SSC CGL Tier 2 Quant Previous Year Paper (Held On: 9 March 2018)

Shortcut Trick

Filter digits: Primes must use digits {1, 3, 7, 9} only; even digits or 5 result in composite numbers when reversed.

Possible primes with these digits: 11, 13, 17, 19, 31, 37, 71, 73, 79, 97.

Verify reversals: 11→11, 13↔31, 17↔71, 37↔73, 79↔97 (all prime). Note 19→91 (7×13).

Total valid numbers: 1 + 2 + 2 + 2 + 2 = 9.

∴ The correct answer is 9 prime numbers.

Alternate Method

Given: Two-digit prime numbers between 10 and 100.

Method: Systematic verification of prime numbers and their reversals.

⇒ List of all prime numbers between 10 to 100: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

⇒ If a prime contains an even digit (2, 4, 6, 8) or 5, its reverse will be divisible by 2 or 5. Such primes are excluded (e.g., 23, 53, etc.).

⇒ Testing primes with digits 1, 3, 7, 9:

⇒ 11 reversed is 11 (Prime)

⇒ 13 reversed is 31 (Prime) & 31 reversed is 13 (Prime)

⇒ 17 reversed is 71 (Prime) & 71 reversed is 17 (Prime)

⇒ 19 reversed is 91 (91 = 13 × 7, Not Prime)

⇒ 37 reversed is 73 (Prime) & 73 reversed is 37 (Prime)

⇒ 79 reversed is 97 (Prime) & 97 reversed is 79 (Prime)

⇒ Valid primes: {11, 13, 17, 31, 37, 71, 73, 79, 97}.

⇒ Total count = 9.

∴ The correct answer is 9 prime numbers.

Additional Information

Emirp Primes

An emirp (prime spelled backwards) is a prime number that results in a different prime when its digits are reversed. Examples include 13, 17, 31, 37, 71, 73, 79, and 97.

Palindromic Primes

A prime number that remains the same when its digits are reversed is called a palindromic prime. 11 is the only two-digit palindromic prime.

Twin Primes

Pairs of prime numbers that have a difference of 2 are called twin primes, such as (3, 5), (5, 7), (11, 13), and (17, 19).

Composite Numbers

Numbers that have more than two factors are composite numbers. 1 is neither prime nor composite.

Shortcut Trick For a perfect cube n3 to be divisible by a prime number 7, the base n must also be divisible by 7. The required range is 1 < n3 < 100,000. Taking cube roots: 1 < n < ³√100,000 ≈ 46.41. Multiples of 7 within the range [2, 46.41] are: 7, 14, 21, 28, 35, and 42. Count of such numbers = 6. ∴ The correct answer is 6.

Alternate Method Given: Perfect cubes between 1 and 100,000 divisible by 7. Property Used: If a prime p divides n3, then p must divide n.

⇒ Let the perfect cube be x3. ⇒ 1 < x3 < 100,000 ⇒ Taking cube root on all sides: ³√1 < x < ³√100,000 ⇒ 1 < x < 46.41 (Since 463 = 97,336 and 473 = 103,823) ⇒ Since x3 is divisible by 7, and 7 is prime, x must be a multiple of 7. ⇒ Possible values for x are multiples of 7 between 1 and 46.41. ⇒ x = 7 × 1, 7 × 2, 7 × 3, 7 × 4, 7 × 5, 7 × 6. ⇒ x = 7, 14, 21, 28, 35, 42. ⇒ Total count = 6. ∴ The correct answer is 6.

Additional Information Divisibility of Powers If a prime number p divides an, then p must divide a. This property is essential for finding bases of powers in a given range. Cube Root Estimation To estimate ³√N, find the nearest cubes. For example, 403 = 64,000 and 503 = 125,000. Since 100,000 is between them, the root is between 40 and 50. Multiples in a Range The number of multiples of k in the range [1, N] is given by the floor value of N/k.

Given:

A = 0.142857142857... = $\frac{1}{7}$

B = 0.16666... = $\frac{1}{6}$

Formula used:

$\frac{A + B}{AB} = \frac{A}{AB} + \frac{B}{AB} = \frac{1}{B} + \frac{1}{A}$

Calculations:

$\frac{A + B}{AB} = \frac{1}{B} + \frac{1}{A}$

⇒ $\frac{1}{1/6} + \frac{1}{1/7}$

⇒ 6 + 7

⇒ 13

∴ The correct answer is option (4).

Given:

A = 0.abcabc...

Formula used:

A = 0.\overline{pqr} = $\frac{pqr}{999}$

Calculations:

A = 0.abcabc...

⇒ A = 0.\overline{abc}

⇒ A = $\frac{abc}{999}$

To obtain an integral value, A must be multiplied by a multiple of 999.

⇒ Option (1): 2997 = 999 × 3

⇒ Option (2): 1000 = (Not a multiple of 999)

⇒ Option (3): 1998 = 999 × 2

⇒ Both 2997 and 1998 result in an integral value.

∴ The correct answer is option (4).

Given:

Series: $1 \frac{1}{2} + 4 \frac{1}{6} + 7 \frac{1}{12} + 10 \frac{1}{20} \dots$

Number of terms (n) = 20

Formula used:

Sum of AP (S_n) = $\frac{n}{2} [2a + (n - 1)d]$

Telescoping series: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

Calculation:

⇒ Sum = $(1 + \frac{1}{1 \times 2}) + (4 + \frac{1}{2 \times 3}) + (7 + \frac{1}{3 \times 4}) + \dots \text{ to 20 terms}$

⇒ Integer sum = $\frac{20}{2} [2 \times 1 + (20 - 1) \times 3]$

⇒ Integer sum = 10 \times [2 + 57] = 590

⇒ Fractional sum = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{20} - \frac{1}{21})

⇒ Fractional sum = 1 - \frac{1}{21} = \frac{20}{21}

⇒ Total sum = 590 + \frac{20}{21}

⇒ Total sum = $\frac{590 \times 21 + 20}{21}$

⇒ Total sum = $\frac{12390 + 20}{21} = \frac{12410}{21}$

∴ The correct answer is option (1).

Shortcut Trick

For any geometric series of the form (1/2¹) + (1/2²) + ... + (1/2ⁿ), the sum is always equal to (2ⁿ − 1) / 2ⁿ.

Here, n = 10, so the sum S = (2¹⁰ − 1) / 2¹⁰.

S = (1024 − 1) / 1024 = 1023 / 1024.

Given 1/k = 1023 / 1024, by reciprocating both sides, we get k = 1024 / 1023.

∴ The correct answer is 1024/1023.

Alternate Method

Given:

Series: (1/2¹) + (1/2²) + (1/2³) + ... + (1/2¹⁰) = 1/k

First term (a) = 1/2

Common ratio (r) = (1/2²) ÷ (1/2¹) = 1/2

Number of terms (n) = 10

Formula Used:

Sum of a Geometric Progression (GP): S_n = a(1 − rⁿ) / (1 − r), where |r| < 1

Calculations:

⇒ 1/k = (1/2) × [1 − (1/2)¹⁰] / [1 − 1/2]

⇒ 1/k = (1/2) × [1 − 1/1024] / (1/2)

⇒ 1/k = 1 − 1/1024

⇒ 1/k = (1024 − 1) / 1024

⇒ 1/k = 1023 / 1024

⇒ k = 1024 / 1023

∴ The correct answer is 1024/1023.

Additional Information

Sum of Infinite GP

If the number of terms is infinite and |r| < 1, the sum is given by S_∞ = a / (1 − r).

Geometric Progression (GP)

A sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio (r).

Powers of 2

Commonly used powers include 2⁹ = 512 and 2¹⁰ = 1024, which are essential for solving binary-based series quickly.

p>Given:

I. $1 \frac{2}{3} + 2 \frac{3}{4} + 3 \frac{4}{5} > 8$

II. $6 \frac{1}{2} - 5 \frac{3}{4} + 4 \frac{1}{4} > 5$

Formula used:

Mixed fraction: $a \frac{b}{c} = a + \frac{b}{c}$

Calculation:

Statement I:

$1 \frac{2}{3} + 2 \frac{3}{4} + 3 \frac{4}{5}$

⇒ $(1 + 2 + 3) + (\frac{2}{3} + \frac{3}{4} + \frac{4}{5})$

⇒ LCM of (3, 4, 5) = 60

⇒ $6 + (\frac{40 + 45 + 48}{60})$

⇒ $6 + \frac{133}{60}$

⇒ $6 + 2 \frac{13}{60} = 8 \frac{13}{60}$

⇒ $8 \frac{13}{60} > 8$ (True)

Statement II:

$6 \frac{1}{2} - 5 \frac{3}{4} + 4 \frac{1}{4}$

⇒ $(6 - 5 + 4) + (\frac{1}{2} - \frac{3}{4} + \frac{1}{4})$

⇒ $5 + (\frac{2 - 3 + 1}{4})$

⇒ $5 + \frac{0}{4} = 5$

⇒ $5 > 5$ (False)

∴ The correct answer is option (1).

Shortcut Trick

3²⁰⁰² is an odd number, so (3²⁰⁰² − 1) and (3²⁰⁰² + 1) are consecutive even integers.

The Highest Common Factor (HCF) of any two consecutive even integers is always 2.

For Statement II, 4^even always ends in 6; therefore, 4⁸⁴ − 1 ends in 5.

Any integer ending in 5 is exactly divisible by 5, making Statement II TRUE.

∴ The correct answer is Only II.

Alternate Method

Given:

I. A = (3²⁰⁰² − 1), B = (3²⁰⁰² + 1)

II. C = (4⁸⁴ − 1)

Formula/Property Used:

⇒ HCF(x, y) must divide their difference |y − x|.

⇒ (aⁿ − bⁿ) is divisible by (a + b) if n is even.

Calculations:

⇒ Difference of terms in Statement I = (3²⁰⁰² + 1) − (3²⁰⁰² − 1) = 2.

⇒ The HCF must be a factor of 2. Since both terms are even, HCF = 2.

⇒ Statement I is FALSE (it claims HCF is 4).

⇒ For Statement II: 4⁸⁴ − 1 = (4²)⁴² − 1⁴² = 16⁴² − 1.

⇒ Any expression (xⁿ − 1) is divisible by (x − 1).

⇒ 16⁴² − 1 is divisible by (16 − 1) = 15.

⇒ Since it is divisible by 15, it is also divisible by 5. Statement II is TRUE.

∴ The correct answer is Only II.

Additional Information

HCF of Consecutive Even Integers

For any integer n, the HCF of consecutive even integers 2n and 2n + 2 is always 2.

Divisibility of Polynomial Forms

The form (xⁿ − aⁿ) is always divisible by (x − a) for all natural numbers n, and by (x + a) if n is an even number.

Unit Digit of 4ⁿ

If n is odd, the unit digit of 4ⁿ is 4. If n is even, the unit digit of 4ⁿ is 6.

Shortcut Trick Statement I: For odd n, (an + bn) is divisible by (a + b). Grouping terms: (199 + 499) + (299 + 399) + 599. ⇒ (1 + 4) = 5 and (2 + 3) = 5. All groups and 599 are divisible by 5. (True) Statement II: Approximate bases to powers of 2. 31 < 25 and 17 > 24. ⇒ 3111 < (25)11 = 255 and 1714 > (24)14 = 256. ⇒ Since 255 < 256, then 3111 < 1714. (False) ∴ The correct answer is Only I. Alternate Method Given: I. 199 + 299 + 399 + 499 + 599 is divisible by 5. II. 3111 > 1714 Method Used: Modular Arithmetic and Power Comparison. Calculations: For Statement I, calculate remainder of each term when divided by 5: ⇒ 199 ≡ 1 (mod 5) ⇒ 299 ≡ (24)24 × 23 ≡ 124 × 8 ≡ 3 (mod 5) [By Fermat's Little Theorem, 24 ≡ 1 mod 5] ⇒ 399 ≡ (34)24 × 33 ≡ 124 × 27 ≡ 2 (mod 5) ⇒ 499 ≡ (−1)99 ≡ −1 ≡ 4 (mod 5) ⇒ 599 ≡ 0 (mod 5) ⇒ Sum of remainders = 1 + 3 + 2 + 4 + 0 = 10 ⇒ 10 ÷ 5 = 2, Remainder = 0. So, Statement I is TRUE. For Statement II, compare using a common base proxy (2): ⇒ 3111 < (32)11 = (25)11 = 255 ⇒ 1714 > (16)14 = (24)14 = 256 ⇒ Since 255 < 256, it follows that 3111 < 1714. So, Statement II is FALSE. ∴ The correct answer is Only I. Additional Information Divisibility Rule for (an + bn) If n is an odd positive integer, then (an + bn) is always divisible by (a + b). If n is even, it is not necessarily divisible by (a + b). Fermat's Little Theorem If p is a prime number and a is not divisible by p, then ap−1 ≡ 1 (mod p). This helps in finding remainders for large powers. Comparison of Exponential Terms To compare ab and cd, try to find a common base or use logarithms. Alternatively, use nearby powers of 2 or 10 as benchmarks. Divisibility of (an − bn) (an − bn) is always divisible by (a − b) for all natural numbers n, and is divisible by (a + b) if n is even.

Shortcut Trick

Identify the power of 2 near the range 60−70, which is 2⁶ = 64.

Rewrite N = 2⁴⁸ − 1 as (2⁶)⁸ − 1.

Since the exponent 8 is even, (xⁿ − 1) is divisible by both (x − 1) and (x + 1).

Here, x = 2⁶ = 64, so N is divisible by (64 − 1) = 63 and (64 + 1) = 65.

Sum of these two numbers = 63 + 65 = 128.

∴ The correct answer is 128.

Alternate Method

Given: N = 2⁴⁸ − 1. We need to find two factors between 60 and 70.

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

⇒ N = 2⁴⁸ − 1

⇒ N = (2²⁴)² − 1² = (2²⁴ − 1)(2²⁴ + 1)

⇒ N = (2¹² − 1)(2¹² + 1)(2²⁴ + 1)

⇒ N = (2⁶ − 1)(2⁶ + 1)(2¹² + 1)(2²⁴ + 1)

⇒ 2⁶ − 1 = 64 − 1 = 63

⇒ 2⁶ + 1 = 64 + 1 = 65

⇒ Since 63 and 65 both lie between 60 and 70, these are the required numbers.

⇒ Sum = 63 + 65 = 128

∴ The correct answer is 128.

Additional Information

Divisibility of aⁿ − bⁿ

The expression aⁿ − bⁿ is always divisible by (a − b) for all natural numbers n. If n is even, it is also divisible by (a + b).

Divisibility of aⁿ + bⁿ

The expression aⁿ + bⁿ is divisible by (a + b) only if n is an odd natural number.

Factors of 2ⁿ − 1

For any expression of the form 2ⁿ − 1, if k is a factor of n, then 2^k − 1 is a factor of 2ⁿ − 1.

Given: I. $\sqrt{625} + \sqrt[4]{1296} + \sqrt{1024} > 90$ II. $\sqrt[3]{(\sqrt{729})} + \sqrt[4]{(\sqrt{256})} = 5$ Formula used: $\sqrt[n]{x}$ = x1/n Calculation: Statement I: ⇒ 25 + 6 + 32 = 63 ⇒ 63 > 90 (False) Statement II: ⇒ $\sqrt[3]{27} + \sqrt[4]{16}$ ⇒ 3 + 2 = 5 ⇒ 5 = 5 (True) ∴ The correct answer is option (2).

Given:

I. $\sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{5} + \sqrt{6} > 10$

II. $\sqrt{10} + \sqrt{12} + \sqrt{14} > 3\sqrt{12}$

Formula used:

Approximate values: $\sqrt{2} \approx 1.414, \sqrt{3} \approx 1.732, \sqrt{5} \approx 2.236, \sqrt{6} \approx 2.449$

(a + b)² = a² + b² + 2ab

Calculation:

For Statement I:

⇒ $\sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{5} + \sqrt{6}$

⇒ 1 + 1.414 + 1.732 + 2 + 2.236 + 2.449

⇒ 10.831 > 10 (True)

For Statement II:

⇒ $\sqrt{10} + \sqrt{12} + \sqrt{14} > 3\sqrt{12}$

⇒ $\sqrt{10} + \sqrt{14} > 2\sqrt{12}$

⇒ $(\sqrt{10} + \sqrt{14})^2 > (2\sqrt{12})^2$

⇒ $10 + 14 + 2\sqrt{10 \times 14} > 4 \times 12$

⇒ $24 + 2\sqrt{140} > 48$

⇒ $2\sqrt{140} > 24$

⇒ $\sqrt{140} > 12$

⇒ 140 > 144 (False)

∴ The correct answer is option (1).

Shortcut Trick

Multiply the given equation y² = y + 7 by y on both sides.

y³ = y² + 7y

Replace y² with the given value (y + 7).

y³ = (y + 7) + 7y = 8y + 7

∴ The correct answer is 8y + 7.

Alternate Method

Given:

y² = y + 7

Method Used:

Recursive substitution to reduce the degree of the polynomial expression.

Calculations:

⇒ y³ = y² × y

⇒ Substituting the given value y² = y + 7 into the expression:

⇒ y³ = (y + 7) × y

⇒ y³ = y² + 7y

⇒ Substituting y² = y + 7 once again to eliminate the squared term:

⇒ y³ = (y + 7) + 7y

⇒ y³ = y + 7 + 7y

⇒ y³ = 8y + 7

∴ The correct answer is 8y + 7.

Additional Information

Laws of Indices

The product rule for exponents states that a^m × aⁿ = a^m+n. In this problem, y² × y¹ = y³.

Polynomial Substitution

When given a lower-degree identity (like y² = y + 7), you can find higher powers by repeatedly substituting the known identity into the expression.

Quadratic Equation Roots

The equation y² − y − 7 = 0 can be solved using the quadratic formula y = (−b ± √(b² − 4ac)) / 2a, but substitution is faster for expression simplification.

Shortcut Trick

By Factor Theorem, if (x − 3) is a factor, then f(3) = 0.

Substitute x = 3: (3 − 2)(3² + 3P + 4) = 0

1 × (9 + 3P + 4) = 0 ⇒ 13 + 3P = 0

3P = −13 ⇒ P = −13/3

∴ The correct answer is −13/3.

Alternate Method

Given: f(x) = (x − 2)(x² + Px + 4) and (x − 3) is a factor.

Formula Used: Factor Theorem states that if (x − a) is a factor of a polynomial f(x), then f(a) = 0.

⇒ Here, (x − 3) is a factor, so a = 3.

⇒ Therefore, f(3) = 0.

⇒ f(3) = (3 − 2)(3² + P × 3 + 4) = 0

⇒ 1 × (9 + 3P + 4) = 0

⇒ 13 + 3P = 0

⇒ 3P = −13

⇒ P = −13/3

∴ The correct answer is −13/3.

Additional Information

Factor Theorem

For any polynomial f(x), (x − a) is a factor if and only if f(a) = 0.

Remainder Theorem

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

Relationship of Roots

If f(x) is a product of linear factors like (x − r₁)(x − r₂), then r₁ and r₂ are the roots of the equation f(x) = 0.

Shortcut Trick

If (x − 1/x) = k, then (x³ − 1/x³) = k³ + 3k.

Given k = 2 ⇒ (x³ − 1/x³) = 2³ + 3(2) = 14.

Using (x³ + 1/x³) = √((x³ − 1/x³)² + 4) = √(14² + 4) = √200 = 10√2.

Value of (x⁶ − 1/x⁶) = (x³ − 1/x³)(x³ + 1/x³) = 14 × 10√2 = 140√2.

∴ The correct answer is 140√2.

Alternate Method

Given: x − (1/x) = 2

Formula Used: x⁶ − 1/x⁶ = (x² − 1/x²)(x⁴ + 1/x⁴ + 1)

⇒ x − 1/x = 2

⇒ Squaring both sides: x² + 1/x² − 2 = 4 ⇒ x² + 1/x² = 6

⇒ Again squaring: (x² + 1/x²)² = 6² ⇒ x⁴ + 1/x⁴ + 2 = 36 ⇒ x⁴ + 1/x⁴ = 34

⇒ We know (x + 1/x)² = (x − 1/x)² + 4 = 2² + 4 = 8 ⇒ x + 1/x = √8 = 2√2

⇒ Now, x² − 1/x² = (x − 1/x)(x + 1/x) = 2 × 2√2 = 4√2

⇒ x⁶ − 1/x⁶ = (x² − 1/x²)(x⁴ + 1/x⁴ + 1) = 4√2 × (34 + 1)

⇒ x⁶ − 1/x⁶ = 4√2 × 35 = 140√2

∴ The correct answer is 140√2.

Additional Information

Difference of Squares and Cubes

The identity a² − b² = (a − b)(a + b) and a³ − b³ = (a − b)(a² + ab + b²) are fundamental for simplifying higher power expressions like x⁶ − 1/x⁶.

Relationship between Sum and Difference

For any non-zero value x, the relation between the sum and difference of reciprocals is given by (x + 1/x)² − (x − 1/x)² = 4.

Power Transformations

If x + 1/x = k, then x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k. These shortcuts allow for quick computation in competitive exams.

Given: x, y, z > 0 3x > 9y 2y > 4z Formula used: (am)n = am × n If am > an and a > 1, then m > n Calculations: 3x > 9y ⇒ 3x > (32)y ⇒ 3x > 32y ⇒ x > 2y 2y > 4z ⇒ 2y > (22)z ⇒ 2y > 22z ⇒ y > 2z Since x, y, z are positive numbers: ⇒ x > 2y > y ⇒ y > 2z > z ⇒ x > y > z ∴ The correct answer is option (1).

Shortcut Trick

For any positive proper fraction x (where 0 < x < 1), the reciprocal 1/x is always greater than 1.

In contrast, the square (x²), sub-multiples (x/2), and square root (√x) of such a fraction will result in values less than 1 (specifically √x < 1).

Given x = 1/8, the value of 1/x = 1 ÷ (1/8) = 8.

Since 8 is the only result greater than 1, it is the largest value.

∴ The correct answer is 8.

Alternate Method

Given: x = 1/8 = 0.125

⇒ Option 1: x/2 = (1/8) ÷ 2 = 1/16 = 0.0625

⇒ Option 2: x² = (1/8) × (1/8) = 1/64 ≈ 0.0156

⇒ Option 3: √x = √(1/8) = 1/(2√2) ≈ 1/2.828 ≈ 0.3535

⇒ Option 4: 1/x = 1 ÷ (1/8) = 8

⇒ Comparing all values: 8 > 0.3535 > 0.0625 > 0.0156

⇒ Clearly, 1/x = 8 is the largest value among the choices.

∴ The correct answer is 8.

Additional Information

Comparison of Proper Fractions

For any fraction x where 0 < x < 1, the following inequality holds true: x² < x < √x < 1.

Reciprocal Property

The reciprocal of a positive proper fraction (1/x) is always greater than 1, whereas the reciprocal of an improper fraction is less than or equal to 1.

Power Behavior

As the exponent n increases, the value of xⁿ decreases if 0 < x < 1, and increases if x > 1.

Shortcut Trick Simplify X: X = (1 + X) ÷ (X + 2) ⇒ X2 + X − 1 = 0. Solve for positive X using quadratic formula: X = (−1 + √5) ÷ 2 = (−2 + 2√5) ÷ 4. Simplify Y: Y = (2 + 2Y) ÷ (2Y + 3) ⇒ 2Y2 + Y − 2 = 0. Solve for positive Y using quadratic formula: Y = (−1 + √17) ÷ 4. Calculate sum: X + Y = (−2 + 2√5 − 1 + √17) ÷ 4 = (2√5 + √17 − 3) ÷ 4. ∴ The correct answer is (2√5 + √17 − 3)/4. Alternate Method Given: X = 1 / (1 + 1 / (1 + X)) Y = 2 / (2 + 1 / (1 + Y)) Formula Used: For a quadratic equation ax2 + bx + c = 0, roots are x = (−b ± √(b2 − 4ac)) / 2a Calculation: For X: ⇒ X = 1 ÷ ((1 + X + 1) ÷ (1 + X)) ⇒ X = (1 + X) ÷ (X + 2) ⇒ X(X + 2) = 1 + X ⇒ X2 + 2X − 1 − X = 0 ⇒ X2 + X − 1 = 0 ⇒ X = (−1 ± √(12 − 4 × 1 × (−1))) ÷ (2 × 1) ⇒ X = (−1 ± √5) ÷ 2 For Y: ⇒ Y = 2 ÷ ((2(1 + Y) + 1) ÷ (1 + Y)) ⇒ Y = 2(1 + Y) ÷ (2Y + 2 + 1) ⇒ Y = (2 + 2Y) ÷ (2Y + 3) ⇒ Y(2Y + 3) = 2 + 2Y ⇒ 2Y2 + 3Y − 2 − 2Y = 0 ⇒ 2Y2 + Y − 2 = 0 ⇒ Y = (−1 ± √(12 − 4 × 2 × (−2))) ÷ (2 × 2) ⇒ Y = (−1 ± √17) ÷ 4 Finding X + Y (taking positive roots for both): ⇒ X + Y = [(−1 + √5) ÷ 2] + [(−1 + √17) ÷ 4] ⇒ X + Y = [2(−1 + √5) + (−1 + √17)] ÷ 4 ⇒ X + Y = (−2 + 2√5 − 1 + √17) ÷ 4 ⇒ X + Y = (2√5 + √17 − 3) ÷ 4 ∴ The correct answer is (2√5 + √17 − 3)/4. Additional Information Quadratic Formula The roots of any quadratic equation ax2 + bx + c = 0 are calculated using x = (−b ± √D) / 2a, where D = b2 − 4ac. Continued Fractions Continued fractions that are self-referential can be solved by substituting the recursive part with the variable itself and solving the resulting algebraic equation. Discriminant and Nature of Roots If D > 0 and not a perfect square, the roots are real, irrational, and distinct. These roots often appear in conjugate pairs like p ± √q.

Given:

P = 2²⁹ × 3²¹ × 5⁸

Q = 2²⁷ × 3²¹ × 5⁸

R = 2²⁶ × 3²² × 5⁸

S = 2²⁵ × 3²² × 5⁹

Formula used:

To compare numbers, divide each by the common factor: 2^a × 3^b × 5^c

Calculation:

Highest Common Factor = 2²⁵ × 3²¹ × 5⁸

Divide P, Q, R, and S by the Common Factor:

⇒ P = $2^{29-25} \times 3^{21-21} \times 5^{8-8}$

⇒ P = 2⁴ = 16

⇒ Q = $2^{27-25} \times 3^{21-21} \times 5^{8-8}$

⇒ Q = 2² = 4

⇒ R = $2^{26-25} \times 3^{22-21} \times 5^{8-8}$

⇒ R = 2¹ × 3¹ = 6

⇒ S = $2^{25-25} \times 3^{22-21} \times 5^{9-8}$

⇒ S = 3¹ × 5¹ = 15

Comparing the simplified values:

⇒ 16 > 15 > 6 > 4

⇒ P > S > R > Q

∴ The correct answer is option (1).

Shortcut Trick

Let x = (A + B) and y = (A − B), then (x − y) = 2B and xy = (A² − B²).

The expression becomes x³ − y³ − 3(x − y)xy, which simplifies to (x − y)³.

Substitute (x − y) = 2B: (2B)³ = 8B³.

Given B = 8, Value = 8 × 8³ = 8⁴ = 4096.

∴ The correct answer is 4096.

Alternate Method

Given: A = 125, B = 8

Formula: (x − y)³ = x³ − y³ − 3xy(x − y)

⇒ Expansion of (A + B)³ = A³ + 3A²B + 3AB² + B³

⇒ Expansion of (A − B)³ = A³ − 3A²B + 3AB² − B³

⇒ Difference (A + B)³ − (A − B)³ = 6A²B + 2B³

⇒ Third term −6B(A² − B²) = −6A²B + 6B³

⇒ Total Expression = (6A²B + 2B³) − 6A²B + 6B³ = 8B³

⇒ Substitute value of B: 8 × (8)³ = 8 × 512 = 4096

∴ The correct answer is 4096.

Additional Information

Algebraic Cube Identity

(a − b)³ = a³ − b³ − 3ab(a − b). This is useful for simplifying expressions involving cubes and differences.

Difference of Sum and Difference Cubes

(a + b)³ − (a − b)³ = 6a²b + 2b³, while (a + b)³ + (a − b)³ = 2a³ + 6ab².

Squares Identity

(a + b)(a − b) = a² − b². Often used to substitute terms in cubic expansions.

Shortcut Trick

Identify prime bases: 125 = 5³ and 243 = 3⁵.

Compare Y^zx = 5³ ⇒ Y = 5 and z^x = 3. Since x, z are natural numbers, z = 3 and x = 1.

Compare Z^yx = 3⁵ ⇒ Z = 3 and y^x = 5. Since x = 1, y = 5.

Required value = 9x + 10y − 18z = 9(1) + 10(5) − 18(3) = 9 + 50 − 54 = 5.

∴ The correct answer is 5.

Alternate Method

Given Equations:

1. X^yz = 1

2. Y^zx = 125

3. Z^yx = 243

Where x, y, and z are natural numbers {1, 2, 3, ...}.

From (2): Y^zx = 5³

⇒ Possible values: (Y = 5, z^x = 3) or (Y = 125, z^x = 1).

Case 1: If z^x = 3, then z = 3 and x = 1 (since 3 is prime).

From (3): Z^yx = 243

⇒ Z^y1 = 3⁵ (Substituting x = 1)

⇒ Z = 3 and y = 5.

Check with (1): X⁵³ = 1 ⇒ X = 1 (Satisfied as X is a natural number).

Now, substitute x = 1, y = 5, z = 3 in the required expression:

⇒ 9x + 10y − 18z

⇒ 9(1) + 10(5) − 18(3)

⇒ 9 + 50 − 54

⇒ 59 − 54

⇒ 5

∴ The correct answer is 5.

Additional Information

Laws of Indices

For any base 'a', a^m = aⁿ ⇒ m = n (if a ≠ 1). Also, a⁰ = 1 and 1ⁿ = 1 for any n.

Prime Factorization

Expressing a composite number as a product of primes helps in solving exponential equations, e.g., 243 = 3 × 3 × 3 × 3 × 3 = 3⁵.

Natural Numbers

Natural numbers (N) are positive integers starting from 1 {1, 2, 3, ...}. This constraint often limits the possible integer solutions for exponents.

Shortcut Trick

Identify a linear combination of the three equations to eliminate variable 'z'.

Target expression 75x + 113y = (Equation 1 + Equation 2 + 12 × Equation 3) ÷ 3.

Substitute values: (20/3 + 17/3 + 12 × 113/9) ÷ 3 = (37/3 + 452/3) ÷ 3.

Final calculation: (489/3) ÷ 3 = 163 ÷ 3 = 163/3.

∴ The correct answer is 163/3.

Alternate Method

Given Equations:

(1) 3x + 6y + 9z = 20/3

(2) 6x + 9y + 3z = 17/3

(3) 18x + 27y − z = 113/9

Multiply Equation (3) by 9 and add to Equation (1) to eliminate 'z':

⇒ (3x + 6y + 9z) + 9(18x + 27y − z) = 20/3 + 9(113/9)

⇒ 3x + 6y + 9z + 162x + 243y − 9z = 20/3 + 113

⇒ 165x + 249y = (20 + 339) ÷ 3 = 359/3 .... (Eq 4)

Multiply Equation (3) by 3 and add to Equation (2) to eliminate 'z':

⇒ (6x + 9y + 3z) + 3(18x + 27y − z) = 17/3 + 3(113/9)

⇒ 6x + 9y + 3z + 54x + 81y − 3z = 17/3 + 113/3

⇒ 60x + 90y = 130/3 .... (Eq 5)

Add Equation 4 and Equation 5:

⇒ (165x + 249y) + (60x + 90y) = 359/3 + 130/3

⇒ 225x + 339y = 489/3 = 163

⇒ 3(75x + 113y) = 163

⇒ 75x + 113y = 163/3

∴ The correct answer is 163/3.

Additional Information

Linear Equations in Three Variables

A system of three linear equations with three variables can be solved by eliminating one variable at a time to reduce the system to two variables, then one.

Method of Elimination

This method involves adding or subtracting equations to cancel out a variable. Multiplying equations by a constant helps align the coefficients for cancellation.

Linear Combination Shortcut

In many competitive exams, the target expression is a specific linear combination (a×Eq1 + b×Eq2 + c×Eq3) of the given equations, designed to eliminate unwanted variables quickly.

​

Shortcut Trick

Inradius r = Area ÷ s, where s = semi-perimeter

s = (12 + 15 + 21)/2 = 24

Area = √[24 × 12 × 9 × 3] = 36√6 cm2

r = 36√6 ÷ 24 = (3√6)/2 cm

∴ The correct answer is (3√6)/2 cm.

Alternate Method

Let AB = 12 cm, BC = 15 cm and CA = 21 cm

Let the inradius be r

Area of triangle ABC using Herons formula

s = (a + b + c)/2 = (12 + 15 + 21)/2 = 24

Area = [s (s - a) (s - b) (s - c)]1/2

⇒ [24 × (24 - 12) × (24 - 15) × (24 - 21)]1/2

⇒ [24 × 12 × 9 × 3]1/2 = 36√6 cm2

Area of the triangle ABC = Sum of area of triangles AOC, COB and AOB

36√6 = ½ × 12 × r + ½ × 15 × r + ½ × 21 × r [∵ Area of the triangle = ½ × base × height]

36√6 = 24r

r = (36√6)/24 = (3√6)/2 cm

Additional Information

Inradius of a Right-angled Triangle

For a right-angled triangle with base b, perpendicular p, and hypotenuse h, the inradius is given by r = (p + b − h) ÷ 2.

Circumradius of a Triangle

The circumradius (R) of any triangle with sides a, b, and c is given by R = (abc) ÷ (4 × Area).

Euler's Theorem in Geometry

The distance (d) between the incenter and circumcenter of a triangle is d = √[R(R − 2r)], where R is circumradius and r is inradius.

Shortcut Trick

Length of median AX = 1/2 √(2AB² + 2AC² − BC²)

Centroid O property: OX = 1/3 × Median AX

⇒ AX = 1/2 √(2 × 12² + 2 × 15² − 18²) = 1/2 √(288 + 450 − 324) = √414 / 2

⇒ OX = (√414 / 2) ÷ 3 = √414 / 6 = (3√46) / 6 = √46 / 2 = (√23)/(√2)

∴ The correct answer is (√23)/(√2).

Alternate Method

Given: AB = 12, BC = 18, AC = 15. AX and BY are medians. O is the centroid.

Formula: Apollonius Theorem, AB² + AC² = 2(AX² + BX²)

⇒ In triangle ABC, X is the midpoint of BC ∴ BX = 18 ÷ 2 = 9

⇒ By Apollonius Theorem: AB² + AC² = 2(AX² + BX²)

⇒ 12² + 15² = 2(AX² + 9²)

⇒ 144 + 225 = 2(AX² + 81)

⇒ 369 = 2AX² + 162

⇒ 2AX² = 369 − 162 = 207

⇒ AX² = 207 ÷ 2 ⇒ AX = √(207/2) = 3√(23/2)

⇒ Since O is the centroid, it divides AX in the ratio AO : OX = 2 : 1

⇒ OX = 1/3 × AX = 1/3 × 3√(23/2) = √(23/2) = (√23)/(√2)

∴ The correct answer is (√23)/(√2).

Additional Information

Apollonius Theorem

For any triangle with sides a, b, c and median m_a to side a: b² + c² = 2(m_a² + (a/2)²).

Centroid Division Property

The centroid (intersection of medians) divides each median in a 2:1 ratio from the vertex to the midpoint of the opposite side.

Sum of Squares of Medians

The relationship between medians (m_a, m_b, m_c) and sides (a, b, c) is given by: 3(a² + b² + c²) = 4(m_a² + m_b² + m_c²).

Shortcut Trick

In any triangle, if the angle bisector is also the median, the triangle must be isosceles (PQ = PR = 12 cm).

The angle bisector PX also acts as the altitude (PX ⊥ QR).

Base QR = 2 × QX = 2 × 3 = 6 cm.

Height PX = √(12² − 3²) = √(144 − 9) = √135 = 3√15 cm.

Area = 1/2 × base × height = 1/2 × 6 × 3√15 = 9√15 cm².

∴ The correct answer is 9√15 cm².

Alternate Method

Given: PQ = 12 cm, QX = 3 cm, PX bisects QR, PX is angle bisector of ∠P.

Formula Used: Internal Angle Bisector Theorem (PQ/PR = QX/XR) and Area = 1/2 × base × height.

⇒ By Angle Bisector Theorem: PQ / PR = QX / XR

⇒ Since PX bisects QR, QX = XR = 3 cm

⇒ 12 / PR = 3 / 3

⇒ PR = 12 cm

⇒ In ΔPQX, using Pythagoras Theorem: PX² + QX² = PQ²

⇒ PX² + 3² = 12²

⇒ PX² + 9 = 144

⇒ PX² = 135 ⇒ PX = √135 = 3√15 cm

⇒ QR = QX + XR = 3 + 3 = 6 cm

⇒ Area of ΔPQR = 1/2 × QR × PX

⇒ Area = 1/2 × 6 × 3√15

⇒ Area = 3 × 3√15 = 9√15 cm²

∴ The correct answer is 9√15 cm².

Additional Information

Internal Angle Bisector Theorem

The bisector of an interior angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle: BD/DC = AB/AC.

Isosceles Triangle Properties

In an isosceles triangle, the altitude, median, and angle bisector drawn from the vertex angle to the base all coincide on the same line segment.

Area by Heron's Formula

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

Shortcut Trick

Given PT : TS : SR = 2 : 1 : 1 and RS = 8 cm ⇒ TS = 8 cm, PT = 16 cm.

In ΔRSU, 6² + 8² = 10² (100 = 100), so ΔRSU is right-angled at S.

Since SU || TQ, ΔRSU ≈ ΔRTQ ⇒ TQ/SU = RT/RS = (8+8)/8 = 2 ⇒ TQ = 12 cm.

In right-angled ΔPTQ, PQ = √(PT² + TQ²) = √(16² + 12²) = 20 cm.

∴ The correct answer is 20 cm.

Alternate Method

Given: PT : TS : SR = 2 : 1 : 1, RS = 8 cm, RU = 10 cm, SU = 6 cm, SU || TQ.

Formula: Pythagoras Theorem: a² + b² = c² | Similarity: Corresponding sides ratio.

⇒ From PT : TS : SR = 2 : 1 : 1 and RS = 8 cm:

⇒ TS = 8 cm and PT = 16 cm.

⇒ In ΔRSU, RS² + SU² = 8² + 6² = 64 + 36 = 100.

⇒ RU² = 10² = 100. Since RS² + SU² = RU², ∠RSU = 90°.

⇒ Since SU || TQ, corresponding ∠RTQ = ∠RSU = 90°.

⇒ ΔRSU ≈ ΔRTQ (By AA Similarity).

⇒ TQ / SU = RT / RS ⇒ TQ / 6 = (8 + 8) / 8 = 16 / 8 = 2.

⇒ TQ = 6 × 2 = 12 cm.

⇒ In right-angled ΔPTQ, ∠PTQ = 90° (Linear pair with ∠RTQ).

⇒ PQ² = PT² + TQ² = 16² + 12² = 256 + 144 = 400.

⇒ PQ = √400 = 20 cm.

∴ The correct answer is 20 cm.

Additional Information

Pythagorean Triplets

Common triplets like (3, 4, 5), (6, 8, 10), and (5, 12, 13) satisfy the equation a² + b² = c² and form right-angled triangles.

Basic Proportionality Theorem (Thales Theorem)

If a line is parallel to one side of a triangle intersecting the other two sides, it divides those sides in the same ratio.

Similarity Ratios

For two similar triangles, the ratio of any two corresponding linear dimensions (sides, altitudes, medians) is always equal.

Shortcut Trick

Identify half-lengths of the chords: a = 20/2 = 10 cm and b = 48/2 = 24 cm.

Identify a Pythagorean triplet involving these values: (10, 24, 26).

Sum of distances from center = 10 + 24 = 34 cm, which matches the given distance.

Thus, the radius (r) of the circle is 26 cm.

Area of circle = π × r² = π × 26² = 676π cm².

∴ The correct answer is 676π cm².

Alternate Method

Given: Chord PQ = 20 cm, Chord RS = 48 cm, Distance between them = 34 cm.

Formula: r² = (chord/2)² + (distance from center)²; Area = πr²

⇒ Let O be the center and x be the distance from center to RS.

⇒ In ▵ORN: r² = x² + (48/2)² = x² + 24² = x² + 576

⇒ Distance from center to PQ = 34 − x

⇒ In ▵OPM: r² = (34 − x)² + (20/2)² = (34 − x)² + 100

⇒ Equating r²: x² + 576 = 1156 − 68x + x² + 100

⇒ 576 = 1256 − 68x

⇒ 68x = 1256 − 576 = 680

⇒ x = 10 cm

⇒ r² = 10² + 576 = 100 + 576 = 676

⇒ Area of circle = πr² = 676π cm²

∴ The correct answer is 676π cm².

Additional Information

Perpendicular Bisector Property

A perpendicular drawn from the center of a circle to a chord always bisects the chord into two equal parts.

Parallel Chords on Same Side

If two parallel chords of length 2a and 2b are on the same side of the center, the distance between them is |√(r²−a²) − √(r²−b²)|.

Area of Circle Formula

The area of a circle with radius r is given by the formula A = πr².

Pythagorean Triplet Logic

For quick calculation in circles, use triplets: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (10,24,26), (20,21,29).

Shortcut Trick

Ratio of Areas = 616 : 154 = 4 : 1

Ratio of Radii (R : r) = √4 : √1 = 2 : 1

Since Area 154 → r = 7 cm, then R = 14 cm

Half tangent length AB = √(R² − r²) = √(14² − 7²) = √147 = 7√3 cm

Full length AC = 2 × AB = 14√3 cm

∴ The correct answer is 14√3 cm.

Alternate Method

Given: Area of larger circle = 616 cm², Area of smaller circle = 154 cm².

Formula Used: Area of circle = πr²; Pythagoras theorem: a² + b² = c².

⇒ Area of larger circle = πR² = 616

⇒ (22/7) × R² = 616

⇒ R² = (616 × 7) ÷ 22 = 28 × 7 = 196

⇒ R = 14 cm

⇒ Area of smaller circle = πr² = 154

⇒ (22/7) × r² = 154

⇒ r² = (154 × 7) ÷ 22 = 7 × 7 = 49

⇒ r = 7 cm

⇒ In right-angled triangle OBA (where OB ⊥ AC at B):

⇒ AB² = OA² − OB² = R² − r²

⇒ AB² = 196 − 49 = 147

⇒ AB = √147 = √(49 × 3) = 7√3 cm

⇒ Since B bisects AC, AC = 2 × AB = 2 × 7√3 = 14√3 cm

∴ The correct answer is 14√3 cm.

Additional Information

Tangent Property

A tangent at any point of a circle is perpendicular to the radius through the point of contact (∠OBA = 90°).

Chord Properties in Concentric Circles

A chord of the larger circle that touches the smaller circle is bisected at the point of contact.

Area and Radius Relationship

The ratio of the areas of two circles is equal to the ratio of the squares of their radii (A₁/A₂ = R²/r²).

Shortcut Trick

By similarity of triangles formed by radii and common tangents, PN/PO = R/r = 5/3.

Difference in ratio parts = 5 − 3 = 2 units, which corresponds to ON = 12 cm.

1 unit = 6 cm ⇒ Distance PN = 5 × 6 = 30 cm.

In right-angled ▵PNY, PY = √(PN² − NY²) = √(30² − 5²).

PY = √(900 − 25) = √875 = 5√35 cm.

∴ The correct answer is 5√35 cm.

Alternate Method

Given: Radius of smaller circle (r) = 3 cm, Radius of larger circle (R) = 5 cm, Distance between centers (ON) = 12 cm.

Formula: For triangles formed by the intersection of common tangents P, ▵POX ~ ▵PNY. Also, in ▵PNY, PN² = PY² + NY².

⇒ Let PO = x. Since ▵POX and ▵PNY are similar:

⇒ OX / NY = PO / PN

⇒ 3 / 5 = x / (x + 12)

⇒ 3(x + 12) = 5x

⇒ 3x + 36 = 5x

⇒ 2x = 36 ⇒ x = 18 cm

⇒ Total length PN = PO + ON = 18 + 12 = 30 cm

⇒ In right-angled ▵PNY, PY² = PN² − NY²

⇒ PY² = 30² − 5² = 900 − 25 = 875

⇒ PY = √875 = √(25 × 35) = 5√35 cm

∴ The correct answer is 5√35 cm.

Additional Information

Direct Common Tangent (DCT)

The length of the direct common tangent is given by √(D² − (R−r)²), where D is the distance between centers.

Transverse Common Tangent (TCT)

The length of the transverse common tangent is given by √(D² − (R+r)²).

Intersection of Tangents

The external common tangents intersect at a point P that divides the line joining the centers externally in the ratio of the radii (R : r).

Shortcut Trick

In a circle, the radius is perpendicular to the tangent at the point of contact ∴ ∠ORX = 90°.

Given ∠XRP = 120°, then ∠ORP = ∠XRP − ∠ORX = 120° − 90° = 30°.

In ΔOPR, OP = OR (radii), so it is an isosceles triangle ⇒ ∠OPR = ∠ORP = 30°.

Exterior angle ∠QOR of ΔOPR is the sum of opposite interior angles ⇒ ∠QOR = ∠OPR + ∠ORP = 30° + 30° = 60°.

∴ The correct answer is 60°.

Alternate Method

Given:

XR is a tangent to the circle at point R.

O is the centre of the circle, and PQ is the diameter.

∠XRP = 120°.

Formula Used:

1. Radius ⊥ Tangent at point of contact (∠ORX = 90°).
2. Angles opposite to equal sides in a triangle are equal.
3. Exterior Angle Theorem: Exterior angle = Sum of two opposite interior angles.

Calculations:

⇒ ∠ORX = 90° (Radius is perpendicular to tangent at contact point R)

⇒ ∠ORP = ∠XRP − ∠ORX

⇒ ∠ORP = 120° − 90° = 30°

⇒ In ΔOPR, OR = OP (Both are radii of the same circle)

⇒ ∠OPR = ∠ORP = 30° (Angles opposite to equal sides are equal)

⇒ ∠QOR is the exterior angle to ΔOPR at vertex O

⇒ ∠QOR = ∠OPR + ∠ORP

⇒ ∠QOR = 30° + 30° = 60°

∴ The correct answer is 60°.

Additional Information

Radius-Tangent Theorem

The tangent at any point of a circle is perpendicular to the radius through the point of contact (∠Radius, Tangent = 90°).

Alternate Segment Theorem

The angle between a tangent and a chord through the point of contact is equal to the angle subtended by the chord in the alternate segment.

Exterior Angle Property

An exterior angle of a triangle is equal to the sum of its two interior opposite angles (∠Ext = ∠Int₁ + ∠Int₂).

Central Angle Property

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

Shortcut Trick

According to the Alternate Segment Theorem, the angle between a tangent and a chord (∠BCX) is equal to the angle subtended by the chord in the alternate segment (∠BAC).

In ΔAOC, OA = OC (Radii), so it is an isosceles triangle.

∠OAC = (180° − ∠AOC) ÷ 2 = (180° − 80°) ÷ 2 = 100° ÷ 2 = 50°.

∠BAC = ∠OAC = 50°.

By theorem, ∠BCX = ∠BAC = 50°.

∴ The correct answer is 50°.

Alternate Method

Given:

∠AOC = 80°

Formula used:

Angle at the centre = 2 × Angle at the circumference standing on the same chord

Calculation:

AC is a chord of the circle.

⇒ ∠ABC = 1/2 × ∠AOC

⇒ ∠ABC = 1/2 × 80°

⇒ ∠ABC = 40°

In △OBC,

⇒ OB = OC = Radius of the circle

⇒ ∠OBC = ∠OCB = 40°

⇒ ∠XCO = 90°

Radius is perpendicular to the tangent at the point of contact.

⇒ ∠BCX = ∠XCO − ∠OCB

⇒ ∠BCX = 90° − 40°

⇒ ∠BCX = 50°

∴ ∠BCX = 50°.

Additional Information

Alternate Segment Theorem

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Tangent-Radius Property

A tangent to a circle is perpendicular to the radius at the point of contact (∠OCX = 90°).

Angle at Centre and Circumference

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

Shortcut TrickTransverse common tangent = $√{d^2 - (r1 + r2)^2}$

Transverse common tangent = $√{24^2 - (4 + 8)^2}$

Transverse common tangent = $√{576 - 144}$ = √432

Transverse common tangent = 12√3

Direct Common Tangent = $√{d^2 - (r1 - r2)^2}$

Direct Common Tangent = $√{24^2 - (8 - 4)^2}$

Direct Common Tangent = $√{576 - 16}$

Direct Common Tangent = √560 

Direct Common Tangent = 4√35

Sum = Transverse common tangent + Direct Common Tangent

Sum = 12√3 + 4√35 = 4(√35 + 3√3)

CD is the transverse common tangent and AB in the direct common tangent

Finding CD:

In triangle OCQ and PDQ

∠OCQ = ∠PDQ = 90° [Tangent is perpendicular to the radius]

∠OQC = ∠PQD [Vertically opposite angles are equal]

∴ Triangle OCQ ∼ PDQ

OC/PD = OQ/PQ = CQ/DQ

8/4 = OQ/PQ

∴ OQ = 2PQ

OP = OQ + PQ = 3PQ

24 = 3PQ

∴ PQ = 8 cm

OQ = 2 × 8 = 16 cm

Applying Pythagoras theorem for triangles OCQ

CQ = √(OQ2 - OC2) = √(162 - 82) = √192 = 8√3 cm

DQ = ½ × CQ = 4√3 cm

∴ CD = CQ + DQ = 12√3 cm

Finding AB:

Let us extend the lines AB and OP and let them meet at M

In triangle MBP and MAO

∠M is common

∠MBP = ∠MAO = 90° [Tangent is perpendicular to the radius]

∴ MBP ∼ MAO

BP/AO = MB/MA = MP/MO

4/8 = MB/MA = MP/MO

∴ MA = 2 MB and MO = 2 MP

OP = MO - MP = MP = 24 cm

Applying Pythagoras theorem to triangle MBP

MB = √(MP2 - PB2)

⇒ √(242 - 42) = √560 = 4√35 cm

MA = 2 MB = 8√35 cm

AB = MA - MB = 4√35 cm

Sum of the lengths = CD + AB = 12√3 + 4√35 = 4(√35 + 3√3) cm

Additional Information

Direct Common Tangent (DCT)

A tangent that lies on the same side of the line segment joining the centers. Formula: √[d² − (R − r)²].

Transverse Common Tangent (TCT)

A tangent that intersects the line segment joining the centers. Formula: √[d² − (R + r)²].

Number of Common Tangents

If d > R + r, there are 4 common tangents (2 DCT, 2 TCT). If d = R + r, there are 3. If R − r < d < R + r, there are 2 (only DCT).

Shortcut Trick

In right-angled triangle ABD, AB = 10 cm and AD = 8 cm. Using Pythagoras triplet (6, 8, 10), BD = 6 cm.

D is a point on BC (since AD ⊥ BC), thus DC = BC − BD = 16 − 6 = 10 cm.

In right-angled triangle ADC, AC = √(AD² + DC²) = √(8² + 10²).

AC = √(64 + 100) = √164 = 2√41 cm.

∴ The correct answer is 2√41 cm.

Alternate Method

Given: AB = 10 cm, BC = 16 cm, Altitude AD = 8 cm where AD ⊥ BC.

Formula Used: Pythagoras Theorem: Hypotenuse² = Perpendicular² + Base²

⇒ In right-angled ΔABD:

⇒ AB² = AD² + BD²

⇒ 10² = 8² + BD²

⇒ 100 = 64 + BD²

⇒ BD² = 100 − 64 = 36

⇒ BD = √36 = 6 cm

⇒ Now, DC = BC − BD

⇒ DC = 16 − 6 = 10 cm

⇒ In right-angled ΔADC:

⇒ AC² = AD² + DC²

⇒ AC² = 8² + 10²

⇒ AC² = 64 + 100 = 164

⇒ AC = √164 = √(4 × 41) = 2√41 cm

∴ The correct answer is 2√41 cm.

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: c² = a² + b².

Common Pythagorean Triplets

Standard sets of integers that satisfy the theorem include (3, 4, 5), (5, 12, 13), (8, 15, 17), and their multiples like (6, 8, 10).

Altitude of a Triangle

An altitude is a line segment through a vertex and perpendicular to the opposite side, dividing the triangle into two right-angled triangles.

Shortcut Trick

For an equilateral triangle of side a, the height h = (√3/2)a.

The side of the largest inscribed square x = (a × h) / (a + h).

Given a = 12 cm, h = (√3/2) × 12 = 6√3 cm.

x = (12 × 6√3) / (12 + 6√3) = 72√3 / (6(2 + √3)) = 12√3 / (2 + √3).

Rationalizing: x = 12√3(2 − √3) = 24√3 − 36 cm.

Area = (24√3 − 36)² = 1728 + 1296 − 1728√3 = 3024 − 1728√3 cm².

∴ The correct answer is 3024 − 1728√3 cm².

Alternate Method

Given: Side of equilateral triangle (a) = 12 cm.

Formula Used: Side of largest square (x) = (Base × Height) / (Base + Height).

⇒ Height of equilateral triangle (h) = (√3/2) × Side = (√3/2) × 12 = 6√3 cm.

⇒ Side of square (x) = (12 × 6√3) / (12 + 6√3).

⇒ x = 72√3 / [6 × (2 + √3)] = 12√3 / (2 + √3).

⇒ x = [12√3 × (2 − √3)] / [(2 + √3) × (2 − √3)].

⇒ x = 24√3 − 36 cm.

⇒ Area of square = x² = (24√3 − 36)².

⇒ Area = (24√3)² + (36)² − 2 × 24√3 × 36.

⇒ Area = (576 × 3) + 1296 − 1728√3.

⇒ Area = 1728 + 1296 − 1728√3 = 3024 − 1728√3 cm².

∴ The correct answer is 3024 − 1728√3 cm².

Additional Information

General Inscribed Square Formula

For any triangle with base 'b' and height 'h', the side of the largest inscribed square resting on that base is x = (b × h) / (b + h).

Equilateral Triangle Properties

Height = (√3/2)a, Area = (√3/4)a², In-radius = a/(2√3), and Circum-radius = a/√3.

Square in Right Triangle

For a right triangle with legs P and B and hypotenuse H, the side of the largest square resting on the legs is x = (P × B) / (P + B).

Shortcut Trick

Ratio of sides = 3 : 1

Ratio of diagonal squared = 3² + 1² = 10 units

Actual diagonal squared = 10² = 100 cm²

Scale factor for area = 100 ÷ 10 = 10

Required Area = (3 × 1) ratio parts × 10 = 30 cm²

∴ The correct answer is 30 cm².

Alternate Method

Given: Ratio PQ : QR = 3 : 1 and Diagonal PR = 10 cm

Formula: Diagonal² = Length² + Width² | Area = Length × Width

⇒ Let side PQ = 3x and side QR = x.

⇒ In right-angled ΔPQR, PR² = PQ² + QR²

⇒ 10² = (3x)² + (x)²

⇒ 100 = 9x² + x²

⇒ 100 = 10x²

⇒ x² = 10

⇒ Area of rectangle PQRS = PQ × QR = 3x × x = 3x²

⇒ Area = 3 × 10 = 30 cm²

∴ The correct answer is 30 cm².

Additional Information

Diagonal of a Rectangle

The diagonal d of a rectangle with length l and breadth b is given by d = √(l² + b²).

Area and Perimeter

Area = Length × Breadth, and Perimeter = 2 × (Length + Breadth).

Pythagorean Theorem

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: a² + b² = c².

Shortcut Trick

In a regular hexagon with side 'a', the area can be divided into 6 equal equilateral triangles of area 'k'.

The area of triangle ACE (formed by connecting alternate vertices) is exactly half the area of the hexagon.

Area of ΔACE = 3 × k

The area of the peripheral triangle AEF = 1 × k (since three such triangles AEF, ABC, and CDE plus ΔACE make the whole hexagon).

Ratio = Area(ΔACE) : Area(ΔAEF) = 3k : 1k = 3 : 1

∴ The correct answer is 3 : 1.

Alternate Method

Given: ABCDEF is a regular hexagon with side length 'a'.

Formula Used: Area of triangle = (1/2) × a × b × sin(θ) | Area of equilateral triangle = (√3/4) × side²

⇒ Interior angle of a regular hexagon = (6 − 2) × 180° ÷ 6 = 120°.

⇒ In ΔAEF, side AF = a, side FE = a, and ∠AFE = 120°.

⇒ Area of ΔAEF = (1/2) × a × a × sin(120°) = (1/2) × a² × (√3/2) = (√3/4)a².

⇒ In ΔAEF, by Cosine Rule, side AE² = a² + a² − 2 × a × a × cos(120°).

⇒ AE² = 2a² − 2a² × (−1/2) = 2a² + a² = 3a² ⇒ AE = a√3.

⇒ ΔACE is an equilateral triangle with side length s = a√3.

⇒ Area of ΔACE = (√3/4) × (a√3)² = (√3/4) × 3a² = (3√3/4)a².

⇒ Ratio = Area(ΔACE) : Area(ΔAEF) = [(3√3/4)a²] : [(√3/4)a²].

⇒ Ratio = 3 : 1.

∴ The correct answer is 3 : 1.

Additional Information

Area of Regular Hexagon

The area of a regular hexagon with side 'a' is given by (3√3/2) × a².

Diagonals of a Hexagon

A regular hexagon has two types of diagonals: Short diagonal = a√3 (e.g., AE) and Long diagonal = 2a (e.g., AD).

Symmetry of ΔACE

Triangle ACE is an equilateral triangle because all its sides are short diagonals of the regular hexagon (AC = CE = EA = a√3).

Interior and Exterior Angles

For a regular n-sided polygon, Interior Angle = [(n−2)×180]/n and Exterior Angle = 360/n.

Shortcut Trick

For equal perimeters, the sum of non-parallel sides in the top part plus the top base must equal the sum of non-parallel sides in the bottom part plus the bottom base.

Let the ratio BE/EC = AF/FD = k. The shared line EF cancels out on both sides.

Perimeter condition: AB + BE + AF = CD + EC + FD

Substitute values: 6 + [k/(k+1)] × (BC + AD) = 18 + [1/(k+1)] × (BC + AD)

6 + [k/(k+1)] × 20 = 18 + [1/(k+1)] × 20

20(k − 1) / (k + 1) = 12 ⇒ 5k − 5 = 3k + 3 ⇒ 2k = 8 ⇒ k = 4.

∴ The correct answer is 4.

Alternate Method

Given: AB = 6 cm, CD = 18 cm, BC = 8 cm, AD = 12 cm, EF ∥ AB.

Let BE/EC = AF/FD = k. This implies:

⇒ BE = 8k / (k + 1) and EC = 8 / (k + 1)

⇒ AF = 12k / (k + 1) and FD = 12 / (k + 1)

Since the perimeter of ABEF = Perimeter of FECD:

⇒ AB + BE + EF + FA = FE + EC + CD + DF

⇒ AB + BE + AF = EC + CD + FD

⇒ 6 + [8k / (k + 1)] + [12k / (k + 1)] = [8 / (k + 1)] + 18 + [12 / (k + 1)]

⇒ 6 + [20k / (k + 1)] = 18 + [20 / (k + 1)]

⇒ [20k / (k + 1)] − [20 / (k + 1)] = 18 − 6

⇒ 20(k − 1) / (k + 1) = 12

⇒ 5(k − 1) = 3(k + 1)

⇒ 5k − 5 = 3k + 3

⇒ 2k = 8

⇒ k = 4

∴ The correct answer is 4.

Additional Information

Length of Parallel Segment EF

If EF divides the non-parallel sides in ratio k:1 (where k = AF/FD), its length is given by EF = (k×CD + 1×AB) / (k + 1).

Area Bisector Parallel Line

A line parallel to the bases that divides a trapezium into two equal areas has a length of √[(AB² + CD²) / 2].

Median of a Trapezium

The segment joining the midpoints of the non-parallel sides is parallel to the bases and its length is the average of the bases: (AB + CD) / 2.

Shortcut Trick

For the minimum area of a circular sheet, the diagonal of the rectangle must be the diameter of the circle.

Radius² (r²) = (Length² + Breadth²) ÷ 4

r² = (42² + 14²) ÷ 4 = (1764 + 196) ÷ 4 = 1960 ÷ 4 = 490

Area = πr² = (22 ÷ 7) × 490 = 22 × 70 = 1540 cm².

∴ The correct answer is 1540 cm².

Alternate Method

Given:

Length of the rectangular sheet (l) = 42 cm

Breadth of the rectangular sheet (b) = 14 cm

Formula Used:

Diagonal of rectangle = √(l² + b²)

Area of circle = π × r²

For the minimum area of the circular sheet, the rectangle should be perfectly inscribed in the circle. Thus, the diagonal of the rectangle serves as the diameter of the circle.

⇒ Diameter (D) = √(42² + 14²)

⇒ D = √(1764 + 196)

⇒ D = √1960 cm

⇒ Radius (r) = D ÷ 2 = (√1960) ÷ 2

⇒ Area of circle = π × r²

⇒ Area = (22 ÷ 7) × [(√1960) ÷ 2]²

⇒ Area = (22 ÷ 7) × (1960 ÷ 4)

⇒ Area = (22 ÷ 7) × 490

⇒ Area = 22 × 70

⇒ Area = 1540 cm²

∴ The correct answer is 1540 cm².

Additional Information

Diagonal of Rectangle

The distance between opposite vertices of a rectangle with length l and breadth b is given by d = √(l² + b²).

Circumcircle of a Rectangle

The radius of a circle circumscribing a rectangle is always half the length of its diagonal: r = d ÷ 2.

Area of Circle

The area enclosed by a circle of radius r is A = πr², where π ≈ 22/7 or 3.14.

Shortcut Trick

In an equilateral triangle of side a, the side of the largest square x = a√3 ÷ (2 + √3).

Large circle radius R = a ÷ √3 and small circle radius r = x ÷ 2.

Ratio of radii r/R = [a√3 ÷ (2(2 + √3))] ÷ [a ÷ √3] = 3 ÷ (2(2 + √3)).

Ratio of Areas = (r/R)² = 9 ÷ [4(7 + 4√3)].

Rationalizing: 9(7 − 4√3) ÷ [4(49 − 48)] = (63 − 36√3) ÷ 4.

∴ The correct answer is (63 − 36√3) : 4.

Alternate Method

Given:

Triangle ABC is equilateral and inscribed in a large circle.

A square is inscribed in the triangle, and a small circle is inscribed in that square.

Formula Used:

Circumradius of equilateral triangle R = a ÷ √3

Side of square in equilateral triangle x = (base × height) ÷ (base + height)

Area of Circle = πr²

Calculations:

Let the side of the equilateral triangle be a.

Height of triangle h = (a√3) ÷ 2

⇒ Side of square x = (a × (a√3 ÷ 2)) ÷ (a + (a√3 ÷ 2))

⇒ x = (a²√3 ÷ 2) ÷ (a(2 + √3) ÷ 2) = a√3 ÷ (2 + √3)

⇒ Radius of small circle r = x ÷ 2 = a√3 ÷ [2(2 + √3)]

⇒ Radius of large circle R = a ÷ √3

⇒ Area of small circle = π × (a√3 ÷ [2(2 + √3)])² = π × [3a² ÷ (4(7 + 4√3))]

⇒ Area of large circle = π × (a ÷ √3)² = π × (a² ÷ 3)

⇒ Ratio = [3a² ÷ (4(7 + 4√3))] ÷ [a² ÷ 3]

⇒ Ratio = 9 ÷ [4(7 + 4√3)]

⇒ Ratio = [9 × (7 − 4√3)] ÷ [4 × (7 + 4√3) × (7 − 4√3)]

⇒ Ratio = (63 − 36√3) ÷ [4 × (49 − 48)] = (63 − 36√3) ÷ 4

∴ The correct answer is (63 − 36√3) : 4.

Additional Information

Circumradius and Inradius

For an equilateral triangle of side a, circumradius R = a/√3 and inradius r = a/(2√3). The ratio of areas of circumcircle to incircle is always 4 : 1.

Square Inscribed in a Triangle

The side of the largest square x that can be inscribed in a triangle with base b and height h (with square base on triangle base) is x = (b × h) / (b + h).

Rationalization of Surds

To rationalize a denominator of the form (a + √b), multiply both numerator and denominator by its conjugate (a − √b) to use the identity (x + y)(x − y) = x² − y².

Shortcut Trick

Sum of TSA of parts = Original TSA + 2 × (Area of Vertical Cut 1) + 2 × (Area of Vertical Cut 2)

Original TSA = 2 × (216√3) + (72 × 24) = 432√3 + 1728 cm²

Area of Cut 1 (through vertices) = 2 × 12 × 24 = 576 cm²

Area of Cut 2 (through midpoints) = 12√3 × 24 = 288√3 cm²

Total = (432√3 + 1728) + 2(576) + 2(288√3) = 2880 + 1008√3 cm²

∴ The correct answer is 2880 + 1008√3 cm².

Alternate Method

Given:

Side of regular hexagon (a) = 12 cm

Height of prism (h) = 24 cm

Formula Used:

Area of Hexagon = (3√3 / 2) × a²

TSA of Prism = 2 × Base Area + Perimeter × Height

Surface Area increase per cut = 2 × Area of the cut face

Calculations:

⇒ Base Area of Hexagon = (3√3 / 2) × 12² = (3√3 / 2) × 144 = 216√3 cm²

⇒ Lateral Surface Area = (6 × 12) × 24 = 72 × 24 = 1728 cm²

⇒ Original TSA = 2 × 216√3 + 1728 = 432√3 + 1728 cm²

⇒ Long diagonal of hexagon = 2 × 12 = 24 cm

⇒ Short diagonal of hexagon = 12√3 cm

⇒ Area of Cut Face 1 (Vertical) = 24 × 24 = 576 cm²

⇒ Area of Cut Face 2 (Vertical) = 12√3 × 24 = 288√3 cm²

⇒ Total Surface Area of 4 parts = Original TSA + 2(Area of Cut 1) + 2(Area of Cut 2)

⇒ Sum of TSA = (432√3 + 1728) + 2(576) + 2(288√3)

⇒ Sum of TSA = 432√3 + 1728 + 1152 + 576√3

⇒ Sum of TSA = 2880 + 1008√3 cm²

∴ The correct answer is 2880 + 1008√3 cm².

Additional Information

Regular Hexagon Properties

A regular hexagon has a long diagonal of 2a and a short diagonal of a√3. Area = (3√3 / 2) × a².

Surface Area Increase on Cutting

Every plane cut through a solid increases the total surface area by exactly twice the cross-sectional area of that cut.

Prism Volume and Area

Volume = Base Area × Height. Lateral Surface Area = Base Perimeter × Height.

Shortcut Trick

Total Surface Area (TSA) of remaining solid = TSA of Cuboid + 4 × (CSA of Cone − Base Area of Cone)

Cuboid TSA = 2(30×32 + 32×40 + 40×30) = 2(960 + 1280 + 1200) = 6880 cm²

Cone Slant Height (l) = √(10.5² + 14²) = √(110.25 + 196) = 17.5 cm

Net change per cone = π × r × l − π × r² = π × r × (l − r) = (22/7) × 10.5 × (17.5 − 10.5) = 33 × 7 = 231 cm²

Total TSA = 6880 + 4 × 231 = 6880 + 924 = 7804 cm²

∴ The correct answer is 7804 cm².

Alternate Method

Given:

Cuboid: Length (L) = 30 cm, Breadth (B) = 32 cm, Height (H) = 40 cm

Cone: Radius (r) = 10.5 cm, Height (h) = 14 cm, Number (n) = 4

Formula Used:

TSA of Cuboid = 2(LB + BH + LH)

CSA of Cone = πrl | Base Area of Cone = πr²

Slant height (l) = √(r² + h²)

Calculations:

⇒ Slant height (l) = √(10.5² + 14²) = √(110.25 + 196) = √306.25 = 17.5 cm

⇒ TSA of cuboid = 2 × (30 × 32 + 32 × 40 + 40 × 30)

⇒ TSA of cuboid = 2 × (960 + 1280 + 1200) = 2 × 3440 = 6880 cm²

⇒ Base Area of 4 cones = 4 × πr² = 4 × (22/7) × 10.5 × 10.5 = 4 × 346.5 = 1386 cm²

⇒ CSA of 4 cones = 4 × πrl = 4 × (22/7) × 10.5 × 17.5 = 4 × 577.5 = 2310 cm²

⇒ TSA of remaining solid = TSA of Cuboid − Base Area of cones + CSA of cones

⇒ TSA = 6880 − 1386 + 2310

⇒ TSA = 6880 + 924 = 7804 cm²

∴ The correct answer is 7804 cm².

Additional Information

Surface Area Modification

When a cavity is removed from a solid, the total surface area increases by the difference between the new internal surface area and the area of the opening: ΔArea = CSA − Base Area.

Cone Relations

Slant height l = √(r² + h²). Volume of cone = (1/3)πr²h. Volume of remaining solid = Volume of cuboid − n × Volume of cone.

Cuboid Formulas

Total Surface Area = 2(lb + bh + hl). Lateral Surface Area = 2h(l + b). Longest diagonal = √(l² + b² + h²).

Shortcut Trick

Inner radius (r) = 3.5 cm and Thickness (t) = 0.7 cm.

Outer radius (R) = 3.5 + 0.7 = 4.2 cm.

Formula for Total Surface Area of hollow cylinder = 2π(R + r)(h + t).

TSA = 2 × (22/7) × (4.2 + 3.5) × (15 + 0.7) = 2 × (22/7) × 7.7 × 15.7.

TSA = 44 × 1.1 × 15.7 = 48.4 × 15.7 = 759.88 cm².

∴ The correct answer is 759.88 cm².

Alternate Method

Given:

Inner radius (r) = 3.5 cm

Thickness (t) = 0.7 cm

Height (h) = 15 cm

Formula Used:

Outer radius (R) = r + t

Total Surface Area (TSA) = Inner CSA + Outer CSA + 2 × Area of base ring

TSA = 2πrh + 2πRh + 2π(R² − r²)

⇒ Outer radius (R) = 3.5 + 0.7 = 4.2 cm

⇒ TSA = 2πh(R + r) + 2π(R + r)(R − r)

⇒ TSA = 2π(R + r) [h + (R − r)]

⇒ TSA = 2 × (22/7) × (4.2 + 3.5) × [15 + (4.2 − 3.5)]

⇒ TSA = 2 × (22/7) × 7.7 × [15 + 0.7]

⇒ TSA = 2 × 22 × 1.1 × 15.7

⇒ TSA = 44 × 1.1 × 15.7

⇒ TSA = 48.4 × 15.7 = 759.88 cm²

∴ The correct answer is 759.88 cm².

Additional Information

Volume of Hollow Cylinder

The capacity of the material used or the space between the walls: V = π(R² − r²)h.

Curved Surface Area (CSA)

The sum of inner and outer lateral surface areas: CSA = 2πRh + 2πrh = 2πh(R + r).

Base Area of Hollow Cylinder

Area of the ring-shaped cross-section: Area = π(R² − r²).

Shortcut Trick

Outer radius R = OCSA ÷ (2 × π × h) = 11880 ÷ (2 × 22/7 × 90) = 21 cm

Inner radius r = √(Volume ÷ (π × h)) = √(55440 ÷ (22/7 × 90)) = √196 = 14 cm

Thickness = Outer Radius (R) − Inner Radius (r) = 21 − 14 = 7 cm

∴ The correct answer is 7 cm.

Alternate Method

Given:

Height (h) = 90 cm

Outer Curved Surface Area (OCSA) = 11880 cm²

Internal Volume (V_in) = 55440 cm³

Formula Used:

Outer Curved Surface Area = 2 × π × R × h

Internal Volume = π × r² × h

Thickness (T) = R − r

Calculations:

Finding Outer Radius (R):

⇒ 2 × (22/7) × R × 90 = 11880

⇒ (44/7) × 90 × R = 11880

⇒ 3960/7 × R = 11880

⇒ R = (11880 × 7) ÷ 3960

⇒ R = 3 × 7 = 21 cm

Finding Inner Radius (r):

⇒ (22/7) × r² × 90 = 55440

⇒ r² = (55440 × 7) ÷ (22 × 90)

⇒ r² = (55440 × 7) ÷ 1980

⇒ r² = 28 × 7 = 196

⇒ r = √196 = 14 cm

Calculating Thickness:

⇒ Thickness = R − r

⇒ Thickness = 21 − 14 = 7 cm

∴ The correct answer is 7 cm.

Additional Information

Volume of Material in Hollow Cylinder

The volume of the material (shell) is given by π × (R² − r²) × h.

Total Surface Area (TSA) of Hollow Cylinder

TSA = Outer CSA + Inner CSA + 2 × Area of Base Ring = 2πRh + 2πrh + 2π(R² − r²).

Base Area of Hollow Cylinder

The area of the circular cross-section (the ring) is π × (R² − r²).

Shortcut Trick

Number of spheres = [R₁³ − r₁³] ÷ [R₂³ − r₂³]

Large Sphere: 6³ − 4³ = 216 − 64 = 152

Small Sphere: 3³ − 2³ = 27 − 8 = 19

Number of spheres = 152 ÷ 19 = 8

∴ The correct answer is 8 spheres.

Alternate Method

Given:

Outer radius of big sphere (R₁) = 6 cm

Inner radius of big sphere (r₁) = 4 cm

Outer radius of small sphere (R₂) = 3 cm

Inner radius of small sphere (r₂) = 2 cm

Formula Used:

Volume of hollow sphere = (4/3) × π × (R³ − r³)

Calculations:

⇒ Let the number of smaller spheres be 'n'.

⇒ Volume of big sphere = n × Volume of small sphere

⇒ (4/3) × π × (6³ − 4³) = n × (4/3) × π × (3³ − 2³)

⇒ (6³ − 4³) = n × (3³ − 2³)

⇒ (216 − 64) = n × (27 − 8)

⇒ 152 = n × 19

⇒ n = 152 ÷ 19

⇒ n = 8

∴ The correct answer is 8 spheres.

Additional Information

Volume of a Solid Sphere

The volume is given by V = (4/3) × π × R³, where R is the radius of the sphere.

Surface Area of a Sphere

The total surface area of a sphere is 4 × π × R².

Hollow Sphere Logic

A hollow sphere's material volume is the difference between the outer volume and the inner volume: (4/3)π(R³ − r³).

Shortcut Trick

Outer Surface Area = Total Cost ÷ Rate = 2464 ÷ 8 = 308 m².

2 × (22/7) × R² = 308 ⇒ R² = 49 ⇒ Outer Radius (R) = 7 m.

Inner Radius (r) = R − thickness = 7 − 3.5 = 3.5 m.

Volume of iron = 2/3 × π × (R³ − r³) = 2/3 × 22/7 × (7³ − 3.5³).

Volume = (44/21) × (343 − 42.875) = (44/21) × 300.125 ≈ 628.83 m³.

∴ The correct answer is 628.83 m³.

Alternate Method

Given:

Thickness (t) = 3.5 m

Cost of painting Outer CSA = ₹2464

Rate of painting = ₹8/m²

Formula Used:

Outer CSA of Hemisphere = 2πR²

Volume of Hollow Hemisphere = 2/3 π(R³ − r³)

Calculations:

⇒ Outer CSA = Total Cost ÷ Rate

⇒ 2 × (22/7) × R² = 2464 ÷ 8

⇒ (44/7) × R² = 308

⇒ R² = (308 × 7) ÷ 44

⇒ R² = 7 × 7 = 49 ⇒ R = 7 m

⇒ Inner radius (r) = R − thickness = 7 − 3.5 = 3.5 m

⇒ Volume of iron = 2/3 × (22/7) × (R³ − r³)

⇒ Volume = 2/3 × 22/7 × (7³ − 3.5³)

⇒ Volume = 44/21 × (343 − 42.875)

⇒ Volume = 44/21 × 300.125

⇒ Volume ≈ 628.833 m³

∴ The correct answer is 628.83 m³.

Additional Information

Hemisphere Surface Areas

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr² for a solid hemisphere.

Volume of Sphere and Hemisphere

Volume of Sphere = 4/3 πr³ and Volume of Hemisphere = 2/3 πr³.

Hollow Sphere Volume

The volume of material in a hollow sphere is 4/3 π(R³ − r³), where R is outer and r is inner radius.

Shortcut Trick

Net Surface Area = TSA of cuboid + 6 × (CSA of hemisphere − Area of circle base)

Area added per face = 2πr² − πr² = πr²

Total added area = 6 × πr² = 6 × 38.5 = 231 cm²

Cuboid TSA = 2(432 + 252 + 336) = 2040 cm²

Final TSA = 2040 + 231 = 2271 cm²

∴ The correct answer is 2271 cm².

Alternate Method

Given: Dimensions of cuboid = 14 cm × 18 cm × 24 cm; Radius of hemisphere (r) = 3.5 cm.

Formula: Total Surface Area (TSA) = 2(lb + bh + hl) + 6(πr²)

⇒ Total Surface Area of original cuboid = 2(24 × 18 + 18 × 14 + 14 × 24)

⇒ Cuboid TSA = 2(432 + 252 + 336) = 2 × 1020 = 2040 cm²

⇒ When a hemisphere is cut out, the circular top area (πr²) is removed from the face, but the curved surface area (2πr²) of the hole is added.

⇒ Net change in surface area for one face = 2πr² − πr² = πr²

⇒ For 6 faces, total area added = 6 × πr²

⇒ Total added area = 6 × (22/7) × 3.5 × 3.5 = 6 × 38.5 = 231 cm²

⇒ Surface area of remaining solid = 2040 + 231 = 2271 cm²

∴ The correct answer is 2271 cm².

Additional Information

Surface Area of Hemisphere

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr².

Volume of Remaining Solid

Volume = Volume of Cuboid − 6 × (Volume of Hemisphere) = (l × b × h) − 6 × (2/3 πr³).

Diagonal of Cuboid

The longest rod that can be placed in a cuboid is given by √(l² + b² + h²).

Shortcut Trick

When a pyramid is cut into equal heights, the ratio of heights from the vertex is 1 : 2 : 3 : 4.

Volume of similar solids is proportional to the cube of their heights: V ∝ h³.

Ratio of volumes of cumulative pyramids from top: 1³ : 2³ : 3³ : 4³ = 1 : 8 : 27 : 64.

Ratio of the four individual parts = 1 : (8 − 1) : (27 − 8) : (64 − 27) = 1 : 7 : 19 : 37.

∴ The correct answer is 1 : 7 : 19 : 37.

Alternate Method

Given: Side of square base = 12 cm, Height (H) = 40 cm.

The pyramid is cut into 4 parts of equal heights, so each part has a height of 40 ÷ 4 = 10 cm.

Formula: Volume of similar pyramid ∝ (Height)³

⇒ Let V₁, V₂, V₃, and V₄ be the volumes of the pyramids formed from the vertex with heights 10, 20, 30, and 40 cm respectively.

⇒ V₁ : V₂ : V₃ : V₄ = (10)³ : (20)³ : (30)³ : (40)³

⇒ V₁ : V₂ : V₃ : V₄ = 1000 : 8000 : 27000 : 64000

⇒ V₁ : V₂ : V₃ : V₄ = 1 : 8 : 27 : 64

⇒ Volume of 1st part (top) = V₁ = 1 unit

⇒ Volume of 2nd part = V₂ − V₁ = 8 − 1 = 7 units

⇒ Volume of 3rd part = V₃ − V₂ = 27 − 8 = 19 units

⇒ Volume of 4th part (bottom) = V₄ − V₃ = 64 − 27 = 37 units

⇒ Ratio of volumes of the four parts = 1 : 7 : 19 : 37

∴ The correct answer is 1 : 7 : 19 : 37.

Additional Information

Volume of Similar Solids

If two similar solids have heights h₁ and h₂, the ratio of their volumes is V₁/V₂ = (h₁/h₂)³.

Surface Area Ratio

The ratio of the lateral or total surface areas of similar solids is equal to the square of the ratio of their corresponding linear dimensions (h₁/h₂)².

Volume of a Frustum

The volume of a frustum of a pyramid is given by V = (h/3) × (A₁ + A₂ + √(A₁ × A₂)), where A₁ and A₂ are areas of the two bases.

Shortcut Trick

Factor out 2 sin 15° cos 15° from the expression.

Expression = 2 sin 15° cos 15° (1 − 2 sin² 15°)

Apply identities: sin 2θ = 2 sin θ cos θ and cos 2θ = 1 − 2 sin² θ.

Expression = sin(2 × 15°) × cos(2 × 15°) = sin 30° cos 30°.

Result = (1/2) × (√3/2) = √3/4.

∴ The correct answer is √3/4.

Alternate Method

Given: 2 sin 15° cos 15° − 4 sin³ 15° cos 15°

Formulas Used:

1. sin 2A = 2 sin A cos A
2. cos 2A = 1 − 2 sin² A
3. sin 2A = (1/2) sin 2A (Used for rewriting sin A cos A)

Calculations:

⇒ Expression = 2 sin 15° cos 15° − 4 sin³ 15° cos 15°

⇒ Take 2 sin 15° cos 15° as common factor:

⇒ = (2 sin 15° cos 15°) × (1 − 2 sin² 15°)

⇒ Substitute sin 2θ for (2 sin θ cos θ) and cos 2θ for (1 − 2 sin² θ):

⇒ = sin(2 × 15°) × cos(2 × 15°)

⇒ = sin 30° × cos 30°

⇒ Substitute standard values: sin 30° = 1/2 and cos 30° = √3/2

⇒ = (1/2) × (√3/2)

⇒ = √3/4

∴ The correct answer is √3/4.

Additional Information

Double Angle Identities

sin 2θ = 2 sin θ cos θ and cos 2θ = cos²θ − sin²θ = 2 cos²θ − 1 = 1 − 2 sin²θ.

Product to Sum Formula

sin A cos B = 1/2 [sin(A + B) + sin(A − B)]. For A = B, sin A cos A = 1/2 sin 2A.

Trigonometric Values of 15°

sin 15° = (√6 − √2)/4 and cos 15° = (√6 + √2)/4.

Triple Angle Identity for Sine

sin 3θ = 3 sin θ − 4 sin³ θ. This is often used in problems involving cubic trigonometric terms.

Shortcut Trick

Directly substitute cos² x ≡ 1 − sin² x and cos² y ≡ 1 − sin² y.

Given sin x = 1/2 ⇒ cos² x = 1 − 1/4 = 3/4

Given sin y = 2/3 ⇒ cos² y = 1 − 4/9 = 5/9

Numerator = 6(3/4) − 4(3/4)² = 18/4 − 9/4 = 9/4

Denominator = 18(5/9) − 27(5/9)² = 10 − 25/3 = 5/3

Value = (9/4) ÷ (5/3) = 27/20

∴ The correct answer is 27/20.

Alternate Method

Given:

sin x = 1/2

sin y = 2/3

Formula Used:

cos² θ = 1 − sin² θ

Calculations:

⇒ cos² x = 1 − (1/2)² = 1 − 1/4 = 3/4

⇒ cos² y = 1 − (2/3)² = 1 − 4/9 = 5/9

⇒ Numerator = 6 cos² x − 4 cos⁴ x = cos² x (6 − 4 cos² x)

⇒ Numerator = (3/4) × (6 − 4 × 3/4) = (3/4) × (6 − 3) = 9/4

⇒ Denominator = 18 cos² y − 27 cos⁴ y = 9 cos² y (2 − 3 cos² y)

⇒ Denominator = 9 × (5/9) × (2 − 3 × 5/9) = 5 × (2 − 5/3) = 5 × (1/3) = 5/3

⇒ Final Value = (9/4) ÷ (5/3) = (9/4) × (3/5) = 27/20

∴ The correct answer is 27/20.

Additional Information

Pythagorean Identities

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

Double Angle Formulas

cos 2θ = cos² θ − sin² θ = 2 cos² θ − 1 = 1 − 2 sin² θ.

Power Reduction

Expressions involving cos⁴ θ can be solved by substituting (cos² θ)² or using half-angle reduction: cos² θ = (1 + cos 2θ) / 2.

Shortcut Trick

Apply the sum-to-product transformation formula: cos C + cos D = 2 cos((C+D)/2) cos((C-D)/2).

The average of the angles is (105° + 15°) ÷ 2 = 60°.

The half-difference of the angles is (105° − 15°) ÷ 2 = 45°.

Result = 2 × cos 60° × cos 45° = 2 × (1/2) × (1/√2) = 1/√2.

∴ The correct answer is 1/√2.

Alternate Method

Given:

Expression = cos 15° + cos 105°

Formula Used:

cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2)

Calculations:

⇒ cos 105° + cos 15°

⇒ 2 cos((105° + 15°)/2) cos((105° − 15°)/2)

⇒ 2 cos(120°/2) cos(90°/2)

⇒ 2 cos 60° cos 45°

⇒ 2 × (1/2) × (1/√2)

⇒ 1 × (1/√2)

⇒ 1/√2

∴ The correct answer is 1/√2.

Additional Information

Trigonometric Sum-to-Product Formulas

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

Values of 15° and 75°

sin 15° = cos 75° = (√3 − 1)/(2√2) and cos 15° = sin 75° = (√3 + 1)/(2√2).

Complementary and Supplementary Angles

cos(90° + θ) = −sin θ and cos(180° − θ) = −cos θ.

Shortcut Trick

Identify angles: sin (A − B) = 1/2 ⇒ A − B = 30° and cos (A + B) = 1/2 ⇒ A + B = 60°.

Solve for A and B: A = (60 + 30)/2 = 45° and B = (60 − 30)/2 = 15°.

Substitute: (sin 45° cos 45°) + (sin² 45° × 1/2 sin 30°) + (cos³ 45° cos 15° tan 45°).

Calculation: 1/2 + (1/2 × 1/4) + (1/(2√2) × (√3 + 1)/(2√2)) = 1/2 + 1/8 + (√3 + 1)/8 = (4 + 1 + √3 + 1)/8 = (6 + √3)/8.

∴ The correct answer is (6 + √3)/8.

Alternate Method

Given: sin (A − B) = 1/2 and cos (A + B) = 1/2.

Formula used: sin 30° = 1/2, cos 60° = 1/2, sin 2θ = 2 sin θ cos θ, cos (X − Y) = cos X cos Y + sin X sin Y.

⇒ A − B = 30° ... (i)

⇒ A + B = 60° ... (ii)

⇒ Adding (i) and (ii): 2A = 90° ⇒ A = 45°

⇒ Subtracting (i) from (ii): 2B = 30° ⇒ B = 15°

⇒ sin A = 1/√2, cos A = 1/√2, tan A = 1

⇒ cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30° = (1/√2 × √3/2) + (1/√2 × 1/2) = (√3 + 1)/(2√2)

⇒ sin 15° cos 15° = 1/2 sin 30° = 1/2 × 1/2 = 1/4

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

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

⇒ Expression = (4 + 1 + √3 + 1) / 8 = (6 + √3) / 8

∴ The correct answer is (6 + √3)/8.

Additional Information

Compound Angle Formulas

sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B.

Trigonometric Values of 15° and 75°

sin 15° = cos 75° = (√3 − 1)/(2√2) and cos 15° = sin 75° = (√3 + 1)/(2√2).

Double Angle Identity

sin 2θ = 2 sin θ cos θ, which implies sin θ cos θ = (1/2) sin 2θ.

Shortcut Trick

Use the allied angle identity: cot (90° + θ) = −tan θ.

Here, θ = 75°, so cot (90° + 75°) = −tan 75°.

The standard value for tan 75° is 2 + √3.

Substitute this value: −(2 + √3) = −2 − √3.

∴ The correct answer is −2 − √3.

Alternate Method

Given: expression cot (90° + 75°)

Formula Used: cot (90° + θ) = −tan θ and tan (A + B) = (tan A + tan B) ÷ (1 − tan A × tan B)

Calculations:

⇒ cot (90° + 75°) = −tan 75°

⇒ tan 75° = tan (45° + 30°)

⇒ tan 75° = (tan 45° + tan 30°) ÷ (1 − tan 45° × tan 30°)

⇒ tan 75° = (1 + 1/√3) ÷ (1 − 1/√3)

⇒ tan 75° = (√3 + 1) ÷ (√3 − 1)

⇒ Rationalizing the denominator: [(√3 + 1)(√3 + 1)] ÷ [(√3 − 1)(√3 + 1)]

⇒ tan 75° = (3 + 1 + 2√3) ÷ (3 − 1) = (4 + 2√3) ÷ 2 = 2 + √3

⇒ cot (90° + 75°) = −tan 75° = −(2 + √3) = −2 − √3

∴ The correct answer is −2 − √3.

Additional Information

Allied Angles Property

For angles in the form (90° ± θ) or (270° ± θ), trigonometric functions change to their co-functions: sin ↔ cos, tan ↔ cot, sec ↔ cosec.

ASTC Rule

In the second quadrant (90° to 180°), only Sine and Cosecant are positive. Tangent and Cotangent are negative.

Common Trig Values

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

Shortcut Trick

Use value substitution: Let A = 90°, B = 0°, and C = 0°, which satisfies the condition A + B + C = 90°.

Substitute in the expression: sin(90°/2) sin[(180° − 0° − 0°)/2] + cos(90°/2) sin[(0° + 0°)/2]

⇒ sin 45° sin 90° + cos 45° sin 0°

⇒ (1 / √2) × 1 + (1 / √2) × 0 = 1 / √2

∴ The correct answer is 1 / √2.

Alternate Method

Given: A + B + C = 90°

Formula Used: sin (90° − θ) = cos θ and sin (X + Y) = sin X cos Y + cos X sin Y

⇒ Given expression: sin (A / 2) sin [(180 − (B + C)) / 2] + cos (A / 2) sin [(B + C) / 2]

⇒ sin (A / 2) sin [90° − (B + C) / 2] + cos (A / 2) sin [(B + C) / 2]

⇒ Using sin (90° − θ) = cos θ:

⇒ sin (A / 2) cos [(B + C) / 2] + cos (A / 2) sin [(B + C) / 2]

⇒ This matches the form sin X cos Y + cos X sin Y, where X = A / 2 and Y = (B + C) / 2

⇒ sin [A / 2 + (B + C) / 2] = sin [(A + B + C) / 2]

⇒ Substituting A + B + C = 90°:

⇒ sin (90° / 2) = sin 45° = 1 / √2

∴ The correct answer is 1 / √2.

Additional Information

Sine Addition Formula

The sine of the sum of two angles is given by sin(X + Y) = sin X cos Y + cos X sin Y.

Complementary Angles

Trigonometric ratios of complementary angles follow the rules: sin(90° − θ) = cos θ and cos(90° − θ) = sin θ.

Standard Trigonometric Values

For the angle 45°, both sin 45° and cos 45° are equal to 1 / √2.

Shortcut Trick

Apply allied angle reductions: cot(90 − x) = tan x, sin(90 − x) = cos x, cot(180 − x) = −cot x, and sin(180 − x) = sin x.

Expression = tan x cos⁴ x − cot x sin⁴ x.

Convert to sin and cos: (sin x / cos x) × cos⁴ x − (cos x / sin x) × sin⁴ x = sin x cos³ x − cos x sin³ x.

Factor out common terms: sin x cos x (cos² x − sin² x) = (1/2) sin 2x × cos 2x.

Using 2 sin A cos A = sin 2A: (1/4) sin 4x.

∴ The correct answer is (sin 4x) / 4.

Alternate Method

Given: cot (90 − x) sin⁴ (90 − x) + cot (180 − x) sin⁴ (180 − x)

Formula Used: Allied angles and double angle identities.

⇒ Term 1: cot (90 − x) sin⁴ (90 − x) = tan x (cos x)⁴ = tan x cos⁴ x

⇒ Term 2: cot (180 − x) sin⁴ (180 − x) = (−cot x) (sin x)⁴ = −cot x sin⁴ x

⇒ Expression = tan x cos⁴ x − cot x sin⁴ x

⇒ Expression = (sin x / cos x) × cos⁴ x − (cos x / sin x) × sin⁴ x

⇒ Expression = sin x cos³ x − cos x sin³ x

⇒ Expression = sin x cos x (cos² x − sin² x)

⇒ Expression = (1/2) × (2 sin x cos x) × (cos² x − sin² x)

⇒ Expression = (1/2) sin 2x cos 2x

⇒ Expression = (1/4) (2 sin 2x cos 2x) = (1/4) sin 4x

∴ The correct answer is (sin 4x) / 4.

Additional Information

Allied Angle Relations

Trigonometric ratios change when using (90 ± θ) or (270 ± θ), where sin ↔ cos, tan ↔ cot, and sec ↔ cosec. Ratios remain the same for (180 ± θ) or (360 ± θ), only the sign changes based on the quadrant.

Double and Multiple Angle Identities

Key formulas include sin 2θ = 2 sin θ cos θ and cos 2θ = cos² θ − sin² θ = 2 cos² θ − 1 = 1 − 2 sin² θ. These allow reducing powers and expanding compound angles like sin 4x.

Fundamental Identities

Recall that sin² θ + cos² θ = 1, tan θ = sin θ / cos θ, and cot θ = cos θ / sin θ. These are essential for simplifying rational trigonometric expressions.

Shortcut Trick

Using trigonometry ratios: For 30°, height : base = 1 : √3. For 45°, total height : base = 1 : 1.

To keep the base constant, total height : base = √3 : √3.

Flag height = Total height − Building height = (√3 − 1) units.

Given: (√3 − 1) units = 4 m ⇒ 1 unit (Building height) = 4 ÷ (√3 − 1).

On rationalizing: 4(√3 + 1) ÷ (3 − 1) = 2(√3 + 1) m.

∴ The correct answer is 2(√3 + 1) m.

Alternate Method

Let the height of the building be H and the distance from point X to the building base be D.

Given: Flag height = 4 m. Total height of (Building + Flag) = H + 4.

Angle of elevation to top of building = 30°. Angle of elevation to top of flag = 45°.

In triangle formed by building top and X:

⇒ tan 30° = H ÷ D

⇒ 1 ÷ √3 = H ÷ D

⇒ D = H√3 --- (Equation 1)

In triangle formed by flag top and X:

⇒ tan 45° = (H + 4) ÷ D

⇒ 1 = (H + 4) ÷ D

⇒ D = H + 4 --- (Equation 2)

Equating D from both equations:

⇒ H√3 = H + 4

⇒ H√3 − H = 4

⇒ H(√3 − 1) = 4

⇒ H = 4 ÷ (√3 − 1)

⇒ H = [4 × (√3 + 1)] ÷ [(√3 − 1) × (√3 + 1)]

⇒ H = [4(√3 + 1)] ÷ (3 − 1)

⇒ H = 4(√3 + 1) ÷ 2

⇒ H = 2(√3 + 1) m

∴ The correct answer is 2(√3 + 1) m.

Additional Information

Trigonometric Ratios

tan θ = Perpendicular ÷ Base. Standard values include tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3.

Angle of Elevation

The angle between the horizontal line of sight and the line from the observer's eye to some object above the horizontal level.

Rationalization

To remove the radical from the denominator, multiply both numerator and denominator by the conjugate of the denominator: (a − √b) conjugate is (a + √b).

Shortcut Trick

The relationship between Height (h) and Distance from base (d) is given by d = h × cot θ.

Given: h = 120 m and θ = 75°.

Standard value: cot 75° = tan 15° = 2 − √3.

Substituting the values: d = 120 × (2 − √3) m.

∴ The correct answer is 120 (2 − √3) m.

Alternate Method

Given: Height of tower (AC) = 120 m, Angle of elevation at point B = 75°.

Formula Used: tan θ = Perpendicular / Base

⇒ tan 75° = 120 / BC

⇒ BC = 120 / tan 75°

⇒ tan 75° = tan(45° + 30°) = (tan 45° + tan 30°) / (1 − tan 45° × tan 30°)

⇒ tan 75° = (1 + 1/√3) / (1 − 1/√3) = (√3 + 1) / (√3 − 1) = 2 + √3

⇒ BC = 120 / (2 + √3)

⇒ BC = [120 / (2 + √3)] × [(2 − √3) / (2 − √3)]

⇒ BC = 120(2 − √3) / (4 − 3) = 120(2 − √3)

∴ The correct answer is 120 (2 − √3) m.

Additional Information

Trigonometric Values for 15° and 75°

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

Height and Distance Relation

For an angle of elevation θ, Distance = Height / tan θ. For θ = 45°, Height = Distance.

Rationalization Rule

To rationalize 1/(a + √b), multiply numerator and denominator by (a − √b) to use the identity (a+b)(a−b) = a² − b².

Shortcut Trick

Height above Mohit's eye level (h) = 60 − 1.8 = 58.2 m.

Distance covered (d) = h × (cot θ₁ − cot θ₂), where θ₁ = 45° and θ₂ = 60°.

d = 58.2 × (cot 45° − cot 60°) = 58.2 × (1 − 1/√3).

d = 58.2 − (58.2/√3) = 19.4 × 3 − 19.4√3.

d = 19.4 (3 − √3) m.

∴ The correct answer is 19.4 (3 − √3) m.

Alternate Method

Given:

Height of building (H) = 60 m

Height of Mohit (h_m) = 1.8 m

Initial angle (θ₁) = 45° and Final angle (θ₂) = 60°

Formula Used:

tan θ = Perpendicular / Base

Calculations:

⇒ Effective height (AB) = Total height − Mohit height = 60 − 1.8 = 58.2 m

⇒ In first position (45°), let base distance be x: tan 45° = 58.2 / x

⇒ 1 = 58.2 / x ⇒ x = 58.2 m

⇒ In second position (60°), let base distance be y: tan 60° = 58.2 / y

⇒ √3 = 58.2 / y ⇒ y = 58.2 / √3 = (58.2 × √3) / 3 = 19.4√3 m

⇒ Distance covered = x − y = 58.2 − 19.4√3 m

⇒ Distance covered = 19.4 (3 − √3) m

∴ The correct answer is 19.4 (3 − √3) m.

Additional Information

Trigonometric Ratios for Heights

tan 45° = 1 and tan 60° = √3. These are the most common values used in height and distance problems.

General Formula for Distance

If an observer moves towards an object and the angle changes from α to β, the distance d = h(cot α − cot β), where h is the relative height.

Angle Observation Rule

As you move closer to a building or tower, the angle of elevation always increases (e.g., from 45° to 60°).

Given :

Total number of bikes manufactured by company D = (2/3) of 300000

⇒ 200000 

Total number of cars manufactured by by company G = (3/4) of 400000

⇒ 300000 

C₁ type cars manufactured by G = (2/6) of 300000 

⇒ 100000 

Required percent = (200000/100000) × 100 

⇒ 200% 

∴ option 3 is the correct choice

Calculations :

C₁ type car manufactured by D = (2/10) × (1/3) × 300000 =  20000

C₁ type car manufactured by E = (3/4) × (1/3) × 280000 = 70000

C₁ type car manufactured by F = (1/2) × (2/4) × 320000 = 80000

C₁ type car manufactured by G = (3/4) × (2/6) × 400000 = 100000

C₁ type car manufactured by H = (1/3) × (1/4) × 480000 = 40000

Total number of cars = 20000 + 70000 + 80000 + 100000 + 40000 = 310000

Average of C₁ type of cars produced by all companies = Total number of cars/5

⇒ 310000/5 

⇒ 62000

∴ average of cars of type C1 will be 62000

Calculations :

Total number of C₃ type car manufactured by company E = (3/4) × (1/3) of 280000 = 70000

Total number of C₃ type car manufactured by company G = (3/4) × (1/6) of 400000 = 50000

Total number of car of C3​ type manufacture by E and G = 70000 + 50000 = 120000 

Total number of B₁ type of bikes manufactured by company H = (2/3) × (2/8) of 480000 = 80000

Required difference = 120000 - 80000 

⇒ 40000 

∴ option 2 is the correct answer