SSC CGL Tier 2 Quant Previous Year Paper (Held On: 16 November 2020)

Shortcut Trick

Net deviation for the middle (13^th) number = 13 × [(52.8 − 54) + (62.2 − 54)]

Net deviation = 13 × [−1.2 + 8.2] = 13 × 7 = 91

Value of 13^th number = 54 + 91 = 145

Sum of remaining 24 numbers = (25 × 54) − 145 = 1350 − 145 = 1205

New Average = 1205 ÷ 24 = 50.208... ≈ 50.2

∴ The correct answer is 50.2.

Alternate Method

Given:

Total numbers (n) = 25

Total Average (A) = 54

Average of first 13 = 52.8

Average of last 13 = 62.2

Formula Used:

Total Sum = Average × Number of terms

Value of repeated term = (Sum of subgroups) − (Total Sum)

Calculations:

⇒ Total Sum of 25 numbers = 25 × 54 = 1350

⇒ Sum of first 13 numbers = 13 × 52.8 = 686.4

⇒ Sum of last 13 numbers = 13 × 62.2 = 808.6

⇒ Sum of (13 + 13) numbers = 686.4 + 808.6 = 1495

⇒ Value of 13^th number = 1495 − 1350 = 145

⇒ Sum of remaining 24 numbers = 1350 − 145 = 1205

⇒ Average of 24 numbers = 1205 ÷ 24

⇒ Average ≈ 50.208

∴ The correct answer is 50.2.

Additional Information

Net Deviation Concept

The sum of deviations of all observations from their arithmetic mean is always zero. If a term is repeated, the net deviation is the value by which the sum increases.

Exclusion Principle

When a value is excluded, the new average = (Original Total Sum − Excluded Value) ÷ (Original Count − 1).

Weighted Average

If groups with different averages are combined, the overall average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Shortcut Trick

Substitute θ = 60° to simplify calculations: sin 60° = √3/2 and cos 60° = 1/2.

First term: sin² 60° / [cos 60° (1 + cos 60°)] = (3/4) / [1/2 × 3/2] = (3/4) / (3/4) = 1.

Second term: (1 + cos 60°) / cos 60° = (3/2) / (1/2) = 3.

Total value: 1 + 3 = 4.

Check options: Option (2) 2sec 60° = 2 × 2 = 4 (Match).

∴ The correct answer is 2secθ.

Alternate Method

Given expression: sin²θ / [cosθ(1 + cosθ)] + (1 + cosθ) / cosθ

Formula used: sin²θ + cos²θ = 1 and secθ = 1/cosθ

⇒ Take LCM: [sin²θ + (1 + cosθ)²] / [cosθ(1 + cosθ)]

⇒ Expand the numerator: [sin²θ + 1 + 2cosθ + cos²θ] / [cosθ(1 + cosθ)]

⇒ Rearrange terms: [(sin²θ + cos²θ) + 1 + 2cosθ] / [cosθ(1 + cosθ)]

⇒ Using identity sin²θ + cos²θ = 1: [1 + 1 + 2cosθ] / [cosθ(1 + cosθ)]

⇒ Simplify numerator: [2 + 2cosθ] / [cosθ(1 + cosθ)]

⇒ Factor out 2: 2(1 + cosθ) / [cosθ(1 + cosθ)]

⇒ Cancel (1 + cosθ): 2 / cosθ = 2secθ

∴ The correct answer is 2secθ.

Additional Information

Pythagorean Identities

Fundamental identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Relations

Reciprocal trigonometric functions are defined as secθ = 1/cosθ, cosecθ = 1/sinθ, and cotθ = 1/tanθ.

Algebraic Expansion

The square of a binomial (a + b)² = a² + 2ab + b² is frequently used to simplify trigonometric fractions.

Shortcut Trick

Area of equilateral triangular base = (√3 / 4) × side²

Base Area = (√3 / 4) × 8 × 8 = 16√3 cm²

Volume of Pyramid = (1/3) × Base Area × Height

Volume = (1/3) × 16√3 × 30√3 = 16 × 10 × 3 = 480 cm³

∴ The correct answer is 480 cm³.

Alternate Method

Given:

Side of equilateral triangular base (a) = 8 cm

Height of the pyramid (h) = 30√3 cm

Formula Used:

Volume of a Pyramid = (1/3) × Area of Base × Vertical Height

Area of Equilateral Triangle = (√3 / 4) × a²

Calculations:

⇒ Area of base = (√3 / 4) × (8)²

⇒ Area of base = (√3 / 4) × 64

⇒ Area of base = 16√3 cm²

⇒ Volume = (1/3) × 16√3 × 30√3

⇒ Volume = (1/3) × 16 × 30 × (√3 × √3)

⇒ Volume = (1/3) × 16 × 30 × 3

⇒ Volume = 16 × 30 = 480 cm³

∴ The correct answer is 480 cm³.

Additional Information

Volume of a Pyramid

The volume is always one-third of the product of the base area and the perpendicular height, regardless of the base shape: V = (1/3) × A_base × h.

Total Surface Area of Pyramid

TSA = Base Area + Lateral Surface Area, where Lateral Surface Area = (1/2) × Perimeter of base × Slant height (l).

Slant Height Relationship

For a right pyramid with a regular polygon base, l² = h² + r², where r is the inradius of the base polygon.

Shortcut Trick

Income Ratio (X : Y) = 5 : 4

Expenditure Ratio (X : Y) = 9 : 7

Given: Income of Y = Expenditure of X. Take LCM of 4 and 9 = 36.

New Income (multiply by 9) → X : Y = 45 : 36

New Expenditure (multiply by 4) → X : Y = 36 : 28

Savings (X : Y) = (45 − 36) : (36 − 28) = 9 : 8

∴ The correct answer is 9 : 8.

Alternate Method

Given:

Ratio of Incomes = 5 : 4

Ratio of Expenditures = 9 : 7

Condition: Income of Y = Expenditure of X

Calculations:

⇒ Let Monthly Income of X and Y be 5a and 4a respectively.

⇒ Let Monthly Expenditure of X and Y be 9b and 7b respectively.

⇒ According to the condition: 4a = 9b

⇒ a/b = 9/4 or a = 9k, b = 4k

⇒ Income of X = 5 × 9k = 45k

⇒ Income of Y = 4 × 9k = 36k

⇒ Expenditure of X = 9 × 4k = 36k

⇒ Expenditure of Y = 7 × 4k = 28k

⇒ Savings of X = Income_X − Expenditure_X = 45k − 36k = 9k

⇒ Savings of Y = Income_Y − Expenditure_Y = 36k − 28k = 8k

⇒ Ratio of Savings = 9k : 8k = 9 : 8

∴ The correct answer is 9 : 8.

Additional Information

Income, Expenditure, and Savings Relationship

The fundamental relation is Savings = Income − Expenditure. If any two are in ratio, a third can be found if a cross-linking condition is given.

Ratio Scaling

To compare two different ratios, identify a common term or a related condition to bring them to a common scale (using LCM).

Percentage Conversion

Savings % = (Savings ÷ Income) × 100. In this case, X's savings % = (9/45) × 100 = 20%.

Shortcut Trick

When a number x divides a, b, and c with the same remainder, x is the HCF of the differences |a−b|, |b−c|, and |c−a|.

Difference 1: 1027 − 955 = 72

Difference 2: 1075 − 1027 = 48

x = HCF(72, 48) = 24

Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Comparing with options, 16 is not a factor of 24.

∴ The correct answer is 16.

Alternate Method

Given: Numbers are 955, 1027, and 1075.

Formula Used: For equal remainders, Divisor x = HCF(|Number₁ − Number₂|, |Number₂ − Number₃|)

⇒ Let x be the HCF of (1027 − 955) and (1075 − 1027).

⇒ x = HCF(72, 48)

⇒ Prime factors of 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²

⇒ Prime factors of 48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3¹

⇒ HCF(72, 48) = 2³ × 3¹ = 8 × 3 = 24

⇒ Now, check the options to find which is NOT a factor of 24:

⇒ 24 ÷ 6 = 4 (Factor)

⇒ 24 ÷ 8 = 3 (Factor)

⇒ 24 ÷ 4 = 6 (Factor)

⇒ 24 ÷ 16 = 1.5 (Not an integer, hence NOT a factor)

∴ The correct answer is 16.

Additional Information

HCF of Differences Property

The greatest number that divides a, b, and c leaving the same remainder r is the HCF of |a−b|, |b−c|, and |c−a|.

Definition of a Factor

A factor is a number that divides another number completely without leaving a remainder. For any number x, all its divisors are its factors.

Finding the Remainder

Once x is found, the common remainder r can be calculated as r = a mod x. In this case, 955 ÷ 24 leaves a remainder of 19.

Given:

$4 \div 12 \text{ of } [3 \div 4 \text{ of } {(4 - 2) \times 6 \div 2}] - 2 \times 6 \div 8 + 3$ = ?

Formula used:

Simplification follows the VBODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction).

Calculation:

$4 \div 12 \text{ of } [3 \div 4 \text{ of } {(4 - 2) \times 6 \div 2}] - 2 \times 6 \div 8 + 3$

⇒ $4 \div 12 \text{ of } [3 \div 4 \text{ of } {2 \times 6 \div 2}] - 12 \div 8 + 3$

⇒ $4 \div 12 \text{ of } [3 \div 4 \text{ of } 6] - \frac{3}{2} + 3$

⇒ $4 \div 12 \text{ of } [3 \div 24] - \frac{3}{2} + 3$

⇒ $4 \div (12 \times \frac{1}{8}) - \frac{3}{2} + 3$

⇒ $4 \div \frac{3}{2} - \frac{3}{2} + 3$

⇒ $\frac{8}{3} - \frac{3}{2} + 3$

⇒ $\frac{16 - 9 + 18}{6} = \frac{25}{6}$

⇒ $4\frac{1}{6}$

∴ The correct answer is option (4).

Shortcut Trick

Let the numerator be N. Denominator D = 2N + 4.

New Numerator = N + 3, New Denominator = (2N + 4) − 3 = 2N + 1.

New fraction = (N + 3) / (2N + 1) = 2/3.

By cross-multiplication: 3N + 9 = 4N + 2 ⇒ N = 7.

Original Denominator = 2(7) + 4 = 18.

Difference = 18 − 7 = 11.

∴ The correct answer is 11.

Alternate Method

Given:

Relationship between numerator (N) and denominator (D): D = 2N + 4

Condition: (N + 3) / (D − 3) = 2/3

Calculations:

⇒ Substitute D = 2N + 4 into the condition:

⇒ (N + 3) / ((2N + 4) − 3) = 2/3

⇒ (N + 3) / (2N + 1) = 2/3

⇒ 3 × (N + 3) = 2 × (2N + 1)

⇒ 3N + 9 = 4N + 2

⇒ 4N − 3N = 9 − 2

⇒ N = 7

⇒ D = 2 × 7 + 4 = 14 + 4 = 18

⇒ Difference = D − N = 18 − 7 = 11

∴ The correct answer is 11.

Additional Information

Proper and Improper Fractions

A fraction where the numerator is less than the denominator is a proper fraction (e.g., 7/18), while one where the numerator is greater is an improper fraction.

Equivalent Fractions

Fractions that represent the same value even though they look different (e.g., 2/3 = 10/15), obtained by multiplying/dividing N and D by the same non-zero number.

Reciprocal of a Fraction

The reciprocal of a fraction a/b is b/a. The product of a fraction and its reciprocal is always 1.

Shortcut Trick

Using the cross-multiplication formula for intersection (x, y):

x = [(−2 × −5) − (3 × −8)] ÷ [3 × 3 − 4 × (−2)] = [10 + 24] ÷ [9 + 8] = 2

y = [(−8 × 4) − (−5 × 3)] ÷ 17 = [−32 + 15] ÷ 17 = −1

Therefore, α = 2 and β = −1.

Required value (2α − β) = 2(2) − (−1) = 4 + 1 = 5.

∴ The correct answer is 5.

Alternate Method

Given Equations:

3x − 2y = 8 ... (i)

4x + 3y = 5 ... (ii)

⇒ To eliminate y, multiply (i) by 3 and (ii) by 2:

⇒ 9x − 6y = 24

⇒ 8x + 6y = 10

⇒ Adding both equations: 17x = 34

⇒ x = 2 ∴ α = 2

⇒ Substitute x = 2 in (i): 3(2) − 2y = 8

⇒ 6 − 2y = 8 ⇒ −2y = 2

⇒ y = −1 ∴ β = −1

⇒ Value of (2α − β) = 2(2) − (−1) = 4 + 1 = 5.

∴ The correct answer is 5.

Additional Information

Nature of Solutions

For equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂, a unique solution exists if a₁/a₂ ≠ b₁/b₂.

Inconsistent Equations

Lines are parallel (no solution) if a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Dependent Equations

Lines are coincident (infinite solutions) if a₁/a₂ = b₁/b₂ = c₁/c₂.

Shortcut Trick

Property: The angle between the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining two angles.

∠DEC = (∠A + ∠B) ÷ 2

80° = (∠A + 82°) ÷ 2

160° = ∠A + 82°

∠A = 160° − 82° = 78°

∴ The correct answer is 78°.

Alternate Method

Given: ∠B = 82°, ∠DEC = 80°. DE and CE are bisectors of ∠D and ∠C respectively.

Let ∠EDC = x and ∠ECD = y. Since DE and CE are bisectors:

⇒ ∠D = 2x and ∠C = 2y

In ▵DEC:

⇒ ∠EDC + ∠ECD + ∠DEC = 180°

⇒ x + y + 80° = 180°

⇒ x + y = 100°

Multiplying by 2:

⇒ 2x + 2y = 200° → ∠D + ∠C = 200°

In quadrilateral ABCD:

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠A + 82° + 200° = 360°

⇒ ∠A + 282° = 360°

⇒ ∠A = 360° − 282°

⇒ ∠A = 78°

∴ The correct answer is 78°.

Additional Information

Angle Sum Property of Quadrilateral

The sum of all interior angles of a quadrilateral is always 360°.

Angle Bisector Relation

In any quadrilateral, the angle between the bisectors of two adjacent angles is half the sum of the other two angles: ∠E = (∠A + ∠B) ÷ 2 if bisectors of ∠C and ∠D meet at E.

Opposite Angle Bisectors

If bisectors of opposite angles ∠A and ∠C meet at P, the angle between them is |∠B − ∠D| ÷ 2.

Shortcut Trick

Find vertices: solve (1) & (2) ⇒ P(1, 2); put y = 0 in (1) ⇒ R(6, 0); put y = 0 in (2) ⇒ Q(3, 0)

Base QR (on x-axis) = 6 − 3 = 3 | Height PM = y-coordinate of P = 2

Area = ½ × Base × Height = ½ × 3 × 2 = 3

∴ The correct answer is 3 unit².

Alternate Method

Given:

2x + 5y - 12 = 0 ------(1)

x + y = 3 ------(2)

y = 0 ------(3)

Concept Used:

On the x-axis, y = 0

Area of a triangle = (1/2) × Base × Height

Calculation:

By solving equations (1) and (2), we get

⇒ x = 1 and y = 2

Point of intersection P = (1, 2)

Putting y = 0 in equation (1), we get

⇒ R(x, y) = (6, 0)

Putting y = 0 in equation (2), we get

⇒ Q(x, y) = (3, 0)

⇒ The vertices of the triangle are (1, 2), (6, 0) and (3, 0)

Here, we can see that ΔPQR is an obtuse-angled triangle.

Therefore, its base will be QR (= 3) and its height will be PM (= 2)

Therefore, Area of the triangle = (1/2) × 3 × 2

⇒ Area of the triangle = 3 square units

∴ The area of the triangle is 3 square units.

Alternate Method

From the figure, we get

Area of ΔPQR = (1/2) × QR × PM

Area of ΔPQR = (3 × 2)/2 = 3 square units

Additional Information

Area of Triangle (Coordinate Geometry)

For vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃), Area = 1/2 |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Intercept Form of a Line

The equation of a line with x-intercept 'a' and y-intercept 'b' is x/a + y/b = 1. This helps in finding vertices quickly when the third line is an axis.

Intersection of Linear Equations

The point where two graphs intersect is the unique solution to the system of linear equations representing those lines.

∴ The correct answer is 3 unit².

Shortcut Trick

Assume Total Capacity of tank = LCM of (12, 15, 10) = 60 units.

Efficiency of A = 60 ÷ 12 = +5 units/min; Efficiency of B = 60 ÷ 15 = +4 units/min.

Combined efficiency (A + B − C) = 60 ÷ 10 = +6 units/min.

Efficiency of C = (Efficiency of A + Efficiency of B) − Combined efficiency.

Efficiency of C = (5 + 4) − 6 = 3 units/min.

Time x = Total Capacity ÷ Efficiency of C = 60 ÷ 3 = 20 minutes.

∴ The correct answer is 20 minutes.

Alternate Method

Given:

Time taken by Pipe A = 12 minutes

Time taken by Pipe B = 15 minutes

Time taken by Pipe C (empty) = x minutes

Combined time (A + B − C) = 10 minutes

Formula Used:

Total Work = Time × Efficiency

Calculations:

⇒ Let efficiency of Pipe C be ‘c’.

⇒ Efficiency(A) + Efficiency(B) − Efficiency(C) = Efficiency(Combined)

⇒ 5 + 4 − c = 6

⇒ 9 − c = 6

⇒ c = 9 − 6 = 3 units/min

⇒ Time taken by Pipe C to empty the tank (x) = Total Work ÷ Efficiency(C)

⇒ x = 60 ÷ 3

⇒ x = 20 minutes

∴ The correct answer is 20 minutes.

Additional Information

Pipes and Cisterns Basics

Filling pipes are considered as having positive efficiency, while emptying pipes (outlets) are considered as having negative efficiency.

Combined Rate Formula

If two filling pipes take A and B hours and an emptying pipe takes C hours, the net work done in one hour is (1/A + 1/B − 1/C).

Work and Wages

Wages are always distributed in the ratio of the efficiency of the pipes or the work done by each pipe, provided they work for the same duration.

Shortcut Trick

LCM of 3, 7, and 11 is 231. Assume the number is the maximum possible 23599.

23599 ÷ 231 ⇒ Remainder = 37.

Exact divisible number = 23599 − 37 = 23562. This gives x = 6 and y = 2.

Required value of 3x − 4y = 3(6) − 4(2) = 18 − 8 = 10.

∴ The correct answer is 10.

Alternate Method

Given: A five-digit number 235xy is divisible by 3, 7, and 11.

Formula: If a number is divisible by coprime factors, it is divisible by their LCM.

⇒ LCM of 3, 7, and 11 = 3 × 7 × 11 = 231.

⇒ For the number to be of the form 235xy, we divide 23599 by 231.

⇒ 23599 = 231 × 102 + 37.

⇒ The required number = 23599 − 37 = 23562.

⇒ By comparing 235xy with 23562, we find x = 6 and y = 2.

⇒ Value = 3x − 4y = 3 × 6 − 4 × 2.

⇒ Value = 18 − 8 = 10.

∴ The correct answer is 10.

Additional Information

Divisibility Rule of 11

A number is divisible by 11 if the difference between the sum of digits at odd positions and the sum of digits at even positions is either 0 or a multiple of 11.

Divisibility Rule of 3

A number is divisible by 3 if the sum of all its digits is divisible by 3. For 23562, 2+3+5+6+2 = 18, which is divisible by 3.

LCM of Prime Numbers

If a set of numbers are all prime (like 3, 7, and 11), their Least Common Multiple (LCM) is simply the product of those numbers.

Given:

Calculation:

Total number of students in B = 2100 × (75.6^° /360^° ) 

⇒ (21 × 756)/36 = 441

The ratio of male and female students in B = 4 : 3

Number of male students in B = 441 × 4/7 = 252

Number of female students in B = 441 - 252 = 189

Total number of students in D = 2100 × (54^° /360^° ) = 315

The ratio of male and female students in D = 8 : 7

Number of male students in D = 315 × 8/15 = 168

Number of female students in D = 315 - 168 = 147

now,

% decrease of female in D that of male students in B = (105/252) × 100 = 41.7%

∴ The % decrease is 41.7%.

Shortcut Trick

Convert all angles to 57° using complementary relation (33° = 90° − 57°).

Expression → tan² 57° + (sin² 57° + cos² 57°) + (cosec² 57° × sin² 57°) + (cosec² 57° × sin² 57°)

Expression → tan² 57° + 1 + 1 + 1 = tan² 57° + 3

Since x = sec 57°, we have tan² 57° = sec² 57° − 1 = x² − 1

Total Value = (x² − 1) + 3 = x² + 2

∴ The correct answer is x² + 2.

Alternate Method

Given: x = sec 57°

Formula used: sin(90° − θ) = cos θ | cos(90° − θ) = sin θ | cot(90° − θ) = tan θ | sec(90° − θ) = cosec θ

Based on x = sec 57°, let base BC = 1 and hypotenuse AC = x. Then perpendicular AB = √(x² − 1).

⇒ cot 33° = tan 57° = AB/BC = √(x² − 1)

⇒ sin 33° = cos 57° = BC/AC = 1/x

⇒ cos 33° = sin 57° = AB/AC = √(x² − 1)/x

⇒ sec 33° = cosec 57° = AC/AB = x/√(x² − 1)

Now, substituting into the expression:

⇒ cot² 33° + (sin² 57° + sin² 33°) + cosec² 57° cos² 33° + sec² 33° sin² 57°

⇒ (√(x² − 1))² + 1 + (x/√(x² − 1))² × (√(x² − 1)/x)² + (x/√(x² − 1))² × (√(x² − 1)/x)²

⇒ (x² − 1) + 1 + 1 + 1

⇒ x² + 2

∴ The correct answer is x² + 2.

Additional Information

Complementary Angle Identities

For any angle θ, sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, and sec(90° − θ) = cosec θ.

Pythagorean Identities

Fundamental relations: sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = cosec² θ.

Reciprocal Relations

Multiplying reciprocal trigonometric ratios gives unity: sin θ × cosec θ = 1, cos θ × sec θ = 1, and tan θ × cot θ = 1.

Shortcut Trick

Calculate ∠B = 180° − (80° + 38°) = 62°.

Using Exterior Angle Theorem in ΔABD: ∠ADC = ∠BAD + ∠B.

Given ∠ADC = 2∠BAD, so 2∠BAD = ∠BAD + 62° ⇒ ∠BAD = 62°.

Substitute to find ∠ADC = 2 × 62° = 124°.

Measure of ∠ADB = 180° − 124° = 56°.

∴ The correct answer is 56°.

Alternate Method

Given: ∠A = 80°, ∠C = 38°, D is on BC such that ∠ADC = 2∠BAD.

Formula: Sum of angles in a triangle = 180°; Exterior angle is the sum of opposite interior angles.

⇒ In ΔABC: ∠B = 180° − (∠A + ∠C)

⇒ ∠B = 180° − (80° + 38°) = 62°

⇒ In ΔABD, ∠ADC is the exterior angle at vertex D.

⇒ ∠ADC = ∠BAD + ∠ABD

⇒ Let ∠BAD = x, then ∠ADC = 2x (given).

⇒ 2x = x + 62°

⇒ x = 62°

⇒ ∠ADC = 2 × 62° = 124°

⇒ Since BDC is a straight line, ∠ADB + ∠ADC = 180°

⇒ ∠ADB = 180° − 124° = 56°

∴ The correct answer is 56°.

Additional Information

Exterior Angle Theorem

The exterior angle of a triangle is equal to the sum of the two opposite interior angles (∠Ext = ∠Opp1 + ∠Opp2).

Angle Sum Property

The sum of all interior angles of a triangle is always 180° (∠A + ∠B + ∠C = 180°).

Supplementary Angles

Angles on a straight line add up to 180°; thus, ∠ADB and ∠ADC are supplementary angles.

Given:

$\frac{7+3\sqrt5}{3+\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt5}$

Formula used:

(a + b)(a - b) = a² - b²

Calculation:

⇒ $\frac{(7+3\sqrt5)(3-\sqrt5) - (7-3\sqrt5)(3+\sqrt5)}{(3+\sqrt5)(3-\sqrt5)}$

⇒ $\frac{(21 - 7\sqrt5 + 9\sqrt5 - 3 \times 5) - (21 + 7\sqrt5 - 9\sqrt5 - 3 \times 5)}{3^2 - (\sqrt5)^2}$

⇒ $\frac{(6 + 2\sqrt5) - (6 - 2\sqrt5)}{9 - 5}$

⇒ $\frac{6 + 2\sqrt5 - 6 + 2\sqrt5}{4}$

⇒ $\frac{4\sqrt5}{4} = \sqrt5$

⇒ $\sqrt5 \approx 2.236$

2.236 lies between 2 and 2.5.

∴ The correct answer is option (4).

Shortcut Trick

Volume of cuboid = 18 × 36 × 72 = 18 × (2 × 18) × (4 × 18) = 8 × 18³.

Since this is melted into 8 cubes, 8 × a³ = 8 × 18³ ⇒ Side of each cube (a) = 18 cm.

Ratio of cuboid dimensions = 18 : 36 : 72 = 1 : 2 : 4.

TSA of cuboid parts = 2 × (1×2 + 2×4 + 4×1) = 28 units.

Sum of LSA of 8 cubes = 8 × 4 × (1)² = 32 units.

Required Ratio = 28 : 32 = 7 : 8.

∴ The correct answer is 7 : 8.

Alternate Method

Given:

Dimensions of cuboid: L = 18 cm, B = 36 cm, H = 72 cm.

Number of cubes formed = 8.

Formula Used:

Volume of Cuboid = L × B × H

Volume of Cube = a³ (where a is side length)

Total Surface Area (TSA) of Cuboid = 2(LB + BH + HL)

Lateral Surface Area (LSA) of Cube = 4a²

Calculations:

⇒ Volume of cuboid = 18 × 36 × 72 = 46656 cm³

⇒ Volume of 8 cubes = Volume of cuboid

⇒ 8 × a³ = 46656

⇒ a³ = 46656 ÷ 8 = 5832

⇒ a = √³5832 = 18 cm

⇒ TSA of cuboid = 2(18×36 + 36×72 + 72×18)

⇒ TSA of cuboid = 2(648 + 2592 + 1296) = 2(4536) = 9072 cm²

⇒ LSA of one cube = 4 × 18² = 4 × 324 = 1296 cm²

⇒ Sum of LSA of 8 cubes = 8 × 1296 = 10368 cm²

⇒ Required Ratio = 9072 : 10368

⇒ Dividing by 1296: 7 : 8

∴ The correct answer is 7 : 8.

Additional Information

Recasting Rule

When a solid is melted and recast into another shape, the total volume remains constant (Volume_Initial = Volume_Final).

Cuboid Formulas

Volume = L × B × H | TSA = 2(LB + BH + HL) | CSA (Lateral Area) = 2H(L + B).

Cube Formulas

Volume = a³ | TSA = 6a² | LSA = 4a² | Diagonal = a√3.

Shortcut Trick

Use the direct identity: a³ + b³ + c³ − 3abc = (a + b + c) × [(a + b + c)² − 3(ab + bc + ca)]

Substitute the given values into the formula: 175 = 7 × [7² − 3(ab + bc + ca)]

⇒ 175 ÷ 7 = 49 − 3(ab + bc + ca)

⇒ 25 = 49 − 3(ab + bc + ca)

⇒ 3(ab + bc + ca) = 49 − 25 = 24

⇒ ab + bc + ca = 8

∴ The correct answer is 8.

Alternate Method

Given:

a + b + c = 7

a³ + b³ + c³ − 3abc = 175

Formula Used:

(i) (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(ii) a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − (ab + bc + ca))

Calculations:

⇒ From (i): a² + b² + c² = (a + b + c)² − 2(ab + bc + ca)

⇒ a² + b² + c² = 7² − 2(ab + bc + ca) = 49 − 2(ab + bc + ca)

⇒ Substitute this into (ii): 175 = 7 × [(49 − 2(ab + bc + ca)) − (ab + bc + ca)]

⇒ 175 ÷ 7 = 49 − 3(ab + bc + ca)

⇒ 25 = 49 − 3(ab + bc + ca)

⇒ 3(ab + bc + ca) = 49 − 25

⇒ 3(ab + bc + ca) = 24

⇒ ab + bc + ca = 24 ÷ 3

⇒ ab + bc + ca = 8

∴ The correct answer is 8.

Additional Information

The Factorized Identity

The expression a³ + b³ + c³ − 3abc is always equal to (a + b + c)(a² + b² + c² − ab − bc − ca).

Special Case: a + b + c = 0

If a + b + c = 0, then the entire right side becomes zero, leading to the property a³ + b³ + c³ = 3abc.

Alternative Form of Identity

The term (a² + b² + c² − ab − bc − ca) can also be written as ½[(a − b)² + (b − c)² + (c − a)²].

Sum of Squares

The relation a² + b² + c² = (a + b + c)² − 2(ab + bc + ca) is frequently used to simplify algebraic problems involving three variables.

Shortcut Trick

Surface Area increase = 4π(2rh + h²) = 110, where h = 2.5 dm is the increase in radius.

4 × (22/7) × (5r + 6.25) = 110 ⇒ 5r + 6.25 = 8.75

5r = 2.5 ⇒ r = 0.5 dm

Volume = (4/3) × (22/7) × (0.5)³ = (4/3) × (22/7) × (1/8) = 11/21 dm³.

∴ The correct answer is 11/21 dm³.

Alternate Method

Given:

Increase in radius (h) = 2.5 dm

Increase in Surface Area = 110 dm²

π = 22/7

Formula Used:

Surface Area (SA) of Sphere = 4πr²

Volume (V) of Sphere = (4/3)πr³

Calculations:

⇒ Increase in Surface Area = 4π(r + 2.5)² − 4πr² = 110

⇒ 4 × (22/7) × [r² + 5r + 6.25 − r²] = 110

⇒ (88/7) × (5r + 6.25) = 110

⇒ 5r + 6.25 = (110 × 7) ÷ 88

⇒ 5r + 6.25 = 770 ÷ 88 = 8.75

⇒ 5r = 8.75 − 6.25 = 2.5

⇒ r = 0.5 dm

⇒ Volume of Sphere (V) = (4/3) × (22/7) × (0.5)³

⇒ V = (4/3) × (22/7) × (1/8)

⇒ V = 88 ÷ 168 = 11/21 dm³

∴ The correct answer is 11/21 dm³.

Additional Information

Properties of a Sphere

Surface Area = 4πr² and Volume = (4/3)πr³. If the radius is multiplied by a factor k, the surface area increases by k² and the volume increases by k³.

Properties of a Hemisphere

For a hemisphere of radius r, the Curved Surface Area (CSA) = 2πr², the Total Surface Area (TSA) = 3πr², and the Volume = (2/3)πr³.

Volume and Surface Area Relationship

The volume of a sphere can also be expressed in terms of its surface area (S) as V = (S/3) × r.

Shortcut Trick

Combine ratios: A : B : C : D = (1×2×2) : (3×2×2) : (3×5×2) : (3×5×3)

Ratio A : B : C : D = 4 : 12 : 30 : 45

Difference between A and C = 30 − 4 = 26 units

26 units = ₹3,510 ⇒ 1 unit = ₹135

Share of D = 45 units = 45 × 135 = ₹6,075

∴ The correct answer is ₹6,075.

Alternate Method

Given:

Ratio of A : B = 1 : 3

Ratio of B : C = 2 : 5

Ratio of C : D = 2 : 3

Difference (C − A) = ₹3,510

To find the combined ratio A : B : C : D:

⇒ A : B = 1 : 3

⇒ B : C = 2 : 5

⇒ Multiply A:B by 2 and B:C by 3 to equate B:

⇒ A : B : C = (1×2) : (3×2) : (5×3) = 2 : 6 : 15

⇒ Now combine A : B : C with C : D = 2 : 3:

⇒ Multiply A:B:C by 2 and C:D by 15 to equate C:

⇒ A : B : C : D = (2×2) : (6×2) : (15×2) : (3×15)

⇒ A : B : C : D = 4 : 12 : 30 : 45

⇒ Let shares be 4x, 12x, 30x, and 45x.

⇒ Difference between C and A = 30x − 4x = 26x

⇒ 26x = 3,510

⇒ x = 3,510 ÷ 26 = 135

⇒ Share of D = 45x = 45 × 135

⇒ Share of D = ₹6,075

∴ The correct answer is ₹6,075.

Additional Information

Combined Ratio Rule

To combine A:B and B:C, find the LCM of common term B. If A:B = x:y and B:C = p:q, then A:B:C = (x×p) : (y×p) : (y×q).

Compound Ratio

The compound ratio of (a:b) and (c:d) is (ac : bd). In this question, A:D is the compound ratio (1×2×2) : (3×5×3) = 4 : 45.

Mean Proportional

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

Shortcut Trick

Volume to be emptied = 2/3 × (2/3 × π × r³) = 4/9 × 22/7 × (10.5)³

Volume in m³ = 4/9 × 22/7 × 1157.625 = 1617 m³

Total litres to empty = 1617 × 1000 = 1,617,000 litres

Time (in hours) = Total litres ÷ (Rate × 3600) = 1,617,000 ÷ (7.7 × 3600)

Time = 1,617,000 ÷ 27720 = 175/3 hours

∴ The correct answer is 175/3 hours.

Alternate Method

Given:

Internal radius (r) = 10.5 m = 21/2 m

Rate of emptying = 7.7 litres/second

Part of tank to empty = 2/3

Formula Used:

Volume of a hemisphere = 2/3 π r³

1 m³ = 1000 Litres

Time = Volume ÷ Rate

Calculations:

⇒ Volume of tank = 2/3 × 22/7 × (21/2) × (21/2) × (21/2)

⇒ Volume of tank = 11 × 21 × 21 ÷ 2 = 2425.5 m³

⇒ Volume to be emptied = 2/3 × 2425.5 = 1617 m³

⇒ Volume in litres = 1617 × 1000 = 1,617,000 litres

⇒ Time (sec) = 1,617,000 ÷ 7.7 = 210,000 seconds

⇒ Time (hours) = 210,000 ÷ 3600

⇒ Time = 2100 ÷ 36 = 175/3 hours

∴ The correct answer is 175/3 hours.

Additional Information

Volume and Surface Area of Sphere

The volume of a full sphere is 4/3 π r³ and its surface area is 4 π r².

Surface Area of Hemisphere

Curved Surface Area (CSA) = 2 π r² | Total Surface Area (TSA) = 3 π r².

Metric Conversion

Always remember: 1 m³ = 1000 litres and 1 litre = 1000 cm³.

Shortcut Trick

Use the direct ratio formula: MP/CP = (100 + Profit%)/(100 − Discount%)

Let CP = 100, then MP = 125 and Profit = 5.5%

125/100 = (100 + 5.5)/(100 − x)

5/4 = 105.5/(100 − x) ⇒ 100 − x = (105.5 × 4) ÷ 5 = 84.4

x = 100 − 84.4 = 15.6%

∴ The correct answer is 15.6%.

Alternate Method

Given:

Marked Price (MP) = CP + 25% of CP

Profit = 5.5% of CP

Discount = x% on MP

Assume Cost Price (CP) = 100

⇒ Marked Price (MP) = 100 + (25/100) × 100 = 125

⇒ Selling Price (SP) = 100 + (5.5/100) × 100 = 105.5

⇒ Discount = MP − SP

⇒ Discount = 125 − 105.5 = 19.5

⇒ Discount Percentage (x) = (Discount ÷ MP) × 100

⇒ x = (19.5 ÷ 125) × 100

⇒ x = 0.156 × 100 = 15.6%

∴ The correct answer is 15.6%.

Additional Information

Markup and Discount Relation

The relationship between CP and MP is given by MP/CP = (100 + Profit%)/(100 − Discount%). If there is a loss, use (100 − Loss%).

Successive Percentage Change

The net change when a value is increased by M% (markup) and then decreased by D% (discount) is given by: Net Profit/Loss = M − D − (M × D)/100.

Mark-up Percentage

Mark-up % is always calculated on the Cost Price (CP), whereas Discount % is always calculated on the Marked Price (MP).

Shortcut Trick

Arc AB (diameter) subtends 180° at the center.

Arc PQ subtends 54° at the center.

For secants AQ and BP intersecting at C outside the circle, ∠PCA = 1/2 × (arc AB − arc PQ).

∠PCA = 1/2 × (180° − 54°) = 1/2 × 126° = 63°.

∴ The correct answer is 63°.

Alternate Method

Given:

AB is the diameter of the circle with centre O.

∠POQ = 54°.

AQ and BP intersect at C.

Formula Used:

1. Angle subtended by an arc at the center is double the angle at the circumference.
2. Angle in a semicircle is 90°.
3. Sum of angles in a triangle is 180°.

Calculations:

⇒ In the circle, ∠POQ = 54° (Angle at center by arc PQ).

⇒ ∠PAQ = 1/2 × ∠POQ = 1/2 × 54° = 27° (Angle at circumference by arc PQ).

⇒ AB is the diameter ∴ ∠APB = 90° (Angle in a semicircle).

⇒ Points B, P, and C lie on the same line, so ∠APC = 180° − ∠APB = 180° − 90° = 90°.

⇒ In ▵PAC, the sum of internal angles is 180°.

⇒ ∠PCA + ∠APC + ∠PAC = 180°.

⇒ ∠PCA + 90° + 27° = 180°.

⇒ ∠PCA + 117° = 180°.

⇒ ∠PCA = 180° − 117° = 63°.

∴ The correct answer is 63°.

Additional Information

Angle in a Semicircle

The angle subtended by a diameter at any point on the circumference of the circle is always 90°.

Central Angle Theorem

The angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle (∠Center = 2 × ∠Circumference).

Secant-Secant Angle Theorem

If two secants intersect outside a circle, the measure of the angle formed is half the difference of the measures of the intercepted arcs.

Angles in the Same Segment

Angles subtended by the same chord (or arc) at different points on the same segment of the circumference are equal.

Shortcut Trick Find the nearest perfect square below 10424. Since 1002 = 10000 and 1022 = 10404, we have x = 10424 − 10404 = 20. Prime factorize x: 20 = 2 × 2 × 5 = 22 × 51. To make it a perfect square, multiply by the unpaired factor 5. ∴ The correct answer is 5.

Alternate Method Given number = 10424 Formula: Perfect Square Number = (Number − Remainder)

Step 1: Calculate the square root of 10424 to find the remainder. ⇒ Using Long Division Method: ⇒ 1002 = 10000 ⇒ 1022 = (100 + 2)2 = 10000 + 4 + 400 = 10404 ⇒ 1032 = (100 + 3)2 = 10000 + 9 + 600 = 10609 Step 2: Since 10404 < 10424 < 10609, the nearest perfect square is 10404. ⇒ x = 10424 − 10404 ⇒ x = 20 Step 3: Find the multiplier for x = 20 to make it a perfect square. ⇒ Prime factorization of 20 = 2 × 2 × 5 ⇒ 20 = 22 × 51 ⇒ To make it a perfect square, all prime factors must have even powers. ⇒ Required multiplier = 5 ∴ The correct answer is 5.

Additional Information Perfect Square Properties A perfect square never ends with the digits 2, 3, 7, or 8. The digital sum of a perfect square is always 1, 4, 7, or 9. Finding the Least Multiplier To find the least number to multiply/divide to make N a perfect square, perform prime factorization and identify factors with odd exponents. Long Division for Square Roots The remainder obtained after performing the long division of a non-perfect square number is the least number to be subtracted to make it a perfect square.

Shortcut Trick

Volume of a cone (V) ∝ r² × h, so Height (h) ∝ V ÷ r².

Ratio of heights h₁ : h₂ = (V₁ ÷ r₁²) : (V₂ ÷ r₂²)

h₁ : h₂ = (8 ÷ 5²) : (9 ÷ 6²)

h₁ : h₂ = (8 ÷ 25) : (9 ÷ 36) = 8/25 : 1/4

h₁ : h₂ = (8 × 4) : (25 × 1) = 32 : 25

∴ The correct answer is 32 : 25.

Alternate Method

Given:

Ratio of radii (r₁ : r₂) = 5 : 6

Ratio of volumes (V₁ : V₂) = 8 : 9

Formula:

Volume of cone (V) = (1/3) × π × r² × h

Calculations:

⇒ V₁ ÷ V₂ = [(1/3) × π × r₁² × h₁] ÷ [(1/3) × π × r₂² × h₂]

⇒ V₁ ÷ V₂ = (r₁² × h₁) ÷ (r₂² × h₂)

⇒ 8 ÷ 9 = (5² × h₁) ÷ (6² × h₂)

⇒ 8 ÷ 9 = (25 × h₁) ÷ (36 × h₂)

⇒ h₁ ÷ h₂ = (8 ÷ 9) × (36 ÷ 25)

⇒ h₁ ÷ h₂ = (8 × 4) ÷ (1 × 25)

⇒ h₁ ÷ h₂ = 32 ÷ 25

∴ The correct answer is 32 : 25.

Additional Information

Volume of a Cone

The volume of a cone is given by V = (1/3) × π × r² × h, where r is the radius and h is the vertical height.

Curved Surface Area (CSA)

The CSA of a cone is π × r × l, where l is the slant height and l = √(r² + h²).

Total Surface Area (TSA)

The TSA of a cone includes the base area: TSA = π × r × l + π × r² = π × r(l + r).

Comparison with Cylinder

The volume of a cone is exactly one-third the volume of a cylinder with the same radius and height.

Shortcut Trick

Convert fractional ratio a : b : c to integers by multiplying by LCM(4, 3, 2) = 12.

a : b : c = (1/4 × 12) : (1/3 × 12) : (1/2 × 12) = 3 : 4 : 6.

Substitute these values into the required ratio: a/b : b/c : c/a = 3/4 : 4/6 : 6/3.

Simplify: 3/4 : 2/3 : 2/1. Multiply by LCM(4, 3) = 12 to remove denominators.

Final Ratio = (3/4 × 12) : (2/3 × 12) : (2 × 12) = 9 : 8 : 24.

∴ The correct answer is 9 : 8 : 24.

Alternate Method

Given:

a : b : c = 1/4 : 1/3 : 1/2

Formula Used:

To simplify a : b : c, multiply each term by the LCM of the denominators. To find the ratio of fractions, multiply by the LCM of their respective denominators.

Calculations:

⇒ LCM of denominators (4, 3, 2) = 12

⇒ a = 1/4 × 12 = 3

⇒ b = 1/3 × 12 = 4

⇒ c = 1/2 × 12 = 6

⇒ Now, calculate a/b : b/c : c/a

⇒ Ratio = 3/4 : 4/6 : 6/3

⇒ Ratio = 3/4 : 2/3 : 2/1

⇒ LCM of denominators (4, 3, 1) = 12

⇒ New Ratio = (3/4 × 12) : (2/3 × 12) : (2/1 × 12)

⇒ New Ratio = 9 : 8 : 24

∴ The correct answer is 9 : 8 : 24.

Additional Information

Simplifying Fractional Ratios

To convert a ratio of fractions x/y : p/q into integers, multiply both terms by the LCM of denominators y and q.

Reciprocal Ratio

The reciprocal ratio of a : b : c is 1/a : 1/b : 1/c. In this question, the given values of a, b, and c are already in a reciprocal-like fractional form.

Compound Ratio

The compound ratio of (a : b) and (c : d) is calculated by multiplying the corresponding terms: (a × c) : (b × d).

Shortcut Trick

Ratio of Expenditure : Savings = 80% : 20% = 4 : 1.

Expenditure increases by 20% and Income increases by 29%.

Let Savings increase by x%.

By weighted average: (4 × 20% + 1 × x%) / (4 + 1) = 29%.

80 + x = 145 ⇒ x = 65%.

∴ The correct answer is 65%.

Alternate Method

Given:

Initial Savings = 20% of Income

Income increase = 29%

Expenditure increase = 20%

Calculations:

⇒ Let Initial Income = ₹100

⇒ Initial Savings = 20% of 100 = ₹20

⇒ Initial Expenditure = 100 − 20 = ₹80

⇒ New Income = 100 + 29% of 100 = ₹129

⇒ New Expenditure = 80 + 20% of 80 = 80 + 16 = ₹96

⇒ New Savings = New Income − New Expenditure = 129 − 96 = ₹33

⇒ Increase in Savings = 33 − 20 = ₹13

⇒ Percentage Increase in Savings = (13 / 20) × 100

⇒ Percentage Increase = 65%

∴ The correct answer is 65%.

Additional Information

Income, Expenditure and Savings Relation

Income = Expenditure + Savings. Any percentage change in income is the weighted average of the percentage changes in expenditure and savings.

Percentage Change Formula

Percentage Change = [(New Value − Old Value) / Old Value] × 100.

Alligation in Income-Expenditure

The ratio of Expenditure to Savings can be found using the alligation of their respective percentage changes against the total income percentage change.

Shortcut Trick

Rate = 10% = 1/10. For 2 years, ratios are (10:11) and (100:121).

Make installments equal: (10:11) × 11 → 110 : 121.

Total Principal = 110 + 100 = 210 units. Total Paid = 121 + 121 = 242 units.

Interest = 242 − 210 = 32 units.

Given 121 units = ₹ 5,808 ⇒ 1 unit = 48.

Total Interest = 32 × 48 = ₹ 1,536.

∴ The correct answer is Rs. 1,536.

Alternate Method

Given:

Rate (R) = 10% p.a.

Time (n) = 2 years

Installment (I) = Rs. 5,808

Formula Used:

Principal (P) = I / (1 + R/100)¹ + I / (1 + R/100)²

Total Interest = (Total Amount Paid) − Principal

⇒ P = 5808 / (1 + 10/100) + 5808 / (1 + 10/100)²

⇒ P = 5808 / (1.1) + 5808 / (1.21)

⇒ P = 5280 + 4800

⇒ P = Rs. 10,080

⇒ Total amount paid = 2 × 5,808 = Rs. 11,616

⇒ Total Interest = 11616 − 10080

⇒ Total Interest = Rs. 1,536

∴ The correct answer is Rs. 1,536.

Additional Information

Installment Formula for CI

For 'n' yearly installments, Principal (P) = I[1/(1+r) + 1/(1+r)² + ... + 1/(1+r)ⁿ] where r is the rate in decimal.

Effective Interest Rate

For two years at rate r%, the effective compound interest rate is r + r + (r×r)/100 %.

CI vs SI Difference

Difference for 2 years = P(R/100)². Difference for 3 years = P(R/100)² × (3 + R/100).

Shortcut Trick

Radius (r) = √(Volume ÷ (π × height))

r = √(392π ÷ 8π) = √49 = 7 cm

Curved Surface Area (CSA) = 2 × π × r × h

CSA = 2 × π × 7 × 8 = 112π cm²

∴ The correct answer is 112π cm².

Alternate Method

Given: Height (h) = 8 cm, Volume (V) = 392π cm³

Formula: Volume of Cylinder = πr²h | Curved Surface Area = 2πrh

⇒ Volume = πr²h

⇒ 392π = π × r² × 8

⇒ r² = 392 ÷ 8

⇒ r² = 49

⇒ r = 7 cm

⇒ Curved Surface Area (CSA) = 2πrh

⇒ CSA = 2 × π × 7 × 8

⇒ CSA = 112π cm²

∴ The correct answer is 112π cm².

Additional Information

Total Surface Area (TSA) of Cylinder

The TSA includes the curved surface and the area of two circular bases: TSA = 2πr(r + h).

Relationship between Volume and CSA

The Volume can be expressed in terms of Curved Surface Area as: Volume = (CSA × r) ÷ 2.

Volume of a Hollow Cylinder

For a hollow cylinder with external radius R and internal radius r: Volume = π(R² − r²)h.

Shortcut Trick

Replace the variables directly with their respective remainders: a → 9, b → 7, and c → 10.

Expression value = 9 + 2(7) + 5(10) = 9 + 14 + 50 = 73.

Final remainder = Remainder of (73 ÷ 13).

Since 13 × 5 = 65, the remainder is 73 − 65 = 8.

∴ The correct answer is 8.

Alternate Method

Given:

a = 13p + 9

b = 13q + 7

c = 13r + 10

Formula used: Rem(X + Y) = [Rem(X) + Rem(Y)] ÷ Divisor

⇒ Expression = (a + 2b + 5c)

⇒ Remainder = [Rem(a) + 2 × Rem(b) + 5 × Rem(c)] mod 13

⇒ Remainder = [9 + 2(7) + 5(10)] mod 13

⇒ Remainder = [9 + 14 + 50] mod 13

⇒ Remainder = 73 mod 13

⇒ 73 = 13 × 5 + 8

⇒ Remainder = 8

∴ The correct answer is 8.

Additional Information

Remainder Addition Rule

The remainder of a sum of numbers is the same as the remainder of the sum of their individual remainders. Formula: (A + B) mod n = [(A mod n) + (B mod n)] mod n.

Remainder Multiplication Rule

The remainder of a product is the same as the remainder of the product of their individual remainders. Formula: (A × B) mod n = [(A mod n) × (B mod n)] mod n.

Negative Remainder Concept

A remainder of (n − k) can also be written as −k. For example, dividing 12 by 13 gives a remainder of 12 or −1.

Given:

Total number of students = 2100

Difference between students in A and B and students in C and D = x

Calculation:

Angle of the students specialising in A and B = 61.2° + 75.6° = 136.8° 

Angle of the students specialising in C and D = 72° + 54° = 126° 

Angle difference = 136.8° – 126° = 10.8°

360° = 2100

⇒ 1° = 2100/360 = 210/36

⇒ 1° = 210/36

⇒ 10.8° = 210/36 × 10.8

⇒ 10.8° = 2268/36

⇒ 10.8° = 63

⇒ x = 63

∴ The total number of students specializing in A and B exceeds the total number of students specializing in C and D by x, which lies between 60 and 65.

Shortcut Trick

Directly substitute the standard trigonometric values: cos 60° = 1/2, sec 30° = 2/√3, tan 45° = 1, tan 60° = √3, sin 30° = 1/2, and cos 45° = 1/√2.

Numerator = 5(1/4) + 4(4/3) − 1 = 5/4 + 16/3 − 1 = (15 + 64 − 12) / 12 = 67/12.

Denominator = 3 − 1/4 − 1/2 = (12 − 1 − 2) / 4 = 9/4.

Value = (67/12) ÷ (9/4) = 67/12 × 4/9 = 67/27.

∴ The correct answer is 67/27.

Alternate Method

Given:

Expression: (5 cos² 60° + 4 sec² 30° − tan² 45°) ÷ (tan² 60° − sin² 30° − cos² 45°)

Formula Used:

Standard values of trigonometric ratios for specific angles.

Calculations:

⇒ Substituting the values into the numerator:

⇒ Numerator = 5 × (1/2)² + 4 × (2/√3)² − (1)²

⇒ Numerator = 5 × (1/4) + 4 × (4/3) − 1

⇒ Numerator = 5/4 + 16/3 − 1

⇒ Numerator = (15 + 64 − 12) / 12 = 67/12

⇒ Substituting the values into the denominator:

⇒ Denominator = (√3)² − (1/2)² − (1/√2)²

⇒ Denominator = 3 − 1/4 − 1/2

⇒ Denominator = (12 − 1 − 2) / 4 = 9/4

⇒ Final Value = (67/12) / (9/4)

⇒ Final Value = 67/12 × 4/9

⇒ Final Value = 67 / (3 × 9) = 67/27

∴ The correct answer is 67/27.

Additional Information

Trigonometric Identities

Key identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Standard Values Table

For 30°: sin = 1/2, cos = √3/2, tan = 1/√3. For 60°: sin = √3/2, cos = 1/2, tan = √3. For 45°: sin = 1/√2, cos = 1/√2, tan = 1.

Complementary Angles

sin(90°−θ) = cosθ, cos(90°−θ) = sinθ, and tan(90°−θ) = cotθ.

Given:

$\frac{25\% \times (50\% \times 30\% \times 150)}{40\% \times 2250}$ = ?

Formula used:

x% of y = (x / 100) × y

Calculation:

$\frac{25\% \times (50\% \times 30\% \times 150)}{40\% \times 2250}$

⇒ $\frac{0.25 \times (0.5 \times 0.3 \times 150)}{0.4 \times 2250}$

⇒ $\frac{0.25 \times 22.5}{900}$

⇒ $\frac{5.625}{900}$

⇒ 0.00625

⇒ 0.625%

∴ The correct answer is option (4).

Shortcut Trick

When AD = AE and ∠BAD = ∠EAC in ΔABC with D and E on BC, then ΔABD ≅ ΔAEC.

By property of congruence: AB = AC and BD = EC.

3x + 1 = 34 ⇒ 3x = 33 ⇒ x = 11.

y + 1 = 9 ⇒ y = 8.

Sum (x + y) = 11 + 8 = 19.

∴ The correct answer is 19.

Alternate Method

Given: AD = AE, ∠BAD = ∠EAC, AB = (3x + 1) cm, BD = 9 cm, AC = 34 cm, EC = (y + 1) cm.

In ΔADE, since AD = AE, the angles opposite to these sides are equal:

⇒ ∠ADE = ∠AED

Now, consider the supplementary angles on line BC:

⇒ ∠ADB = 180° − ∠ADE

⇒ ∠AEC = 180° − ∠AED

⇒ ∠ADB = ∠AEC (Since ∠ADE = ∠AED)

In ΔABD and ΔAEC:

1. AD = AE (Given)
2. ∠BAD = ∠EAC (Given)
3. ∠ADB = ∠AEC (Proven above)

⇒ ΔABD ≅ ΔAEC (By ASA congruence criterion)

By CPCT (Corresponding Parts of Congruent Triangles):

⇒ AB = AC

⇒ 3x + 1 = 34

⇒ 3x = 33 ⇒ x = 11

⇒ BD = EC

⇒ 9 = y + 1 ⇒ y = 8

⇒ x + y = 11 + 8 = 19

∴ The correct answer is 19.

Additional Information

Isosceles Triangle Properties

In any triangle, if two sides are equal, the angles opposite to them are also equal. If AD = AE, then ∠ADE = ∠AED.

ASA Congruence Rule

Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding parts of another triangle.

CPCT Theorem

Corresponding Parts of Congruent Triangles are equal. Once congruence is proved, all remaining corresponding sides and angles are equal.

Shortcut Trick

Centroid G divides medians AD and BE in a 2 : 1 ratio.

GD = 1/3 × AD = 1/3 × 18 = 6 cm and BG = 2/3 × BE = 2/3 × 12 = 8 cm.

Since medians intersect at 90°, ΔBGD is a right-angled triangle.

BD = √(BG² + GD²) = √(8² + 6²) = 10 cm.

D is the midpoint of BC, so DC = BD = 10 cm.

∴ The correct answer is 10 cm.

Alternate Method

Given: AD = 18 cm, BE = 12 cm, AD ⊥ BE.

Formula Used: Centroid divides median in 2 : 1 ratio; Pythagoras Theorem: a² + b² = c².

⇒ In ΔABC, AD and BE are medians intersecting at centroid G.

⇒ BG = (2/3) × BE = (2/3) × 12 = 8 cm.

⇒ GD = (1/3) × AD = (1/3) × 18 = 6 cm.

⇒ In right-angled ΔBGD, by Pythagoras theorem:

⇒ BD² = BG² + GD²

⇒ BD² = 8² + 6²

⇒ BD² = 64 + 36 = 100

⇒ BD = √100 = 10 cm.

⇒ Since D is the midpoint of BC, DC = BD = 10 cm.

∴ The correct answer is 10 cm.

Additional Information

Centroid Ratio Property

The centroid G of a triangle divides each median in the ratio 2 : 1 from the vertex to the midpoint of the opposite side.

Perpendicular Medians Relation

If two medians m_a and m_b intersect at 90°, then the relationship between the sides is a² + b² = 5c².

Apollonius' Theorem

In any ΔABC, if AD is a median, then AB² + AC² = 2(AD² + BD²).

Centroid and Area

The three medians of a triangle divide it into six smaller triangles of equal area.

Shortcut Trick

Central angle = 2 × inscribed angle on same arc

∠ABD = ∠AOD/2 = 32°/2 = 16° | ∠CDB = ∠COB/2 = 26°/2 = 13°

In ▵BPD: ∠DPB = 180° − (16° + 13°) = 151°

∠APD = 180° − ∠DPB = 180° − 151° = 29°

∴ ∠APD lies between 26° and 30°

Alternate Method

Given:

In a circle with center O, chords AB and CD intersect at P

∠AOD = 32° and ∠COB = 26° 

Concept used:

The angle made by a chord at the center of a circle is double to the angle made at the circumference of the circle

Calculation:

A circle can be drawn based on the given information as -

As per the question ∠AOD = 32° ; ∠COB = 26° 

Then, ∠ABD = 16° (∵ angle on circumference by the base chord AD)

Also, ∠CDB = 13° (∵ angle on circumference by the base chord BC)

Now, consider ΔBPD 

∠DPB = 180° - (13° + 16° ) 

∠DPB = 151°

∠DPB + ∠APD = 180° (Supplementary angles)

∠APD = 180° - 151° = 29°

⇒ ∠APD will be of 29° (∵ ∠APC + ∠APD = 180°)

∴ ∠APD lies between 26° and 30°

Additional Information

Intersecting Chords (Exterior)

If two chords intersect at a point P outside the circle, the angle formed is half the positive difference of the intercepted arcs: ∠P = 1/2 × |Arc1 − Arc2|.

Angle at Center Theorem

The angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the remaining part of the circle.

Chords Intersection Product Rule

If two chords AB and CD intersect at P (inside or outside), then AP × PB = CP × PD.

Shortcut Trick

Total chocolate = 8% of 50 = 4 litres.

Total new volume = 50 + 10 = 60 litres.

New chocolate percentage = (4 ÷ 60) × 100 = 20/3 = 6 ²⁄₃%.

∴ The correct answer is 6 ²⁄₃%.

Alternate Method

Given:

Initial volume of drink = 50 litres

Percentage of chocolate = 8%

Volume of milk added = 10 litres

Formula Used:

Percentage of ingredient = (Volume of ingredient ÷ Total volume) × 100

Calculations:

⇒ Pure chocolate in the initial 50 litres = (8 ÷ 100) × 50 = 4 litres.

⇒ Total volume of the new drink = 50 + 10 = 60 litres.

⇒ Percentage of chocolate in new drink = (4 ÷ 60) × 100

⇒ New percentage = (1 ÷ 15) × 100 = 100/15

⇒ New percentage = 20/3 = 6 ²⁄₃%

∴ The correct answer is 6 ²⁄₃%.

Additional Information

Mixture Dilution Rule

When a solvent (like milk) is added to a mixture, the quantity of the solute (like chocolate) remains constant. The formula used is C₁V₁ = C₂V₂ where C is concentration and V is volume.

Percentage Change in Concentration

If the volume increases, the concentration of the solute decreases in inverse proportion. For example, if volume increases by 20% (60/50 = 1.2), the concentration becomes 1/1.2 of the original.

Fraction to Percentage Conversion

Standard fractions are useful for quick solving: 1/15 ≈ 6.66% or 6 ²⁄₃%.

Shortcut Trick

From the first condition, A − 5 = B + 5 ⇒ A − B = 10.

Check the given ratio 19 : 17 where the difference is 2 units.

Equating: 2 units = 10 ⇒ 1 unit = 5.

Then, A = 19 × 5 = 95 and B = 17 × 5 = 85.

Verify second condition: (95 + 25) : (85 − 25) = 120 : 60 = 2 : 1. Correct.

∴ The correct answer is 19 : 17.

Alternate Method

Given:

Case 1: 5 children from A join B → A − 5 = B + 5

Case 2: 25 children from B join A → A + 25 = 2 × (B − 25)

Formula Used:

Solving simultaneous linear equations: Express one variable in terms of the other and substitute.

Calculations:

⇒ A − 5 = B + 5

⇒ A − B = 10 ⇒ A = B + 10 ... (i)

⇒ A + 25 = 2(B − 25)

⇒ A + 25 = 2B − 50

⇒ A − 2B = −75 ... (ii)

Substitute (i) in (ii):

⇒ (B + 10) − 2B = −75

⇒ −B + 10 = −75

⇒ −B = −85 ⇒ B = 85

⇒ A = 85 + 10 = 95

⇒ Ratio A : B = 95 : 85

⇒ Dividing by 5: 19 : 17

∴ The correct answer is 19 : 17.

Additional Information

Ratio and Proportion

A ratio compares two quantities by division. If a : b = c : d, then a × d = b × c.

Linear Equations

Word problems involving two variables can be solved by translating sentences into equations like ax + by = c and solving for the unknowns.

Difference Principle

When a quantity x moves from A to B and results in equality, the initial difference between A and B is exactly 2x.

Shortcut Trick

Profit = Capital × Time ⇒ Time = Profit ÷ Capital

Ratio of Time (x : y : z) = (Profit_A ÷ Capital_A) : (Profit_B ÷ Capital_B) : (Profit_C ÷ Capital_C)

Ratio (x : y : z) = (45 ÷ 5) : (42 ÷ 7) : (28 ÷ 4)

Ratio (x : y : z) = 9 : 6 : 7

∴ The correct answer is 9 : 6 : 7.

Alternate Method

Given:

Ratio of Capitals (A : B : C) = 5 : 7 : 4

Ratio of Investment Times (A : B : C) = x : y : z

Ratio of Profits (A : B : C) = 45 : 42 : 28

Formula:

Profit = Capital × Time

According to the partnership principle:

⇒ Profit of A = Capital_A × Time_A ⇒ 45 = 5 × x ⇒ x = 45 ÷ 5 = 9

⇒ Profit of B = Capital_B × Time_B ⇒ 42 = 7 × y ⇒ y = 42 ÷ 7 = 6

⇒ Profit of C = Capital_C × Time_C ⇒ 28 = 4 × z ⇒ z = 28 ÷ 4 = 7

⇒ x : y : z = 9 : 6 : 7

∴ The correct answer is 9 : 6 : 7.

Additional Information

Partnership Profit Distribution

In a partnership, profits are shared in the ratio of the product of capital and the time period of investment: P₁ : P₂ = (C₁ × T₁) : (C₂ × T₂).

Calculation of Capital Ratio

If the profit ratio and time ratio are known, the capital ratio can be found using: C₁ : C₂ = (P₁ ÷ T₁) : (P₂ ÷ T₂).

Simple vs. Compound Partnership

Simple partnership involves capitals invested for the same time. Compound partnership involves capitals invested for different time periods.

Shortcut Trick

According to Angle Bisector Theorem, the ratio of sides is equal to the ratio of base segments: PQ : PR = QS : SR.

Side ratio → 16 : 22.4 = 160 : 224 = 5 : 7.

Sum of ratio parts = 5 + 7 = 12 units.

Total base QR = 9.6 cm ⇒ 12 units = 9.6 cm ⇒ 1 unit = 0.8 cm.

Length of SR = 7 units = 7 × 0.8 = 5.6 cm.

∴ The correct answer is 5.6 cm.

Alternate Method

Given:

In ΔPQR, PS is the internal bisector of ∠P.

PQ = 16 cm, PR = 22.4 cm, QR = 9.6 cm.

Formula Used:

Internal Angle Bisector Theorem: PQ ÷ PR = QS ÷ SR

Calculations:

⇒ Let SR = x cm, then QS = (9.6 − x) cm.

⇒ According to Angle Bisector Theorem: PQ ÷ PR = QS ÷ SR

⇒ 16 ÷ 22.4 = (9.6 − x) ÷ x

⇒ 160 ÷ 224 = (9.6 − x) ÷ x

⇒ 5 ÷ 7 = (9.6 − x) ÷ x

⇒ 5 × x = 7 × (9.6 − x)

⇒ 5x = 67.2 − 7x

⇒ 5x + 7x = 67.2

⇒ 12x = 67.2

⇒ x = 67.2 ÷ 12

⇒ x = 5.6 cm

∴ The correct answer is 5.6 cm.

Additional Information

Internal Angle Bisector Theorem

It states that an angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

External Angle Bisector Theorem

If the external bisector of ∠A of ΔABC intersects BC produced at E, then AB ÷ AC = BE ÷ CE.

Length of Angle Bisector

The length of the internal bisector PS is given by PS² = (PQ × PR) − (QS × SR).

Given:

$\frac{1}{x + \frac{1}{y + \frac{2}{z + \frac{1}{4}}}} = \frac{29}{79}$

Formula used:

Continued fraction expansion by reciprocal and division.

Calculations:

⇒ $x + \frac{1}{y + \frac{2}{z + \frac{1}{4}}} = \frac{79}{29}$

⇒ $\frac{79}{29} = 2 + \frac{21}{29}$

⇒ x = 2

⇒ $\frac{1}{y + \frac{2}{z + \frac{1}{4}}} = \frac{21}{29}$

⇒ $y + \frac{2}{z + \frac{1}{4}} = \frac{29}{21} = 1 + \frac{8}{21}$

⇒ y = 1

⇒ $\frac{2}{z + \frac{1}{4}} = \frac{8}{21}$

⇒ $\frac{1}{z + \frac{1}{4}} = \frac{4}{21}$

⇒ $z + \frac{1}{4} = \frac{21}{4} = 5 + \frac{1}{4}$

⇒ z = 5

⇒ Value = 2x + 3y - z

⇒ 2 × 2 + 3 × 1 - 5

⇒ 4 + 3 - 5

⇒ 2

∴ The correct answer is option (3).

Shortcut Trick

Scale the second case by 2: 5 km Upstream + 10 km Downstream = 50 minutes.

Compare with the first case: 5 km Upstream + 7.5 km Downstream = 45 minutes.

The difference of 2.5 km Downstream takes 5 minutes (50 − 45), so 5 km Downstream takes 10 minutes.

From the scaled second case, 5 km Upstream takes 50 − 10 = 40 minutes, which means 1 km Upstream takes 8 minutes? No, let's re-evaluate from the original second case.

In Case 2: 2.5 km Upstream + (5 km Downstream = 10 min) = 25 minutes → 2.5 km Upstream takes 15 minutes.

Rate Upstream = 15 ÷ 2.5 = 6 minutes per km. For 6 km Upstream = 6 × 6 = 36 minutes.

∴ The correct answer is 36 minutes.

Alternate Method

Given:

Case 1: 5 km Upstream + 7.5 km Downstream = 45 minutes

Case 2: 2.5 km Upstream + 5 km Downstream = 25 minutes

Formula Used: Time = Distance ÷ Speed

Let the time taken to travel 1 km upstream be 'u' and 1 km downstream be 'd'.

⇒ 5u + 7.5d = 45 — (i)

⇒ 2.5u + 5d = 25 — (ii)

Multiply equation (ii) by 1.5 to match 'd' coefficients:

⇒ 3.75u + 7.5d = 37.5 — (iii)

Subtract equation (iii) from equation (i):

⇒ (5u − 3.75u) + (7.5d − 7.5d) = 45 − 37.5

⇒ 1.25u = 7.5

⇒ u = 7.5 ÷ 1.25

⇒ u = 6 minutes/km

This 'u' is the time required to travel 1 km upstream.

⇒ Time for 6 km upstream = 6 × u

⇒ Time = 6 × 6 = 36 minutes

∴ The correct answer is 36 minutes.

Additional Information

Upstream and Downstream Speed

If boat speed is B and stream speed is S, then Downstream Speed (D) = B + S and Upstream Speed (U) = B − S.

Boat and Stream Speed from U and D

Boat speed in still water (B) = (D + U) ÷ 2 and Stream speed (S) = (D − U) ÷ 2.

Average Speed for Round Trip

Average Speed = (D × U) ÷ B = (B² − S²) ÷ B when traveling equal distances upstream and downstream.

Given:

$3\frac{1}{5} \div 4\frac{1}{2} \text{ of } 5\frac{1}{3} + \frac{1}{8} \div \frac{1}{2} \text{ of } \frac{1}{4} - \frac{1}{4} ( \frac{1}{2} \div \frac{1}{8} \times \frac{1}{4} )$

Formula used:

BODMAS rule

Calculation:

$\frac{16}{5} \div ( \frac{9}{2} \times \frac{16}{3} ) + \frac{1}{8} \div ( \frac{1}{2} \times \frac{1}{4} ) - \frac{1}{4} ( \frac{1}{2} \times 8 \times \frac{1}{4} )$

⇒ $\frac{16}{5} \div 24 + \frac{1}{8} \div \frac{1}{8} - \frac{1}{4} (1)$

⇒ $\frac{2}{15} + 1 - \frac{1}{4}$

⇒ $\frac{2}{15} + \frac{3}{4}$

⇒ $\frac{8 + 45}{60} = \frac{53}{60}$

∴ The correct answer is option (1).

Shortcut Trick

Equating total work: (2M + 7W) × 28 = (6M + 16W) × 11

56M + 196W = 66M + 176W ⇒ 10M = 20W ⇒ 1 Man = 2 Women.

Total work units = (2 × 2 + 7) × 28 = 11 × 28 = 308 units.

Efficiency of 5 Men + 4 Women = 5 × 2 + 4 = 14 units/day.

Required time = 308 ÷ 14 = 22 days.

∴ The correct answer is 22 days.

Alternate Method

Given:

Time taken by (2M + 7W) = 28 days

Time taken by (6M + 16W) = 11 days

Formula Used:

Total Work = Efficiency × Time

⇒ 2M + 7W = 11 --- (i)

⇒ 6M + 16W = 28 --- (ii)

⇒ Multiply eq. (i) by 3: 6M + 21W = 33 --- (iii)

⇒ Subtracting eq. (ii) from eq. (iii): (6M − 6M) + (21W − 16W) = 33 − 28

⇒ 5W = 5 ⇒ Efficiency of 1 Woman = 1 unit/day

⇒ Substituting W = 1 in eq. (i): 2M + 7(1) = 11 ⇒ 2M = 4 ⇒ Efficiency of 1 Man = 2 units/day

⇒ Efficiency of 5 Men + 4 Women = 5(2) + 4(1) = 10 + 4 = 14 units/day

⇒ Days required = Total Work ÷ Total Efficiency = 308 ÷ 14 = 22 days

∴ The correct answer is 22 days.

Additional Information

M-D-H Rule

(M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂ where M is persons, D is days, H is hours, and W is work.

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete the work (E ∝ 1/T).

Work and Wages

When multiple people work together, wages are distributed in the ratio of their efficiencies or the actual work done by each.

Given:

$\dfrac{27 \times (0.25)^3 + 125 \times (0.05)^3}{(0.75)^2 - 0.25 \times 0.5}$

Formula used:

a³ + b³ = (a + b)(a² - ab + b²)

Calculation:

⇒ $\dfrac{(3 \times 0.25)^3 + (5 \times 0.05)^3}{0.75^2 - 0.25 \times (0.75 - 0.25)}$

⇒ $\dfrac{0.75^3 + 0.25^3}{0.75^2 - 0.75 \times 0.25 + 0.25^2}$

⇒ $\dfrac{(0.75 + 0.25)(0.75^2 - 0.75 \times 0.25 + 0.25^2)}{0.75^2 - 0.75 \times 0.25 + 0.25^2}$

⇒ 0.75 + 0.25

⇒ 1

∴ The correct answer is option (2).

Shortcut Trick

Secant-secant power of point: PA × PB = PC × PD, where PC = OP − r, PD = OP + r

8 × 21 = (17 − r)(17 + r) = 289 − r²

168 = 289 − r² ⇒ r² = 121 ⇒ r = 11

∴ The correct answer is 11 cm.

Alternate Method

Given:

OP = 17 cm

PA = 8 cm

PB = 21 cm

Concept Used:

For two intersecting secants from an external point:

PA × PB = PC × PD

Calculation:

⇒ Let the radius of the circle = x cm

⇒ PC = 17 - x

⇒ PD = 17 + x

⇒ PA × PB = PC × PD

⇒ 8 × 21 = (17 - x)(17 + x)

⇒ 168 = 289 - x2

⇒ x2 = 289 - 168

⇒ x2 = 121

⇒ x = √121

⇒ x = 11

∴ The radius of the circle is 11 cm.

Additional Information

Power of a Point

For any line through P intersecting the circle at A and B, the product PA × PB is constant and equals d² − r², where d is the distance from P to the centre.

Tangent-Secant Theorem

If a tangent PT and a secant PAB are drawn from point P, then PT² = PA × PB.

Intersecting Chords Theorem

If two chords AB and CD intersect inside a circle at point P, then PA × PB = PC × PD.

Shortcut Trick

For meeting point problems, Ratio of Speeds S_A/S_B = √(T_B/T_A), where T_A and T_B are times taken after crossing.

Given: S_A = 33 km/h, T_A = 5 ⁴/₉ = 49/9 hours, T_B = 9 hours.

33 / S_B = √(9 ÷ (49/9)) = √(81/49) = 9/7.

S_B = (33 × 7) ÷ 9 = 77/3 = 25 ²/₃ km/h.

∴ The correct answer is 25 ²/₃ km/h.

Alternate Method

Given:

Speed of A (S_A) = 33 km/h

Time taken by A after meeting (T_A) = 5 ⁴/₉ hr = 49/9 hr

Time taken by B after meeting (T_B) = 9 hr

Formula used:

S_A / S_B = √(T_B / T_A)

Calculations:

⇒ S_A / S_B = √(T_B ÷ T_A)

⇒ 33 / S_B = √(9 ÷ (49/9))

⇒ 33 / S_B = √((9 × 9) / 49)

⇒ 33 / S_B = √(81 / 49)

⇒ 33 / S_B = 9 / 7

⇒ S_B = (33 × 7) ÷ 9

⇒ S_B = 231 ÷ 9 = 77 / 3

⇒ S_B = 25 ²/₃ km/h

∴ The correct answer is 25 ²/₃ km/h.

Additional Information

Meeting Time Formula

The time taken by both objects to meet after starting simultaneously is given by t = √(T_A × T_B), where T_A and T_B are times taken after crossing.

Distance Ratio

The ratio of the distances covered by A and B before they meet is equal to the ratio of their speeds: D_A / D_B = S_A / S_B.

Total Distance

Total Distance D = (S_A + S_B) × t, where t is the time elapsed until the meeting point.

Shortcut Trick

Substitute a standard angle to solve quickly. Let A = 45°.

At 45°, tan 45° = 1, cot 45° = 1, sec 45° = √2, and cosec 45° = √2.

Expression = (1² + 1² − 2) − (√2)² × (√2)²

= (1 + 1 − 2) − (2 × 2) = 0 − 4 = −4.

∴ The correct answer is −4.

Alternate Method

Given: (tan² A + cot² A − 2) − sec² A cosec² A

Formula used: sec² A cosec² A = sec² A + cosec² A

⇒ sec² A cosec² A = (1 + tan² A) + (1 + cot² A)

⇒ sec² A cosec² A = tan² A + cot² A + 2

⇒ Expression = (tan² A + cot² A − 2) − (tan² A + cot² A + 2)

⇒ Expression = tan² A + cot² A − 2 − tan² A − cot² A − 2

⇒ Expression = −2 − 2 = −4

∴ The correct answer is −4.

Additional Information

Fundamental Trigonometric Identities

1 + tan² A = sec² A and 1 + cot² A = cosec² A. Also, sin² A + cos² A = 1.

Special Sum-to-Product Identity

The relation sec² A + cosec² A = sec² A × cosec² A is derived by converting terms to sin and cos.

Algebraic Form of Identities

(tan A − cot A)² = tan² A + cot² A − 2 and (tan A + cot A)² = tan² A + cot² A + 2.

Shortcut Trick

Ratio of CP to MP is given by: CP : MP = (100 − Discount%) : (100 + Profit%)

CP : MP = (100 − 16) : (100 + 18) = 84 : 118

Given CP = 84 units = Rs. 840 → 1 unit = Rs. 10

Marked Price (MP) = 118 units = 118 × 10 = Rs. 1,180

∴ The correct answer is Rs. 1,180.

Alternate Method

Given:

Cost Price (CP) = Rs. 840

Profit% = 18%

Discount% = 16%

Formula Used:

Selling Price (SP) = CP × (100 + Profit%) ÷ 100

Marked Price (MP) = SP × 100 ÷ (100 − Discount%)

Calculations:

⇒ SP = 840 × (118 ÷ 100)

⇒ SP = 840 × 1.18 = Rs. 991.20

⇒ MP = 991.20 × 100 ÷ (100 − 16)

⇒ MP = 991.20 × 100 ÷ 84

⇒ MP = 99120 ÷ 84

⇒ MP = Rs. 1,180

∴ The correct answer is Rs. 1,180.

Additional Information

Mark-up Percentage

It is the percentage by which the cost price is increased to reach the marked price: Mark-up% = [(MP − CP) ÷ CP] × 100.

Successive Discounts

If two discounts d₁% and d₂% are given, the effective discount is: d₁ + d₂ − (d₁ × d₂) ÷ 100.

Net Profit with Markup and Discount

If an item is marked up by M% and then a discount of D% is offered, the Net Profit% = M − D − (M × D) ÷ 100.

Shortcut Trick

Effective Markup = +40%

Net Profit/Loss = [∑(Quantity Ratio × Individual Profit/Loss %)]

For 60% goods (3/5): Profit % = 140 × 0.9 − 100 = 126 − 100 = 26%

For 40% goods (2/5): Profit/Loss % = 140 × 0.5 − 100 = 70 − 100 = −30%

Overall % = (3/5 × 26) + (2/5 × −30) = 15.6 − 12 = 3.6%

∴ The correct answer is Profit 3.6%.

Alternate Method

Assume total Cost Price (CP) of 100 units = ₹10000 (at ₹100 per unit).

Given: Markup = 40% above CP.

⇒ Marked Price (MP) per unit = 100 + 40% of 100 = ₹140

⇒ Quantity 1 (60% of 100) = 60 units

⇒ Selling Price 1 (SP₁) = MP − 10% discount = 140 − (0.10 × 140) = ₹126 per unit

⇒ Total SP₁ = 60 × 126 = ₹7560

⇒ Quantity 2 (40% of 100) = 40 units

⇒ Selling Price 2 (SP₂) = MP − 50% discount = 140 − (0.50 × 140) = ₹70 per unit

⇒ Total SP₂ = 40 × 70 = ₹2800

⇒ Total Selling Price = 7560 + 2800 = ₹10360

⇒ Total Profit = Total SP − Total CP = 10360 − 10000 = ₹360

⇒ Profit % = (360 ÷ 10000) × 100 = 3.6%

∴ The correct answer is Profit 3.6%.

Additional Information

Marked Price and Discount Relation

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

Successive Discounts

If two successive discounts of a% and b% are given, the equivalent single discount is (a + b − ab/100)%.

Effective Profit Percentage

When parts of goods are sold at different profit/loss rates, the overall profit % is the weighted average of individual profit/loss percentages based on the quantity ratio.

Shortcut Trick

Distance of pole from hill (d) = Height ÷ tan 60° = 240 ÷ √3 = 80√3 m.

Vertical gap between hill top and pole top = d × tan 30° = 80√3 × (1/√3) = 80 m.

Height of pole (h) = Total height − Gap = 240 − 80 = 160 m.

Required difference = |h − d| = |160 − 80√3| = 80 (2 − √3) m.

∴ The correct answer is 80 (2 − √3) m.

Alternate Method

Given:

Height of Hill (AB) = 240 m

Angle of depression to pole top (α) = 30°

Angle of depression to pole bottom (β) = 60°

Let height of pole CD be h and distance BD be d.

In ▵ABD (Hill bottom):

⇒ tan 60° = AB ÷ BD

⇒ √3 = 240 ÷ d

⇒ d = 240 ÷ √3 = 80√3 m.

In ▵ACE (where E is point on AB at same level as C):

⇒ AE = AB − h = 240 − h

⇒ tan 30° = AE ÷ EC

⇒ 1 ÷ √3 = (240 − h) ÷ d

⇒ 1 ÷ √3 = (240 − h) ÷ (80√3)

⇒ 80 = 240 − h

⇒ h = 160 m.

Difference between pole height (h) and distance (d):

⇒ Difference = h − d = 160 − 80√3

⇒ Difference = 80 (2 − √3) m.

∴ The correct answer is 80 (2 − √3) m.

Additional Information

Trigonometric Ratios for Angles

Key values used: tan 30° = 1/√3 and tan 60° = √3. These ratios relate the vertical height (perpendicular) to the horizontal base.

Angle of Depression

The angle of depression is equal to the angle of elevation from the target point back to the observer due to the property of alternate interior angles between parallel horizontal lines.

Standard 30-60-90 Ratio

In a right triangle with angles 30° and 60°, the sides opposite to these angles are in the ratio 1 : √3 : 2 (hypotenuse).

Given:

Break up (degree wise) of students in terms of specialization in different areas (A, B, C, D & E) in an MBA program.

Concept used:

Percentage.

Calculation:

E - C = 97.2 - 72

E - C = 25.2

Now E is what percent more than C

⇒ $\frac{{252}}{{720}} \times 100$

∴ 35%

Shortcut Trick

In any regular polygon, Each Exterior Angle = 360° ÷ n.

Given n = 16 sides.

Exterior Angle = 360° ÷ 16 = 22.5°.

Each Interior Angle = 180° − Exterior Angle.

Interior Angle = 180° − 22.5° = 157.5° = 157 &frac{1}{2}°.

∴ The correct answer is 157 &frac{1}{2}°.

Alternate Method

Given:

Number of sides (n) = 16

Formula Used:

Each Interior Angle of a regular polygon = [(n − 2) × 180°] ÷ n

Calculations:

⇒ Sum of all interior angles = (16 − 2) × 180°

⇒ Sum = 14 × 180° = 2520°

⇒ Measure of each interior angle = 2520° ÷ 16

⇒ Each interior angle = 157.5°

⇒ 157.5° = 157 &frac{1}{2}°

∴ The correct answer is 157 &frac{1}{2}°.

Additional Information

Sum of Exterior Angles

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

Interior and Exterior Angle Relation

At any vertex of a polygon, the interior angle and the exterior angle are supplementary: Interior + Exterior = 180°.

Number of Diagonals

The number of diagonals in a polygon with n sides is given by the formula: [n × (n − 3)] ÷ 2.

Regular Polygon Definition

A polygon is regular if all its sides are of equal length and all its interior angles are of equal measure.

Shortcut Trick

The relationship between the angle at the incentre (∠BOC) and the vertex angle (∠BAC) is:

∠BOC = 90° + (∠BAC ÷ 2)

Given ∠BOC = 135°

135° = 90° + (∠BAC ÷ 2)

45° = ∠BAC ÷ 2 ⇒ ∠BAC = 90°

∴ The correct answer is 90°.

Alternate Method

Given:

In ΔABC, O is the incentre.

Measure of ∠BOC = 135°

Formula:

Angle at Incentre ∠BOC = 90° + (∠A ÷ 2)

Calculations:

⇒ ∠BOC = 90° + (∠BAC ÷ 2)

⇒ 135° = 90° + (∠BAC ÷ 2)

⇒ 135° − 90° = ∠BAC ÷ 2

⇒ 45° = ∠BAC ÷ 2

⇒ ∠BAC = 45° × 2

⇒ ∠BAC = 90°

∴ The correct answer is 90°.

Additional Information

Incentre Definition

The incentre is the point of intersection of the internal angle bisectors of a triangle and is equidistant from all sides.

Excentre Angle Formula

The angle formed by the external bisectors of ∠B and ∠C is ∠BEC = 90° − ∠A/2.

Circumcentre Angle Formula

The angle subtended by a side at the circumcentre is double the angle at the opposite vertex: ∠BPC = 2 × ∠A.

Orthocentre Angle Formula

The angle formed at the orthocentre H is supplementary to the vertex angle: ∠BHC = 180° − ∠A.

Shortcut Trick

Since DE || BC, ΔADE is similar to ΔABC.

Ratio of areas of similar triangles = (Ratio of corresponding sides)².

Area(ΔADE) : Area(ΔABC) = 6² : 7² = 36 : 49.

Area of trapezium BCED = Area(ΔABC) − Area(ΔADE) = 49 − 36 = 13.

Required Ratio = Area(ΔADE) : Area(BCED) = 36 : 13.

∴ The correct answer is 36 : 13.

Alternate Method

Given:

In ΔABC, DE || BC and DE : BC = 6 : 7.

Formula Used:

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Calculations:

In ΔADE and ΔABC:

∠ADE = ∠ABC (Corresponding angles as DE || BC)

∠AED = ∠ACB (Corresponding angles as DE || BC)

∴ ΔADE ≈ ΔABC (By AA similarity criterion)

⇒ Area(ΔADE) / Area(ΔABC) = (DE / BC)²

⇒ Area(ΔADE) / Area(ΔABC) = (6 / 7)² = 36 / 49

Let Area(ΔADE) = 36k and Area(ΔABC) = 49k

⇒ Area of trapezium BCED = Area(ΔABC) − Area(ΔADE)

⇒ Area of trapezium BCED = 49k − 36k = 13k

⇒ Ratio = Area(ΔADE) : Area of trapezium BCED

⇒ Ratio = 36k : 13k = 36 : 13

∴ The correct answer is 36 : 13.

Additional Information

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio: AD/DB = AE/EC.

Area Ratio of Similar Figures

For any two similar 2D figures, the ratio of their areas is equal to the square of the ratio of any two corresponding linear dimensions (sides, medians, heights, or perimeters).

Mid-point Theorem

If D and E are the midpoints of AB and AC, then DE || BC and DE = 1/2 × BC. In this case, Area(ΔADE) : Area(BCED) = 1 : 3.

Given:

$0.\overline {57} - 0.4\overline {32} + 0.3\bar 5$

Formula used:

$0.\overline{ab} = \frac{ab}{99}$

$0.a\overline{bc} = \frac{abc - a}{990}$

$0.a\overline{b} = \frac{ab - a}{90}$

Calculation:

$0.\overline{57} = \frac{57}{99} = \frac{570}{990}$

$0.4\overline{32} = \frac{432 - 4}{990} = \frac{428}{990}$

$0.3\overline{5} = \frac{35 - 3}{90} = \frac{32}{90} = \frac{352}{990}$

⇒ $\frac{570}{990} - \frac{428}{990} + \frac{352}{990}$

⇒ $\frac{570 - 428 + 352}{990}$

⇒ $\frac{142 + 352}{990}$

⇒ $\frac{494}{990}$

⇒ $\frac{498 - 4}{990} = 0.4\overline{98}$

∴ The correct answer is option (1).