Quantitative Aptitude PYQs

Shortcut Trick

Let Father's age 10 years ago be x. Then Son's age was x/2 and Daughter's age was x/4.

Present ages: Father = (x + 10), Son = (x/2 + 10), Daughter = (x/4 + 10).

Given: Present Father = Son + Daughter ⇒ (x + 10) = (x/2 + 10) + (x/4 + 10).

⇒ x + 10 = 3x/4 + 20 ⇒ x/4 = 10 ⇒ x = 40.

Present Father = 50. Since Total = Father + (Son + Daughter) and Father = Son + Daughter, Total = 2 × Father = 2 × 50 = 100.

∴ The correct answer is 100 years.

Alternate Method

Let the present ages of the Father, Son, and Daughter be F, S, and D respectively.

Given:

1. F = S + D

2. 10 years ago: (F − 10) = 2(S − 10) and (F − 10) = 4(D − 10)

From the second condition:

⇒ F − 10 = 2S − 20 ⇒ 2S = F + 10 ⇒ S = (F + 10) ÷ 2

⇒ F − 10 = 4D − 40 ⇒ 4D = F + 30 ⇒ D = (F + 30) ÷ 4

Substitute S and D in the first equation:

⇒ F = [(F + 10) ÷ 2] + [(F + 30) ÷ 4]

Multiply the entire equation by 4 to clear denominators:

⇒ 4F = 2(F + 10) + (F + 30)

⇒ 4F = 2F + 20 + F + 30

⇒ 4F = 3F + 50

⇒ F = 50 years

Calculating S and D:

⇒ S = (50 + 10) ÷ 2 = 30 years

⇒ D = (50 + 30) ÷ 4 = 20 years

Total Age:

⇒ Total = F + S + D = 50 + 30 + 20 = 100 years.

∴ The correct answer is 100 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals remains constant at any point in time (past, present, or future).

Time Shift in Ages

If the current age is x, then age n years ago was (x − n) and age n years hence will be (x + n).

Ratio and Units Method

When ratios at two different times are given, ensure the ratio difference between the periods is scaled to match the actual time gap.

Shortcut Trick

Ratio of numbers = 2 : 3

Product of ratio parts = 2 × 3 = 6 units²

Given product of LCM and HCF = 294

1 unit² = 294 ÷ 6 = 49 → 1 unit = √49 = 7

Sum of numbers = (2 + 3) × 7 = 5 × 7 = 35

∴ The correct answer is 35.

Alternate Method

Given:

Ratio of two positive numbers = 2 : 3

Product of LCM and HCF = 294

Formula Used:

Product of two numbers = LCM × HCF

Calculations:

⇒ Let the two positive numbers be 2x and 3x.

⇒ According to the property: (2x) × (3x) = 294

⇒ 6x² = 294

⇒ x² = 294 ÷ 6

⇒ x² = 49

⇒ x = √49 = 7

⇒ The numbers are: 2 × 7 = 14 and 3 × 7 = 21

⇒ Sum of the two numbers = 14 + 21 = 35

∴ The correct answer is 35.

Additional Information

Product Property of HCF and LCM

For any two numbers a and b, the relationship is always: a × b = HCF(a, b) × LCM(a, b).

Ratio and HCF

If two numbers are in the ratio x : y, then the numbers can be written as Hx and Hy, where H is the Highest Common Factor (HCF).

LCM of Ratio Numbers

If the numbers are Hx and Hy (with x and y being coprime), then their LCM is H × x × y.

Shortcut Trick

Relative speed in same direction = 62 − 44 = 18 km/hr.

Convert speed to m/s: 18 × (5/18) = 5 m/s.

Total distance to cross = 130 + 120 = 250 m.

Time taken = Total Distance ÷ Relative Speed = 250 ÷ 5 = 50 sec.

∴ The correct answer is 50 sec.

Alternate Method

Given:

Length of the faster train (L₁) = 130 m

Length of the slower train (L₂) = 120 m

Speed of the faster train (S₁) = 62 km/hr

Speed of the slower train (S₂) = 44 km/hr

Formula Used:

Relative Speed (same direction) = S₁ − S₂

Time = Total Distance ÷ Relative Speed

Calculations:

⇒ Total distance to be covered = 130 + 120 = 250 m

⇒ Relative speed of trains = 62 − 44 = 18 km/hr

⇒ Convert relative speed to m/s: 18 × (5/18) = 5 m/s

⇒ Time taken by the faster train = 250 ÷ 5 = 50 sec

The final calculated time is 50 sec, which corresponds to option (2).

∴ The correct answer is 50 sec.

Additional Information

Relative Speed Concepts

When two objects move in the same direction, Relative Speed = S₁ − S₂. When they move in opposite directions, Relative Speed = S₁ + S₂.

Unit Conversion

To convert speed from km/hr to m/s, multiply by 5/18. To convert from m/s to km/hr, multiply by 18/5.

Train Crossing Distance

When a train crosses another train or a platform, the total distance covered is always the sum of their respective lengths.

Shortcut Trick

Total Work (LCM of 10, 20, 30) = 60 units.

Efficiency of (A+B) = 6, (B+C) = 3, and Pipe B = 2.

Efficiency of Pipe A = 6 − 2 = 4; Efficiency of Pipe C = 3 − 2 = 1.

Combined Efficiency of (A+C) = 4 + 1 = 5 units/hr.

Time taken = 60 ÷ 5 = 12 hr.

∴ The correct answer is 12 hr.

Alternate Method

Given:

Time taken by (A + B) = 10 hr

Time taken by (B + C) = 20 hr

Time taken by Pipe B = 30 hr

Formula Used:

Efficiency = Total Work ÷ Time Taken

Calculations:

⇒ Efficiency of Pipe A = Efficiency of (A+B) − Efficiency of B

⇒ Efficiency of Pipe A = 6 − 2 = 4 units/hr

⇒ Efficiency of Pipe C = Efficiency of (B+C) − Efficiency of B

⇒ Efficiency of Pipe C = 3 − 2 = 1 unit/hr

⇒ Combined Efficiency of Pipe A and Pipe C = 4 + 1 = 5 units/hr

⇒ Time taken by A and C together = Total Work ÷ Combined Efficiency

⇒ Time = 60 ÷ 5 = 12 hr

∴ The correct answer is 12 hr.

Additional Information

Pipes and Cisterns Concept

Filling pipes are considered to have positive efficiency, while emptying (leakage) pipes have negative efficiency.

Efficiency-Time Relationship

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

Combined Work Formula

If two pipes can fill a tank in x and y hours respectively, they together fill it in (x × y) ÷ (x + y) hours.

Shortcut Trick

LCM of the ratio parts (3, 7, and 2) = 3 × 7 × 2 = 42.

Relationship: HCF = Actual LCM ÷ LCM of ratio parts.

HCF = 5376 ÷ 42 = 128.

∴ The correct answer is 128.

Alternate Method

Given: Ratio of three numbers = 3 : 7 : 2 and their LCM = 5376.

Formula: LCM(ax, bx, cx) = x × LCM(a, b, c), where x is the HCF.

Let the Highest Common Factor (HCF) be x.

The three numbers can be represented as 3x, 7x, and 2x.

⇒ LCM(3x, 7x, 2x) = 5376

⇒ x × LCM(3, 7, 2) = 5376

⇒ Since 3, 7, and 2 are prime numbers and have no common factors, their LCM = 3 × 7 × 2 = 42.

⇒ 42 × x = 5376

⇒ x = 5376 ÷ 42

⇒ x = 128

∴ The correct answer is 128.

Additional Information

Relationship between HCF and LCM

For any two numbers a and b, HCF(a, b) × LCM(a, b) = a × b. Note: This rule does not directly apply to more than two numbers.

LCM of Ratios

If numbers are in the ratio a : b : c, then their LCM = HCF × LCM(a, b, c).

Co-prime Numbers

If numbers are co-prime (pairwise), their HCF is 1 and their LCM is equal to the product of the numbers.

HCF and LCM Properties

The LCM of a set of integers is always exactly divisible by their HCF.

Shortcut Trick

Identify a = 25, b = −7, and c = −18. First, check the sum: 25 + (−7) + (−18) = 0.

According to the algebraic identity, if a + b + c = 0, then a³ + b³ + c³ = 3abc.

Required value = 3 × 25 × (−7) × (−18) = 75 × 126 = 9450.

∴ The correct answer is 9450.

Alternate Method

Given: a = 25, b = −7, c = −18

Formula Used: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

⇒ Check the sum of terms: a + b + c = 25 + (−7) + (−18)

⇒ a + b + c = 25 − 25 = 0

⇒ Substituting the zero sum into the identity: a³ + b³ + c³ − 3abc = 0 × (a² + b² + c² − ab − bc − ca)

⇒ a³ + b³ + c³ = 3abc

⇒ Value = 3 × 25 × (−7) × (−18)

⇒ Value = 75 × 126 = 9450

∴ The correct answer is 9450.

Additional Information

Conditional Algebraic Identity

When the sum of three variables a + b + c = 0, the expression a³ + b³ + c³ simplifies directly to 3abc.

General Cubic Formula

The complete identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), useful for simplifying complex symmetric expressions.

Alternate Expansion

The term (a² + b² + c² − ab − bc − ca) can also be written as ½[(a − b)² + (b − c)² + (c − a)²].

Shortcut Trick

Monthly Savings = 16000 − (6000 + 1500) = ₹8500

Savings for 11 months (as 1 month's savings is spent on birthday) = 8500 × 11

Total Yearly Savings = 93500

∴ The correct answer is ₹93500.

Alternate Method

Given:

Monthly Salary = ₹16000

House Rent = ₹6000

Monthly Bills = ₹1500

Formula: Savings = Income − Expenditure

Step-by-step Calculation:

⇒ Total Monthly Expenditure = House Rent + Bills

⇒ Total Monthly Expenditure = 6000 + 1500 = ₹7500

⇒ Monthly Savings = Salary − Expenditure

⇒ Monthly Savings = 16000 − 7500 = ₹8500

⇒ Number of months in a year = 12

⇒ Savings are spent in 1 month (birthday month), so net savings occur for = 12 − 1 = 11 months

⇒ Total Savings in a year = Monthly Savings × 11

⇒ Total Savings in a year = 8500 × 11

⇒ Total Savings in a year = ₹93500

Therefore, the correct option is (3).

∴ The correct answer is ₹93500.

Additional Information

Income-Expenditure Relationship

Income is always the sum of Expenditure and Savings. If any two are known, the third can be calculated using Income − Expenditure = Savings.

Annual Calculation

To convert monthly data to annual data, multiply by 12. If a specific period involves extra spending or no savings, adjust the multiplier accordingly (e.g., use 11 instead of 12).

Savings Percentage

Savings % = (Savings ÷ Income) × 100. In this case, monthly savings % = (8500 ÷ 16000) × 100 = 53.125%.

Shortcut Trick

Simple Interest (SI) for 4 years = ₹3,200 → SI for 1 year = 3,200 ÷ 4 = ₹800.

Since the Rate is 10% per annum, 10% of Principal = ₹800 → Principal (100%) = ₹8,000.

Simple Interest for 6 years = 800 × 6 = ₹4,800.

Total Amount payable = Principal + SI for 6 years = 8,000 + 4,800 = ₹12,800.

∴ The correct answer is ₹12,800.

Alternate Method

Given: Time (T₁) = 4 years, Rate (R) = 10%, Simple Interest (SI₁) = ₹3,200, Time (T₂) = 6 years.

Formula: Simple Interest = (P × R × T) ÷ 100 | Amount = Principal + Simple Interest

⇒ 3200 = (P × 10 × 4) ÷ 100

⇒ 3200 = (P × 40) ÷ 100

⇒ 40P = 3200 × 100

⇒ P = 320000 ÷ 40 = ₹8,000

⇒ Simple Interest for 6 years (SI₂) = (8000 × 10 × 6) ÷ 100

⇒ SI₂ = (8000 × 60) ÷ 100 = ₹4,800

⇒ Amount after 6 years (A) = Principal (P) + SI₂

⇒ A = 8000 + 4800 = ₹12,800

∴ The correct answer is ₹12,800.

Additional Information

Total Interest as Percentage

Total Simple Interest % = Rate × Time. For 4 years at 10%, it is 40% of the principal.

Amount Calculation Shortcut

Amount % = 100% + (Rate × Time)%. After 6 years at 10%, Amount = 160% of Principal.

Constant Rate and Principal

When Principal and Rate are constant, Simple Interest is directly proportional to Time (SI₁/T₁ = SI₂/T₂).

Given:

8² + √8² − √49 − 6

Formula used:

Square of a number: a² = a × a

Square Root property: √a² = a

Calculations:

8² + √8² − √49 − 6

⇒ 64 + 8 − 7 − 6 (Since 8² = 64 and √49 = 7)

⇒ 72 − 7 − 6

⇒ 65 − 6

⇒ 59

∴ The correct answer is option (2).

Shortcut Trick

Calculate unit price for both Cost Price (CP) and Selling Price (SP).

CP of 1 kg = 900 ÷ 20 = ₹45

SP of 1 kg = 918 ÷ 18 = ₹51

Profit Percentage = (SP − CP) ÷ CP × 100

Profit % = (6 ÷ 45) × 100 = 2/15 × 100 = 40/3% = 13(1/3)%

∴ The correct answer is 13(1/3)%.

Alternate Method

Given:

Total Cost Price (CP) of 20 kg sugar = ₹900

Total Selling Price (SP) of 18 kg sugar = ₹918

Formula used:

Profit = Selling Price − Cost Price

Profit Percentage = (Profit ÷ Cost Price) × 100

Calculations:

⇒ Cost Price (CP) of 1 kg sugar = 900 ÷ 20 = ₹45

⇒ Selling Price (SP) of 1 kg sugar = 918 ÷ 18 = ₹51

⇒ Profit per kg = 51 − 45 = ₹6

⇒ Profit Percentage = (6 ÷ 45) × 100

⇒ Profit Percentage = (2/15) × 100 = 200/15

⇒ Profit Percentage = 40/3 = 13(1/3)%

∴ The correct answer is 13(1/3)%.

Additional Information

Profit and Loss Basics

Profit occurs when SP > CP. Profit % is always calculated on the CP unless otherwise stated.

Fraction to Percentage Conversion

The fraction 1/15 is approximately equal to 6.66%. Thus, 2/15 = 2 × 6.66% = 13.33% or 13(1/3)%.

Unitary Method

To compare prices of different quantities, always find the value of a single unit (1 kg, 1 item, etc.) to simplify calculations.

Given:

324 × 24 = 7776

Formula used:

Dividend ÷ Divisor = Quotient

Calculations:

From the given data:

⇒ 7776 ÷ 24 = 324

We need to find the value of:

⇒ 7.776 ÷ 2.4

Multiplying both the dividend and divisor by 10 to simplify the decimal:

⇒ (7.776 × 10) ÷ (2.4 × 10) = 77.76 ÷ 24

Expressing 77.76 as a fraction:

⇒ 77.76 ÷ 24 = (7776 ÷ 100) ÷ 24

Rearranging the terms:

⇒ (7776 ÷ 24) ÷ 100

Substituting the known value 7776 ÷ 24 = 324:

⇒ 324 ÷ 100 = 3.24

∴ The correct answer is option (1).

Shortcut Trick

Apply the alligation method using the average weights of boys (65) and girls (45) with the class mean (57).

Ratio of Boys to Girls = (57 − 45) : (65 − 57) = 12 : 8

Simplifying the ratio: 12 : 8 = 3 : 2 (Boys : Girls)

The question asks for the ratio of Girls to Boys, which is 2 : 3.

∴ The correct answer is 2:3.

Alternate Method

Given:

Average weight of boys = 65 kg

Average weight of girls = 45 kg

Average weight of the whole class = 57 kg

Formula Used:

Total Weight = (Average Weight × Number of Students)

Let the number of boys be 'B' and girls be 'G'.

⇒ Total weight of class = 57 × (B + G)

⇒ Also, Total weight = 65B + 45G

⇒ 57B + 57G = 65B + 45G

⇒ 57G − 45G = 65B − 57B

⇒ 12G = 8B

⇒ G ÷ B = 8 ÷ 12

⇒ G : B = 2 : 3

∴ The correct answer is 2:3.

Additional Information

Alligation Rule

It is a rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price: (Quantity of Cheaper) ÷ (Quantity of Dearer) = (Price of Dearer − Mean Price) ÷ (Mean Price − Price of Cheaper).

Weighted Average Formula

The average of a combined group is given by A_m = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where n represents the quantity and A represents the average of each group.

Shortcut Trick

Total amount paid using the successive multiplier method:

Net Selling Price = 900 × (90/100) × (95/100)

Net Selling Price = 900 × 0.9 × 0.95

Net Selling Price = 810 × 0.95 = ₹769.50

∴ The correct answer is ₹769.50.

Alternate Method

Given: Marked Price (MP) = ₹900, First Discount = 10%, Additional Discount = 5%.

Step-by-step calculation:

⇒ First discounted price = 900 − (10% of 900)

⇒ First discounted price = 900 − 90 = ₹810

⇒ Final amount paid = 810 − (5% of 810)

⇒ Final amount paid = 810 − 40.50 = ₹769.50

∴ The correct answer is ₹769.50.

Additional Information

Single Equivalent Discount

For two successive discounts of a% and b%, the single equivalent discount is (a + b − ab/100)%.

Marked Price (MP) and Selling Price (SP)

Selling Price is calculated as SP = MP × (100 − Discount%)/100. For successive discounts, multiply each reduction factor.

Successive Discounts with Fraction

If discounts are 10% (1/10) and 5% (1/20), the price becomes 9/10 and 19/20 of the original value respectively.

Shortcut Trick

Initial radius ratio R₁ : R₂ = 2 : 3

New R₁ = 2 × 150% = 3 units

New R₂ = 3 × 120% = 3.6 units

New Radius Ratio = 3 : 3.6 = 5 : 6

Volume Ratio = (Radius Ratio)³ = 5³ : 6³ = 125 : 216

∴ The correct answer is 125 : 216.

Alternate Method

Given:

Ratio of radii R₁ : R₂ = 2 : 3

Increase in R₁ = 50%

Increase in R₂ = 20%

Formula Used:

Volume of a sphere (V) = (4/3) × π × R³

Calculations:

Let the initial radii be R₁ = 2x and R₂ = 3x

⇒ New radius R'₁ = 2x + 50% of 2x = 2x + x = 3x

⇒ New radius R'₂ = 3x + 20% of 3x = 3x + 0.6x = 3.6x

⇒ Ratio of new volumes V'₁ : V'₂ = ((4/3)π(R'₁)³) : ((4/3)π(R'₂)³)

⇒ V'₁ : V'₂ = (R'₁)³ : (R'₂)³

⇒ V'₁ : V'₂ = (3x)³ : (3.6x)³

⇒ V'₁ : V'₂ = 27 : 46.656

⇒ Simplify the ratio: (3 / 3.6)³ = (30 / 36)³

⇒ (5 / 6)³ = 125 : 216

∴ The correct answer is 125 : 216.

Additional Information

Volume of Sphere

The volume of a sphere is given by V = (4/3)πr³, where r is the radius.

Surface Area of Sphere

The total surface area of a sphere is S = 4πr².

Ratio Property

If the ratio of the radii of two spheres is a : b, then the ratio of their surface areas is a² : b² and the ratio of their volumes is a³ : b³.

Shortcut Trick

Total Effective CP = 8000 + 500 = ₹8500

Net multiplier for 20% profit and 25% loss = 1.20 × 0.75 = 0.9

Final Price = 8500 × 0.9 = ₹7650

∴ The correct answer is ₹7650.

Alternate Method

Given:

Purchase price of mobile phone = ₹8000

Amount spent on repairs = ₹500

Profit percentage = 20%

Loss percentage of the buyer = 25%

Formula Used:

Total CP = Cost Price + Overhead Expenses

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

Calculations:

⇒ Total CP for the trader = 8000 + 500 = ₹8500

⇒ Selling price of the trader (SP₁) = 8500 + 20% of 8500

⇒ SP₁ = 8500 + (20/100) × 8500 = 8500 + 1700 = ₹10200

⇒ This SP₁ becomes the Cost Price for the first buyer (CP₂ = ₹10200)

⇒ Final Selling Price (SP₂) = 10200 − 25% of 10200

⇒ SP₂ = 10200 − (25/100) × 10200 = 10200 − 2550 = ₹7650

∴ The correct answer is ₹7650.

Additional Information

Overhead Expenses

Any additional costs like repairs, transportation, or installation must be added to the initial purchase price to determine the actual Cost Price.

Successive Percentage Changes

When a value changes by x% and then y% successively, the net percentage change is [x + y + (xy/100)]%.

Successive Multiplication Factor

Final Value = Initial Value × (1 ± r₁/100) × (1 ± r₂/100) × ... where 'r' is the percentage profit (+) or loss (−).

Shortcut Trick

Check options based on the given conditions.

Condition 1: Denominator − Numerator = 8. For 27/35, 35 − 27 = 8 (Satisfied).

Condition 2: (Numerator + 3) / (Denominator + 5) = 3/4.

⇒ (27 + 3) / (35 + 5) = 30 / 40 = 3/4 (Satisfied).

∴ The correct answer is 27/35.

Alternate Method

Given:

Numerator = Denominator − 8

New Numerator = Numerator + 3

New Denominator = Denominator + 5

New Fraction = 3/4

Calculations:

Let the denominator of the original fraction be x.

⇒ Numerator = x − 8

⇒ Original Fraction = (x − 8) / x

According to the question:

⇒ ((x − 8) + 3) / (x + 5) = 3 / 4

⇒ (x − 5) / (x + 5) = 3 / 4

By cross-multiplication:

⇒ 4 × (x − 5) = 3 × (x + 5)

⇒ 4x − 20 = 3x + 15

⇒ 4x − 3x = 15 + 20

⇒ x = 35

Now, find the numerator:

⇒ Numerator = 35 − 8 = 27

⇒ Original Fraction = 27/35

∴ The correct answer is 27/35.

Additional Information

Proper and Improper Fractions

A Proper Fraction is one where the numerator is less than the denominator (e.g., 27/35). An Improper Fraction has a numerator greater than or equal to the denominator.

Equivalent Fractions

Fractions that represent the same value even though they look different, obtained by multiplying or dividing the numerator and denominator by the same non-zero number.

Cross-Multiplication Rule

In an equation like a/b = c/d, the property a × d = b × c is used to solve for unknown variables in linear equations involving fractions.

Shortcut Trick

Distance between two objects on the same side = Height × (cot θ_smaller − cot θ_larger)

Distance = 75 × (cot 30° − cot 45°)

Distance = 75 × (√3 − 1) = 75 × (1.732 − 1)

Distance = 75 × 0.732 = 54.9 m

∴ The correct answer is 54.9 m.

Alternate Method

Given:

Height of the lighthouse (AB) = 75 m

Angles of depression of two ships = 30° and 45°

Since the ships are on the same side, the angles of elevation from the ships to the top are also 30° and 45°.

Calculations:

⇒ In right ΔABC (closer ship at C):

⇒ tan 45° = AB ÷ BC

⇒ 1 = 75 ÷ BC ⇒ BC = 75 m

⇒ In right ΔABD (farther ship at D):

⇒ tan 30° = AB ÷ BD

⇒ 1 ÷ √3 = 75 ÷ BD ⇒ BD = 75√3 m

⇒ Distance between ships (CD) = BD − BC

⇒ CD = 75√3 − 75 = 75(√3 − 1)

⇒ CD = 75 × (1.732 − 1) = 75 × 0.732

⇒ CD = 54.9 m

∴ The correct answer is 54.9 m.

Additional Information

Angle of Depression

The angle formed by the line of sight with the horizontal line for an object below the horizontal level.

Trigonometric Ratios

For height and distance problems, tan θ = Perpendicular ÷ Base and cot θ = Base ÷ Perpendicular are most commonly used.

General Formula for Same Side

Distance d = h × |cot α − cot β|, where h is height and α, β are angles of elevation/depression.

Given:

((13/21) / (13/10)) ÷ ((5/9 × 18/15) + 5/7) + 10/6 ÷ (29/15 of 15/6)

Formula used:

The VBODMAS rule (Vinculum, Bracket, Of, Division, Multiplication, Addition, Subtraction) is used to simplify the expression.

Calculations:

((13/21) × (10/13)) ÷ ((1/3 × 2/1) + 5/7) + 10/6 ÷ (29/15 × 15/6)

⇒ (10/21) ÷ (2/3 + 5/7) + 10/6 ÷ 29/6

⇒ (10/21) ÷ ((14 + 15) / 21) + (10/6 × 6/29)

⇒ (10/21) ÷ (29/21) + 10/29

⇒ (10/21 × 21/29) + 10/29

⇒ 10/29 + 10/29

⇒ 20/29

∴ The correct answer is option (2).

Shortcut Trick

Using the Net Deviation method where the Average is 15.

Deviations of given numbers from 15: (10 − 15) = −5, (20 − 15) = +5, (25 − 15) = +10.

Sum of known deviations = −5 + 5 + 10 = +10.

To keep the average 15, the total deviation must be 0; thus, the fourth number deviation = −10.

Fourth number = 15 − 10 = 5.

∴ The correct answer is 5.

Alternate Method

Given:

Average of 4 numbers = 15

Three numbers are 10, 20, and 25.

Formula Used:

Average = (Sum of all observations) ÷ (Total number of observations)

Calculations:

⇒ Let the fourth number be x.

⇒ Average = (10 + 20 + 25 + x) ÷ 4

⇒ 15 = (55 + x) ÷ 4

⇒ 15 × 4 = 55 + x

⇒ 60 = 55 + x

⇒ x = 60 − 55

⇒ x = 5

∴ The correct answer is 5.

Additional Information

Sum of Observations

The total sum is always equal to Average × Number of items (Sum = A × n).

Net Deviation Rule

The algebraic sum of deviations of all observations from their arithmetic mean is always zero.

Change in Average

If a constant k is added to or subtracted from every observation, the new average will be (Old Average ± k).

Shortcut Trick

Let the Marked Price (MP) be 100 units.

Cost Price (CP) for Ravi = 100 − 25% of 100 = 75 units.

Ravi sells it at 10% profit ⇒ Selling Price (SP) = 75 × 110 ÷ 100 = 82.5 units.

Given SP = ₹1,320 ⇒ 82.5 units = 1320.

1 unit = 1320 ÷ 82.5 = 16.

Marked Price = 100 units = 100 × 16 = ₹1,600.

∴ The correct answer is Rs. 1600.

Alternate Method

Given: Discount = 25%, Profit = 10%, Selling Price (SP) = ₹1,320.

Formulas: CP = SP × 100 ÷ (100 + Profit%); MP = CP × 100 ÷ (100 − Discount%)

⇒ First, find the Cost Price (CP) for Ravi: CP = 1320 × 100 ÷ 110 = ₹1,200.

⇒ This CP is the price after a 25% discount on the Marked Price (MP).

⇒ Let MP be x. Then, 75% of x = 1200.

⇒ x = (1200 × 100) ÷ 75 = 16 × 100 = ₹1,600.

∴ The correct answer is Rs. 1600.

Additional Information

Marked Price & Discount

The Marked Price (MP) is the price printed on an item. Discount is always calculated on the MP to arrive at the Selling Price (SP).

Relation between CP and MP

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This direct ratio helps find MP when CP and profit/discount percentages are known.