Search Previous Year Questions

Shortcut Trick

Apply the Successive Percentage formula: Net % change = a + b + (ab/100)

Here, Mark-up (a) = +40 and Discount (b) = −25

Net change = 40 − 25 + [40 × (−25) ÷ 100]

Net change = 15 − 10 = 5%

Positive result indicates Profit.

∴ The correct answer is 5% profit.

Alternate Method

Given:

Mark-up Percentage = 40%

Discount Percentage = 25%

Calculations:

⇒ Let the Cost Price (CP) of the article be 100.

⇒ Marked Price (MP) = CP + 40% of CP = 100 + 40 = 140.

⇒ Discount = 25% of MP = (25/100) × 140 = 35.

⇒ Selling Price (SP) = MP − Discount = 140 − 35 = 105.

⇒ Profit = SP − CP = 105 − 100 = 5.

⇒ Profit Percentage = (Profit ÷ CP) × 100 = (5 ÷ 100) × 100 = 5%.

∴ The correct answer is 5% profit.

Additional Information

CP-MP Relation Formula

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This is extremely useful for finding any one unknown variable when others are given.

Markup Percentage

Markup% = [(Marked Price − Cost Price) ÷ Cost Price] × 100. It represents how much the price is increased over the cost before selling.

Successive Discounts

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

Shortcut Trick

Ratio of areas: (Base₁ × Height₁) ÷ (Base₂ × Height₂) = Area₁ ÷ Area₂

Given: Area₂ = 2 × Area₁ ⇒ Area₁ : Area₂ = 1 : 2

(18 × 20) ÷ (24 × H₂) = 1 ÷ 2

360 ÷ (24 × H₂) = 1 ÷ 2

H₂ = (360 × 2) ÷ 24 = 720 ÷ 24 = 30 cm

∴ The correct answer is 30 cm.

Alternate Method

Given:

Base of first triangle (b₁) = 18 cm

Height of first triangle (h₁) = 20 cm

Base of second triangle (b₂) = 24 cm

Area of second triangle (A₂) = 2 × Area of first triangle (A₁)

Formula Used:

Area of a triangle = (1 ÷ 2) × base × height

Calculations:

⇒ Area₁ = (1 ÷ 2) × 18 × 20

⇒ Area₁ = 9 × 20 = 180 cm²

⇒ Area₂ = 2 × Area₁ = 2 × 180 = 360 cm²

⇒ 360 = (1 ÷ 2) × 24 × h₂

⇒ 360 = 12 × h₂

⇒ h₂ = 360 ÷ 12

⇒ h₂ = 30 cm

∴ The correct answer is 30 cm.

Additional Information

Heron's Formula

When all three sides a, b, and c are known, Area = √[s(s−a)(s−b)(s−c)], where s is the semi-perimeter (a+b+c)/2.

Equilateral Triangle Area

For a triangle with all sides equal to 'a', the area is given by (√3 ÷ 4) × a².

Right-Angled Triangle

The area is (1 ÷ 2) × base × perpendicular, where the two sides forming the 90° angle act as base and height.

Shortcut Trick

Net efficiency = Efficiency of Tap 1 + Efficiency of Tap 2 − Efficiency of Tap 3.

Since Tap 1 fills in 4 hr and Tap 3 empties in 4 hr, their rates cancel each other out completely.

The net filling rate is solely determined by Tap 2, which takes 16 hr to fill the full cistern.

Time to fill 1/4 of the cistern = 1/4 × 16 hr = 4 hr.

∴ The correct answer is 4 hr.

Alternate Method

Given: Filling time for Tap A = 4 hr, Tap B = 16 hr, and Emptying time for Tap C = 4 hr.

Formula Used: Total Work = L.C.M. of time taken | Efficiency = Total Work ÷ Time.

⇒ Total efficiency when all taps are open = 4 + 1 − 4 = 1 unit/hr

⇒ Required capacity to fill = 1/4 × Total Capacity = 1/4 × 16 = 4 units

⇒ Time taken = Required Capacity ÷ Total Efficiency

⇒ Time taken = 4 units ÷ 1 unit/hr = 4 hr

∴ The correct answer is 4 hr.

Additional Information

Efficiency of Emptying Pipe

The efficiency of an outlet or emptying pipe is always taken as negative because it reduces the total volume of the cistern.

Net Work Formula

If pipe A fills in x hr and pipe B empties in y hr, the net part filled in 1 hr is (1/x − 1/y). If y > x, the tank will eventually fill.

Fractional Filling

To find the time for a fractional part of the tank, multiply the total time required for a full tank by that specific fraction.

Shortcut Trick

By the Remainder Theorem, if a polynomial is divisible by (x − 5), substituting x = 5 must yield zero.

Put x = 5 in the expression: 5³ − 3(5²) − 7(5) − k = 0

125 − 75 − 35 − k = 0

15 − k = 0 ⇒ k = 15

∴ The correct answer is 15.

Alternate Method

Given:

Polynomial P(x) = x³ − 3x² − 7x − k

Divisor = (x − 5)

Formula Used:

According to the Factor Theorem, if (x − a) is a factor of P(x), then P(a) = 0.

Calculations:

⇒ For the expression to be divisible by (x − 5), the remainder must be 0.

⇒ P(5) = 0

⇒ (5)³ − 3 × (5)² − 7 × (5) − k = 0

⇒ 125 − 3 × 25 − 35 − k = 0

⇒ 125 − 75 − 35 − k = 0

⇒ 50 − 35 − k = 0

⇒ 15 − k = 0

⇒ k = 15

∴ The correct answer is 15.

Additional Information

Remainder Theorem

When a polynomial f(x) is divided by (x − a), the remainder is always f(a).

Factor Theorem

A linear polynomial (x − a) is a factor of a polynomial f(x) if and only if f(a) = 0.

Divisibility Rule for Polynomials

If P(x) is divisible by Q(x), then every root of Q(x) must also be a root of P(x).

Shortcut Trick

Given cos A = 4/5, we identify the Pythagorean triplet (3, 4, 5) where Base = 4, Hypotenuse = 5, and Perpendicular = 3.

Then sin A = 3/5, tan A = 3/4, and sec A = 5/4.

Substitute values into the expression: ((3/5)² − 5×(4/5)² + (5/4)²) ÷ (3/4)².

Numerator = (9/25 − 80/25 + 25/16) = −71/25 + 25/16 = (−1136 + 625)/400 = −511/400.

Result = (−511/400) × (16/9) = −511/(25×9) = −511/225.

∴ The correct answer is −511/225.

Alternate Method

Given: cos A = 4/5

Formula Used: In a right-angled triangle, cos A = Base/Hypotenuse. By Pythagoras theorem, Perpendicular² = Hypotenuse² − Base². Also, sin A = P/H, tan A = P/B, and sec A = H/B.

Calculations:

⇒ Let Base (B) = 4 and Hypotenuse (H) = 5.

⇒ Perpendicular (P) = √(5² − 4²) = √(25 − 16) = √9 = 3.

⇒ sin A = P/H = 3/5

⇒ tan A = P/B = 3/4

⇒ sec A = H/B = 5/4

⇒ Value = (sin² A − 5 cos² A + sec² A) / tan² A

⇒ Value = ((3/5)² − 5×(4/5)² + (5/4)²) / (3/4)²

⇒ Value = (9/25 − 5×(16/25) + 25/16) / (9/16)

⇒ Value = (9/25 − 80/25 + 25/16) / (9/16) = (−71/25 + 25/16) / (9/16)

⇒ Value = [(−71×16 + 25×25) / (25×16)] ÷ (9/16)

⇒ Value = [(−1136 + 625) / 400] ÷ (9/16) = (−511/400) × (16/9)

⇒ Value = −511 / (25 × 9) = −511/225.

∴ The correct answer is −511/225.

Additional Information

Trigonometric Identities

Standard identities include sin² A + cos² A = 1, 1 + tan² A = sec² A, and 1 + cot² A = cosec² A.

Pythagorean Triplets

Common integer sets satisfying a² + b² = c² like (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25) are extremely useful in exam time-saving.

Reciprocal Ratios

The reciprocal ratios are defined as sec A = 1/cos A, cosec A = 1/sin A, and cot A = 1/tan A.

Shortcut Trick

Simplify the first bracket: {15/42 × 3/15} = 3/42 = 1/14.

Simplify the second bracket: (2/3 + 5/14) = (28 + 15)/42 = 43/42.

Divide the results: (1/14) ÷ (43/42) = 1/14 × 42/43 = 3/43.

Simplify the final term using BODMAS: 1/2 ÷ (43/6) = 1/2 × 6/43 = 3/43.

Combine: 3/43 + 3/43 = 6/43.

∴ The correct answer is 6/43.

Alternate Method

Given: { (15/42) ÷ (15/3) } ÷ ( (3/10) × (20/9) + (5/14) ) + (3/6) ÷ (43/20) of (20/6)

Formula Used: VBODMAS (Vinculum, Brackets, Of, Division, Multiplication, Addition, Subtraction)

⇒ Step 1 (Solve Brackets): { 15/42 × 3/15 } ÷ ( 2/3 + 5/14 ) + 3/6 ÷ ( 43/20 of 20/6 )

⇒ Step 2 (Simplify Fractions): { 1/14 } ÷ ( 43/42 ) + 1/2 ÷ ( 43/20 × 20/6 )

⇒ Step 3 (Process 'Of' and Divisions): (1/14 × 42/43) + 1/2 ÷ 43/6

⇒ Step 4 (Multiplication and Reciprocal): 3/43 + (1/2 × 6/43)

⇒ Step 5 (Final Addition): 3/43 + 3/43 = 6/43

∴ The correct answer is 6/43.

Additional Information

VBODMAS Rule

This rule defines the precedence of operations: Vinculum (Bar), Brackets (), Of, Division (÷), Multiplication (×), Addition (+), and Subtraction (−).

Fraction Reciprocal Rule

To divide by a fraction, multiply by its reciprocal: (a/b) ÷ (c/d) = (a/b) × (d/c).

'Of' vs Multiplication

The operator 'of' behaves as multiplication but holds a higher priority than division and regular multiplication in the order of operations.

Shortcut Trick

The ratio of the perimeter of an outer square to the inner square (formed by joining midpoints) is always √2 : 1.

Given, Perimeter of inner square = 40 cm.

Perimeter of outer square = Perimeter of inner square × √2

Perimeter of outer square = 40 × √2 = 40√2 cm.

∴ The correct answer is 40√2 cm.

Alternate Method

Given:

Perimeter of inner square = 40 cm

Formula Used:

Perimeter of a square = 4 × Side

In a right-angled triangle, Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular²

Calculations:

⇒ Side of the inner square (s) = 40 ÷ 4 = 10 cm

⇒ Let the side of the outer square be a. Since the inner vertices are midpoints, the side of the inner square forms a triangle with legs a/2.

⇒ By Pythagoras Theorem: (a/2)² + (a/2)² = 10²

⇒ a²/4 + a²/4 = 100

⇒ 2a²/4 = 100

⇒ a²/2 = 100 ⇒ a² = 200

⇒ a = √200 = 10√2 cm

⇒ Perimeter of outer square = 4 × a = 4 × 10√2 = 40√2 cm

∴ The correct answer is 40√2 cm.

Additional Information

Area Ratio of Midpoint Squares

The area of a square formed by joining the midpoints of another square is exactly half (1/2) the area of the original square.

Successive Midpoint Squares

If the process is repeated, the perimeters form a Geometric Progression (GP) with a common ratio of 1/√2.

Diagonal Relationship

The side of the inner square is equal to half the diagonal of the outer square: s = d/2 = (a√2)/2 = a/√2.

Shortcut Trick

Price Ratio = 200 : 180 → 10 : 9

Reduction in units = 10 − 9 = 1 unit

Percentage Decrease = (1 ÷ 10) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Initial shirt's price = ₹200

Final shirt's price = ₹180

Formula Used:

Percentage Decrease = (Decrease ÷ Original Value) × 100

Calculations:

⇒ Price Decrease = ₹200 − ₹180 = ₹20

⇒ Percentage Decrease = (20 ÷ 200) × 100

⇒ Percentage Decrease = (1 ÷ 10) × 100

⇒ Percentage Decrease = 10%

∴ The correct answer is 10%.

Additional Information

Percentage Change Formula

Percentage change = ((New Value − Old Value) ÷ Old Value) × 100. A negative result indicates a decrease, while a positive result indicates an increase.

Fraction to Percentage Conversion

Common fractions: 1/2 = 50%, 1/4 = 25%, 1/5 = 20%, 1/10 = 10%, and 1/20 = 5%.

Successive Percentage Change

If a value changes by a% and then by b%, the effective change is (a + b + ab/100)%.

Shortcut Trick

Fourth proportional m = (Product of middle terms) ÷ (First term)

⇒ m = (6 × 8) ÷ 4 = 12

Third proportional n = (Square of second term) ÷ (First term)

⇒ n = (12 × 12) ÷ 8 = 144 ÷ 8 = 18

∴ The correct answer is 18.

Alternate Method

Given:

Numbers for fourth proportional: 4, 6, 8, and m

Numbers for third proportional: 8, m, and n

Formula Used:

1. If a, b, c, d are in proportion, then a/b = c/d
2. If a, b, c are in continued proportion, then a/b = b/c (where c is the third proportional)

Calculations:

Step 1: Finding m

⇒ 4 ÷ 6 = 8 ÷ m

⇒ 4 × m = 6 × 8

⇒ 4 × m = 48

⇒ m = 12

Step 2: Finding n

⇒ 8 ÷ 12 = 12 ÷ n

⇒ 8 × n = 12 × 12

⇒ 8 × n = 144

⇒ n = 144 ÷ 8 = 18

∴ The correct answer is 18.

Additional Information

Fourth Proportional

For four numbers a, b, c, and d to be in proportion, the relation is a/b = c/d, which gives d = (b × c) / a.

Third Proportional

For three numbers a, b, and c to be in continued proportion, the relation is a/b = b/c, which gives c = b² / a.

Mean Proportional

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

Shortcut Trick

First Number = 2 × 50 = 100.

Product of numbers = HCF × LCM.

100 × Second Number = 25 × 500.

Second Number = 12500 ÷ 100 = 125.

∴ The correct answer is 125.

Alternate Method

Given:

H.C.F. = 25

L.C.M. = 500

First number when divided by 2 gives quotient = 50 and remainder = 0.

Formula Used:

Dividend = (Divisor × Quotient) + Remainder

Product of two numbers = H.C.F. × L.C.M.

Calculations:

⇒ First number (N₁) = (2 × 50) + 0

⇒ N₁ = 100

⇒ N₁ × N₂ = 25 × 500

⇒ 100 × N₂ = 12500

⇒ N₂ = 12500 ÷ 100

⇒ N₂ = 125

∴ The correct answer is 125.

Additional Information

Product Rule

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

Highest Common Factor (HCF)

The HCF is the greatest number that divides each of the two or more numbers without leaving a remainder.

Least Common Multiple (LCM)

The LCM of two or more numbers is the smallest number (not zero) that is a multiple of all the numbers.

Division Algorithm

The rule states that Dividend = (Divisor × Quotient) + Remainder, where 0 ≤ Remainder < Divisor.