Search Previous Year Questions

Shortcut Trick

Use the Successive Percentage formula: Net % Change = a + b + (a × b) ÷ 100

Here, both first and second year increases are 5%, so a = 5 and b = 5.

Net Increase = 5 + 5 + (5 × 5) ÷ 100

Net Increase = 10 + 0.25 = 10.25%

∴ The correct answer is 10.25%.

Alternate Method

Assume the Initial Salary of the employee is 100 units.

⇒ Let initial salary = 100

⇒ Salary after 1^st year = 100 + (5% of 100) = 100 + 5 = 105

⇒ Salary after 2^nd year = 105 + (5% of 105) = 105 + 5.25 = 110.25

⇒ Total Percentage Increase = ((110.25 − 100) ÷ 100) × 100 = 10.25%

∴ The correct answer is 10.25%.

Additional Information

Successive Percentage Increase/Decrease

When a value changes by x% and then y%, the net change is given by [x + y + (xy/100)]%. Use positive signs for increase and negative signs for decrease.

Multiplying Factor Method

For a 5% increase, the multiplying factor is 1.05. For two successive 5% increases, the final value is Initial × 1.05 × 1.05 = Initial × 1.1025.

Successive Discounts

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

Shortcut Trick

Ratio of numbers = 4 : 5 → LCM of ratio parts = 4 × 5 = 20 units

Given LCM = 240 ⇒ 20 units = 240 ⇒ 1 unit = 12 (this is the HCF)

Sum of ratio parts = 4 + 5 = 9 units

Sum of numbers = 9 units × 12 = 108

∴ The correct answer is 108.

Alternate Method

Given: Ratio = 4 : 5 and LCM = 240.

⇒ Let the two numbers be 4x and 5x (where x is the HCF).

⇒ LCM of 4x and 5x = 4 × 5 × x = 20x

⇒ 20x = 240

⇒ x = 240 ÷ 20 = 12

⇒ First number = 4 × 12 = 48

⇒ Second number = 5 × 12 = 60

⇒ Sum of the numbers = 48 + 60 = 108

∴ The correct answer is 108.

Additional Information

Product of Two Numbers

For any two numbers, Product of numbers = HCF × LCM.

HCF and LCM Relation

The HCF (Highest Common Factor) is always a factor of the LCM (Least Common Multiple).

LCM of Ratios

If two numbers are in the ratio a : b, then their LCM = a × b × HCF and their Sum = (a + b) × HCF.

Shortcut Trick

Ratio of goods = 40 : (25% of 60) : (100 − 40 − 15) = 40 : 15 : 45 = 8 : 3 : 9.

Net change % = (Sum of weight × Value) ÷ Total weight

Net change % = (8 × 10 + 3 × 15 − 9 × 29) ÷ 20

Net change % = (80 + 45 − 261) ÷ 20 = −136 ÷ 20 = −6.8%.

Negative sign indicates loss, so x = 6.8.

∴ The correct answer is 6.8.

Alternate Method

Given:

Part 1 = 40% of goods, Profit = 10%

Part 2 = 25% of remaining (60%) = 15% of total, Profit = 15%

Part 3 = Rest (100% − 40% − 15%) = 45% of total, Loss = 29%

Formula Used:

Overall Profit/Loss % = (Σ Percentage Part × Profit/Loss %) ÷ 100

Calculations:

⇒ Let total quantity be 100 units and CP per unit be 1.

⇒ Total CP = 100

⇒ Profit from 40 units = 40 × 10% = 4

⇒ Profit from 15 units = 15 × 15% = 2.25

⇒ Loss from 45 units = 45 × 29% = 13.05

⇒ Net Profit/Loss = 4 + 2.25 − 13.05

⇒ Net Profit/Loss = 6.25 − 13.05 = −6.8

⇒ Overall Loss % = (6.8 ÷ 100) × 100 = 6.8%

⇒ Comparing with overall loss of x%, we get x = 6.8.

∴ The correct answer is 6.8.

Additional Information

Weighted Average Profit/Loss

When goods are divided into different parts sold at different profit/loss rates, the overall result is the weighted average: Net % = (Σ n_i × r_i) ÷ (Σ n_i).

Relation between Loss and x

If a problem states an 'overall loss of x%', the value of x is always positive. The 'loss' keyword itself accounts for the negative sign of the net result.

Successive Percentage vs Weighted Average

Weighted average is used for different parts of a whole, whereas successive percentage change [x + y + (xy/100)] is used when changes occur on the same value one after another.

Shortcut Trick

Identify the scaling factor between the first two terms: 4.8 ÷ 1.2 = 4.

Since the terms are in proportion, the same factor applies to the third and fourth terms.

Fourth proportional = 7.15 × 4 = 28.6.

∴ The correct answer is 28.6.

Alternate Method

Given:

First term (a) = 1.2

Second term (b) = 4.8

Third term (c) = 7.15

Formula Used:

If a, b, c, and d are in proportion (a : b :: c : d), then a × d = b × c.

Calculations:

⇒ Let the fourth proportional be d.

⇒ 1.2 : 4.8 :: 7.15 : d

⇒ 1.2 ÷ 4.8 = 7.15 ÷ d

⇒ 1 ÷ 4 = 7.15 ÷ d

⇒ d = 7.15 × 4

⇒ d = 28.6

∴ The correct answer is 28.6.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is given by √(a × b).

Third Proportional

If a, b, and c are in continued proportion (a : b :: b : c), then c is the third proportional, where c = b² ÷ a.

Product Rule of Proportion

In any proportion, the Product of Extremes (first and fourth terms) is always equal to the Product of Means (second and third terms).

Shortcut Trick

Directly multiply 1.5 × 0.2 to get 0.3.

Divide the result by the denominator 0.3.

Simplification: 0.3 ÷ 0.3 = 1

∴ The correct answer is 1.

Alternate Method

Given:

Expression: ((1.5 × 0.2)) / 0.3

Formula Used:

Follow the BODMAS order of operations, starting with multiplication in the numerator followed by division.

Calculations:

⇒ (1.5 × 0.2) ÷ 0.3

⇒ 0.3 ÷ 0.3

⇒ 1

∴ The correct answer is 1.

Additional Information

BODMAS Rule

The rule defines the correct sequence: Brackets, Of/Orders, Division, Multiplication, Addition, and Subtraction.

Decimal Multiplication

Multiply digits as whole numbers (15 × 2 = 30) and then shift the decimal point left by the total number of decimal places (1 + 1 = 2) to get 0.30.

Division of Decimals

Multiplying both the numerator and denominator by a power of 10 (like 10 or 100) can simplify the division by removing decimals.

Shortcut Trick

For a rectangular garden of length L and width W with a path of uniform width w outside:

Outer Perimeter = 2 × (L + W) + 8w

⇒ Outer Perimeter = 2 × (25 + 18) + 8 × 2

⇒ Outer Perimeter = 2 × 43 + 16 = 86 + 16 = 102 m

∴ The correct answer is 102 m.

Alternate Method

Given:

Length of garden (L) = 25 m

Width of garden (W) = 18 m

Width of path (w) = 2 m (outside)

When a path is constructed outside, it adds to the dimensions on both sides:

⇒ Outer Length = L + 2w = 25 + 2 × 2 = 29 m

⇒ Outer Width = W + 2w = 18 + 2 × 2 = 22 m

The perimeter of a rectangle is given by the formula: 2 × (Length + Width)

⇒ Perimeter of outer edge = 2 × (29 + 22)

⇒ Perimeter of outer edge = 2 × 51 = 102 m

∴ The correct answer is 102 m.

Additional Information

Area of Outer Path

The area of a path built outside a rectangle is calculated as: Area = (L + 2w)(W + 2w) − LW = 2w(L + W + 2w).

Area of Inner Path

If the path is built inside the garden, the Inner Dimensions become (L − 2w) and (W − 2w). Area = 2w(L + W − 2w).

Perimeter of Inner Path

The perimeter of the inner edge (when path is inside) is: Perimeter = 2(L + W − 4w) or 2 × (L + W) − 8w.

Shortcut Trick

Total work (LCM of 12 and 36) = 36 units.

Efficiency of A = 3 units/day; Efficiency of B = 1 unit/day.

Combined efficiency = 3 + 1 = 4 units/day.

Work done in 4 days = 4 × 4 = 16 units.

Remaining work = 36 − 16 = 20 units.

Days taken by B = 20 ÷ 1 = 20 days.

∴ The correct answer is 20 days.

Alternate Method

Given:

Time taken by A = 12 days

Time taken by B = 36 days

Working duration together = 4 days

Calculations:

⇒ Efficiency of A + B = 3 + 1 = 4 units/day

⇒ Work done by A and B in 4 days = 4 × 4 = 16 units

⇒ Remaining Work = Total Work − Completed Work

⇒ Remaining Work = 36 − 16 = 20 units

⇒ Time taken by B = Remaining Work ÷ Efficiency of B

⇒ Time taken by B = 20 ÷ 1 = 20 days

∴ The correct answer is 20 days.

Additional Information

Work-Efficiency Relation

Total Work = Efficiency × Time. Efficiency is inversely proportional to the time taken to complete a task.

Pipes and Cisterns

Similar logic applies where efficiency of inlet pipes is positive and outlet/leakage efficiency is negative.

Work and Wages

Wages are distributed in the ratio of work done or the ratio of efficiencies when workers work for the same duration.

Shortcut Trick

Apply the BODMAS rule sequence: Brackets → Multiplication → Addition.

Solve inner parentheses first: (26 − 11 + 6) = 21.

Multiply by −9: −9 × 21 = −189.

Final calculation: 30 + (−189) = 30 − 189 = −159.

∴ The correct answer is −159.

Alternate Method

Given: 30 + [{−9 × (26 − 11 + 6)}]

Method Used: BODMAS/VBODMAS Rule for order of operations.

⇒ Step 1: Simplify the value inside the small brackets ().

⇒ (26 − 11 + 6) = 15 + 6 = 21

⇒ Step 2: Substitute this value back into the braces {}.

⇒ {−9 × 21} = −189

⇒ Step 3: Now the expression is 30 + [−189].

⇒ 30 − 189 = −159

∴ The correct answer is −159.

Additional Information

BODMAS Rule

It specifies the order: Brackets, Orders (powers, roots, etc.), Division, Multiplication, Addition, and Subtraction.

Types of Brackets

The order of opening brackets is Bar/Vinculum, then Parentheses ( ), then Braces { }, and finally Square Brackets [ ].

Integer Multiplication Rules

Multiplying a positive number by a negative number always results in a negative product (+ × − = −).

Shortcut Trick

When an inlet pipe takes T₁ hours and with a leak it takes T₂ hours:

Time taken by leak alone to empty = (T₁ × T₂) ÷ (T₂ − T₁)

Time = (10 × 12) ÷ (12 − 10)

Time = 120 ÷ 2 = 60 hr

∴ The correct answer is 60 hr.

Alternate Method

Given:

Filling time of Pipe A = 10 hr

Filling time with leakage (A + Leak) = 12 hr

Formula Used:

Total Capacity = LCM of individual times

Efficiency = Total Capacity ÷ Time

⇒ Efficiency of Pipe A = 6 units/hr

⇒ Net efficiency with leak = 5 units/hr

⇒ Efficiency of Leak alone = 6 − 5 = 1 unit/hr

⇒ Time taken by leak to empty the tank = Total Work ÷ Leak Efficiency

⇒ Time = 60 ÷ 1 = 60 hr

∴ The correct answer is 60 hr.

Additional Information

Negative Work Concept

In pipes and cisterns, an outlet pipe or a leak performs negative work, which reduces the overall efficiency of the filling system.

Net Efficiency Rule

If multiple pipes operate, the net efficiency is the algebraic sum of individual efficiencies (positive for inlet, negative for outlet).

Capacity and Time Relation

The time taken to fill or empty is inversely proportional to the efficiency: Time = Work ÷ Efficiency.

Shortcut Trick

Total interest earned = ₹74,263 − ₹70,000 = ₹4,263.

Total interest percentage = (4,263 ÷ 70,000) × 100 = 6.09%.

Using successive percentage for 3%: 3 + 3 + (3 × 3)/100 = 6.09%.

This matches for a time period of 2 years.

∴ The correct answer is 2 years.

Alternate Method

Given:

Principal (P) = ₹70,000

Amount (A) = ₹74,263

Rate (R) = 3% p.a. compounded annually.

The formula for Amount (A) under Compound Interest is:

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

⇒ 74263 = 70000(1 + 3/100)^X

⇒ 74263 ÷ 70000 = (1 + 0.03)^X

⇒ 10609 ÷ 10000 = (1.03)^X

⇒ 1.0609 = (1.03)^X

⇒ (1.03)² = (1.03)^X

⇒ X = 2 years.

∴ The correct answer is 2 years.

Additional Information

Compound Interest Formula

The amount is given by A = P(1 + r/n)^nt, where n is the number of times interest is compounded per year.

Effective Interest Rate

For compounding annually, the effective rate for 2 years at r% is r + r + (r×r)/100.

Difference between CI and SI

For 2 years, the difference between Compound Interest and Simple Interest is CI − SI = P(R/100)².