is 1:1.

Additional Information Ratio Comparison A ratio compares two quantities of the same kind. It is typically expressed in its simplest form by dividing both components by their highest common factor. Tabular Data Interpretation When verifying conditions such as 'less than a value', you only aggregate the frequencies of the categories strictly falling under that boundary condition.

Shortcut Trick

Profit Ratio = Ratio of (Capital × Time)

Since the time is the same (1 year), Profit Ratio = Capital Ratio

A : B : C = 40,000 : 60,000 : 80,000 = 2 : 3 : 4

B's share = [3 ÷ (2 + 3 + 4)] × 55,000 = (3 ÷ 9) × 55,000

B's share = 55,000 ÷ 3 = 18,333.33

∴ The correct answer is ₹18,333.33.

Alternate Method

Given:

A's investment = ₹40,000

B's investment = ₹60,000

C's investment = ₹80,000

Total time for each = 1 year

Total profit = ₹55,000

Formula Used:

Profit Ratio = (Investment of A × Time) : (Investment of B × Time) : (Investment of C × Time)

Share of a partner = (Partner's ratio part ÷ Total ratio parts) × Total Profit

₹40k A

₹60k B

₹80k C

Profit Ratio = 2 : 3 : 4

Calculations:

⇒ Ratio of profits of A, B, and C = 40000 : 60000 : 80000

⇒ Ratio of profits = 4 : 6 : 8

⇒ Simplifying the ratio further, we get A : B : C = 2 : 3 : 4

⇒ Sum of ratio parts = 2 + 3 + 4 = 9 parts

⇒ B's share in profit = (3 parts ÷ 9 parts) × Total Profit

⇒ B's share = (1 ÷ 3) × 55,000

⇒ B's share = 55,000 ÷ 3 = 18,333.33

∴ The correct answer is ₹18,333.33.

Additional Information

Simple Partnership

When all partners invest their capital for the same time period, the profits or losses are distributed exactly in the ratio of their original investments.

Compound Partnership

When partners invest their capital for different time periods, the profit sharing ratio is determined by multiplying each partner's capital by their respective time period (C₁T₁ : C₂T₂).

Sleeping Partner

A partner who only invests capital but does not actively participate in the management of the business. They still receive a share of the profit proportional to their investment unless explicitly stated otherwise.

Shortcut Trick

Total Capital = 60,000 + 90,000 = 1,50,000

Total Interest = 10% of 1,50,000 = 15,000

Remaining Profit = 25,000 − 15,000 = 10,000

Profit Ratio = 60,000 : 90,000 = 2 : 3

A's Total Share = (10% of 60,000) + (2/5 × 10,000) = 6,000 + 4,000 = 10,000

∴ The correct answer is ₹10,000.

Alternate Method

₹60,000 Ratio: 2 A

₹90,000 Ratio: 3 B

Given:

A's investment = ₹60,000

B's investment = ₹90,000

Total profit including interest = ₹25,000

Rate of interest on capital = 10% per annum

Formula Used:

Total Interest = Interest on A's capital + Interest on B's capital

Profit Ratio = (A's Investment × Time) : (B's Investment × Time)

Calculation:

⇒ Total investment = 60,000 + 90,000 = ₹1,50,000

⇒ Total interest to be paid = 10% of 1,50,000 = ₹15,000

⇒ Remaining profit = 25,000 − 15,000 = ₹10,000

⇒ Ratio of investments A:B = 60,000 : 90,000 = 2 : 3

⇒ A's share in remaining profit = (2 ÷ 5) × 10,000 = ₹4,000

⇒ A's interest amount = 10% of 60,000 = ₹6,000

⇒ A's total share = A's interest + A's share in profit

⇒ A's total share = 6,000 + 4,000 = ₹10,000

∴ The correct answer is ₹10,000.

Additional Information

Partnership Profit Division

Profits in a partnership are divided in the ratio of the product of capital invested and the time period of investment. Ratio = (C₁ × T₁) : (C₂ × T₂).

Active vs Sleeping Partner

An active partner manages the business and usually receives a fixed salary or commission from the profit before the remaining profit is distributed. A sleeping partner only invests capital.

Interest on Capital in Partnership

If the partnership deed specifies interest on capital, it is paid out of the total profit first. The leftover profit is then distributed in the standard capital ratio.

Shortcut Trick Since the number of months is the same (3) for all products, the product with the highest total sales will have the highest average. By simple observation of the table values: Sales of Product C (150, 160, 155) are strictly greater than those of Product A (100, 120, 110) and Product B (90, 85, 95) in every month. Thus, Product C clearly has the maximum sum and therefore the highest average without needing full calculations. ∴ The correct answer is Product C.

Alternate Method Given: Sales of Product A for Jan, Feb, Mar are 100, 120, and 110 respectively. Sales of Product B for Jan, Feb, Mar are 90, 85, and 95 respectively. Sales of Product C for Jan, Feb, Mar are 150, 160, and 155 respectively. Formula Used: Average = Total sum of values ÷ Number of values Calculations: ⇒ Average units sold for Product A = (100 + 120 + 110) ÷ 3 ⇒ Average for Product A = 330 ÷ 3 = 110 ⇒ Average units sold for Product B = (90 + 85 + 95) ÷ 3 ⇒ Average for Product B = 270 ÷ 3 = 90 ⇒ Average units sold for Product C = (150 + 160 + 155) ÷ 3 ⇒ Average for Product C = 465 ÷ 3 = 155 ⇒ Comparing the averages: 155 > 110 > 90 Product C clearly has the highest average. ∴ The correct answer is Product C.

Additional Information

1. Average and Sum Relationship For datasets with the same number of elements, the dataset with the largest sum will always have the largest arithmetic mean. This saves valuable calculation time in exams.
2. Data Comparison by Observation In data interpretation problems, if every element of Set X is greater than the corresponding elements of Sets Y and Z, no calculation is required to deduce that Set X has the highest sum and average.
3. Percentage Change A concept often requested along with averages in tabular data is percentage growth: % Change = ((New Value − Old Value) ÷ Old Value) × 100.

Shortcut Trick

Summing the three given results evaluates to 2 times the sum of the three numbers.

⇒ 2 × (Sum of three numbers) = 180 + 168 + 150 = 498

⇒ Sum of three numbers = 498 ÷ 2 = 249

⇒ Average of three numbers = 249 ÷ 3 = 83

∴ The correct answer is 83.

Alternate Method

Given:

Three numbers where the average of any two added to the third gives 180, 168, and 150 respectively.

Formula Used:

Average = Sum of all observations ÷ Total number of observations

a + b + 2c = 360

2a + b + c = 336

a + 2b + c = 300

4(a + b + c) = 996

Average = 83

Calculations:

Let the original three numbers be a, b, and c.

⇒ (a + b) ÷ 2 + c = 180

⇒ a + b + 2c = 360     ... (Equation 1)

⇒ (b + c) ÷ 2 + a = 168

⇒ 2a + b + c = 336     ... (Equation 2)

⇒ (c + a) ÷ 2 + b = 150

⇒ a + 2b + c = 300     ... (Equation 3)

Adding Equation 1, Equation 2, and Equation 3 together:

⇒ 4a + 4b + 4c = 360 + 336 + 300

⇒ 4(a + b + c) = 996

⇒ a + b + c = 249

The average of the three numbers is their sum divided by 3:

⇒ Average = (a + b + c) ÷ 3

⇒ Average = 249 ÷ 3 = 83

∴ The correct answer is 83.

Additional Information

Properties of Average

If a constant value is added, subtracted, multiplied, or divided by each observation in a dataset, the new average also changes exactly by the same arithmetic operation and constant.

Weighted Average

If quantities are divided into multiple groups with respective averages, the combined group average is evaluated as (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Average of Arithmetic Progression

For any sequence of numbers operating with a common difference, the average is strictly derived as (First Term + Last Term) ÷ 2, which also represents the exact middle term.

Shortcut Trick

Total items for students = 100 × (15% of 100) = 100 × 15 = 1500

Total items for chaperones = 10 × (25% of 100) = 10 × 25 = 250

Total items distributed = 1500 + 250 = 1750

∴ The correct answer is 1750 items.

Alternate Method

Given:

Total number of students = 100

Total number of chaperones = 10

Items per student's snack pack = 15% of total students

Items per chaperone's snack pack = 25% of total students

Formula Used:

Value = (Percentage ÷ 100) × Base Value

Students Count = 100 Items/pack = 15% of 100 = 15 Total = 100 × 15 = 1500

Chaperones Count = 10 Items/pack = 25% of 100 = 25 Total = 10 × 25 = 250

Total Snack Items = 1500 + 250 = 1750

Calculations:

⇒ Items in each student's pack = 15% of 100 = (15 ÷ 100) × 100 = 15 items

⇒ Total items given to 100 students = 100 × 15 = 1500 items

⇒ Items in each chaperone's pack = 25% of 100 = (25 ÷ 100) × 100 = 25 items

⇒ Total items given to 10 chaperones = 10 × 25 = 250 items

⇒ Total items distributed = Total student items + Total chaperone items

⇒ Total items distributed = 1500 + 250 = 1750

∴ The correct answer is 1750 items.

Additional Information

Percentage Calculation Rule

Percentage is fundamentally a fraction based on 100. To find x% of a quantity Y, the formula is (x ÷ 100) × Y. Always identify the correct base from the question text carefully before calculating.

Successive Distribution

When distinct items are given to multiple groups, compute the quantities for each group individually according to their independent rules, and then sum up the results.

Common Percentage Ratios

Memorizing basic percentage-to-fraction conversions helps solve equations faster in exams: 25% = 1/4, 20% = 1/5, 15% = 3/20, and 10% = 1/10.

Shortcut Trick

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

For the next 6 months (half-year), the rate becomes = 10% ÷ 2 = 5%

Net effective rate for 2.5 years = 21 + 5 + (21 × 5)/100 = 26 + 1.05 = 27.05%

Compound Interest = 27.05% of 10000 = 2705

∴ The correct answer is ₹2,705.

Alternate Method

Given:

Principal (P) = ₹10,000

Rate (R) = 10% per annum

Time (T) = 2 years 6 months = 2.5 years

Formula Used:

Amount (A) = P × [1 + R/100]ⁿ × [1 + (R × fraction of year)/100]

Compound Interest (CI) = A − P

Start ₹10k

Year 1 ₹11k

Year 2 ₹12.1k

Year 2.5 ₹12705

+10% (₹1000)

+10% (₹1100)

+5% (₹605)

Total CI = 1000 + 1100 + 605 = 2705

Calculations:

⇒ Full years (n) = 2, Fraction of year = 6/12 = 1/2 year

⇒ A = 10000 × [1 + 10/100]² × [1 + (10 × 1/2)/100]

⇒ A = 10000 × (11/10) × (11/10) × (105/100)

⇒ A = 10000 × (121/100) × (21/20)

⇒ A = 100 × 121 × (21/20)

⇒ A = 5 × 121 × 21

⇒ A = 121 × 105 = 12705

⇒ CI = 12705 − 10000 = 2705

∴ The correct answer is ₹2,705.

Additional Information

Effective Rate Method

For 2 years, the effective compound interest rate is calculated directly as A + B + (A × B)/100.

Half-Yearly Compounding

When interest is compounded half-yearly, the applied rate is halved (R/2) and the time periods are doubled (2T).

SI vs CI Difference

The exact difference between Compound Interest and Simple Interest for a 2-year period is given by the formula P(R/100)².

Shortcut Trick

Fraction value of 12% = 3/25 ⇒ Multiplying factor = 28/25.

For 2 years, Amount / Principal = (28/25)² = 784/625.

Given Amount = 784 parts = ₹3136 ⇒ 1 part = 4.

Principal = 625 parts = 625 × 4 = ₹2500.

140% of Principal = 1.4 × 2500 = ₹3500.

∴ The correct answer is ₹3500.

Alternate Method

Given:

Rate (R) = 12% per annum

Time (T) = 2 years

Amount (A) = ₹3136

Formula Used:

A = P(1 + R/100)^T

P = ? Year 1 ₹3136

+12% CI +12% CI

Equivalent to 25.44% Total Interest (Multiplier: 1.2544)

Calculation:

⇒ 3136 = P(1 + 12/100)²

⇒ 3136 = P(112/100)²

⇒ 3136 = P(28/25)²

⇒ 3136 = P × (784/625)

⇒ P = (3136 × 625) ÷ 784

⇒ P = 4 × 625

⇒ P = ₹2500

⇒ Required value = 140% of P

⇒ Required value = (140 ÷ 100) × 2500

⇒ Required value = 1.4 × 2500 = 3500

∴ The correct answer is ₹3500.

Additional Information

Effective Rate of Compound Interest

For 2 years at R%, the effective CI rate is given by R + R + (R × R)/100. For 12%, it is 12 + 12 + 1.44 = 25.44%.

Difference between CI and SI for 2 years

The difference between compound interest and simple interest for 2 years is calculated using the formula: Difference = P(R/100)².

Half-Yearly Compounding

When interest is compounded half-yearly, the rate is halved (R ÷ 2) and the time periods are doubled (2T).