Search Previous Year Questions

Given:

30 + [{-9 × (26 − 11 + 3)}]

Formula used:

The BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction) is used to simplify the expression.

Calculations:

30 + [{-9 × (26 − 11 + 3)}]

⇒ 30 + [{-9 × (15 + 3)}]

⇒ 30 + [{-9 × 18}]

⇒ 30 + (−162)

⇒ 30 − 162

⇒ −132

∴ The correct answer is option (4).

Shortcut Trick

Sum of ratio squares = 3² + 5² + 7² = 9 + 25 + 49 = 83 parts.

Value of 1 part² = 2075 ÷ 83 = 25.

Value of 1 part = √25 = 5.

Middle number (5 parts) = 5 × 5 = 25.

∴ The correct answer is 25.

Alternate Method

Given:

Ratio of three positive numbers = 3 : 5 : 7

Sum of their squares = 2075

Calculations:

⇒ Let the three positive numbers be 3x, 5x, and 7x.

⇒ Sum of their squares = (3x)² + (5x)² + (7x)² = 2075

⇒ 9x² + 25x² + 49x² = 2075

⇒ 83x² = 2075

⇒ x² = 2075 ÷ 83

⇒ x² = 25

⇒ x = √25 = 5 (Since numbers are positive)

⇒ The middle number = 5x = 5 × 5 = 25

∴ The correct answer is 25.

Additional Information

Ratio and Proportion Basics

A ratio compares two quantities. If a : b = c : d, then a × d = b × c. To find actual values from a ratio, always multiply by a common constant x.

Duplicate Ratio

The duplicate ratio of a : b is a² : b². When squares of numbers in a ratio are given, the ratio of those squares is the duplicate ratio.

Mean Proportional

The mean proportional between two numbers a and b is √(a × b). This is a common property used in geometric progression and ratio problems.

Shortcut Trick

Ratio (8 years ago) G : R = 4 : 1 (Difference = 3 units)

Ratio (6 years later) G : R = 2 : 1 (Difference = 1 unit)

Make Difference equal by multiplying second ratio by 3 → 6 : 3

Change in units = 6 − 4 = 2 units | Total years gap = 8 + 6 = 14 years

2 units = 14 years ⇒ 1 unit = 7 years | G (8 years ago) = 4 × 7 = 28 years

∴ The correct answer is 36 years.

Alternate Method

Given:

Age of Mr. Gupta (8 years ago) = 4 × age of Rohit (8 years ago)

Age of Mr. Gupta (after 6 years) = 2 × age of Rohit (after 6 years)

Let Rohit’s age 8 years ago be x.

⇒ Mr. Gupta’s age 8 years ago = 4x

⇒ Present ages: Rohit = x + 8, Mr. Gupta = 4x + 8

⇒ Ages after 6 years: Rohit = x + 8 + 6 = x + 14, Mr. Gupta = 4x + 8 + 6 = 4x + 14

According to the question:

⇒ 4x + 14 = 2 × (x + 14)

⇒ 4x + 14 = 2x + 28

⇒ 4x − 2x = 28 − 14

⇒ 2x = 14

⇒ x = 7

⇒ Present Age of Mr. Gupta = 4x + 8 = 4 × 7 + 8 = 28 + 8 = 36 years

∴ The correct answer is 36 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals always remains constant regardless of the number of years passed or preceding.

Ratio Balancing Method

To solve age problems quickly, ensure the difference between ratio parts is equal for both time periods by multiplying with suitable constants.

Linear Equation Formulation

Always define variables for the past or future to simplify the present age expression: Present = Past + t or Present = Future − t.

Shortcut Trick

Net discount for Option 2 = 20 + 15 − (20 × 15) ÷ 100 = 32%

Discount for Option 1 = 30%

Difference in discount percentage = 32% − 30% = 2%

Amount difference = 2% of ₹8,000 = ₹160

Since 32% > 30%, Option 2 provides the greater discount.

∴ The correct answer is Option 2 and by ₹160.

Alternate Method

Given:

Marked Price (MP) = ₹8,000

Option 1: Single discount = 30%

Option 2: Successive discounts = 20% and 15%

Calculations:

⇒ Discount in Option 1 = 30% of 8,000 = (30 ÷ 100) × 8,000 = ₹2,400

⇒ Successive discount formula = x + y − (xy ÷ 100)

⇒ Effective discount % for Option 2 = 20 + 15 − (20 × 15 ÷ 100) = 35 − 3 = 32%

⇒ Discount in Option 2 = 32% of 8,000 = (32 ÷ 100) × 8,000 = ₹2,560

⇒ Difference in discount amounts = ₹2,560 − ₹2,400 = ₹160

⇒ Comparing both, Option 2 offers a discount that is ₹160 more than Option 1.

∴ The correct answer is Option 2 and by ₹160.

Additional Information

Successive Discount Formula

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

Marked Price and Selling Price

Discount is always calculated on the Marked Price (MP). The price after deduction is the Selling Price (SP), where SP = MP × (100 − Discount%)/100.

Three Successive Discounts

If three discounts a%, b%, and c% are given, first find the equivalent of a and b, then use that result with c using the same formula.

Shortcut Trick

Value of 4 extra ₹500 notes = 4 × 500 = ₹2,000.

Remaining Total Amount = 24,800 − 2,000 = ₹22,800.

Combined value of one ₹100 note and one ₹500 note = 100 + 500 = ₹600.

Number of ₹100 notes = 22,800 ÷ 600 = 38.

∴ The correct answer is 38.

Alternate Method

Given: Total Amount = ₹24,800; ₹500 notes = 4 + ₹100 notes.

⇒ Let the number of ₹100 notes be x.

⇒ Number of ₹500 notes = x + 4.

⇒ Total Value = (x × 100) + ((x + 4) × 500) = 24,800.

⇒ 100x + 500x + 2,000 = 24,800.

⇒ 600x + 2,000 = 24,800.

⇒ 600x = 24,800 − 2,000 = 22,800.

⇒ x = 22,800 ÷ 600 = 38.

∴ The correct answer is 38.

Additional Information

Currency Denomination Principle

The Total Value of currency is calculated by the sum of (Value of one note × Number of notes) for each denomination.

Linear Equations in One Variable

A Linear Equation involves solving for an unknown variable by balancing constants and variables on either side of the equals sign.

Unitary Method Adjustment

In problems where one quantity is more than another, subtracting the extra value first allows the remaining amount to be divided by the sum of unit values.

Shortcut Trick

Number of sides (n) = (2 × Area) ÷ (Side × Apothem)

n = (2 × 247.5) ÷ (11 × 9)

n = 495 ÷ 99

n = 5

∴ The correct answer is 5.

Alternate Method

Given:

Side length (s) = 11 cm

Area (A) = 247.5 cm²

Perpendicular distance from centre to side (apothem, a) = 9 cm

Formula Used:

Area of a regular polygon = (1/2) × Perimeter × Apothem

Where Perimeter (P) = Number of sides (n) × Side length (s)

Calculations:

⇒ Area = (1/2) × (n × s) × a

⇒ 247.5 = (1/2) × (n × 11) × 9

⇒ 247.5 = (99/2) × n

⇒ 247.5 = 49.5 × n

⇒ n = 247.5 ÷ 49.5

⇒ n = 5

∴ The correct answer is 5.

Additional Information

Interior Angle of a Regular Polygon

The measure of each interior angle is given by [(n − 2) × 180°] ÷ n, where n is the number of sides.

Sum of Exterior Angles

For any convex polygon, the sum of all exterior angles is always 360°.

Relation between Side and Apothem

For a regular polygon, the side length (s) and apothem (a) are related by: s = 2 × a × tan(180°/n).

Shortcut Trick

Discount = 10630 − 9248.10 = 1381.90

Discount % = (Discount ÷ Marked Price) × 100

Discount % = (1381.90 ÷ 10630) × 100 = 13%

∴ The correct answer is 13%.

Alternate Method

Given:

Marked Price (MP) of the article = ₹10,630

Selling Price (SP) of the article = ₹9248.10

Formula Used:

Discount = Marked Price − Selling Price

Discount Percentage = (Discount ÷ Marked Price) × 100

Calculations:

⇒ Discount = 10630 − 9248.10

⇒ Discount = 1381.90

⇒ Discount % = (1381.90 ÷ 10630) × 100

⇒ Discount % = 138190 ÷ 10630

⇒ Discount % = 13%

∴ The correct answer is 13%.

Additional Information

Marked Price (List Price)

The price at which an article is listed or labeled by the retailer or manufacturer is called its Marked Price.

Successive Discounts

If two discounts d₁% and d₂% are given, the single equivalent discount is [d₁ + d₂ − (d₁ × d₂) ÷ 100]%.

Relationship between CP and MP

The ratio of Cost Price to Marked Price is given by: CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%).

Shortcut Trick

Since the height H remains constant, the Volume of a cylinder is directly proportional to the square of its radius R (V ∝ R²).

Original Radius = R ⇒ Original Volume ∝ R².

New Radius = R/2 ⇒ Volume of one new cylinder ∝ (R/2)² = R²/4.

Total Volume of 3 new cylinders = 3 × (R²/4) = 3/4 of the original volume.

Unused fraction of Volume = 1 − 3/4 = 1/4.

∴ The correct answer is 1/4.

Alternate Method

Given: Original radius = R, height = H. New radius = R/2, height = H, and number of new cylinders = 3.

Formula Used: Volume of a Cylinder = π × r² × h

⇒ Volume of original cylinder (V₁) = πR²H

⇒ Volume of one new cylinder (v) = π × (R/2)² × H = π × (R²/4) × H = πR²H/4

⇒ Total volume of 3 new cylinders (V₂) = 3 × (πR²H/4) = 3πR²H/4

⇒ Unused Volume = V₁ − V₂ = πR²H − 3πR²H/4

⇒ Unused Volume = (4πR²H − 3πR²H) / 4 = πR²H/4

⇒ Fraction of original volume remaining unused = (Unused Volume) ÷ (Original Volume)

⇒ Fraction = (πR²H/4) ÷ (πR²H) = 1/4

∴ The correct answer is 1/4.

Additional Information

Volume of a Cylinder

The amount of space occupied by a cylinder is given by V = πr²h, where r is the radius and h is the height.

Scaling Effect on Volume

If the radius is changed by a factor of 'k' and the height remains constant, the volume changes by a factor of k².

Total Surface Area of Cylinder

The total surface area including the top and bottom bases is given by 2πr(h + r).

Shortcut Trick

Identify the direct algebraic relationship: (Difference)² = (Sum)² − 4 × (Product).

Substitute the given values: (Difference)² = 88² − 4 × 1927

(Difference)² = 7744 − 7708 = 36

Difference = √36 = 6

∴ The correct answer is 6.

Alternate Method

Given:

Sum of two numbers (x + y) = 88

Product of two numbers (xy) = 1927

Formula Used:

(x − y)² = (x + y)² − 4xy

Calculations:

⇒ (x − y)² = (88)² − 4 × (1927)

⇒ Square of 88 = (90 − 2)² = 8100 − 360 + 4 = 7744

⇒ 4 × 1927 = 7708

⇒ (x − y)² = 7744 − 7708

⇒ (x − y)² = 36

⇒ x − y = √36

⇒ x − y = 6

∴ The correct answer is 6.

Additional Information

Algebraic Identity

The relation (x + y)² − (x − y)² = 4xy is a fundamental identity used to find the difference when sum and product are known.

Quadratic Roots Property

For a quadratic equation t² − (Sum)t + (Product) = 0, the roots are the two numbers. The difference between roots is given by √D/a where D = b² − 4ac.

Squaring Shortcut

To square numbers near 100 like 88: (100 − 12)² = 10000 − 2400 + 144 = 7744.

Shortcut Trick

Let Total Work (Capacity) = LCM(10, 15, 30) = 30 units.

Efficiencies: Pipe A = +3, Pipe B = +2, Pipe C = −1. Combined = +4 units/hr.

Work done by all three in 6 hr = 4 × 6 = 24 units.

Remaining Work = 30 − 24 = 6 units.

Efficiency of (B + C) = 2 − 1 = 1 unit/hr. Additional time = 6 ÷ 1 = 6 hr.

Total time = 6 (initial) + 6 (extra) = 12 hr.

∴ The correct answer is 12 hr.

Alternate Method

Given: Pipe A fills in 10 hr, Pipe B fills in 15 hr, and Pipe C empties in 30 hr.

Total Work is the LCM of 10, 15, and 30.

⇒ Combined Efficiency (A + B + C) = 3 + 2 − 1 = 4 units/hr

⇒ Work completed by all pipes in first 6 hours = 4 × 6 = 24 units

⇒ Remaining capacity to be filled = 30 − 24 = 6 units

⇒ Since Pipe A is closed, remaining pipes (B and C) fill the rest.

⇒ Combined Efficiency of (B + C) = 2 − 1 = 1 unit/hr

⇒ Time taken to fill remaining 6 units = 6 ÷ 1 = 6 hr

⇒ Total Time taken to fill the cistern = Initial time + Extra time = 6 hr + 6 hr = 12 hr

∴ The correct answer is 12 hr.

Additional Information

Inlet and Outlet Pipes

The rate of work of an inlet pipe (filling) is taken as positive, while that of an outlet pipe (emptying) is taken as negative efficiency.

Net Work Done

If multiple pipes work together, the net efficiency is the algebraic sum of their individual efficiencies (e.g., Net E = E₁ + E₂ − E₃).

Time and Efficiency Relation

Total Capacity = Efficiency × Time. If the net efficiency is negative, the cistern will eventually be emptied rather than filled.