Search Previous Year Questions

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.