Search Previous Year Questions

Shortcut Trick

Total Work (LCM of 15, 20, 6) = 60 units.

Efficiencies: A = 4, B = 3, (A+B+C) = 10.

Efficiency of C = 10 − (4 + 3) = 3 units.

C's share = (Efficiency of C ÷ Total Efficiency) × Total Amount.

C's share = (3 ÷ 10) × 18,000 = ₹ 5,400.

∴ The correct answer is ₹ 5,400.

Alternate Method

Given: A = 15 days, B = 20 days, A+B+C = 6 days, Total Amount = ₹ 18,000.

Formula: Efficiency = Total Work ÷ Time; Share of money ∝ Efficiency.

Combined efficiency of A + B + C = 10 units/day

⇒ Efficiency of C = Total Efficiency − (Efficiency of A + Efficiency of B)

⇒ Efficiency of C = 10 − (4 + 3) = 3 units/day

⇒ Ratio of shares A : B : C = 4 : 3 : 3

⇒ C's share = (3 ÷ 10) × 18,000

⇒ C's share = 3 × 1,800 = 5,400

∴ The correct answer is ₹ 5,400.

Additional Information

Work and Wages

Wages are distributed in the ratio of work done. If all persons work for the same duration, wages are distributed in the ratio of their efficiencies.

Efficiency and Time

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

Combined Work Formula

If A, B, and C work together, their combined efficiency is the sum of their individual efficiencies: E_total = E_A + E_B + E_C.

Shortcut Trick

Sum of three values = 50 × 3 = 150.

Let 1^st value = x, then 3^rd value = 2x and 2^nd value = 2x − 30.

Total sum = x + (2x − 30) + 2x = 5x − 30.

5x − 30 = 150 ⇒ 5x = 180 ⇒ x = 36.

∴ The correct answer is 36.

Alternate Method

Given:

Average of 3 values = 50

1^st value = 1/2 × 3^rd value

3^rd value = 2^nd value + 30

Formula Used:

Sum of observations = Average × Number of observations

Calculations:

⇒ Total sum = 50 × 3 = 150

⇒ Let the 3^rd value be 2x.

⇒ 1^st value = 1/2 × 2x = x

⇒ 2^nd value = 3^rd value − 30 = 2x − 30

⇒ x + (2x − 30) + 2x = 150

⇒ 5x − 30 = 150

⇒ 5x = 180

⇒ x = 36

∴ The correct answer is 36.

Additional Information

Average Concept

The average (arithmetic mean) is calculated by dividing the sum of all observations by the total number of observations: Avg = Σx ÷ n.

Total Sum Rule

If the average of n numbers is A, then their sum is always n × A. This is the starting point for most average-based word problems.

Algebraic Translation

When comparing three values, express all unknowns in terms of a single variable (usually the one referenced most) to form a linear equation.

Shortcut Trick

Divisibility by 11: Sum of digits at odd positions − Sum of digits at even positions = 0 or 11k.

Sum of digits at odd positions (3 + 8) = 11.

Sum of digits at even positions (P + 2) = P + 2.

Difference: 11 − (P + 2) = 0 ⇒ 9 − P = 0 ⇒ P = 9.

∴ The correct answer is 9.

Alternate Method

Given:

Number = 28P3

Divisible by 11

Formula Used:

A number is divisible by 11 if the difference of the sum of digits at odd places and the sum of digits at even places is 0, 11, 22, etc.

Calculations:

⇒ Sum of digits at odd places (from right) = 3 + 8 = 11

⇒ Sum of digits at even places (from right) = P + 2

⇒ Difference = 11 − (P + 2)

⇒ 11 − P − 2 = 0 (for the smallest multiple of 11)

⇒ 9 − P = 0

⇒ P = 9

∴ The correct answer is 9.

Additional Information

Divisibility by 11 Rule

A number is divisible by 11 if the alternating sum of its digits is divisible by 11. Example: For abcd, (d + b) − (c + a) = 0 or 11k.

Divisibility by 3 and 9

A number is divisible by 3 (or 9) if the sum of all its digits is divisible by 3 (or 9).

Divisibility by 4

A number is divisible by 4 if the last two digits of the number are divisible by 4 or are both zeros.

Shortcut Trick

Difference of squares of two consecutive odd/even numbers = 4 × (Middle number).

Middle even number = 24 ÷ 4 = 6.

The sum of the two numbers is twice the middle number.

Sum = 2 × 6 = 12.

∴ The correct answer is 12.

Alternate Method

Given: Difference of squares of two consecutive odd numbers = 24.

Formula Used: a² − b² = (a + b)(a − b)

Let the two consecutive odd numbers be x and (x + 2).

⇒ (x + 2)² − x² = 24

⇒ (x + 2 + x)(x + 2 − x) = 24

⇒ (2x + 2) × 2 = 24

⇒ 2x + 2 = 24 ÷ 2

⇒ 2x + 2 = 12

Since the sum of the numbers is x + (x + 2) = 2x + 2:

∴ The correct answer is 12.

Additional Information

Consecutive Odd/Even Numbers

Any two consecutive odd or even numbers always have a common difference of 2. If the first number is n, the next is n + 2.

Difference of Squares of Consecutive Odd/Even Numbers

The difference (n+2)² − n² simplifies to 4n + 4, which is exactly 4 × (n + 1), where (n + 1) is the even/odd number between them.

Sum of Consecutive Numbers

The sum of any two consecutive odd numbers is always a multiple of 4, while the sum of two consecutive even numbers is a multiple of 2 but not necessarily 4.

Shortcut Trick

Let the side of the square be S.

Area of square = Area of rectangle ⇒ S² = (S/2 + 9) × (S/2)

Divide both sides by S: S = (S/2 + 9) ÷ 2

Multiply by 2: 2S = S/2 + 9 ⇒ 1.5S = 9

S = 9 ÷ 1.5 = 6 m.

∴ The correct answer is 6 m.

Alternate Method

Given:

Area of square park = Area of rectangle

Breadth of rectangle (B) = Side of square (S) / 2

Length of rectangle (L) = Breadth + 9 m

Formula Used:

Area of square = Side²

Area of rectangle = Length × Breadth

Calculations:

⇒ Let the side of the square be S m.

⇒ Breadth of rectangle (B) = S/2 m

⇒ Length of rectangle (L) = (S/2 + 9) m

⇒ Area of square = Area of rectangle

⇒ S² = (S/2 + 9) × (S/2)

⇒ S² = S²/4 + 9S/2

⇒ Multiply the entire equation by 4:

⇒ 4S² = S² + 18S

⇒ 3S² − 18S = 0

⇒ 3S(S − 6) = 0

⇒ S = 6 m (Since side cannot be zero)

∴ The correct answer is 6 m.

Additional Information

Area and Perimeter of Square

Area = side² and Perimeter = 4 × side. The diagonal of a square is side × √2.

Area and Perimeter of Rectangle

Area = length × breadth and Perimeter = 2 × (length + breadth). The diagonal is √(length² + breadth²).

Unit Conversion

Always ensure all dimensions (length, breadth, side) are in the same unit (m, cm, etc.) before equating areas.

Shortcut Trick

Align all decimal numbers to the same length by adding trailing zeros.

0.7000 + 0.0700 + 0.0070 + 0.0007

Summing each decimal place: 0.7 in the 1^st place, 7 in the 2^nd, 7 in the 3^rd, and 7 in the 4^th.

Total = 0.7777.

∴ The correct answer is 0.7777.

Alternate Method

Given expression: 0.0007 + 0.07 + 0.007 + 0.7

To add decimals, arrange them vertically by aligning the decimal points:

⇒ 0.7000

⇒ 0.0700

⇒ 0.0070

⇒ 0.0007

⇒ ____________

⇒ Sum = 0.7777

∴ The correct answer is 0.7777.

Additional Information

Decimal Addition Rule

Always align the decimal points vertically. If a number has fewer decimal places than others, add trailing zeros to balance the length for easier calculation.

BODMAS in Simplification

For mixed operations, follow the order: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Place Value of Decimals

The positions to the right of the decimal point represent tenths (10⁻¹), hundredths (10⁻²), thousandths (10⁻³), and ten-thousandths (10⁻⁴).

Shortcut Trick

Ratio of Income : Expenditure = 7 : 6

Saving unit = Income unit − Expenditure unit = 7 − 6 = 1 unit

Given: 7 units (Income) = ₹3500

⇒ 1 unit (Saving) = 3500 ÷ 7 = ₹500

∴ The correct answer is ₹500.

Alternate Method

Given:

Income : Expenditure = 7 : 6

Total Income = ₹3500

Formula Used:

Saving = Income − Expenditure

⇒ Let Income = 7x and Expenditure = 6x

⇒ 7x = 3500

⇒ x = 3500 ÷ 7

⇒ x = 500

⇒ Saving = Income − Expenditure

⇒ Saving = 7x − 6x = 1x

⇒ Saving = 1 × 500 = ₹500

∴ The correct answer is ₹500.

Additional Information

Income-Expenditure Relationship

Income is the sum of Expenditure and Savings (Income = Expenditure + Savings).

Saving Percentage

It is the ratio of saving to total income expressed as a percentage: (Saving ÷ Income) × 100.

Ratio Multiplier

If A : B is the ratio and the value of A is given, the multiplier is Value of A ÷ Ratio unit of A.

Shortcut Trick

Apply the unitary formula: (Price₁ / (Quantity₁ × Selling%₁)) = (Price₂ / (Quantity₂ × Selling%₂))

Here, Selling%₁ = 100 − 20 = 80 and Selling%₂ = 100 + 60 = 160

⇒ (3600 / (36 × 80)) = (3200 / (n × 160))

⇒ 100 / 80 = 20 / n

⇒ n = (20 × 80) / 100 = 16

∴ The correct answer is 16 bags.

Alternate Method

Given:

Initial Selling Price (SP₁) = ₹3600 for 36 bags

Initial Loss = 20% | Target Profit = 60%

Target Selling Price (SP₂) = ₹3200

Formula Used:

CP = SP × [100 / (100 − Loss%)]

New SP = CP × [(100 + Profit%) / 100]

⇒ Initial SP per bag = 3600 ÷ 36 = ₹100

⇒ Cost Price (CP) per bag = 100 × (100 / 80) = ₹125

⇒ New SP per bag for 60% profit = 125 × (160 / 100) = ₹200

⇒ Number of bags for ₹3200 = 3200 ÷ 200

⇒ Number of bags = 16

∴ The correct answer is 16 bags.

Additional Information

Successive Percentage for SP

If an item is sold at multiple profit/loss stages, the final SP = CP × (1 ± r₁/100) × (1 ± r₂/100).

Marked Price and Discount

The relationship between CP and MP is given by: CP/MP = (100 − Discount%)/(100 + Profit%).

Dishonest Dealer Concept

Profit% = [Error / (True Value − Error)] × 100 when the dealer uses a false weight.

Shortcut Trick

Apply BODMAS: Bracket → Division → Addition → Subtraction.

Solve bracket and division first: 19 × 2 = 38 and 37 ÷ 1 = 37.

New expression: 83 + 38 − 44 + 37 + 26 − 21.

Group terms: (83 + 38 + 37 + 26) − (44 + 21) = 184 − 65.

Final calculation: 184 − 65 = 119.

∴ The correct answer is 119.

Alternate Method

Given: 83 + (19 × 2) − 44 + 37 ÷ 1 + 26 − 21

Formula: Follow BODMAS (Bracket, Of, Division, Multiplication, Addition, Subtraction).

⇒ 83 + 38 − 44 + 37 ÷ 1 + 26 − 21

⇒ 83 + 38 − 44 + 37 + 26 − 21

⇒ 121 − 44 + 37 + 26 − 21

⇒ 77 + 37 + 26 − 21

⇒ 114 + 26 − 21

⇒ 140 − 21

⇒ 119

∴ The correct answer is 119.

Additional Information

BODMAS / VBODMAS Rule

Defines the order of operations: Vinculum (Bar), Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Identity Element of Division

Dividing any number by 1 results in the original number itself (x ÷ 1 = x).

Left-to-Right Rule

Operations like addition and subtraction have the same priority and can be solved from left to right or by grouping.

Total marks of student 3 and 5 = 10 + 7 = 17

Total marks of student 4 and 1 = 6 + 8 = 14

∴ Required difference = 17 – 14 = 3