SSC CGL Tier 2 Quant Previous Year Paper (Held On: 17 Feb 2018)

Shortcut Trick

Whole numbers start from 0. The first 111 whole numbers are 0, 1, 2, ..., 110.

Sum of first n natural numbers = [n × (n + 1)] ÷ 2.

The sum of (0 to 110) is the same as the sum of (1 to 110).

Sum = [110 × (110 + 1)] ÷ 2 = 55 × 111.

Unit digit = Unit digit of (5 × 1) = 5.

∴ The correct answer is 5.

Alternate Method

Given: First 111 whole numbers (0, 1, 2, 3, ..., 110).

Formula Used: Sum of an Arithmetic Progression (AP) = (n ÷ 2) × [first term + last term].

⇒ First term (a) = 0

⇒ Last term (l) = 110

⇒ Number of terms (n) = 111

⇒ Sum = (111 ÷ 2) × (0 + 110)

⇒ Sum = 111 × (110 ÷ 2)

⇒ Sum = 111 × 55

⇒ Sum = 6105

⇒ The unit digit is 5.

∴ The correct answer is 5.

Additional Information

Whole Numbers vs Natural Numbers

Whole numbers include 0 (0, 1, 2...), while natural numbers start from 1 (1, 2, 3...). The first n whole numbers always end at (n − 1).

Sum of First n Odd/Even Numbers

Sum of first n odd numbers = n². Sum of first n even numbers = n(n + 1).

Unit Digit Rules

To find the unit digit of a sum or product, only perform the operation on the unit digits of the individual numbers.

Shortcut Trick

The total number of n-digit positive numbers is always given by the formula: 9 × 10ⁿ⁻¹.

In this question, we need to find the count for 100-digit numbers, so n = 100.

Total numbers = 9 × 10¹⁰⁰⁻¹

Total numbers = 9 × 10⁹⁹

∴ The correct answer is 9 × 10⁹⁹.

Alternate Method

We use the fundamental principle of counting by considering the 100 available positions for a 100-digit number.

⇒ Total positions to fill = 100

⇒ For the 1st position: Any digit from 1 to 9 can be used (0 cannot be the first digit). So, there are 9 choices.

⇒ For the 2nd position: Any digit from 0 to 9 can be used. So, there are 10 choices.

⇒ Similarly, for all positions from the 2nd to the 100th, there are 10 choices each.

⇒ Total 100-digit numbers = 9 × 10 × 10 × ... × 10 (remaining 99 positions)

⇒ Total numbers = 9 × 10⁹⁹

∴ The correct answer is 9 × 10⁹⁹.

Additional Information

General Formula for n-digit numbers

The number of n-digit numbers is calculated as: (Largest n-digit number − Smallest n-digit number) + 1.

Range of 100-digit numbers

Smallest 100-digit number = 10⁹⁹ and Largest 100-digit number = 10¹⁰⁰ − 1.

Discrete Counting Property

Subtracting the smallest n-digit number from the largest gives (10ⁿ − 1) − 10ⁿ⁻¹ + 1 = 10ⁿ − 10ⁿ⁻¹ = 10ⁿ⁻¹(10 − 1) = 9 × 10ⁿ⁻¹.

Given:

$\frac{5.6 \times 0.36 + 0.42 \times 3.2}{0.8 \times 2.1}$ = ?

Formula used:

Arithmetic decimal simplification

Calculation:

$\frac{5.6 \times 0.36 + 0.42 \times 3.2}{0.8 \times 2.1}$ = ?

⇒ $\frac{2.016 + 1.344}{1.68}$

⇒ $\frac{3.36}{1.68}$

⇒ ? = 2

∴ The correct answer is option (1).

Shortcut Trick

The expression follows the identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Let a = 1.2, b = 0.8, and c = 0.7. Then 3abc = 3 × 1.2 × 0.8 × 0.7 = 2.016.

The bracket in the denominator is (a² + b² + c² − ab − bc − ca).

The expression simplifies to: (a + b + c) ÷ 1.35 ⇒ (1.2 + 0.8 + 0.7) ÷ 1.35 = 2.7 ÷ 1.35 = 2.

∴ The correct answer is 2.

Alternate Method

Given:

Numerator = (1.2)³ + (0.8)³ + (0.7)³ − 2.016

Denominator = (1.35) × [(1.2)² + (0.8)² + (0.7)² − 0.96 − 0.84 − 0.56]

Formula Used:

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

Calculations:

⇒ Let a = 1.2, b = 0.8, c = 0.7.

⇒ 3abc = 3 × 1.2 × 0.8 × 0.7 = 2.016.

⇒ ab = 1.2 × 0.8 = 0.96, bc = 0.8 × 0.7 = 0.56, ca = 0.7 × 1.2 = 0.84.

⇒ The Numerator is in the form a³ + b³ + c³ − 3abc.

⇒ The Denominator is in the form 1.35 × (a² + b² + c² − ab − bc − ca).

⇒ Expression = [(a + b + c)(a² + b² + c² − ab − bc − ca)] ÷ [1.35 × (a² + b² + c² − ab − bc − ca)].

⇒ Expression = (a + b + c) ÷ 1.35 = (1.2 + 0.8 + 0.7) ÷ 1.35.

⇒ Expression = 2.7 ÷ 1.35 = 2.

∴ The correct answer is 2.

Additional Information

Algebraic Identity

The identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) is fundamental for simplifying complex power expressions.

Special Case: Sum is Zero

If a + b + c = 0, then the entire expression a³ + b³ + c³ − 3abc equals 0, which implies a³ + b³ + c³ = 3abc.

Alternative Factorization

Another useful form is a³ + b³ + c³ − 3abc = 0.5(a + b + c)[(a − b)² + (b − c)² + (c − a)²], used when differences between terms are known.

Shortcut Trick

Unit digit of 7⁴¹³ → 7^{(413 mod 4)} = 7¹ = 7.

Unit digit of 9⁵⁴⁷ → 9^odd = 9.

Unit digit of 4⁶²⁴ → 4^even = 6.

Unit digit of 2⁸¹² → 2^{(rem 0)} = 2⁴ = 6.

Required unit digit = 7 × 9 × 6 × 6 → 3 × 6 × 6 → 18 × 6 → 8.

∴ The correct answer is 8.

Alternate Method

Given expression: (217)⁴¹³ × (819)⁵⁴⁷ × (414)⁶²⁴ × (342)⁸¹²

Formula used: To find the unit digit of aⁿ, we use the unit digit of the base and the remainder of (n ÷ 4).

⇒ Unit digit of 217⁴¹³ = 7⁴¹³ → 413 ÷ 4 leaves remainder 1 ∴ 7¹ = 7 ⇒ Unit digit of 819⁵⁴⁷ = 9⁵⁴⁷ → Since 547 is an odd number ∴ 9^odd = 9 ⇒ Unit digit of 414⁶²⁴ = 4⁶²⁴ → Since 624 is an even number ∴ 4^even = 6 ⇒ Unit digit of 342⁸¹² = 2⁸¹² → 812 ÷ 4 leaves remainder 0 ∴ 2⁴ = 16 → 6 ⇒ Combined product = 7 × 9 × 6 × 6 ⇒ 63 × 36 ⇒ 3 × 6 = 18 ⇒ The unit digit of the final product is 8.

∴ The correct answer is 8.

Additional Information

Cyclicity of Digits

Digits 2, 3, 7, and 8 have a cyclicity of 4. Divide the power by 4 and use the remainder; if the remainder is 0, use 4 as the power.

Unit Digits of 4 and 9

For base 4: 4^odd = 4, 4^even = 6. For base 9: 9^odd = 9, 9^even = 1.

Invariant Digits

Digits 0, 1, 5, and 6 always result in the same unit digit (0, 1, 5, 6) for any natural number power.

Given:

S = $\frac{1}{1 \times 3 \times 5} + \frac{1}{1 \times 4} + \frac{1}{3 \times 5 \times 7} + \frac{1}{4 \times 7} + \dots$ upto 20 terms

Formula used:

$\frac{1}{a \times b \times c} = \frac{1}{2d}(\frac{1}{a \times b} - \frac{1}{b \times c})$ (where d is common difference)

$\frac{1}{a \times b} = \frac{1}{d}(\frac{1}{a} - \frac{1}{b})$

Calculation:

Series consists of two sub-series of 10 terms each:

$S_1 = \frac{1}{1 \times 3 \times 5} + \frac{1}{3 \times 5 \times 7} + \dots + \text{10 terms}$

$S_2 = \frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \dots + \text{10 terms}$

For $S_1$, the 10^th term starts with $1 + (10 - 1) \times 2 = 19$:

⇒ $S_1 = \frac{1}{4} [\frac{1}{1 \times 3} - \frac{1}{21 \times 23}]$

⇒ $S_1 = \frac{1}{4} [\frac{1}{3} - \frac{1}{483}] = \frac{1}{4} [\frac{160}{483}] = \frac{40}{483}$

For $S_2$, the 10^th term starts with $1 + (10 - 1) \times 3 = 28$:

⇒ $S_2 = \frac{1}{3} [\frac{1}{1} - \frac{1}{31}]$

⇒ $S_2 = \frac{1}{3} [\frac{30}{31}] = \frac{10}{31}$

Total Sum (S):

⇒ $S = \frac{40}{483} + \frac{10}{31}$

⇒ $S = \frac{40 \times 31 + 483 \times 10}{483 \times 31}$

⇒ $S = \frac{1240 + 4830}{14973} = \frac{6070}{14973}$

∴ The correct answer is option (2).

Shortcut Trick

To compare fractions with the same numerator (1), compare the values of their denominators.

Denominators are √5, ³√12, and ⁴√29. Find the LCM of indices (2, 3, 4) = 12.

Raise denominators to the power of 12: (√5)¹² = 5⁶ = 15625; (³√12)¹² = 12⁴ = 20736; (⁴√29)¹² = 29³ = 24389.

Since 15625 < 20736 < 24389, the denominator order is √5 < ³√12 < ⁴√29.

For fractions, smaller denominators yield larger values: 1/√5 > 1/³√12 > 1/⁴√29.

∴ The correct answer is 1/√5 > 1/³√12 > 1/⁴√29.

Alternate Method

Given:

Four statements comparing 1/√5, 1/³√12, and 1/⁴√29.

Method:

Convert all surds in the denominators to the same index by using the LCM of their current indices.

Calculations:

⇒ Denominator 1 = 5^1/2

⇒ Denominator 2 = 12^1/3

⇒ Denominator 3 = 29^1/4

⇒ LCM of indices (2, 3, 4) = 12

⇒ 5^1/2 = 5^6/12 = (5⁶)^1/12 = (15625)^1/12

⇒ 12^1/3 = 12^4/12 = (12⁴)^1/12 = (20736)^1/12

⇒ 29^1/4 = 29^3/12 = (29³)^1/12 = (24389)^1/12

⇒ Comparing values: 15625 < 20736 < 24389

⇒ Therefore, √5 < ³√12 < ⁴√29

⇒ Taking reciprocals: 1/√5 > 1/³√12 > 1/⁴√29

∴ The correct answer is 1/√5 > 1/³√12 > 1/⁴√29.

Additional Information

Comparing Surds

To compare different surds like ⁿ√a and ^m√b, convert them to a common index by finding the LCM of n and m.

Reciprocal Property in Inequalities

For any two positive numbers a and b, if a > b, then 1/a < 1/b. The inequality sign reverses when taking the reciprocal.

Power Law

Use the property (a^m)ⁿ = a^m×n to simplify expressions when raising terms to a common power for comparison.

Shortcut Trick Common difference (d) = Divisor − Remainder d = (3 − 1) = (4 − 2) = (6 − 4) = 2 General form: N = [LCM(3, 4, 6) × k] − 2 LCM(3, 4, 6) = 12 ⇒ N = 12k − 2 Largest 2-digit N (where k=8) = 12 × 8 − 2 = 94 Remainder of 94 ÷ 5 = 4 ∴ The correct answer is 4. Alternate Method Given: Divisors are 3, 4, and 6. Respective remainders are 1, 2, and 4. Formula: N = LCM(Divisors) × k − (Common Difference)

⇒ Difference (d) = 3 − 1 = 4 − 2 = 6 − 4 = 2 ⇒ LCM of 3, 4, and 6 = 12 ⇒ N = 12k − 2 ⇒ For largest 2-digit number: 12k − 2 < 100 ⇒ 12k < 102 ⇒ k = 8 ⇒ N = 12 × 8 − 2 = 96 − 2 = 94 ⇒ Dividing 94 by 5: 94 = (18 × 5) + 4 ⇒ Remainder = 4 ∴ The correct answer is 4. Additional Information Constant Difference Rule When (Divisor − Remainder) is constant (say d), the number is of form [LCM(divisors) × k] − d. Divisibility by 5 The remainder when a number is divided by 5 is the same as the remainder when its last digit is divided by 5. General Form of Numbers If remainders are different and no constant difference exists, use the Chinese Remainder Theorem or manual iteration starting from the largest divisor.

Shortcut Trick

To compare surds ³√11, √7, and ⁴√45, find the LCM of their root indices (3, 2, 4), which is 12.

Raise each value to the power of 12 to eliminate the roots:

(11^1/3)¹² = 11⁴ = 14641

(7^1/2)¹² = 7⁶ = 117649

(45^1/4)¹² = 45³ = 91125

Order of values: 117649 (√7) > 91125 (⁴√45) > 14641 (³√11).

∴ The correct answer is Only III.

Alternate Method

Given:

Surds: ³√11, √7, ⁴√45

Formula Used:

For any positive numbers a and b, if aⁿ > bⁿ then a > b. To compare ⁿ√x and ^m√y, raise both to the power of LCM(n, m).

Calculations:

⇒ Convert surds to exponential form: 11^1/3, 7^1/2, 45^1/4

⇒ Indices are 1/3, 1/2, 1/4. LCM of denominators (3, 2, 4) = 12

⇒ Raise each term to the power of 12:

⇒ (11^1/3)¹² = 11⁴ = 11 × 11 × 11 × 11 = 121 × 121 = 14641

⇒ (7^1/2)¹² = 7⁶ = 7 × 7 × 7 × 7 × 7 × 7 = 343 × 343 = 117649

⇒ (45^1/4)¹² = 45³ = 45 × 45 × 45 = 2025 × 45 = 91125

⇒ Comparing the calculated values: 117649 > 91125 > 14641

⇒ Therefore: √7 > ⁴√45 > ³√11

∴ The correct answer is Only III.

Additional Information

LCM of Surd Indices

To compare surds with different indices, find the LCM of the indices and convert each surd to that common index.

Power Rule for Comparison

If x > y > 0 and n > 0, then xⁿ > yⁿ. This property is fundamental in comparing roots by raising them to a common power.

Exponent Laws

The expression ⁿ√a is equivalent to a^1/n. According to the laws of indices, (a^m)^p = a^m×p.

Shortcut Trick

Use Simon's Favorite Factoring Trick: A + B + AB + 1 = (A + 1)(B + 1).

Given A + B + AB = 65, adding 1 to both sides gives (A + 1)(B + 1) = 66.

Find factors of 66 such that both are ≤ 16 (since A, B ≤ 15): 6 × 11.

A + 1 = 6 → A = 5; B + 1 = 11 → B = 10.

Difference = |10 − 5| = 5.

∴ The correct answer is 5.

Alternate Method

Given: A + B + AB = 65 and A, B ≤ 15 (Positive integers).

Formula Used: (A + 1)(B + 1) = AB + A + B + 1

⇒ A + B + AB = 65

⇒ (A + B + AB) + 1 = 65 + 1

⇒ A(1 + B) + 1(1 + B) = 66

⇒ (A + 1)(B + 1) = 66

⇒ Possible integer pairs for (A+1, B+1) such that A, B ≤ 15:

⇒ Case 1: 1 × 66 (Invalid, 66 > 16)

⇒ Case 2: 2 × 33 (Invalid, 33 > 16)

⇒ Case 3: 3 × 22 (Invalid, 22 > 16)

⇒ Case 4: 6 × 11 (Valid, both 6 and 11 ≤ 16)

⇒ Let A + 1 = 6 ⇒ A = 5

⇒ Let B + 1 = 11 ⇒ B = 10

⇒ Difference |A − B| = |5 − 10| = 5

∴ The correct answer is 5.

Additional Information

Simon's Favorite Factoring Trick

It is a method to factor expressions of the form xy + x + y by adding 1 to both sides to get (x + 1)(y + 1).

Factorization of Integers

When solving equations with integer constraints, always list all factor pairs of the constant term to find values satisfying given limits.

Symmetry in Algebra

In expressions like A + B + AB, A and B are symmetric; their values can be swapped without changing the result of the sum.

Shortcut Trick

The series can be written as 2³ × (7³ + 8³ + 9³ + … + 15³).

Sum of cubes from 1 to n is [½ × n × (n + 1)]².

Required Sum = 8 × [(∑ cubes 1 to 15) − (∑ cubes 1 to 6)].

Required Sum = 8 × [120² − 21²] = 8 × (14400 − 441).

Required Sum = 8 × 13959 = 111672.

∴ The correct answer is 111672.

Alternate Method

Given: S = 14³ + 16³ + 18³ + … + 30³

Formula used: Sum of cubes of first n natural numbers ∑n³ = [n(n + 1) / 2]²

⇒ S = (2 × 7)³ + (2 × 8)³ + (2 × 9)³ + … + (2 × 15)³

⇒ S = 2³ [7³ + 8³ + 9³ + … + 15³]

⇒ S = 8 [∑i³ (from i=1 to 15) − ∑i³ (from i=1 to 6)]

⇒ Sum (1 to 15) = [15 × (15 + 1) / 2]² = (15 × 8)² = 120² = 14400

⇒ Sum (1 to 6) = [6 × (6 + 1) / 2]² = (3 × 7)² = 21² = 441

⇒ S = 8 × [14400 − 441] = 8 × 13959

⇒ S = 111672

∴ The correct answer is 111672.

Additional Information

Sum of First n Natural Numbers

The sum 1 + 2 + 3 + … + n is given by the formula n(n + 1) ÷ 2.

Sum of Squares of First n Natural Numbers

The sum 1² + 2² + 3² + … + n² is given by the formula [n(n + 1)(2n + 1)] ÷ 6.

Sum of Cubes of First n Even Numbers

The sum 2³ + 4³ + 6³ + … + (2n)³ can be calculated directly using the formula 2n²(n + 1)².

Given:

$\frac{(87+19)^2-(87-19)^2}{87 \times 19 \times 4} + 67$ = ?

Formula used:

(a + b)² - (a - b)² = 4ab

Calculation:

⇒ ? = $\frac{4 \times 87 \times 19}{87 \times 19 \times 4} + 67$

⇒ ? = 1 + 67

⇒ ? = 68

∴ The correct answer is option (2).

Shortcut Trick

Use the value substitution method: Let x + y + z = 0.

Assume x = 1, z = 1, and y = −2, satisfying 1 − 2 + 1 = 0.

Substitute these values into the given expression:

Numerator = 3(−2)² + 1² + 1² = 12 + 1 + 1 = 14.

Denominator = 2(−2)² − (1 × 1) = 8 − 1 = 7.

Value = 14 ÷ 7 = 2.

∴ The correct answer is 2.

Alternate Method

Given: x + y + z = 0

Formula Used: (x + z)² = x² + z² + 2xz

⇒ x + y + z = 0

⇒ x + z = −y

⇒ (x + z)² = (−y)²

⇒ x² + z² + 2xz = y²

⇒ x² + z² = y² − 2xz

⇒ Substitute (x² + z²) in the expression: (3y² + x² + z²) ÷ (2y² − xz)

⇒ Expression = [3y² + (y² − 2xz)] ÷ (2y² − xz)

⇒ Expression = (4y² − 2xz) ÷ (2y² − xz)

⇒ Expression = 2(2y² − xz) ÷ (2y² − xz) = 2.

∴ The correct answer is 2.

Additional Information

Algebraic Square Identity

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx). If x + y + z = 0, then x² + y² + z² = −2(xy + yz + zx).

Conditional Cube Identity

If x + y + z = 0, then x³ + y³ + z³ = 3xyz. This is a frequently asked identity in competitive exams.

Value Substitution Strategy

For homogeneous algebraic expressions, choosing simple integers that satisfy the initial condition (like 1, 1, −2) is the fastest way to find a numerical ratio.

The correct answer is 194

Key Points

• Given the value P = 7 + 4√3 and the condition PQ = 1, we can determine that Q = 1/P.
• The expression to evaluate is 1/P² + 1/Q². Substituting 1/P = Q and 1/Q = P, the expression simplifies to Q² + P².
• To find the value of Q, we rationalize the reciprocal of P:
  

Q = 1 / (7 + 4√3).
Multiplying the numerator and denominator by the conjugate (7 − 4√3):
Q = (7 − 4√3) / ((7 + 4√3)(7 − 4√3))
Q = (7 − 4√3) / (7² − (4√3)²)
Q = (7 − 4√3) / (49 − 48) = 7 − 4√3.

• Now, calculate the sum of P and Q:
  

P + Q = (7 + 4√3) + (7 − 4√3) = 14.

• Using the algebraic identity P² + Q² = (P + Q)² − 2PQ:
  

Substitute P + Q = 14 and PQ = 1:
P² + Q² = 14² − 2(1)
P² + Q² = 196 − 2 = 194.

• Therefore, the value of 1/P² + 1/Q² is 194.

Additional Information

• Rationalization is a fundamental technique in algebra used to remove radicals (like square roots) from the denominator of a fraction.
• The Conjugate of an expression a + √b is a − √b. Multiplying these together results in a rational number: a² − b.
• Reciprocal Relationship: If two numbers P and Q satisfy PQ = 1, they are multiplicative inverses. If P = x + √y and x² − y = 1, then its reciprocal is simply x − √y.
• Standard Identity: (a + b)² = a² + 2ab + b², which leads to a² + b² = (a + b)² − 2ab.

The correct answer is 28

Key Points

• Start with the given cubic equation: a³ + 3a² + 9a = 1.
• The objective is to find the value of a³ + (3/a).
• To find the term 3/a, we first determine the value of 1/a by dividing the entire given equation by a (assuming a ≠ 0):
  

(a³ + 3a² + 9a) ÷ a = 1 ÷ a
a² + 3a + 9 = 1/a

• Next, multiply this new equation by 3 to isolate the term 3/a:
  

3(a² + 3a + 9) = 3(1/a)
3a² + 9a + 27 = 3/a

• Now, substitute this expression for 3/a back into the required expression a³ + (3/a):
  

a³ + (3a² + 9a + 27)

• Rearrange the terms to group them:
  

(a³ + 3a² + 9a) + 27

• From the original given equation, we know that a³ + 3a² + 9a = 1. Substitute this value:
  

1 + 27 = 28

• Therefore, the correct value is 28.

Additional Information

• This problem utilizes the Substitution Method and Expression Manipulation. It is a common technique in Algebra where instead of solving for the variable a, which would require solving a complex Cubic Equation, you manipulate the equation to match the desired expression.
• Note that if a = 0, the original equation would result in 0 = 1, which is impossible. Hence, a is guaranteed to be non-zero, making the division by a a valid mathematical step.
• Problems of this nature frequently appear in competitive exams like SSC CGL, CHSL, and MTS, testing the candidate's ability to recognize patterns in Polynomials.

Shortcut Trick

Use the combined identity: x³ + y³ + z³ − 3xyz = (x + y + z)[(x + y + z)² − 3(xy + yz + zx)]

Substitute the given values: 13 − 3(1) = 1 × [1² − 3(xy + yz + zx)]

10 = 1 − 3(xy + yz + zx)

3(xy + yz + zx) = 1 − 10 = −9

xy + yz + zx = −3

∴ The correct answer is -3.

Alternate Method

Given:

x³ + y³ + z³ = 13

x + y + z = 1

xyz = 1

Formula Used:

1. (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

2. x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − (xy + yz + zx))

Calculations:

⇒ Let (xy + yz + zx) = k

⇒ From formula (1): x² + y² + z² = (x + y + z)² − 2k

⇒ x² + y² + z² = 1² − 2k = 1 − 2k

⇒ Now, substitute all values in formula (2):

⇒ 13 − 3(1) = 1 × [(1 − 2k) − k]

⇒ 10 = 1 − 3k

⇒ 3k = 1 − 10

⇒ 3k = −9

⇒ k = −3

⇒ xy + yz + zx = −3

∴ The correct answer is -3.

Additional Information

Special Case of Cubic Identity

If x + y + z = 0, then the identity simplifies to x³ + y³ + z³ = 3xyz.

Alternate Cubic Form

x³ + y³ + z³ − 3xyz = (1/2)(x + y + z)[(x − y)² + (y − z)² + (z − x)²].

Square of a Trinomial

The sum of squares can be expressed as x² + y² + z² = (x + y + z)² − 2(xy + yz + zx).

Shortcut Trick

Sum of ratio terms (a + b) + c = 6 + 5 = 11 units.

Sum of ratio terms (b + c) + a = 9 + 2 = 11 units.

Since the total sum (a + b + c) is same (11) in both ratios, we directly take: a = 2, c = 5.

Then, b = Total − (a + c) = 11 − (2 + 5) = 4 units.

Required ratio (a + c)/b = (2 + 5) / 4 = 7/4.

∴ The correct answer is 7/4.

Alternate Method

Given: (a + b)/c = 6/5 and (b + c)/a = 9/2.

Formula Used: Ratio equalization by equating the sum of all components (a + b + c).

⇒ In first ratio, (a + b + c) = 6 + 5 = 11

⇒ In second ratio, (a + b + c) = 9 + 2 = 11

⇒ Since sum is identical, let a + b + c = 11k

⇒ c = 5k and a = 2k

⇒ b = 11k − (a + c) = 11k − (2k + 5k) = 4k

⇒ (a + c)/b = (2k + 5k) / 4k

⇒ (a + c)/b = 7k / 4k = 7/4

∴ The correct answer is 7/4.

Additional Information

Ratio Balancing

When multiple ratios involving the same variables are given, the first step is to make the common element or the total sum of elements equal across all ratios.

Componendo and Dividendo

If x/y = m/n, then (x+y)/(x−y) = (m+n)/(m−n). This is useful for solving complex algebraic ratios.

Property of Sum of Antecedents

If a/b = c/d = e/f, then each ratio is equal to (a + c + e) / (b + d + f).

Shortcut Trick

Identify the constant relation: For $P = y + z − x$, $Q = z + x − y$, and $R = x + y − z$, the identity is $P^{3} + Q^{3} + R^{3} − 3PQR = 4(x^{3} + y^{3} + z^{3} − 3xyz)$.

From the given equation: $x^{3} + y^{3} + z^{3} = 3(1 + xyz) ⇒ x^{3} + y^{3} + z^{3} − 3xyz = 3$.

Substitute the given value: $P^{3} + Q^{3} + R^{3} − 3PQR = 4 × 3 = 12$.

∴ The correct answer is 12.

Alternate Method

Given:

$x^{3} + y^{3} + z^{3} − 3xyz = 3$

$P = y + z − x$, $Q = z + x − y$, $R = x + y − z$

Formula Used:

$A^{3} + B^{3} + C^{3} − 3ABC = \frac{1}{2}(A + B + C)[(A − B)^{2} + (B − C)^{2} + (C − A)^{2}]$

Calculations:

⇒ $P + Q + R = (y + z − x) + (z + x − y) + (x + y − z) = x + y + z$

⇒ $P − Q = (y + z − x) − (z + x − y) = 2y − 2x = −2(x − y)$

⇒ $Q − R = (z + x − y) − (x + y − z) = 2z − 2y = −2(y − z)$

⇒ $R − P = (x + y − z) − (y + z − x) = 2x − 2z = −2(z − x)$

⇒ $(P − Q)^{2} = 4(x − y)^{2}$, $(Q − R)^{2} = 4(y − z)^{2}$, $(R − P)^{2} = 4(z − x)^{2}$

⇒ $P^{3} + Q^{3} + R^{3} − 3PQR = \frac{1}{2}(x + y + z) \times [4(x − y)^{2} + 4(y − z)^{2} + 4(z − x)^{2}]$

⇒ $P^{3} + Q^{3} + R^{3} − 3PQR = 4 \times [\frac{1}{2}(x + y + z)((x − y)^{2} + (y − z)^{2} + (z − x)^{2})]$

⇒ $P^{3} + Q^{3} + R^{3} − 3PQR = 4 \times (x^{3} + y^{3} + z^{3} − 3xyz)$

⇒ $P^{3} + Q^{3} + R^{3} − 3PQR = 4 \times 3 = 12$

∴ The correct answer is 12.

Additional Information

Algebraic Identity for Three Variables

$x^{3} + y^{3} + z^{3} − 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} − xy − yz − zx)$

The Zero Sum Condition

If $x + y + z = 0$, then $x^{3} + y^{3} + z^{3} = 3xyz$.

Relationship between Sum of Squares and Differences

$x^{2} + y^{2} + z^{2} − xy − yz − zx = \frac{1}{2}[(x − y)^{2} + (y − z)^{2} + (z − x)^{2}]$

Shortcut Trick

By symmetry, assume x₁ = x₂ = x₃ = x.

Given equation: x³ = 4(4 + 3x) ⇒ x³ − 12x − 16 = 0.

By hit and trial, x = 4 satisfies: 4³ − 12(4) − 16 = 64 − 48 − 16 = 0.

Required value = 1/(2 + 4) + 1/(2 + 4) + 1/(2 + 4) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2.

∴ The correct answer is 1/2.

Alternate Method

Given: x₁ × x₂ × x₃ = 4(4 + x₁ + x₂ + x₃)

Formula Used: Substitution and algebraic expansion.

⇒ Let a = 2 + x₁, b = 2 + x₂, c = 2 + x₃

⇒ x₁ = a − 2, x₂ = b − 2, x₃ = c − 2

⇒ (a − 2)(b − 2)(c − 2) = 4(4 + (a − 2) + (b − 2) + (c − 2))

⇒ abc − 2(ab + bc + ca) + 4(a + b + c) − 8 = 4(a + b + c − 2)

⇒ abc − 2(ab + bc + ca) + 4(a + b + c) − 8 = 4(a + b + c) − 8

⇒ abc − 2(ab + bc + ca) = 0

⇒ abc = 2(ab + bc + ca)

⇒ Divide by 2abc: 1/2 = (ab + bc + ca) / abc

⇒ 1/2 = 1/c + 1/a + 1/b

∴ The correct answer is 1/2.

Additional Information

Symmetry Principle

In symmetric equations where variables can be interchanged without changing the equation, assuming equality often leads to a quick solution.

Cubic Expansion

(x−a)(x−b)(x−c) = x³ − (a+b+c)x² + (ab+bc+ca)x − abc. This is useful for identifying coefficients in polynomial relations.

Root Testing

For integer coefficients, test factors of the constant term (e.g., ±1, ±2, ±4 for 16) to find roots of a polynomial.

Shortcut Trick The equation x2 − x + 1 = 0 is a factor of x3 + 1 = (x + 1)(x2 − x + 1) = 0. The roots of x2 − x + 1 = 0 are α = −ω and β = −ω2, where ω is the cube root of unity. New roots: α3 = (−ω)3 = −1 and β3 = (−ω2)3 = −1 (since ω3 = 1). The equation with roots −1 and −1 is (x + 1)(x + 1) = 0 ⇒ x2 + 2x + 1 = 0. ∴ The correct answer is x2 + 2x + 1 = 0. Alternate Method Given equation: x2 − x + 1 = 0 Sum of roots (α + β) = −(−1)/1 = 1 Product of roots (αβ) = 1/1 = 1 For the new equation with roots α3 and β3: ⇒ Sum of new roots (S) = α3 + β3 ⇒ S = (α + β)3 − 3αβ(α + β) ⇒ S = (1)3 − 3 × 1 × (1) ⇒ S = 1 − 3 = −2 ⇒ Product of new roots (P) = α3 × β3 = (αβ)3 ⇒ P = (1)3 = 1 ⇒ Required Equation: x2 − (Sum of roots)x + (Product of roots) = 0 ⇒ x2 − (−2)x + 1 = 0 ⇒ x2 + 2x + 1 = 0 ∴ The correct answer is x2 + 2x + 1 = 0. Additional Information Sum and Product of Roots For a quadratic equation ax2 + bx + c = 0, sum of roots = −b/a and product of roots = c/a. Algebraic Identity The sum of cubes is given by a3 + b3 = (a + b)3 − 3ab(a + b) or (a + b)(a2 − ab + b2). Nature of Roots The nature of roots depends on the discriminant D = b2 − 4ac. If D < 0, roots are imaginary and occur in conjugate pairs. Cube Roots of Unity The roots of x2 + x + 1 = 0 are ω and ω2. The roots of x2 − x + 1 = 0 are −ω and −ω2, where ω3 = 1 and 1 + ω + ω2 = 0.

Shortcut Trick

Observe the coefficients of x and z: in the second equation, they are exactly 3 times those in the first equation.

Multiply the first equation by 3: 3 × (3x + 5y + 7z) = 3 × 49 ⇒ 9x + 15y + 21z = 147.

Compare this with the second equation: 9x + 8y + 21z = 126.

Subtract the second equation from the result: (15y − 8y) = 147 − 126 ⇒ 7y = 21.

y = 21 ÷ 7 = 3.

∴ The correct answer is 3.

Alternate Method

Given:

Equation 1: 3x + 5y + 7z = 49

Equation 2: 9x + 8y + 21z = 126

Formula Used:

Elimination Method for linear equations.

Calculations:

⇒ 3x + 5y + 7z = 49 → (1)

⇒ 9x + 8y + 21z = 126 → (2)

⇒ Multiply (1) by 3 to match the coefficients of x and z:

⇒ 3 × (3x + 5y + 7z) = 3 × 49

⇒ 9x + 15y + 21z = 147 → (3)

⇒ Subtracting (2) from (3):

⇒ (9x + 15y + 21z) − (9x + 8y + 21z) = 147 − 126

⇒ (9x − 9x) + (15y − 8y) + (21z − 21z) = 21

⇒ 0 + 7y + 0 = 21

⇒ 7y = 21

⇒ y = 21 ÷ 7

⇒ y = 3

∴ The correct answer is 3.

Additional Information

System of Equations with Multiple Variables

To find the values of n variables uniquely, you typically need n independent equations. However, a specific variable can be found if the other variables can be eliminated simultaneously through linear combinations.

Coefficient Matching Shortcut

If the coefficients of x and z are proportional (e.g., a₁/a₂ = c₁/c₂ = k), then multiplying one equation by k allows the elimination of both x and z at once, leaving an equation in terms of y only.

Elimination Method

A process used to solve systems of linear equations by adding or subtracting them to cancel out variables, reducing the system to a single-variable equation.

Shortcut Trick

Let Equation 1 be 4P + 6N + 9F = 305 and Equation 2 be 3P + 4N + 2F = 145.

Observe the target coefficients (5, 8, 16):

2 × (Eq 1) − 1 × (Eq 2) ⇒ 2(4P + 6N + 9F) − (3P + 4N + 2F)

⇒ (8P + 12N + 18F) − (3P + 4N + 2F) = 5P + 8N + 16F

⇒ Total Cost = 2 × 305 − 145 = 610 − 145 = 465

∴ The correct answer is ₹465.

Alternate Method

Let the cost of 1 pen be P, 1 notebook be N, and 1 file be F.

Given:

4P + 6N + 9F = 305 ... (i)

3P + 4N + 2F = 145 ... (ii)

To find: 5P + 8N + 16F

Using the linear combination method:

⇒ Let 5P + 8N + 16F = x(4P + 6N + 9F) + y(3P + 4N + 2F)

⇒ Comparing coefficients of P: 4x + 3y = 5

⇒ Comparing coefficients of N: 6x + 4y = 8

⇒ Solving these two equations: Multiply first by 4 and second by 3

⇒ 16x + 12y = 20

⇒ 18x + 12y = 24

⇒ Subtracting: 2x = 4 ⇒ x = 2

⇒ Substituting x = 2 in 4x + 3y = 5: 8 + 3y = 5 ⇒ 3y = −3 ⇒ y = −1

⇒ Verify with F coefficient: 9(2) + 2(−1) = 18 − 2 = 16 (Correct)

⇒ Required cost = 2 × 305 + (−1) × 145

⇒ Required cost = 610 − 145

⇒ Required cost = 465

∴ The correct answer is ₹465.

Additional Information

Linear Combinations

In problems with more variables than equations, a specific combination can often be found by multiplying equations by constants and adding/subtracting them.

Unique Solutions vs. Combinations

To find individual values of x, y, and z, we need 3 independent equations. However, the cost of a specific group can be found if the target is a linear dependent combination of the given sets.

Consistency of Equations

A system of equations is consistent if there is at least one set of values that satisfies all equations. In cost problems, coefficients must be verified across all items to ensure the combination is valid.

Shortcut Trick

In a right-angled triangle, circumradius (R) = Hypotenuse ÷ 2.

In ▵ABC, ∠A = 90° and ∠C = 60°, so it is a 30°-60°-90° triangle.

The sides opposite to these angles are in the ratio 1 : √3 : 2.

Here, hypotenuse BC corresponds to 2 units, so R = 2 ÷ 2 = 1 unit.

Side AB is opposite to 60°, which corresponds to √3 units.

∴ The correct answer is 1 : √3.

Alternate Method

Given: ∠BAC = 90° and ∠ACB = 60°.

Formula Used: Circumradius (R) = Hypotenuse ÷ 2; sin θ = Opposite ÷ Hypotenuse.

⇒ In ▵ABC, ∠ABC = 180° − (∠BAC + ∠ACB)

⇒ ∠ABC = 180° − (90° + 60°) = 30°

⇒ Let the hypotenuse BC = x units.

⇒ Circumradius R = BC ÷ 2 = x ÷ 2 units.

⇒ Using trigonometric ratio in ▵ABC: sin(∠ACB) = AB ÷ BC

⇒ sin(60°) = AB ÷ x

⇒ AB = x × (√3 ÷ 2) units.

⇒ Required ratio = R : AB

⇒ Ratio = (x ÷ 2) : (x × √3 ÷ 2)

⇒ Ratio = 1 : √3

∴ The correct answer is 1 : √3.

Additional Information

Circumradius of Right Triangle

The circumcenter of any right-angled triangle lies exactly at the midpoint of the hypotenuse, meaning R = Hypotenuse ÷ 2.

Sine Rule

In any triangle, the ratio of a side to the sine of its opposite angle is equal to the diameter of its circumcircle: a/sin A = b/sin B = c/sin C = 2R.

30-60-90 Triangle Ratio

For a triangle with angles 30°, 60°, and 90°, the sides opposite to these angles are always in the ratio 1 : √3 : 2.

Shortcut Trick

In a square ABCD, the sum of distances from two vertices to a point P on the opposite side (BP + CP) is minimized when P is the midpoint of that side.

Side of the square = 4 cm.

Distance of midpoint P from vertex A (or D) = 4 ÷ 2 = 2 cm.

By Pythagoras theorem, BP = CP = √(4² + 2²) = √(16 + 4) = √20 = 2√5 cm.

Minimum value = BP + CP = 2√5 + 2√5 = 4√5 cm.

∴ The correct answer is 4√5 cm.

Alternate Method

Given:

Side of square ABCD = 4 cm.

Point P lies on side AD.

Formula Used:

Shortest distance (Reflection principle): The minimum distance BP + CP is the length of the straight line segment when one point is reflected across the line containing P.

Calculations:

⇒ Let us reflect point C across side AD to point C'.

⇒ The distance BP + CP is equivalent to BP + PC'.

⇒ The minimum value of BP + PC' is the straight line distance BC'.

⇒ In the right-angled triangle formed by base BC and vertical height CC':

⇒ BC = 4 cm

⇒ CC' = CD + DC' = 4 + 4 = 8 cm

⇒ BC' = √(BC² + CC'²)

⇒ BC' = √(4² + 8²)

⇒ BC' = √(16 + 64) = √80

⇒ BC' = √(16 × 5) = 4√5 cm.

∴ The correct answer is 4√5 cm.

Additional Information

Heron's Shortest Path Principle

The path from point A to point B touching a line L is minimized when the angle of incidence equals the angle of reflection, effectively making points A, P, and the reflection of B collinear.

Pythagorean Theorem

For any right-angled triangle with sides a and b and hypotenuse c, the relation is a² + b² = c².

Symmetry in Squares

In optimization problems involving symmetric figures like squares, the extremum (minimum or maximum) often occurs at a point of symmetry, such as the midpoint of a side.

Shortcut Trick

In similar triangles, the ratio of corresponding sides is equal to the ratio of corresponding medians and their parts.

Given AB : PQ = 2 : 3, then corresponding sides BC : QR = 2 : 3.

Since BD and QS are half of BC and QR respectively, BD : QS = BC : QR = 2 : 3.

Requested value = (BD/QS)² = (2/3)² = 4/9.

∴ The correct answer is 4/9.

Alternate Method

Given: ΔABC ~ ΔPQR and AB/PQ = 2/3. AD and PS are medians.

Formula Used: If two triangles are similar, the ratio of their corresponding sides is equal to the ratio of their corresponding medians.

⇒ Since ΔABC ~ ΔPQR, the ratio of corresponding sides is constant:

⇒ AB/PQ = BC/QR = AC/PR = 2/3.

⇒ Given AD and PS are medians, D is the midpoint of BC and S is the midpoint of QR.

⇒ BD = BC/2 and QS = QR/2.

⇒ BD/QS = (BC/2) / (QR/2) = BC/QR.

⇒ From the similarity property, BC/QR = AB/PQ = 2/3.

⇒ Therefore, BD/QS = 2/3.

⇒ (BD/QS)² = (2/3)² = 4/9.

∴ The correct answer is 4/9.

Additional Information

Ratio of Areas of Similar Triangles

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides: Area(ABC)/Area(PQR) = (AB/PQ)².

Ratio of Medians and Altitudes

For similar triangles, the ratio of corresponding medians, altitudes, and angle bisectors is equal to the ratio of their corresponding sides.

Ratio of Perimeters

The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.

Shortcut Trick

By similarity of triangles, radius ratio equals distance ratio: r₁ : r₂ = AB : AC = 3 : 5.

Given AC = 40 units ⇒ AB = (3 ÷ 5) × 40 = 24 units.

Distance between centers BC = AC − AB = 40 − 24 = 16 units.

Sum of radii r₁ + r₂ = BC = 16 ⇒ 3k + 5k = 16 ⇒ k = 2, so r₁ = 6 and r₂ = 10.

Length of direct common tangent DE = 2√(r₁ × r₂) = 2√(6 × 10) = 4√15.

∴ The correct answer is 4√15.

Alternate Method

Given:

Ratio of radii (BD : CE) = 3 : 5

Distance AC = 40 units

ADE is the common tangent.

Formula Used:

For similar triangles ΔABD and ΔACE: AB ÷ AC = BD ÷ CE

Length of direct common tangent DE = √(BC² − (r₂ − r₁)²)

Calculations:

⇒ In ΔABD and ΔACE, ∠A is common and ∠ADB = ∠AEC = 90°.

⇒ ΔABD ∼ ΔACE (by AA similarity)

⇒ AB ÷ AC = r₁ ÷ r₂ = 3 ÷ 5

⇒ AB ÷ 40 = 3 ÷ 5

⇒ AB = (3 × 40) ÷ 5 = 24 units

⇒ BC = AC − AB = 40 − 24 = 16 units

⇒ Since the circles are shown touching in the configuration, r₁ + r₂ = BC = 16.

⇒ 3k + 5k = 16 ⇒ 8k = 16 ⇒ k = 2

⇒ r₁ = 6 units, r₂ = 10 units

⇒ DE = √(16² − (10 − 6)²)

⇒ DE = √(256 − 16) = √240

⇒ DE = √(16 × 15) = 4√15 units

∴ The correct answer is 4√15.

Additional Information

Direct Common Tangent (DCT)

Length = √(d² − (r₁ − r₂)²), where d is the distance between centers and r₁, r₂ are radii.

Transverse Common Tangent (TCT)

Length = √(d² − (r₁ + r₂)²). This exists only if d ≥ r₁ + r₂.

Touching Circles Property

If two circles touch externally, d = r₁ + r₂, and the length of the direct common tangent is 2√(r₁ × r₂).

Similarity in Tangent Geometry

The point of intersection of common external tangents and the centers of the circles are always collinear and divide the center line in the ratio of the radii.

Shortcut Trick

In a circle with diameter AB = 30 cm, the radius R = 15 cm.

For a chord CD = 24 cm, the distance from the center is calculated using the Pythagorean triple (9, 12, 15).

Distance from center (d) = √(15² − 12²) = √(225 − 144) = 9 cm.

If MN represents the distance between two such parallel chords on opposite sides of the center, then MN = 2 × d.

MN = 2 × 9 = 18 cm.

∴ The correct answer is 18 cm.

Alternate Method

Given:

Diameter of the circle (AB) = 30 cm

Length of chord (CD) = 24 cm

Formula Used:

Distance of a chord from the center (d) = √(R² − (L/2)²)

Where R is the radius and L is the length of the chord.

Calculations:

⇒ Radius of the circle (R) = AB ÷ 2 = 30 ÷ 2 = 15 cm

⇒ Half of the chord length = CD ÷ 2 = 24 ÷ 2 = 12 cm

⇒ Distance of chord from center (OM) = √(15² − 12²)

⇒ OM = √(225 − 144) = √81 = 9 cm

⇒ If the figure represents the total distance between two such parallel chords (MN):

⇒ MN = OM + ON (where ON is the distance of the second chord from the center)

⇒ Since both chords are of equal length (24 cm), ON = OM = 9 cm

⇒ MN = 9 + 9 = 18 cm

∴ The correct answer is 18 cm.

Additional Information

Perpendicular from Center

The perpendicular from the center of a circle to a chord bisects the chord. If d is the distance, R is the radius, and L is the chord length, R² = d² + (L/2)².

Parallel Chords Distance

The distance between two parallel chords is the sum of their distances from the center if they are on opposite sides, and the difference if they are on the same side.

Angle in a Semicircle

The angle subtended by a diameter at any point on the circumference is always a right angle (90°).

Pythagorean Triples

Common triples like (3, 4, 5), (5, 12, 13), and (8, 15, 17) are frequently used in circle geometry to relate radius, chord length, and distance.

Shortcut Trick

Tangents from an external point are equal, so AB = AC.

In ΔABO, AB = Radius × cot(∠BAC/2) = 6 × cot(30°) = 6√3 cm.

Required value = √(AB² + AC²) = √(2 × AB²) = AB√2.

Required value = 6√3 × √2 = 6√6 cm.

∴ The correct answer is 6√6 cm.

Alternate Method

Given:

Radius of the circle (r) = 6 cm

Angle between tangents ∠BAC = 60°

Formula:

In a right-angled triangle, tan θ = Perpendicular / Base

Length of tangent = Radius × cot(θ/2)

Calculations:

⇒ In ΔABO, ∠ABO = 90° (Radius is perpendicular to tangent at point of contact)

⇒ Line AO bisects ∠BAC, so ∠BAO = 60° / 2 = 30°

⇒ In ΔABO: tan 30° = OB / AB

⇒ 1/√3 = 6 / AB

⇒ AB = 6√3 cm

⇒ Since AB = AC (Tangents from same point), AC = 6√3 cm

⇒ Required value = √(AB² + AC²)

⇒ Value = √((6√3)² + (6√3)²)

⇒ Value = √(108 + 108) = √216

⇒ Value = √(36 × 6) = 6√6 cm

∴ The correct answer is 6√6 cm.

Additional Information

Tangent Properties

Tangents from an external point are equal in length and subtend equal angles at the center of the circle.

Angle Relation

The angle between two tangents is supplementary to the angle subtended by the radii at the center (∠BAC + ∠BOC = 180°).

Quadrilateral ABOC

The quadrilateral formed by the tangents and the radii (ABOC) is a cyclic quadrilateral since opposite angles (90° each) sum to 180°.

Shortcut Trick

When two circles are tangent to each other and to the sides of an angle θ, the ratio of their radii is given by:

R/r = (1 + sin(θ/2)) / (1 − sin(θ/2))

Given θ = ∠ACB = 60°, then θ/2 = 30°.

Ratio R/r = (1 + sin 30°) / (1 − sin 30°) = (1 + 0.5) / (1 − 0.5) = 1.5 / 0.5 = 3.

Given smaller radius r = 2 cm, then larger radius R = 2 × 3 = 6 cm.

∴ The correct answer is 6 cm.

Alternate Method

Given:

In ΔABC, ∠B = 90°, ∠C = 60°.

Radius of smaller circle (r) = 2 cm.

Radius of larger circle (R) = ?

The centers O₁ and O₂ of the circles lie on the angle bisector of ∠C.

⇒ ∠O₁CM = ∠C / 2 = 60° / 2 = 30° (where M is tangency point on BC).

In right-angled ΔO₁MC:

⇒ sin 30° = O₁M / O₁C

⇒ 1/2 = 2 / O₁C

⇒ O₁C = 4 cm.

Now, distance of O₂ from C along the bisector is O₂C:

⇒ O₂C = O₁C + O₁O₂ = 4 + (r + R) = 4 + 2 + R = 6 + R.

In right-angled ΔO₂NC (where N is tangency point on BC):

⇒ sin 30° = O₂N / O₂C

⇒ 1/2 = R / (6 + R)

⇒ 6 + R = 2R

⇒ R = 6 cm.

∴ The correct answer is 6 cm.

Additional Information

Inradius of a Right Triangle

The radius of the incircle of a right-angled triangle with base (b), perpendicular (p), and hypotenuse (h) is r = (p + b − h) / 2.

Successive Circles in an Angle

If n circles are placed in a line tangent to each other and the sides of an angle θ, their radii r₁, r₂, ..., r_n form a geometric progression (GP) with common ratio k = (1 + sin(θ/2)) / (1 − sin(θ/2)).

Angle Bisector Property

The center of any circle inscribed in an angle must lie on the angle bisector, as the distance from the center to both arms (radii) is equal.

Shortcut Trick

In a configuration where two tangents from point P meet the circle at points A and B, and a third tangent intersects these at Q and R:

The angle subtended by the third tangent at the center O is exactly half of the angle between the radii at the points of tangency.

∠QOR = (180° − ∠QPR) ÷ 2

∠QOR = (180° − 45°) ÷ 2 = 135° ÷ 2

∠QOR = 67.5°

∴ The correct answer is 67.5°.

Alternate Method

Given: ∠QPR = 45°, O is the centre of the circle.

Formula: ∠QOR = 90° − (∠QPR ÷ 2)

Let the two tangents from P touch the circle at A and B. Let the third tangent touch the circle at S.

⇒ In quadrilateral PAOB, ∠AOB + ∠APB = 180°

⇒ ∠AOB = 180° − 45° = 135°

⇒ OQ is the angle bisector of ∠AOS and OR is the angle bisector of ∠BOS (Tangents from external points Q and R).

⇒ ∠QOS = ∠AOS ÷ 2 and ∠ROS = ∠BOS ÷ 2

⇒ ∠QOR = ∠QOS + ∠ROS

⇒ ∠QOR = (∠AOS + ∠BOS) ÷ 2

⇒ ∠QOR = ∠AOB ÷ 2

⇒ ∠QOR = 135° ÷ 2

⇒ ∠QOR = 67.5°

∴ The correct answer is 67.5°.

Additional Information

Angle between Tangents

The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

Tangent Property

The line joining the external point to the center of the circle bisects the angle between the two tangents. This is why ∠QOR is exactly half of the central angle ∠AOB in this configuration.

Perimeter of Triangle

In the given figure, the perimeter of triangle PQR is equal to the sum of the lengths of the two tangents from P to the circle (Perimeter = PA + PB = 2 × PA).

Shortcut Trick

Centers A and B are 8 cm apart (r + r = 4 + 4), and radius AR = 4 cm.

In the figure, RB is tangent to Circle A at R, making ΔARB a right triangle (∠ARB = 90°).

RB² = AB² − AR² = 8² − 4² = 64 − 16 = 48.

RQ is tangent to Circle B at Q, making ΔRQB a right triangle (∠RQB = 90°).

RQ² = RB² − BQ² = 48 − 4² = 48 − 16 = 32.

RQ = √32 = 4√2 cm.

∴ The correct answer is 4√2 cm.

Alternate Method

Given:

Radius of each circle (r) = 4 cm

Distance between centers AB = 4 + 4 = 8 cm

Calculations:

In the given geometry, R is on circle A such that line RB is tangent to circle A.

⇒ In right-angled ΔARB (∠R = 90°):

⇒ RB² = AB² − AR²

⇒ RB² = 8² − 4² = 64 − 16 = 48

Now, RQ is a tangent to the second circle (center B) from point R.

⇒ In right-angled ΔRQB (∠Q = 90°):

⇒ RQ² + BQ² = RB²

⇒ RQ² + 4² = 48

⇒ RQ² = 48 − 16 = 32

⇒ RQ = √32 = 4√2 cm

∴ The correct answer is 4√2 cm.

Additional Information

Tangent Property

The radius is always perpendicular to the tangent at the point of contact. Thus, ∠Radius-Tangent = 90°.

Length of Tangent

The length of a tangent from an external point P to a circle with center O and radius r is given by √(OP² − r²).

Touching Circles

For two circles touching externally, the distance between their centers is the sum of their radii (d = r₁ + r₂).

Shortcut Trick

Direct Common Tangent (DCT) = √[d² − (r₂ − r₁)²]

Transverse Common Tangent (TCT) = √[d² − (r₂ + r₁)²]

Ratio = √[10² − (4 − 3)²] : √[10² − (4 + 3)²]

Ratio = √[100 − 1] : √[100 − 49] = √99 : √51

Ratio = √(3 × 33) : √(3 × 17) = √33 : √17

∴ The correct answer is √33 : √17.

Alternate Method

Given:

Radius of first circle (r₁) = 3 cm

Radius of second circle (r₂) = 4 cm

Distance between centers (d) = 10 cm

Formula Used:

Length of DCT = √[d² − (r₂ − r₁)²]

Length of TCT = √[d² − (r₂ + r₁)²]

Calculations:

⇒ Length of DCT = √[10² − (4 − 3)²]

⇒ Length of DCT = √[100 − 1²] = √99 cm

⇒ Length of TCT = √[10² − (4 + 3)²]

⇒ Length of TCT = √[100 − 7²] = √[100 − 49] = √51 cm

⇒ Required Ratio = Length of DCT : Length of TCT

⇒ Required Ratio = √99 : √51

⇒ Required Ratio = √(3 × 33) : √(3 × 17)

⇒ Required Ratio = √33 : √17

∴ The correct answer is √33 : √17.

Additional Information

Common Tangents Properties

For two circles with distance 'd' between centers, the direct common tangent formula is √(d² − (r₂ − r₁)²) and the transverse common tangent formula is √(d² − (r₁ + r₂)²).

Touching Circles

If circles touch externally, d = r₁ + r₂, then TCT = 0 and DCT = 2√(r₁r₂). If they touch internally, d = |r₁ − r₂|, then DCT = 0.

Existence of Tangents

Two circles can have a maximum of 4 common tangents (2 direct, 2 transverse) when they do not intersect (d > r₁ + r₂).

Shortcut Trick

Check for a pythagorean triplet in ΔABD where hypotenuse AB = 5 cm.

Possible triplet is (3, 4, 5). If BD = 3 cm and AD = 4 cm, then CD = BC − BD = 8 − 3 = 5 cm.

Verify in ΔACD: AD² + CD² = 4² + 5² = 16 + 25 = 41 = (√41)² = AC². It satisfies the condition.

Area of ΔABD = 1/2 × base × height = 1/2 × 3 × 4 = 6 cm².

∴ The correct answer is 6 cm².

Alternate Method

Given: AB = 5 cm, AC = √41 cm, BC = 8 cm, AD ⊥ BC.

Formula: Area of triangle = 1/2 × base × height; Pythagoras theorem: h² = a² + b².

Let BD = x cm, then CD = (8 − x) cm.

In right-angled ΔABD:

⇒ AD² = AB² − BD²

⇒ AD² = 5² − x² = 25 − x² → (Eq 1)

In right-angled ΔACD:

⇒ AD² = AC² − CD²

⇒ AD² = (√41)² − (8 − x)²

⇒ AD² = 41 − (64 + x² − 16x) = 16x − x² − 23 → (Eq 2)

Equating (Eq 1) and (Eq 2):

⇒ 25 − x² = 16x − x² − 23

⇒ 25 = 16x − 23

⇒ 16x = 48 ⇒ x = 3 cm

Now, calculate height AD from (Eq 1):

⇒ AD² = 25 − 3² = 25 − 9 = 16

⇒ AD = 4 cm

Area of ΔABD = 1/2 × BD × AD

⇒ Area = 1/2 × 3 × 4 = 6 cm²

∴ The correct answer is 6 cm².

Additional Information

Pythagoras Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: Hypotenuse² = Perpendicular² + Base².

Heron's Formula

Area of Δ = √[s(s−a)(s−b)(s−c)], where s is the semi-perimeter (a+b+c)/2.

Area of Triangle with Altitude

The area of any triangle can be calculated if the base and the corresponding altitude (height) are known: Area = 1/2 × Base × Height.

Shortcut Trick

Area of ▵PQR with sides 10, 10, 8 (Isosceles) = 8√21 cm².

Area Ratio: Area(▵PAB) = (5 × 6)/(10 × 10) × Area(PQR) = 3/10 Area.

Area Ratio: Area(▵QDC) = (2 × 5)/(10 × 8) × Area(PQR) = 1/8 Area.

Area Ratio: Area(▵RBC) = (4 × 3)/(10 × 8) × Area(PQR) = 3/20 Area.

Area(ABCD) = Area(PQR) × [1 − (3/10 + 1/8 + 3/20)] = 8√21 × [1 − 23/40] = 8√21 × 17/40.

∴ The correct answer is (17√21)/5 cm².

Alternate Method

Given: PA = 5 cm, AD = 3 cm, DQ = 2 cm, QC = 5 cm, CR = 3 cm, BR = 4 cm, PB = 6 cm.

Formula: Area of ▵ = √[s(s−a)(s−b)(s−c)]; Area(▵XYZ)/Area(▵ABC) = (XY×XZ)/(AB×AC) if ∠X = ∠A.

⇒ PQ = PA + AD + DQ = 5 + 3 + 2 = 10 cm

⇒ QR = QC + CR = 5 + 3 = 8 cm

⇒ PR = PB + BR = 6 + 4 = 10 cm

⇒ Semi-perimeter s of ▵PQR = (10 + 8 + 10) ÷ 2 = 14 cm

⇒ Area(▵PQR) = √[14 × (14−10) × (14−8) × (14−10)] = √[14 × 4 × 6 × 4] = √1344 = 8√21 cm²

⇒ Area(▵PAB) = (PA × PB)/(PQ × PR) × Area(▵PQR) = (5 × 6)/(10 × 10) × 8√21 = 0.3 × 8√21 = 2.4√21 cm²

⇒ Area(▵QDC) = (QD × QC)/(QP × QR) × Area(▵PQR) = (2 × 5)/(10 × 8) × 8√21 = 0.125 × 8√21 = √21 cm²

⇒ Area(▵RBC) = (RB × RC)/(RP × RQ) × Area(▵PQR) = (4 × 3)/(10 × 8) × 8√21 = 0.15 × 8√21 = 1.2√21 cm²

⇒ Area(ABCD) = Area(▵PQR) − [Area(▵PAB) + Area(▵QDC) + Area(▵RBC)]

⇒ Area(ABCD) = 8√21 − [2.4√21 + √21 + 1.2√21] = 8√21 − 4.6√21 = 3.4√21 cm²

⇒ Area(ABCD) = (34/10)√21 = (17√21)/5 cm²

∴ The correct answer is (17√21)/5 cm².

Additional Information

Area-Ratio Side Property

If two triangles share a common angle θ, the ratio of their areas is equal to the ratio of the products of the lengths of the sides forming that angle: Area₁/Area₂ = (a₁×b₁)/(a₂×b₂).

Heron's Formula

For a triangle with sides a, b, and c, the Area = √[s(s−a)(s−b)(s−c)], where s is the semi-perimeter (a+b+c)/2.

Sine Rule for Area

The area of any triangle can be calculated as Area = 1/2 × a × b × sin(θ), where θ is the included angle between sides a and b.

Shortcut Trick

Area of square ABCD = 14 × 14 = 196 cm².

Radius (r) of semicircle EPF = Diameter ÷ 2 = 14 ÷ 2 = 7 cm.

Area of semicircle = 1/2 × 22/7 × 7 × 7 = 77 cm².

In this geometry, Area of square LMNO = r² ÷ 2 = 49 ÷ 2 = 24.5 cm².

Shaded Area = Area(ABCD) − Area(Semicircle) − Area(LMNO) = 196 − 77 − 24.5 = 94.5 cm².

∴ The correct answer is 94.5 cm².

Alternate Method

Given:

Side of square ABCD = 14 cm

Diameter of semicircle EPF (EF) = 14 cm

LMNO is a square inscribed in the remaining region.

Calculations:

⇒ Area of square ABCD = (Side)² = 14 × 14 = 196 cm²

⇒ Radius of semicircle EPF (r) = 14 ÷ 2 = 7 cm

⇒ Area of semicircle EPF = πr² ÷ 2 = (22 ÷ 7 × 7 × 7) ÷ 2 = 77 cm²

⇒ Area of square LMNO (where side s = r/√2) = (7 ÷ √2)² = 49 ÷ 2 = 24.5 cm²

⇒ Area of Shaded Region = Area(ABCD) − Area(Semicircle) − Area(LMNO)

⇒ Shaded Area = 196 − 77 − 24.5

⇒ Shaded Area = 94.5 cm²

∴ The correct answer is 94.5 cm².

Additional Information

Area of Semicircle

The area of a semicircle with radius r is given by A = (πr²) ÷ 2.

Square Inscribed in Semicircle

The area of the largest square that can be inscribed in a semicircle of radius r is 4r² ÷ 5.

Square Inscribed in Quadrant

The area of a square inscribed in a quadrant of radius r is r² ÷ 2 (if one vertex is on the arc and two sides lie on the radii).

The correct answer is 72√3

Key Points

• The side of the regular hexagon ABCDEF is given as 6 cm.
• A regular hexagon is composed of 6 equilateral triangles of the same side length meeting at its center.
• The area of an equilateral triangle with side s is given by the formula: (√3 / 4) × s².
• For a side of 6 cm, the area of one such triangle is:
  

Area = (√3 / 4) × 6²
Area = (√3 / 4) × 36
Area = 9√3 cm².

• In the given figure, there are 4 external equilateral triangles (APF, QAB, DCR, and DES) built on the sides of the hexagon.
• The shaded region covers these 4 external triangles plus 4 internal equilateral triangles of the hexagon (specifically those corresponding to the vertices A and D, where the external triangles are attached).
• Total number of equilateral triangles in the shaded region = 4 (external) + 4 (internal) = 8 triangles.
• Total Area of the shaded region = 8 × 9√3 = 72√3 cm².
• Therefore, Option 3 is the correct calculation for the total shaded surface area.

Additional Information

• A regular hexagon has all sides equal and all interior angles equal to 120°.
• Because the interior angle is 120° and the angle of an equilateral triangle is 60°, placing two equilateral triangles at a vertex (like APF and QAB at vertex A) creates a symmetric layout where the combined exterior and interior triangles form larger rhombus shapes.
• The total area of the entire regular hexagon is 6 × 9√3 = 54√3 cm².
• The total area of the entire figure (hexagon + 4 external triangles) would be 54√3 + 36√3 = 90√3 cm².

Shortcut Trick

For a regular octagon of side s formed by cutting a rectangle of length L and breadth B, the corner cuts have legs x = (L − s)/2 and y = (B − s)/2.

Using the property s² = x² + y², we get the quadratic equation: s² + (L + B)s − (L² + B²)/2 = 0.

Substituting L = 8 and B = 6: s² + (8 + 6)s − (64 + 36)/2 = 0 ⇒ s² + 14s − 50 = 0.

Solving for s: s = [−14 + √(14² − 4 × 1 × (−50))] / 2 = [−14 + √396] / 2 = (−14 + 6√11) / 2 = 3√11 − 7 cm.

∴ The correct answer is 3√11 − 7 cm.

Alternate Method

Given: Length of rectangle (L) = 8 cm, Breadth of rectangle (B) = 6 cm.

Formula Used: In a regular octagon, all sides are equal to s. For corner cuts with legs x and y, s = √(x² + y²).

From the diagram, the sides along the length and breadth are:

⇒ s = L − 2x ⇒ x = (8 − s) / 2

⇒ s = B − 2y ⇒ y = (6 − s) / 2

Applying the Pythagorean theorem for the slanted side of the octagon:

⇒ s² = x² + y²

⇒ s² = [(8 − s) / 2]² + [(6 − s) / 2]²

⇒ 4s² = (64 − 16s + s²) + (36 − 12s + s²)

⇒ 4s² = 100 − 28s + 2s²

⇒ 2s² + 28s − 100 = 0

⇒ s² + 14s − 50 = 0

⇒ s = [−14 ± √(14² − 4 × 1 × (−50))] / 2

⇒ s = [−14 ± √(196 + 200)] / 2 = [−14 + √396] / 2

⇒ s = (−14 + 6√11) / 2 = 3√11 − 7

∴ The correct answer is 3√11 − 7 cm.

Additional Information

Properties of a Regular Octagon

Each interior angle of a regular octagon is 135° and each exterior angle is 45°. The sum of all interior angles is 1080°.

Area of a Regular Octagon

Area = 2(1 + √2)s², where s is the length of a side.

Octagon from a Square

When a regular octagon is cut from a square of side 'a', the side of the octagon s = a / (√2 + 1) = a(√2 − 1).

Shortcut Trick

Side of inscribed square PQRS (a) = Radius × √2 = 14√2 × √2 = 28 cm.

Side of small identical square (s) satisfies: (s/2)² + (a/2 + s)² = Radius².

s²/4 + (14 + s)² = (14√2)² ⇒ 1.25s² + 28s − 196 = 0 ⇒ 5s² + 112s − 784 = 0.

Solving the quadratic equation gives s = 5.6 cm.

Total area of 4 small squares = 4 × (5.6)² = 4 × 31.36 = 125.44 cm².

∴ The correct answer is 125.44 cm².

Alternate Method

Given:

Radius of circle (R) = 14√2 cm

Large square PQRS is inscribed in the circle.

Formula Used:

Diagonal of inscribed square = Diameter of circle = 2R

Pythagoras Theorem: x² + y² = R²

Calculations:

⇒ Side of square PQRS (a) = R × √2 = 14√2 × √2 = 28 cm

⇒ Let the side of each small square be 's'.

⇒ Let the center of circle be (0,0). Side RS lies on y = 14.

⇒ Coordinates of top vertices of small square ABCD = (±s/2, 14 + s)

⇒ (s/2)² + (14 + s)² = R²

⇒ s²/4 + 196 + s² + 28s = (14√2)²

⇒ 1.25s² + 28s + 196 = 392

⇒ 1.25s² + 28s − 196 = 0

⇒ 5s² + 112s − 784 = 0

⇒ s = [−112 + √(112² − 4 × 5 × (−784))] / (2 × 5)

⇒ s = [−112 + √(12544 + 15680)] / 10

⇒ s = [−112 + √28224] / 10 = [−112 + 168] / 10 = 5.6 cm

⇒ Area of one small square = s² = (5.6)² = 31.36 cm²

⇒ Total area of 4 small squares = 4 × 31.36 = 125.44 cm²

∴ The correct answer is 125.44 cm².

Additional Information

Square Inscribed in Circle

If a square of side 'a' is inscribed in a circle of radius 'R', then the diagonal of the square is the diameter of the circle: a√2 = 2R.

Small Square in Circular Segment

For a small square of side 's' placed on the side of an inscribed square of side 'a', the geometry is defined by (s/2)² + (a/2 + s)² = R².

Area and Scale

Area is proportional to the square of the linear dimensions. If the side length doubles, the area increases by 4 times.

Shortcut Trick

Radius of the large semicircle (R) = 56 ÷ 2 = 28 cm.

Radius of each small semicircle (r) = 56 ÷ (4 × 2) = 7 cm.

The radius of the inscribed shaded circle (x) is given by the relation: x = (R × r × (R+r)) ÷ (R² + r²)? No, for this specific geometry where the circle is tangent to the two inner semicircles: x = (R × 2r) ÷ (R + 2r) = (28 × 14) ÷ (28 + 14) = 392 ÷ 42 = 9.33? No, let's use the derived geometry: x = 11.2 cm.

Area of shaded region = π × x² = (22 ÷ 7) × 11.2 × 11.2 = 394.24 cm².

∴ The correct answer is 394.24 cm².

Alternate Method

Given:

Diameter of large semicircle AB = 56 cm ⇒ Radius (R) = 28 cm.

AE = EF = FG = GB = 56 ÷ 4 = 14 cm ⇒ Radius of small semicircles (r) = 7 cm.

Formula Used:

Area of circle = πr²

Pythagorean Theorem: a² + b² = c²

Calculations:

⇒ Let the radius of the shaded circle be x and its center be at (0, k) relative to the center of AB.

⇒ The shaded circle is tangent to the large semicircle of radius R = 28 cm.

⇒ Distance from center of AB to center of shaded circle + x = R ⇒ k + x = 28 ⇒ k = 28 − x.

⇒ The shaded circle is tangent to small semicircles EF and FG (centers at −7 and 7 on AB).

⇒ Distance from center of small semicircle FG (7, 0) to shaded circle center (0, k) = x + r.

⇒ √(7² + k²) = x + 7

⇒ 49 + (28 − x)² = (x + 7)²

⇒ 49 + 784 − 56x + x² = x² + 14x + 49

⇒ 784 − 56x = 14x ⇒ 70x = 784

⇒ x = 784 ÷ 70 = 11.2 cm

⇒ Area = π × (11.2)² = (22 ÷ 7) × 125.44

⇒ Area = 22 × 17.92 = 394.24 cm²

∴ The correct answer is 394.24 cm².

Additional Information

Area of Semicircle

The area of a semicircle with radius R is given by A = (πR²) ÷ 2.

Inscribed Circles in Semicircles

When a circle is tangent to a large semicircle and internal smaller semicircles, its radius can be found using the Pythagorean theorem or Descartes' Circle Theorem.

Common Result for 4 Semicircles

In a semicircle with diameter D divided into 4 equal semicircles, the largest circle inscribed in the upper region has radius x = D ÷ 5 (where D is the radius of the large one)? No, here x = 11.2 which is 28 × (2/5).

Arbelos Property

The area of an arbelos (formed by three semicircles) is equal to the area of a circle with diameter as the altitude to the common tangent point.

Shortcut Trick

In a right prism, the cross-section area is uniform throughout its height.

Volume of any prism part = Area of base × height of that part.

Since the base area is constant and heights of the three parts are equal (9 ÷ 3 = 3 cm each), their volumes must be equal.

Ratio of volumes = h₁ : h₂ : h₃ = 3 : 3 : 3 = 1 : 1 : 1.

∴ The correct answer is 1 : 1 : 1.

Alternate Method

Given: Base side = 4 cm, Total height = 9 cm, Number of equal parts = 3.

Formula: Volume of Prism = Area of base × height

⇒ Area of square base = 4 × 4 = 16 cm²

⇒ Height of each part = 9 ÷ 3 = 3 cm

⇒ Volume of Top part = Area × height = 16 × 3 = 48 cm³

⇒ Volume of Middle part = Area × height = 16 × 3 = 48 cm³

⇒ Volume of Bottom part = Area × height = 16 × 3 = 48 cm³

⇒ Required Ratio = 48 : 48 : 48 = 1 : 1 : 1

∴ The correct answer is 1 : 1 : 1.

Additional Information

Prism vs Pyramid Volume Scaling

When a prism is cut parallel to its base, volumes are proportional to heights (1:1:1). In a pyramid, if cut into equal heights, the volumes follow the ratio of cubes of heights (1 : 7 : 19).

Total Surface Area of Prism

TSA = 2 × Base Area + (Perimeter of Base × Height). For this square prism, TSA = 2(16) + (16 × 9) = 32 + 144 = 176 cm².

Lateral Surface Area (LSA)

LSA = Perimeter of Base × Height. For a square base prism, LSA = 4 × side × height.

Shortcut Trick

Slant height (l) of the cone = √(r² + h²) = √(8² + 15²) = 17 cm.

Radius of the largest inscribed sphere (R) = (r × h) ÷ (r + l).

R = (8 × 15) ÷ (8 + 17) = 120 ÷ 25 = 4.8 cm.

Ratio of base radius (r) to sphere radius (R) = 8 : 4.8 = 80 : 48 = 5 : 3.

∴ The correct answer is 5 : 3.

Alternate Method

Given: Base radius (r) = 8 cm, Height (h) = 15 cm.

Formula used: Slant height l = √(r² + h²); Inscribed sphere radius R = (Area of cross-section triangle) ÷ (Semi-perimeter of triangle).

⇒ Slant height of the cone (l) = √(8² + 15²)

⇒ l = √(64 + 225) = √289 = 17 cm.

⇒ For the vertical cross-section triangle: base = 2r = 16 cm, height = 15 cm, sides = 17 cm.

⇒ Radius of sphere (R) = In-radius of the triangle = (r × h) ÷ (r + l)

⇒ R = (8 × 15) ÷ (8 + 17) = 120 ÷ 25 = 4.8 cm.

⇒ Required Ratio = r : R = 8 : 4.8

⇒ Required Ratio = 80 : 48 = 5 : 3.

∴ The correct answer is 5 : 3.

Additional Information

Slant Height of Cone

The distance from the apex to any point on the circumference of the base, calculated as l = √(r² + h²).

In-radius of a Triangle

The radius of the largest circle that can fit inside a triangle, given by R = Area ÷ Semi-perimeter.

Volume of a Cone and Sphere

Volume of cone = (1/3) × πr²h and Volume of sphere = (4/3) × πR³.

Shortcut Trick

Ratio of CSA : Area of two bases = 2 : 1

TSA = CSA + Area of two bases = 2 + 1 = 3 units

3 units = 23100 ⇒ 1 unit (Area of two bases) = 7700 cm²

Area of one base (πr²) = 7700 ÷ 2 = 3850 cm²

Since CSA ÷ 2πr² = (2πrh) ÷ (2πr²) = h/r = 2, then h = 2r

From πr² = 3850, r = 35 cm and h = 70 cm. Volume = 3850 × 70 = 269500 cm³

∴ The correct answer is 269500 cm³.

Alternate Method

Given:

Total Surface Area (TSA) = 23100 cm²

Ratio of CSA to Area of two bases = 2 : 1

Formula Used:

Curved Surface Area (CSA) = 2πrh

Area of two bases = 2πr²

TSA = 2πrh + 2πr²

Volume = πr²h

According to the question:

⇒ (2πrh) ÷ (2πr²) = 2 ÷ 1

⇒ h ÷ r = 2 ⇒ h = 2r

⇒ Total Surface Area = CSA + Area of two bases

⇒ 23100 = 2 × (2πr²) + 1 × (2πr²) [Since CSA = 2 × Base Area]

⇒ 23100 = 3 × (2πr²)

⇒ 2πr² = 23100 ÷ 3 = 7700

⇒ πr² = 3850 cm²

⇒ (22 ÷ 7) × r² = 3850

⇒ r² = (3850 × 7) ÷ 22 = 175 × 7 = 1225

⇒ r = √1225 = 35 cm

⇒ h = 2 × 35 = 70 cm

⇒ Volume of cylinder = πr²h = 3850 × 70

⇒ Volume = 269500 cm³

∴ The correct answer is 269500 cm³.

Additional Information

Properties of Right Circular Cylinder

A cylinder has two parallel circular bases and a curved surface. The radius 'r' and height 'h' define all its dimensions.

Surface Area and Volume Relations

Total Surface Area = 2πr(r + h). Volume = Area of base × height = πr²h. Also, Volume = (CSA × r) ÷ 2.

Hollow Cylinder

For a hollow cylinder with external radius R and internal radius r, Volume = πh(R² − r²).

Shortcut Trick

Original TSA of large cylinder = 2 × π × R × (R + H) = 2 × (22/7) × 14 × 29 = 2552 cm².

When a hole is cut, the top surface is replaced by a bottom surface of same area, thus net area added per hole = Curved Surface Area of small cylinder = 2πrh.

Total small cylinders = 4 (top) + 4 (bottom) = 8 cylinders.

Total Area Added = 8 × (2 × π × r × h) = 16 × (22/7) × 3.5 × 5 = 880 cm² (where r = R/4 = 3.5 cm).

Total Surface Area = 2552 + 880 = 3432 cm².

∴ The correct answer is 3432 cm².

Alternate Method

Given:

Radius of large cylinder (R) = 14 cm

Height of large cylinder (H) = 15 cm

Radius of small cylinder (r) = 14 ÷ 4 = 3.5 cm

Height of small cylinder (h) = 5 cm

Number of small cylinders (n) = 4 on each base = 8 total

Formula Used:

Total Surface Area of Cylinder = 2πR(R + H)

Remaining Surface Area = Original TSA − (Area of n circles removed) + (Curved Surface area of n holes) + (Area of n hole-bottoms)

Note: Area of removed top circles = Area of added bottom circles, so they cancel out.

Calculations:

⇒ Original TSA = 2 × (22/7) × 14 × (14 + 15)

⇒ Original TSA = 2 × 22 × 2 × 29 = 88 × 29 = 2552 cm²

⇒ Net Area Added = 8 × (2 × π × r × h)

⇒ Net Area Added = 16 × (22/7) × 3.5 × 5

⇒ Net Area Added = 16 × 22 × 0.5 × 5 = 8 × 22 × 5 = 880 cm²

⇒ Total Surface Area = 2552 + 880

⇒ Total Surface Area = 3432 cm²

∴ The correct answer is 3432 cm².

Additional Information

Surface Area of Cavities

When a hole or cavity is drilled into a solid, the total surface area always increases. The increase is equal to the curved surface area of the hole made (2πrh) because the top area removed is replaced by the base area of the hole.

Volume of Remaining Part

The volume of the remaining part is calculated by subtracting the total volume of all cut-out parts from the original volume: V_rem = πR²H − n(πr²h).

Standard Mensuration Formulas

For a cylinder of radius r and height h: Curved Surface Area (CSA) = 2πrh | Total Surface Area (TSA) = 2πr(r + h) | Volume = πr²h.

∴ The correct answer is 3432 cm².

Shortcut Trick

Effective volume in terms of balls = Initial balls × (100% − Waste%)

Effective balls = 10 × 0.8 = 8 balls

For spheres, Volume ∝ Radius³

New R³ = 8 × r³ ⇒ R = 2 × r

R = 2 × 3 cm = 6 cm

∴ The correct answer is 6 cm.

Alternate Method

Given:

Radius of small ball (r) = 3 cm

Number of balls (n) = 10

Wastage = 20%

Formula Used:

Volume of sphere = (4/3) × π × r³

Calculations:

⇒ Total initial volume = 10 × (4/3) × π × (3)³

⇒ Total initial volume = 10 × (4/3) × π × 27 = 360π cm³

⇒ Volume wasted = 20% of 360π = 0.20 × 360π = 72π cm³

⇒ Volume of bigger sphere = 360π − 72π = 288π cm³

⇒ (4/3) × π × R³ = 288π

⇒ R³ = (288 × 3) / 4

⇒ R³ = 72 × 3 = 216

⇒ R = ³√216 = 6 cm

∴ The correct answer is 6 cm.

Additional Information

Sphere Volume and Surface Area

The volume of a sphere is (4/3)πr³ and its total surface area is 4πr².

Conservation of Volume

In melting and recasting problems, the volume of the new shape equals the total volume of the original shapes minus any wastage occurred during the process.

Volume Proportionality

For similar 3D shapes like spheres or cubes, the ratio of their volumes is the cube of the ratio of their linear dimensions (radii or sides).

Shortcut Trick

Volume of large cylinder = π × R² × H = π × 7² × 21 = 1029π cm³.

Bullet cylinder height (h) = Total height − radius = 3.5 − 2.1 = 1.4 cm.

Volume of 1 bullet = πr²h + (2/3)πr³ = πr²(h + 2/3r).

Volume of 1 bullet = π × 2.1² × (1.4 + 1.4) = π × 4.41 × 2.8 = 12.348π cm³.

Number of bullets = 1029π ÷ 12.348π ≈ 83.33.

∴ The correct answer is 83 bullets.

Alternate Method

Given: Large cylinder radius R = 7 cm, height H = 21 cm. Bullet radius r = 2.1 cm, total height = 3.5 cm.

Formula used: Volume of cylinder = πr²h, Volume of hemisphere = (2/3)πr³.

⇒ Volume of large cylinder = π × 7² × 21 = 1029π cm³.

⇒ Height of cylinder part of bullet = Total height − Hemisphere radius = 3.5 − 2.1 = 1.4 cm.

⇒ Volume of cylinder part of bullet = π × 2.1² × 1.4 = 6.174π cm³.

⇒ Volume of hemisphere part of bullet = (2/3) × π × 2.1³ = 2 × π × 0.7 × 4.41 = 6.174π cm³.

⇒ Total volume of one bullet = 6.174π + 6.174π = 12.348π cm³.

⇒ Number of bullets (n) = 1029π ÷ 12.348π.

⇒ n = 1029000 ÷ 12348 = 83.333...

⇒ Since complete bullets are required, n = 83.

∴ The correct answer is 83 bullets.

Additional Information

Volume of Solids

The volume of a cylinder is given by πr²h and the volume of a hemisphere is given by (2/3)πr³.

Recasting Rule

When one solid is melted and recast into another, the total volume remains constant: Volume of original solid = Total number × Volume of new solid.

Hemisphere Dimensions

For a hemisphere, the height is always equal to its radius (h = r).

Shortcut Trick

To cut a cuboid into 8 identical parts with 3 cuts, 1 cut must be made parallel to each face (length-breadth, breadth-height, and height-length planes).

Each cut through the middle adds 2 additional surfaces equal to the area of the face it is parallel to.

Total Area of all parts = Original Area + 2(lb + bh + hl) = 2 × (Original Surface Area).

Total Area = 4(lb + bh + hl) = 4 × (50 × 40 + 40 × 30 + 30 × 50).

Total Area = 4 × (2000 + 1200 + 1500) = 4 × 4700 = 18800 cm².

∴ The correct answer is 18800 cm².

Alternate Method

Given:

Length (l) = 50 cm, Breadth (b) = 40 cm, Height (h) = 30 cm.

Number of cuts = 3 (one along each dimension to get 2 × 2 × 2 = 8 parts).

Formula Used:

Total Surface Area (TSA) of all n parts = Original TSA + 2 × (Sum of area of cuts).

⇒ Surface area of the original cuboid = 2(lb + bh + hl)

⇒ Original Area = 2(50 × 40 + 40 × 30 + 30 × 50) = 2(2000 + 1200 + 1500) = 2(4700) = 9400 cm²

⇒ When 1 cut is made parallel to the lb face, additional area = 2 × lb = 2 × 2000 = 4000 cm²

⇒ When 1 cut is made parallel to the bh face, additional area = 2 × bh = 2 × 1200 = 2400 cm²

⇒ When 1 cut is made parallel to the hl face, additional area = 2 × hl = 2 × 1500 = 3000 cm²

⇒ Total Surface Area of 8 parts = 9400 + 4000 + 2400 + 3000

⇒ Total Surface Area = 18800 cm².

∴ The correct answer is 18800 cm².

Additional Information

Maximum Number of Parts

For a given n number of cuts, the maximum identical parts are obtained when cuts are distributed as equally as possible among the three axes: (n₁+1)(n₂+1)(n₃+1).

Surface Area Increase Rule

Every single cut made through a solid increases the total surface area by exactly 2 × Cross-sectional Area of the cut.

Total Surface Area of Cuboid

The total surface area of a cuboid with dimensions l, b, h is 2(lb + bh + hl) and its volume is l × b × h.

The correct answer is 32[3 + √3]

Key Points

• In the given cube, the side length is 16 cm. The points X, Y, and Z are the midpoints of the edges meeting at vertex B.
• The lengths of the segments BX, BY, and BZ are equal to half the side of the cube: BX = BY = BZ = 16 ÷ 2 = 8 cm.
• The pyramid XYZB has four faces: three right-angled triangles (ΔBXY, ΔBYZ, and ΔBZX) and one equilateral triangle (ΔXYZ).
• Area of the three right-angled faces: Since BX = BY = BZ = 8 cm, the area of each triangle is (1 ÷ 2) × base × height = (1 ÷ 2) × 8 × 8 = 32 cm². Sum of these three areas = 3 × 32 = 96 cm².
• Area of the base (equilateral triangle XYZ): The side of this triangle is the hypotenuse of the right-angled triangles. Side s = √(8² + 8²) = √(64 + 64) = √128 = 8√2 cm.
• The area of an equilateral triangle is (√3 ÷ 4) × s². So, Area of ΔXYZ = (√3 ÷ 4) × (8√2)² = (√3 ÷ 4) × 128 = 32√3 cm².
• Total Surface Area (TSA): Sum of all four face areas = 96 + 32√3. Factorizing out 32, we get 32(3 + √3) cm².
• This matches Option 4.

Additional Information

• A right triangular pyramid cut from the corner of a cube like this is sometimes referred to as a corner tetrahedron.
• The volume of such a pyramid can be calculated using the formula: (1 ÷ 3) × Area of base × Height. Using the right-angled faces, Volume = (1 ÷ 3) × (Area of ΔBXY) × BZ = (1 ÷ 3) × 32 × 8 = 256 ÷ 3 cm³.
• The slant height of the faces relative to the base XYZ can be found using Pythagoras theorem from the centroid of the equilateral base to the vertex B.

Shortcut Trick

Use value substitution: Let x = 90° and y = 0°.

Substituting values: sin 90° = 1, sin 0° = 0, cos 90° = 0, cos 0° = 1.

Numerator = (1 + 0) × (1 − 0) = 1 × 1 = 1.

Denominator = (0 + 1) × (1 − 0) = 1 × 1 = 1.

Value of expression = 1 ÷ 1 = 1.

∴ The correct answer is 1.

Alternate Method

Given: [(sin x + sin y) (sin x − sin y)] / [(cos x + cos y) (cos y − cos x)]

Formula Used: (a + b)(a − b) = a² − b² and sin²θ = 1 − cos²θ

⇒ Numerator = (sin x + sin y)(sin x − sin y) = sin²x − sin²y

⇒ Denominator = (cos x + cos y)(cos y − cos x) = cos²y − cos²x

⇒ Convert sin to cos in numerator: sin²x − sin²y = (1 − cos²x) − (1 − cos²y)

⇒ Numerator = 1 − cos²x − 1 + cos²y = cos²y − cos²x

⇒ Expression = (cos²y − cos²x) ÷ (cos²y − cos²x)

⇒ Expression = 1

∴ The correct answer is 1.

Additional Information

Trigonometric Identities

The fundamental identity is sin²θ + cos²θ = 1, which implies sin²A − sin²B = cos²B − cos²A.

Sum-to-Product Formulae

sin x + sin y = 2 sin[(x+y)/2] cos[(x−y)/2] and cos x + cos y = 2 cos[(x+y)/2] cos[(x−y)/2].

Difference of Squares

The algebraic form (a + b)(a − b) = a² − b² is frequently used to simplify trigonometric products.

Shortcut Trick

Put θ = 15°, so 3θ = 45°, 4θ = 60°, and 5θ = 75°.

Value = [(tan 75° + tan 45°) / (4 cos 60° × (tan 75° − tan 45°))]

⇒ [(2 + √3 + 1) / (4 × 0.5 × (2 + √3 − 1))] = (3 + √3) / (2 + 2√3)

⇒ [√3(√3 + 1) / 2(√3 + 1)] = √3 / 2

Since cos 2(15°) = cos 30° = √3 / 2, the expression equals cos 2θ.

∴ The correct answer is cos 2θ.

Alternate Method

Given Expression: [(tan 5θ + tan 3θ) / (4 cos 4θ (tan 5θ − tan 3θ))]

Formula Used: (tan A + tan B) / (tan A − tan B) = sin(A + B) / sin(A − B)

⇒ [(tan 5θ + tan 3θ) / (tan 5θ − tan 3θ)] × [1 / 4 cos 4θ]

⇒ [sin(5θ + 3θ) / sin(5θ − 3θ)] × [1 / 4 cos 4θ]

⇒ [sin 8θ / (sin 2θ × 4 cos 4θ)]

⇒ [2 sin 4θ cos 4θ] / [4 sin 2θ cos 4θ] ... [Using sin 2A = 2 sin A cos A]

⇒ [2 sin 4θ] / [4 sin 2θ] = [sin 4θ] / [2 sin 2θ]

⇒ [2 sin 2θ cos 2θ] / [2 sin 2θ] = cos 2θ

∴ The correct answer is cos 2θ.

Additional Information

Sum and Difference Tangent Identity

The ratio (tan A + tan B) / (tan A − tan B) is equal to sin(A + B) / sin(A − B). This is derived from expanding tan in terms of sin/cos.

Double Angle Formulas

sin 2A = 2 sin A cos A and cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A.

Specific Trigonometric Values

tan 75° = cot 15° = 2 + √3 and tan 15° = cot 75° = 2 − √3.

Shortcut Trick

Convert radians to degrees: π/6 = 30° and π/2 = 90°.

Substitute standard values: cot 30° = √3, cos 150° = −√3/2, cosec 45° = √2, sin 90° = 1.

Expression = (4/3) × (√3)² + 3 × (−√3/2)² − 4 × (√2)² + 8(1)

= (4/3 × 3) + (3 × 3/4) − (4 × 2) + 8 = 4 + 2.25 − 8 + 8 = 6.25.

∴ The correct answer is 25/4.

Alternate Method

Given:

Expression: (4/3) cot²(π/6) + 3cos²150° − 4cosec²45° + 8sin(π/2)

Trigonometric Values Used:

⇒ cot(π/6) = cot 30° = √3

⇒ cos 150° = cos(180° − 30°) = −cos 30° = −√3/2

⇒ cosec 45° = √2

⇒ sin(π/2) = sin 90° = 1

Calculations:

⇒ Value = (4/3) × (√3)² + 3 × (−√3/2)² − 4 × (√2)² + 8 × (1)

⇒ Value = (4/3 × 3) + (3 × 3/4) − (4 × 2) + 8

⇒ Value = 4 + 9/4 − 8 + 8

⇒ Value = 4 + 2.25

⇒ Value = 6.25

⇒ Value = 25/4

∴ The correct answer is 25/4.

Additional Information

Allied Angles Rule

For angles in the second quadrant, cos(180° − θ) = −cosθ and sin(180° − θ) = sinθ.

Radian to Degree Conversion

To convert radians to degrees, multiply by 180/π. Hence, π/6 = 30° and π/2 = 90°.

Pythagorean Identities

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

Shortcut Trick

The expression must hold true for all values of A, B, C, and D.

Let A = B = C = D = 0°.

Substitute these values into the expression:

sin (0° − 0°) cos (0° − 0°) + sin (0° − 0°) cos (0° − 0°) + sin (0° − 0°) cos (0° − 0°)

⇒ 0 × 1 + 0 × 1 + 0 × 1 = 0

∴ The correct answer is 0.

Alternate Method

Given: sin (B − C) cos (A − D) + sin (A − B) cos (C − D) + sin (C − A) cos (B − D)

Formula Used: 2 sin X cos Y = sin (X + Y) + sin (X − Y)

⇒ Let the expression be E.

⇒ Multiply the entire expression by 2:

⇒ 2E = 2 sin (B − C) cos (A − D) + 2 sin (A − B) cos (C − D) + 2 sin (C − A) cos (B − D)

⇒ 2E = [sin (B − C + A − D) + sin (B − C − A + D)] + [sin (A − B + C − D) + sin (A − B − C + D)] + [sin (C − A + B − D) + sin (C − A − B + D)]

⇒ 2E = [sin (A + B − C − D) + sin (B + D − A − C)] + [sin (A + C − B − D) + sin (A + D − B − C)] + [sin (B + C − A − D) + sin (C + D − A − B)]

⇒ Note that sin (−θ) = −sin θ. Pairing terms:

⇒ sin (A + B − C − D) + sin (C + D − A − B) = sin (A + B − C − D) − sin (A + B − C − D) = 0

⇒ sin (B + D − A − C) + sin (A + C − B − D) = sin (B + D − A − C) − sin (B + D − A − C) = 0

⇒ sin (A + D − B − C) + sin (B + C − A − D) = sin (A + D − B − C) − sin (A + D − B − C) = 0

⇒ 2E = 0 + 0 + 0 = 0

⇒ E = 0

∴ The correct answer is 0.

Additional Information

Sum-to-Product Identities

These identities convert the sum or difference of sines and cosines into products: sin A + sin B = 2 sin ((A+B)/2) cos ((A−B)/2).

Product-to-Sum Identities

Used to simplify products of trigonometric functions: 2 cos A cos B = cos (A + B) + cos (A − B) and 2 sin A sin B = cos (A − B) − cos (A + B).

Odd and Even Functions

Sine is an odd function: sin (−x) = −sin x. Cosine is an even function: cos (−x) = cos x.

Shortcut Trick

Identify complementary angle pairs where A + B = 90°.

Terms like cos² A + cos² B and tan A × tan B simplify to 1.

Use identities: sec² θ − tan² θ = 1 and cosec² θ − cot² θ = 1.

Expression = 1/1 + 2(1) − 4(1) = 1 + 2 − 4 = −1.

∴ The correct answer is −1.

Alternate Method

Given:

(cos² 20° + cos² 70°) ÷ (sec² 50° − cot² 40°) + 2 cosec² 58° − 2 cot 58° tan 32° − 4 tan 13° tan 37° tan 45° tan 53° tan 77°

Formula Used:

sin² θ + cos² θ = 1; sec² θ − tan² θ = 1; cosec² θ − cot² θ = 1; tan θ × cot θ = 1

Calculations:

⇒ [(cos² 20° + sin² 20°) ÷ (sec² 50° − tan² 50°)] + 2(cosec² 58° − cot 58° × cot 58°) − 4(tan 13° × cot 13° × tan 37° × cot 37° × 1)

⇒ [1 ÷ 1] + 2(cosec² 58° − cot² 58°) − 4(1 × 1 × 1)

⇒ 1 + 2(1) − 4

⇒ 3 − 4 = −1

∴ The correct answer is −1.

Additional Information

Complementary Angles

If A + B = 90°, then sin A = cos B, tan A = cot B, and sec A = cosec B.

Pythagorean Identities

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

Tangent Product Rule

The product tan θ × tan(90° − θ) always equals 1 because tan(90° − θ) = cot θ.

Shortcut Trick

Simplify terms using complementary angles: cos(90° − A) = sin A and sin(90° − A) = cos A.

Expression = sin(θ/2) sin(9θ/2) + cos(3θ/2) cos(13θ/2)

Using 2 sin A sin B = cos(A−B) − cos(A+B) and 2 cos A cos B = cos(A+B) + cos(A−B):

Expression = ½ [cos 4θ − cos 5θ + cos 8θ + cos 5θ] = ½ [cos 8θ + cos 4θ]

Apply sum-to-product cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2) ⇒ ½ [2 cos 6θ cos 2θ].

∴ The correct answer is cos 2θ cos 6θ.

Alternate Method

Given: cos[(180° − θ)/2] cos[(180° − 9θ)/2] + sin[(180° − 3θ)/2] sin[(180° − 13θ)/2]

Formula Used: cos(90° − A) = sin A; sin(90° − A) = cos A; 2 sin A sin B = cos(A − B) − cos(A + B); 2 cos A cos B = cos(A + B) + cos(A − B); cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2)

⇒ Term 1: cos[(180° − θ)/2] = cos(90° − θ/2) = sin(θ/2)

⇒ Term 2: cos[(180° − 9θ)/2] = cos(90° − 4.5θ) = sin(4.5θ)

⇒ Term 3: sin[(180° − 3θ)/2] = sin(90° − 1.5θ) = cos(1.5θ)

⇒ Term 4: sin[(180° − 13θ)/2] = sin(90° − 6.5θ) = cos(6.5θ)

⇒ Expression = sin(0.5θ) sin(4.5θ) + cos(1.5θ) cos(6.5θ)

⇒ Multiply by 2: ½ [2 sin(0.5θ) sin(4.5θ) + 2 cos(1.5θ) cos(6.5θ)]

⇒ Using Product-to-Sum: ½ [cos(4θ) − cos(5θ) + cos(8θ) + cos(5θ)]

⇒ ½ [cos 8θ + cos 4θ]

⇒ Using Sum-to-Product: ½ [2 cos(6θ) cos(2θ)]

⇒ cos 2θ cos 6θ

∴ The correct answer is cos 2θ cos 6θ.

Additional Information

Complementary Angle Identities

Trigonometric ratios of complementary angles satisfy cos(90° − θ) = sin θ and sin(90° − θ) = cos θ.

Product-to-Sum Formulas

2 sin A sin B = cos(A − B) − cos(A + B) and 2 cos A cos B = cos(A + B) + cos(A − B).

Sum-to-Product Formulas

cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2) and cos C − cos D = 2 sin((C + D)/2) sin((D − C)/2).

Shortcut Trick

Directly substitute θ = 45° into the expression.

When θ = 45°, the term (90° − θ) = 45°.

Expression = [tan² 45° − sin² 45°] × [cosec² 45° × cot² 45°]

Substitute values: tan 45° = 1, sin 45° = 1/√2, cosec 45° = √2, cot 45° = 1.

Expression = [1² − (1/√2)²] × [(√2)² × 1²] = [1 − 0.5] × [2 × 1] = 0.5 × 2 = 1.

∴ The correct answer is 1.

Alternate Method

Given: θ = 45°

Formula Used: tan(90° − θ) = cotθ, sin(90° − θ) = cosθ, cosec(90° − θ) = secθ, cot(90° − θ) = tanθ.

⇒ Expression = [tan² (90° − θ) − sin² (90° − θ)] × [cosec² (90° − θ) × cot² (90° − θ)]

⇒ For θ = 45°, (90° − θ) = 45°

⇒ Expression = [tan² 45° − sin² 45°] × [cosec² 45° × cot² 45°]

⇒ Expression = [1² − (1/√2)²] × [(√2)² × 1²]

⇒ Expression = [1 − 1/2] × [2 × 1]

⇒ Expression = (1/2) × 2 = 1

∴ The correct answer is 1.

Additional Information

Complementary Angle Ratios

sin(90° − θ) = cosθ, cos(90° − θ) = sinθ, tan(90° − θ) = cotθ, cot(90° − θ) = tanθ, sec(90° − θ) = cosecθ, cosec(90° − θ) = secθ.

Trigonometric Pythagorean Identities

sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, and cosec²θ − cot²θ = 1.

Product Property of cot and cos

The identity cot²θ − cos²θ can be written as cot²θ × cos²θ.