SSC CGL Tier 2 Quant Previous Year Paper (Held On: 19 Feb 2018)

Shortcut Trick

The sum of the series 1 + 11 + 111 + ... up to n terms (where n ≤ 9) results in the number 123...n.

Here, the series has 9 terms (ending at 111,111,111), so N = 123456789.

Sum of digits of N = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9.

Using the formula for sum of first n natural numbers: [n × (n + 1)] ÷ 2.

Sum = (9 × 10) ÷ 2 = 45.

∴ The correct answer is 45.

Alternate Method

Given: N = 1 + 11 + 111 + 1111 + 11111 + 111111 + 1111111 + 11111111 + 111111111

Let us align the numbers vertically to find the sum N:

⇒ Total terms = 9

⇒ Sum at units place = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9

⇒ Sum at tens place = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8

⇒ Sum at hundreds place = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7

⇒ Continuing this logic, the number N = 123456789

⇒ Sum of digits of N = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

⇒ Sum = (1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5

⇒ Sum = 10 + 10 + 10 + 10 + 5 = 45

∴ The correct answer is 45.

Additional Information

Sum of first n natural numbers

The sum of the first n natural numbers is given by the formula S = [n(n + 1)] ÷ 2.

Repunit Series Sum

For a series of repunits (numbers with only digit 1), if the number of terms n ≤ 9, the sum is always the integer 123...n. If n > 9, carry-over digits will affect the pattern.

Sum of squares and cubes

Sum of squares ∑n² = [n(n + 1)(2n + 1)] ÷ 6 and Sum of cubes ∑n³ = [[n(n + 1)] ÷ 2]².

Shortcut Trick

The series can be grouped into pairs: (1 + 3) + (4 + 5) + (7 + 7) + (10 + 9) + ...

New paired series: 4 + 9 + 14 + 19 + ... which forms an Arithmetic Progression (AP).

For this AP, first term (a) = 4, common difference (d) = 5, and number of terms (n) = 40 ÷ 2 = 20.

Sum = n/2 [2a + (n − 1)d] ⇒ 20/2 [2 × 4 + (20 − 1) × 5]

Sum = 10 [8 + 95] = 10 × 103 = 1030.

∴ The correct answer is 1030.

Alternate Method

Given series: 1 + 3 + 4 + 5 + 7 + 7 + 10 + 9 + ... up to 40 terms.

This series is a combination of two alternating Arithmetic Progressions (APs).

Formula Used: Sum of n terms (S_n) = (n/2) × [2a + (n − 1)d]

⇒ Total terms = 40. Therefore, each AP has 40 ÷ 2 = 20 terms.

⇒ For Series 1 (1, 4, 7...): a₁ = 1, d₁ = 3, n = 20

⇒ S₂₀₍₁₎ = (20/2) × [2(1) + (20 − 1)3] = 10 × [2 + 57] = 590

⇒ For Series 2 (3, 5, 7...): a₂ = 3, d₂ = 2, n = 20

⇒ S₂₀₍₂₎ = (20/2) × [2(3) + (20 − 1)2] = 10 × [6 + 38] = 440

⇒ Total Sum = 590 + 440 = 1030

∴ The correct answer is 1030.

Additional Information

General Term of an AP

The n-th term of an arithmetic progression is given by T_n = a + (n − 1)d, where 'a' is the first term and 'd' is the common difference.

Arithmetic Progression (AP) Sum Formula

The sum can also be calculated as S_n = (n/2) × (a + l), where 'l' is the last term of the series.

Alternating Series Logic

When a series does not show a single pattern, check for alternating terms (1st, 3rd, 5th...) or group terms in pairs/triplets to find a consistent AP or GP.

Shortcut Trick

Term 1: 1 ÷ 0.2 = 5

Term 2: 1 ÷ 0.02 = 50

Term 3: 1 ÷ 0.002 = 500

Pattern: Each term is 10 × the previous term.

Sum of 9 terms: 5 + 50 + 500 + ... (9 terms) = 555,555,555.

∴ The correct answer is 555555555.

Alternate Method

Given: Series is 1/0.2 + 1/0.02 + 1/0.002 + ... upto 9 terms.

Formula: Sum of a Geometric Progression, S_n = a(rⁿ − 1) ÷ (r − 1)

⇒ T₁ = 1 ÷ 0.2 = 10 ÷ 2 = 5 ⇒ T₂ = 1 ÷ 0.02 = 100 ÷ 2 = 50 ⇒ T₃ = 1 ÷ 0.002 = 1000 ÷ 2 = 500 ⇒ Here, first term (a) = 5, common ratio (r) = 10, number of terms (n) = 9. ⇒ S₉ = 5 × (10⁹ − 1) ÷ (10 − 1) ⇒ S₉ = 5 × (1000000000 − 1) ÷ 9 ⇒ S₉ = 5 × 999999999 ÷ 9 ⇒ S₉ = 5 × 111111111 ⇒ S₉ = 555555555

∴ The correct answer is 555555555.

Additional Information

Geometric Progression (GP)

A sequence where the ratio of any two consecutive terms is constant, known as the common ratio (r).

Sum of n terms in GP

The formula is S_n = a(rⁿ − 1) / (r − 1) for r > 1 and S_n = a(1 − rⁿ) / (1 − r) for r < 1.

Decimal to Fraction Conversion

To convert a decimal like 0.002 to a fraction, divide the significant digits by 10^p, where p is the number of decimal places.

Given:

$\frac{3.6 \times 1.62 + 0.48 \times 3.6}{1.8 \times 0.8 + 10.8 \times 0.3 - 2.16}$ = ?

Formula used:

Distributive property: (a × b) + (a × c) = a × (b + c)

Calculation:

$\frac{3.6 \times (1.62 + 0.48)}{1.44 + 3.24 - 2.16}$

⇒ $\frac{3.6 \times 2.1}{4.68 - 2.16}$

⇒ $\frac{7.56}{2.52}$

⇒ 3

∴ The correct answer is option 4.

Shortcut Trick

Successively invert the fraction and subtract the integer part (1) to find the nested values.

1. Invert 5/8 → 8/5 = 1 + 3/5
2. Invert 3/5 → 5/3 = 1 + 2/3
3. Invert 2/3 → 3/2 = 1 + 1/2

Comparing 1 + 1/2 with the last nested term 1 + 1/x, we get 1/x = 1/2 ⇒ x = 2.

∴ The correct answer is 2.

Alternate Method

Given: 1 / (1 + 1 / (1 + 1 / (1 + 1/x))) = 5/8

Formula used: If 1 / A = B, then A = 1 / B.

⇒ 1 + 1 / (1 + 1 / (1 + 1/x)) = 8/5

⇒ 1 / (1 + 1 / (1 + 1/x)) = 8/5 − 1 = 3/5

⇒ 1 + 1 / (1 + 1/x) = 5/3

⇒ 1 / (1 + 1/x) = 5/3 − 1 = 2/3

⇒ 1 + 1/x = 3/2

⇒ 1/x = 3/2 − 1 = 1/2

⇒ x = 2

∴ The correct answer is 2.

Additional Information

Continued Fractions

A continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number.

Reciprocal Property

For any fraction a/b, its reciprocal is 1 ÷ (a/b) = b/a. This property is used to 'unpeel' nested fractions from the top level down.

Euclidean Algorithm Connection

The coefficients in a finite continued fraction (like the 1s in this problem) are the quotients obtained when applying the Euclidean algorithm to the numerator and denominator.

Shortcut Trick

Observe the reciprocal pairs: (1 + 1/2) × (1 − 1/3) = 3/2 × 2/3 = 1.

Similarly, (1 + 1/4) × (1 − 1/5) = 1 and (1 + 1/6) × (1 − 1/7) = 1.

All pairs cancel out, leaving only the term (1 + 1/8) = 9/8.

Comparing 9/8 = 1 + 1/x ⇒ 1 + 1/8 = 1 + 1/x.

∴ The correct answer is 8.

Alternate Method

Given equation:

(1 + 1/2)(1 + 1/4)(1 + 1/6)(1 + 1/8)(1 − 1/3)(1 − 1/5)(1 − 1/7) = 1 + 1/x

Convert each bracketed term into an improper fraction:

⇒ (3/2) × (5/4) × (7/6) × (9/8) × (2/3) × (4/5) × (6/7) = 1 + 1/x

Group terms to facilitate cancellation:

⇒ [(3/2) × (2/3)] × [(5/4) × (4/5)] × [(7/6) × (6/7)] × (9/8) = 1 + 1/x

Simplifying the products of reciprocals:

⇒ 1 × 1 × 1 × 9/8 = 1 + 1/x

⇒ 9/8 = 1 + 1/x

⇒ 1 + 1/8 = 1 + 1/x

⇒ 1/x = 1/8

⇒ x = 8

∴ The correct answer is 8.

Additional Information

Reciprocal Identity

A fraction multiplied by its reciprocal always equals 1, i.e., (a/b) × (b/a) = 1.

BODMAS Rule

Always solve terms within brackets first before performing multiplication across the series.

Series Cancellation

In telescoping products, look for terms in the numerator of one fraction that match terms in the denominator of another to simplify quickly.

Shortcut Trick

Identify the common difference between factors in the denominator: 7 − 3 = 4, 11 − 7 = 4, ..., 903 − 899 = 4.

Direct formula for sum: Sum = (1 ÷ difference) × (1 ÷ first term − 1 ÷ last term)

Sum = (1 ÷ 4) × (1 ÷ 3 − 1 ÷ 903)

Sum = (1 ÷ 4) × (300 ÷ 903) = 75 ÷ 903 = 25 ÷ 301.

∴ The correct answer is 25/301.

Alternate Method

Given series: S = 1/(3 × 7) + 1/(7 × 11) + 1/(11 × 15) + ... + 1/(899 × 903)

Formula used: 1/(a × b) = [1/(b − a)] × (1/a − 1/b)

⇒ Each term can be written as: 1/4 × (1/3 − 1/7) + 1/4 × (1/7 − 1/11) + ... + 1/4 × (1/899 − 1/903)

⇒ S = 1/4 × [(1/3 − 1/7) + (1/7 − 1/11) + (1/11 − 1/15) + ... + (1/899 − 1/903)]

⇒ S = 1/4 × [1/3 − 1/7 + 1/7 − 1/11 + 1/11 − ... − 1/899 + 1/899 − 1/903]

⇒ S = 1/4 × [1/3 − 1/903]

⇒ S = 1/4 × [(301 − 1) ÷ 903]

⇒ S = 1/4 × [300 ÷ 903]

⇒ S = 75 ÷ 903

⇒ S = 25/301

∴ The correct answer is 25/301.

Additional Information

Telescoping Series

A telescoping series is a series where most of the terms cancel each other out, leaving only the first and last parts of the expansion.

General Sum Formula

For a series ∑ 1/[n × (n + d)], the sum of n terms is (1/d) × [1/first − 1/last].

Partial Fraction Decomposition

1/[n(n+k)] = (1/k) × [1/n − 1/(n+k)]. This is the core principle used to break down the terms in such mathematical series.

Shortcut Trick

The unit digit of any number n⁵ is always the same as the unit digit of n¹.

This is because the cyclicity of all digits is 4, and 5 ≡ 1 (mod 4).

Thus, unit digit of 1⁵ + 2⁵ + … + 20⁵ is same as 1 + 2 + … + 20.

Sum of first 20 natural numbers = (20 × 21) ÷ 2 = 210.

The unit digit of 210 is 0.

∴ The correct answer is 0.

Alternate Method

Given: Sum S = 1⁵ + 2⁵ + 3⁵ + … + 20⁵

Concept: The unit digit of a number raised to a power repeats in a cycle of 4. For any digit n, the unit digit of n⁵ is the same as n^{5 − 4} = n¹.

⇒ Required unit digit = Unit digit of (1⁵ + 2⁵ + 3⁵ + … + 20⁵)

⇒ Unit digit = (Unit digit of 1¹ + Unit digit of 2¹ + … + Unit digit of 20¹) mod 10

⇒ Sum of first 20 natural numbers = [n × (n + 1)] ÷ 2

⇒ Sum = [20 × (20 + 1)] ÷ 2

⇒ Sum = [20 × 21] ÷ 2 = 10 × 21 = 210

⇒ The unit digit of 210 is 0.

∴ The correct answer is 0.

Additional Information

Cyclicity of Digits

Digits 0, 1, 5, 6 have a cyclicity of 1 (unit digit remains same for any power). Digits 4 and 9 have a cyclicity of 2. Digits 2, 3, 7, 8 have a cyclicity of 4.

Power Reduction Rule

To find the unit digit of x^y, divide y by 4. If remainder is r, unit digit is x^r. If remainder is 0, use unit digit of x⁴.

Sum of Powers Formula

Sum of first n natural numbers = n(n + 1) ÷ 2. Sum of squares = n(n + 1)(2n + 1) ÷ 6. Sum of cubes = [n(n + 1) ÷ 2]².

Shortcut Trick

Sum of three primes (38) is even → At least one prime must be even (2).

Let z = 2, then x + y = 38 − 2 = 36.

To maximize x, test smallest primes for y: 3, 5, 7...

If y = 3, x = 33 (Composite); if y = 5, x = 31 (Prime).

∴ The correct answer is 31.

Alternate Method

Given: x, y, z are prime numbers and x + y + z = 38.

Formula/Property: Sum of 3 odd numbers is always odd. Since 38 is even, one of the primes must be the only even prime, 2.

⇒ Let z = 2

⇒ x + y + 2 = 38

⇒ x + y = 36

⇒ Testing for maximum x (implies minimum prime y):

⇒ If y = 2, x = 34 (Not prime)

⇒ If y = 3, x = 33 (Not prime, divisible by 3)

⇒ If y = 5, x = 31 (Prime number)

⇒ Max value of x = 31

∴ The correct answer is 31.

Additional Information

Even Prime Number

2 is the only prime number that is even. All other prime numbers are odd.

Sum Parity Rules

Odd + Odd = Even | Odd + Even = Odd | Odd + Odd + Odd = Odd.

Definition of Prime

A number greater than 1 with exactly two factors: 1 and itself.

Shortcut Trick

Smallest three-digit number is 100, which is even (composite).

Next number 101 is the smallest three-digit prime number (N = 101).

Nearest multiple of 13 to 101 is 13 × 7 = 91.

Remainder = 101 − 91 = 10.

∴ The correct answer is 10.

Alternate Method

Given: N is the smallest three-digit prime number.

Method: Identify N and apply the division algorithm: N = (13 × Quotient) + Remainder.

⇒ Smallest three-digit numbers are 100, 101, 102, ...

⇒ 100 is divisible by 2, 4, 5, 10, 20, 25, 50.

⇒ 101 is not divisible by 2, 3, 5, or 7 (primes less than √101 ≈ 10.05).

⇒ ∴ N = 101.

⇒ Now, divide 101 by 13:

⇒ 101 ÷ 13

⇒ 13 × 7 = 91

⇒ Remainder = 101 − 91 = 10.

∴ The correct answer is 10.

Additional Information

Prime Numbers

A natural number greater than 1 that has no positive divisors other than 1 and itself. 2 is the only even prime number.

Testing for Primality

To check if N is prime, test divisibility by all primes less than or equal to √N. If none divide N, then N is prime.

Divisibility Rule for 13

Multiply the last digit by 4 and add it to the remaining number. If the result is divisible by 13, the original number is also divisible by 13.

Shortcut Trick Find the largest integer less than √45109 and the smallest integer greater than √261. Since 162 = 256 and 172 = 289, √261 is approximately 16.15. Smallest integer → 17. Since 2122 = 44944 and 2132 = 45369, √45109 is approximately 212.38. Largest integer → 212. Total numbers = 212 − 17 + 1 = 196. ∴ The correct answer is 196.

Alternate Method Given: Lower limit = √261 Upper limit = √45109

Calculations: ⇒ For lower limit: 162 = 256 and 172 = 289. ⇒ 16 < √261 < 17. The first natural number greater than √261 is 17. ⇒ For upper limit: Let's estimate √45109. Since 2102 = 44100. ⇒ 2122 = (210 + 2)2 = 44100 + 840 + 4 = 44944. ⇒ 2132 = (210 + 3)2 = 44100 + 1260 + 9 = 45369. ⇒ 212 < √45109 < 213. The last natural number smaller than √45109 is 212. ⇒ Total natural numbers from 17 to 212 = (Last number − First number) + 1 ⇒ Count = (212 − 17) + 1 ⇒ Count = 195 + 1 = 196. ∴ The correct answer is 196.

Additional Information Square Root Estimation To find integers between two roots √A and √B, find the perfect squares immediately above A and immediately below B. Number of Terms in AP For a sequence of consecutive natural numbers from 'a' to 'b', the number of terms is given by (b − a + 1). Perfect Square Identification A number ending in 2, 3, 7, or 8 cannot be a perfect square. √45109 is not an integer as it ends in 9 but 2122 ends in 4 and 2132 ends in 9.

Given:

$\sqrt{121} + \sqrt{12321} + \sqrt{1234321} + \sqrt{123454321}$

Formula used:

11² = 121

111² = 12321

1111² = 1234321

11111² = 123454321

Calculation:

⇒ $\sqrt{121}$ = 11

⇒ $\sqrt{12321}$ = 111

⇒ $\sqrt{1234321}$ = 1111

⇒ $\sqrt{123454321}$ = 11111

⇒ Sum = 11 + 111 + 1111 + 11111

⇒ Sum = 12344

∴ The correct answer is option (3).

Shortcut Trick Use the value-putting method by assuming p = √3√4, q = 0, and r = 0. Check given condition: p3 + q3 + r3 − 3pqr = (√3√4)3 + 0 + 0 − 0 = 4. (Condition satisfied) Calculate new variables: a = q + r = 0, b = r + p = √3√4, c = p + q = √3√4. Evaluate target: a3 + b3 + c3 − 3abc = 03 + (√3√4)3 + (√3√4)3 − 3(0)(√3√4)(√3√4) = 0 + 4 + 4 − 0 = 8. ∴ The correct answer is 8. Alternate Method Given: p3 + q3 + r3 − 3pqr = 4 and a = q + r, b = r + p, c = p + q Formula Used: x3 + y3 + z3 − 3xyz = (1/2)(x + y + z)[(x − y)2 + (y − z)2 + (z − x)2]

⇒ Calculate sum: a + b + c = (q + r) + (r + p) + (p + q) = 2(p + q + r) ⇒ Calculate differences: (a − b) = (q + r) − (r + p) = (q − p) ⇒ (b − c) = (r + p) − (p + q) = (r − q) ⇒ (c − a) = (p + q) − (q + r) = (p − r) ⇒ Target = (1/2)(2(p + q + r))[(q − p)2 + (r − q)2 + (p − r)2] ⇒ Target = 2 × [(1/2)(p + q + r)((p − q)2 + (q − r)2 + (r − p)2)] ⇒ Target = 2 × (p3 + q3 + r3 − 3pqr) ⇒ Target = 2 × 4 = 8 ∴ The correct answer is 8. Additional Information Algebraic Identity The expression x3 + y3 + z3 − 3xyz is equivalent to (x + y + z)(x2 + y2 + z2 − xy − yz − zx). Zero Sum Case If x + y + z = 0, then the entire expression x3 + y3 + z3 − 3xyz becomes zero, implying x3 + y3 + z3 = 3xyz. Squared Difference Form An alternative representation is (1/2)(x + y + z)[(x − y)2 + (y − z)2 + (z − x)2], which is useful for proving the expression is non-negative when x, y, z > 0.

Shortcut Trick From x2 + x − 1 = 0, sum α + β = −1 and product αβ = −1. Using Newton's sum Sn = αn + βn where Sn = −Sn−1 + Sn−2. S1 = −1, S2 = (−1)2 − 2(−1) = 3. S3 = −S2 + S1 = −3 − 1 = −4; S4 = −S3 + S2 = 4 + 3 = 7; S5 = −S4 + S3 = −7 − 4 = −11. New sum = −11, New product = (αβ)5 = (−1)5 = −1. ∴ The correct answer is x2 + 11x − 1 = 0.

Alternate Method Given equation: x2 + x − 1 = 0 Roots: α and β Formula used: Sum of roots = −b/a and Product of roots = c/a ⇒ α + β = −1/1 = −1 ⇒ αβ = −1/1 = −1 ⇒ α2 + β2 = (α + β)2 − 2αβ = (−1)2 − 2(−1) = 1 + 2 = 3 ⇒ α3 + β3 = (α + β)3 − 3αβ(α + β) = (−1)3 − 3(−1)(−1) = −1 − 3 = −4 ⇒ New Sum (S) = α5 + β5 = (α2 + β2)(α3 + β3) − α2β2(α + β) ⇒ S = (3)(−4) − (−1)2(−1) = −12 − (1)(−1) = −12 + 1 = −11 ⇒ New Product (P) = α5β5 = (αβ)5 = (−1)5 = −1 ⇒ Required Equation: x2 − Sx + P = 0 ⇒ x2 − (−11)x + (−1) = 0 ⇒ x2 + 11x − 1 = 0 ∴ The correct answer is x2 + 11x − 1 = 0.

Additional Information Sum and Product of Roots For a quadratic equation ax2 + bx + c = 0, sum α + β = −b/a and product αβ = c/a. Newton's Identities If Sn = αn + βn, then for the equation ax2 + bx + c = 0, the relation is aSn + bSn−1 + cSn−2 = 0. Nature of Roots The nature is determined by discriminant D = b2 − 4ac. If D > 0 roots are real and distinct, if D = 0 roots are real and equal.

Shortcut Trick

The sum of two natural numbers x + y = 2017 is an odd number.

For a sum to be odd, one number must be even and the other must be odd.

Property: (−1)^even = 1 and (−1)^odd = −1.

Required value = (−1)^even + (−1)^odd = 1 + (−1) = 0.

∴ The correct answer is 0.

Alternate Method

Given: x + y = 2017 where x and y are natural numbers.

Formula used: (−1)ⁿ = 1 if n is even, and (−1)ⁿ = −1 if n is odd.

⇒ Sum x + y = 2017 (an odd number).

⇒ According to parity rules, an odd sum implies one addend is even and the other is odd.

⇒ If x is even, then y = 2017 − x must be odd.

⇒ If x is odd, then y = 2017 − x must be even.

⇒ The required expression is (−1)^x + (−1)^y.

⇒ (−1)^even + (−1)^odd = 1 + (−1) = 0.

∴ The correct answer is 0.

Additional Information

Parity of Sums

The sum of two integers is odd if and only if one integer is even and the other is odd. (Even + Odd = Odd, Even + Even = Even, Odd + Odd = Even).

Powers of Negative Unity

The expression (−1)ⁿ equals 1 if n is an even integer and equals −1 if n is an odd integer.

Natural Numbers

Natural numbers are the set of positive integers {1, 2, 3, ...}. Since x + y = 2017, neither x nor y can be zero.

Shortcut Trick

If x + 1/x = k, then x² + 1/x² = k² − 2 and x⁴ + 1/x⁴ = (x² + 1/x²)² − 2.

Given k = (√3 + 1)/2 ⇒ k² = (3 + 1 + 2√3) / 4 = (4 + 2√3) / 4 = (2 + √3) / 2.

x² + 1/x² = (2 + √3)/2 − 2 = (√3 − 2)/2.

x⁴ + 1/x⁴ = [(√3 − 2)/2]² − 2 = (3 + 4 − 4√3)/4 − 2 = (7 − 4√3 − 8)/4 = (−4√3 − 1)/4.

∴ The correct answer is (−4√3 − 1)/4.

Alternate Method

Given: x + (1/x) = (√3 + 1)/2

Formula Used: (a + b)² = a² + b² + 2ab

⇒ [x + (1/x)]² = [(√3 + 1)/2]²

⇒ x² + (1/x²) + 2 = (3 + 1 + 2√3) / 4

⇒ x² + (1/x²) = (4 + 2√3) / 4 − 2 = (2 + √3) / 2 − 2

⇒ x² + (1/x²) = (2 + √3 − 4) / 2 = (√3 − 2) / 2

⇒ [x² + (1/x²)]² = [(√3 − 2) / 2]²

⇒ x⁴ + (1/x⁴) + 2 = (3 + 4 − 4√3) / 4

⇒ x⁴ + (1/x⁴) = (7 − 4√3) / 4 − 2 = (7 − 4√3 − 8) / 4

⇒ x⁴ + (1/x⁴) = (−4√3 − 1) / 4

∴ The correct answer is (−4√3 − 1)/4.

Additional Information

Reciprocal Power Shortcuts

For any value x + 1/x = k, the even powers are found by repeating the operation k² − 2. For x³ + 1/x³, use k³ − 3k.

Negative Reciprocal Case

If x − 1/x = k, then x² + 1/x² = k² + 2. Note that the sign between terms changes based on the identity expansion.

Power 5 Calculation

To find x⁵ + 1/x⁵, use the product of the square and cubic forms: (x² + 1/x²)(x³ + 1/x³) − (x + 1/x).

Shortcut Trick

Given equation: a + a² + a³ = 1

Divide the entire equation by 'a': 1 + a + a² = 1/a

We need to find the value of: a³ + 1/a

Substitute the value of 1/a: a³ + (1 + a + a²)

Group the terms: (a + a² + a³) + 1 = 1 + 1 = 2

∴ The correct answer is 2.

Alternate Method

Given: a + a² + a³ − 1 = 0

Formula Used: Algebraic manipulation by isolation and substitution.

⇒ a + a² + a³ = 1 ... (Eq. 1)

⇒ To find the value of 1/a, divide (Eq. 1) by 'a' on both sides:

⇒ (a + a² + a³) / a = 1 / a

⇒ (a/a) + (a²/a) + (a³/a) = 1/a

⇒ 1 + a + a² = 1/a ... (Eq. 2)

⇒ The required expression is a³ + 1/a

⇒ Substituting the value of 1/a from (Eq. 2) into the expression:

⇒ a³ + (1 + a + a²)

⇒ Rearranging the terms: (a + a² + a³) + 1

⇒ From (Eq. 1), we know that a + a² + a³ = 1

⇒ Value = 1 + 1 = 2

∴ The correct answer is 2.

Additional Information

Algebraic Rearrangement

When faced with equations involving higher powers of a variable and their reciprocals, always look for ways to relate the reciprocal (1/x) to the polynomial by dividing the main equation by 'x'.

Geometric Progression in Polynomials

Equations of the form a + a² + a³ + ... + aⁿ follow the sum of a geometric series where the first term is 'a' and the common ratio is 'a'. The sum can be represented as a(1 − aⁿ) / (1 − a).

Substitution Method

If a specific combination of terms appears in the given equation and the required expression, substitute the known numerical value of that group directly to simplify the calculation.

Shortcut Trick

Rewrite the given equations: a − b = 1/a, b − c = 1/b, and c − a = 1/c.

The required expression (1/ab) + (1/bc) + (1/ca) is equal to (a − b)(b − c) + (b − c)(c − a) + (c − a)(a − b).

Expanding this cyclic sum, we get: (ab + bc + ca) − (a² + b² + c²).

From the original relations, a(a − b) = 1, b(b − c) = 1, and c(c − a) = 1, adding them gives (a² + b² + c²) − (ab + bc + ca) = 3.

The required value is the negative of this sum: −3.

∴ The correct answer is -3.

Alternate Method

Given Equations:

1. a − (1/a) = b ⇒ a − b = 1/a

2. b − (1/b) = c ⇒ b − c = 1/b

3. c − (1/c) = a ⇒ c − a = 1/c

To find: (1/ab) + (1/bc) + (1/ca)

⇒ (1/a)(1/b) + (1/b)(1/c) + (1/c)(1/a)

Substituting the values of 1/a, 1/b, and 1/c:

⇒ (a − b)(b − c) + (b − c)(c − a) + (c − a)(a − b)

⇒ (ab − ac − b² + bc) + (bc − ba − c² + ca) + (ca − cb − a² + ab)

⇒ (ab − ba + ab) + (bc + bc − cb) + (−ac + ca + ca) − (a² + b² + c²)

⇒ ab + bc + ca − (a² + b² + c²)

From the original equations:

⇒ a² − 1 = ab ⇒ a² − ab = 1

⇒ b² − 1 = bc ⇒ b² − bc = 1

⇒ c² − 1 = ca ⇒ c² − ca = 1

Adding these three equations:

⇒ (a² + b² + c²) − (ab + bc + ca) = 1 + 1 + 1 = 3

∴ ab + bc + ca − (a² + b² + c²) = −3.

∴ The correct answer is -3.

Additional Information

Cyclic Algebraic Identities

For variables in a cycle, expressions like (a−b)(b−c) + (b−c)(c−a) + (c−a)(a−b) always simplify to −(a² + b² + c² − ab − bc − ca).

Sum of Squared Differences

The identity a² + b² + c² − ab − bc − ca is equal to 1/2 [(a−b)² + (b−c)² + (c−a)²].

Reciprocal Difference Property

If x − 1/x = y, then x² − xy = 1. This property is useful for solving systems with symmetric reciprocal terms.

Shortcut Trick For any quadratic equation Ax2 + Bx + C = 0, if the sum of coefficients A + B + C = 0, then one root is always 1. Here, a(b − c) + b(c − a) + c(a − b) = ab − ac + bc − ab + ac − bc = 0. Since roots are equal, both roots must be 1. ∴ Product of roots (C/A) = 1 × 1 = 1. c(a − b) / a(b − c) = 1 ⇒ ac − bc = ab − ac ⇒ 2ac = ab + bc. Dividing throughout by abc, we get: 2/b = 1/c + 1/a. ∴ The correct answer is 2/b = (1/a) + (1/c).

Alternate Method Given Equation: a(b − c)x2 + b(c − a)x + c(a − b) = 0 This is of the form Ax2 + Bx + C = 0, where: A = a(b − c) B = b(c − a) C = c(a − b) Calculating the sum of coefficients: ⇒ A + B + C = a(b − c) + b(c − a) + c(a − b) ⇒ A + B + C = ab − ac + bc − ab + ac − bc = 0 When the sum of coefficients is zero, the roots are 1 and C/A. Given that the roots are equal: ⇒ 1 = C/A ⇒ 1 = [c(a − b)] / [a(b − c)] ⇒ a(b − c) = c(a − b) ⇒ ab − ac = ac − bc ⇒ ab + bc = 2ac Dividing both sides by abc: ⇒ (ab / abc) + (bc / abc) = 2ac / abc ⇒ 1/c + 1/a = 2/b ∴ The correct answer is 2/b = (1/a) + (1/c).

Additional Information Nature of Roots (Discriminant) For any quadratic equation ax2 + bx + c = 0, the discriminant D = b2 − 4ac determines the nature. If D = 0, roots are real and equal. If D > 0, roots are real and distinct. Harmonic Progression (HP) Three numbers a, b, c are in HP if their reciprocals 1/a, 1/b, 1/c are in Arithmetic Progression (AP). This implies 2/b = 1/a + 1/c. Sum and Product of Roots For Ax2 + Bx + C = 0, the sum of roots is −B/A and the product of roots is C/A.

Shortcut Trick

Assume b = 0 to satisfy the given equation easily.

⇒ √(a² + 0 + 0) + √(a² + 0 − 0) = 1

⇒ |a| + |a| = 1 ⇒ 2a = 1 ⇒ a = 1/2

⇒ a² = 1/4

Evaluate (1 − a²)(1 − b²) = (1 − 1/4)(1 − 0) = 3/4.

∴ The correct answer is 3/4.

Alternate Method

Given: √(a² + b² + ab) + √(a² + b² − ab) = 1

Formula Used: (X + Y)(X − Y) = X² − Y²

⇒ Let X = √(a² + b² + ab) and Y = √(a² + b² − ab)

⇒ X + Y = 1 ... (i)

⇒ X² − Y² = (a² + b² + ab) − (a² + b² − ab) = 2ab

⇒ (X + Y)(X − Y) = 2ab ⇒ 1 × (X − Y) = 2ab ⇒ X − Y = 2ab ... (ii)

⇒ Adding (i) and (ii): 2X = 1 + 2ab ⇒ X = (1 + 2ab) / 2

⇒ Squaring both sides: X² = (1 + 4a²b² + 4ab) / 4

⇒ a² + b² + ab = 1/4 + a²b² + ab

⇒ a² + b² − a²b² = 1/4

⇒ (1 − a²)(1 − b²) = 1 − a² − b² + a²b² = 1 − (a² + b² − a²b²)

⇒ (1 − a²)(1 − b²) = 1 − 1/4 = 3/4

∴ The correct answer is 3/4.

Additional Information

Algebraic Substitution

When an algebraic equation must hold for any values of the variables, substituting simple constants like 0 or 1 can bypass long derivations.

Conjugate Surds

Multiplying the sum of two radicals (√A + √B) by their difference (√A − √B) results in the rational expression (A − B).

Difference of Squares

The identity x² − y² = (x − y)(x + y) is fundamental for simplifying complex radical equations by eliminating the square root terms.

Shortcut Trick

Simplify equations: 3x + 4y = 29 and 4x − 3y = −3.

Equalize coefficients: (3x + 4y = 29) × 3 ⇒ 9x + 12y = 87.

(4x − 3y = −3) × 4 ⇒ 16x − 12y = −12.

Add results: 25x = 75 ⇒ x = 3. Substitute x: 12 − 3y = −3 ⇒ y = 5.

Required value: 5(3) − 3(5) − 9 = 15 − 15 − 9 = −9.

∴ The correct answer is −9.

Alternate Method

Given Equations:

1. 3x + 4y − 11 = 18

2. 8x − 6y + 12 = 6

Method: Elimination Method

⇒ 3x + 4y = 18 + 11 ⇒ 3x + 4y = 29 ... (i)

⇒ 8x − 6y = 6 − 12 ⇒ 8x − 6y = −6 ⇒ 4x − 3y = −3 ... (ii)

⇒ Multiply (i) by 3: 9x + 12y = 87 ... (iii)

⇒ Multiply (ii) by 4: 16x − 12y = −12 ... (iv)

⇒ Adding (iii) and (iv): 25x = 75 ⇒ x = 3

⇒ Substituting x = 3 in (ii): 4(3) − 3y = −3

⇒ 12 − 3y = −3 ⇒ 3y = 15 ⇒ y = 5

⇒ Value of 5x − 3y − 9 = 5(3) − 3(5) − 9

⇒ 15 − 15 − 9 = −9

∴ The correct answer is −9.

Additional Information

Simultaneous Linear Equations

A pair of linear equations in two variables a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 can be solved by elimination, substitution, or cross-multiplication.

Consistency Conditions

Unique solution exists if a₁/a₂ ≠ b₁/b₂. No solution exists if a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Infinite solutions exist if a₁/a₂ = b₁/b₂ = c₁/c₂.

Substitution Method

Isolate one variable in one equation and substitute its expression into the second equation to find the value of the other variable.

Shortcut Trick

To find a + c, we can find the value of b and subtract it from the total sum (a + b + c).

From Eq(2) − 3 × Eq(1): −7b + 2c = −1

From Eq(3) − 7 × Eq(1): −18b − 20c = −14/3

Solving these two equations gives b = 1/6.

Since a + b + c = 7/12, then a + c = 7/12 − 1/6 = 7/12 − 2/12 = 5/12.

∴ The correct answer is 5/12.

Alternate Method

Given:

Equation (1): a + b + c = 7/12

Equation (2): 3a − 4b + 5c = 3/4 = 9/12

Equation (3): 7a − 11b − 13c = −7/12

Formula Used:

Method of elimination for solving a system of linear equations.

Calculations:

Eliminate b using Equation (1) and (2):

⇒ 4 × (a + b + c) + (3a − 4b + 5c) = 4 × (7/12) + 9/12

⇒ 4a + 4b + 4c + 3a − 4b + 5c = 28/12 + 9/12

⇒ 7a + 9c = 37/12 — (Equation 4)

Eliminate b using Equation (1) and (3):

⇒ 11 × (a + b + c) + (7a − 11b − 13c) = 11 × (7/12) + (−7/12)

⇒ 11a + 11b + 11c + 7a − 11b − 13c = 77/12 − 7/12

⇒ 18a − 2c = 70/12

⇒ 9a − c = 35/12 — (Equation 5)

From Equation 5, c = 9a − 35/12. Substitute this into Equation 4:

⇒ 7a + 9 × (9a − 35/12) = 37/12

⇒ 7a + 81a − 315/12 = 37/12

⇒ 88a = 352/12

⇒ 88a = 88/3 ⇒ a = 1/3

Substitute a = 1/3 back into Equation 5:

⇒ 9 × (1/3) − c = 35/12

⇒ 3 − c = 35/12

⇒ c = 3 − 35/12 = (36 − 35)/12 = 1/12

Final calculation for a + c:

⇒ a + c = 1/3 + 1/12 = 4/12 + 1/12 = 5/12

∴ The correct answer is 5/12.

Additional Information

System of Linear Equations

A system of three linear equations in three variables can have a unique solution, infinitely many solutions, or no solution based on the determinant of the coefficient matrix.

Elimination Method

This method involves adding or subtracting equations to systematically remove variables until a single-variable equation is reached.

Cramer’s Rule

A formulaic approach using determinants where x_i = det(A_i) ÷ det(A), useful for small systems of equations with non-zero determinants.

Shortcut Trick

In ΔPQS, PQ = PS ⇒ ∠PSQ = (180° − 40°) ÷ 2 = 70°.

In ΔPSR, PS = SR ⇒ Exterior ∠PSQ = ∠SPR + ∠SRP.

Since PS = SR, ∠SPR = ∠SRP = 70° ÷ 2 = 35°.

Total ∠QPR = ∠QPS + ∠SPR = 40° + 35° = 75°.

∴ The correct answer is 75°.

Alternate Method

Given: PQ = PS = SR and ∠QPS = 40°.

Formula: Angle sum property of triangle = 180°; Isosceles triangle angles opposite to equal sides are equal.

⇒ In ΔPQS, PQ = PS

⇒ ∠PQS = ∠PSQ

⇒ ∠PQS + ∠PSQ + ∠QPS = 180°

⇒ 2 × ∠PSQ + 40° = 180°

⇒ 2 × ∠PSQ = 140° ⇒ ∠PSQ = 70°

⇒ In ΔPSR, PS = SR

⇒ ∠SPR = ∠SRP

⇒ ∠PSQ = ∠SPR + ∠SRP (Exterior angle property)

⇒ 70° = 2 × ∠SPR ⇒ ∠SPR = 35°

⇒ ∠QPR = ∠QPS + ∠SPR

⇒ ∠QPR = 40° + 35° = 75°

∴ The correct answer is 75°.

Additional Information

Isosceles Triangle Property

In a triangle, if two sides are equal, the angles opposite to those sides are always equal.

Exterior Angle Theorem

An exterior angle of a triangle is equal to the sum of its two opposite interior angles.

Angle Sum Property

The sum of all three interior angles of a triangle is 180°.

Shortcut Trick

Length of median PD = (1/2)√(2PQ² + 2PR² − QR²)

PD = (1/2)√(2(30²) + 2(50²) − 36²) = (1/2)√(1800 + 5000 − 1296) = (1/2)√5504

PD = (1/2) × 8√86 = 4√86 cm

Centroid C divides median PD in ratio 2 : 1, where CD is the part from the midpoint to the centroid.

CD = (1/3) × PD = (1/3) × 4√86 = (4√86)/3 cm

∴ The correct answer is (4√86)/3 cm.

Alternate Method

Given: PQ = 30 cm, PR = 50 cm, QR = 36 cm. D is the midpoint of QR. C is the centroid.

Formula Used: Apollonius' Theorem: PQ² + PR² = 2(PD² + (QR/2)²)

In ΔPQR, D is the midpoint of QR, so QD = DR = 36 ÷ 2 = 18 cm.

⇒ By Apollonius' Theorem: PQ² + PR² = 2(PD² + QD²)

⇒ 30² + 50² = 2(PD² + 18²)

⇒ 900 + 2500 = 2(PD² + 324)

⇒ 3400 = 2(PD² + 324)

⇒ 1700 = PD² + 324

⇒ PD² = 1700 − 324 = 1376

⇒ PD = √1376 = √(16 × 86) = 4√86 cm

⇒ Centroid C divides PD in the ratio PC : CD = 2 : 1.

⇒ CD = (1/3) × PD

⇒ CD = (1/3) × 4√86 = (4√86)/3 cm

∴ The correct answer is (4√86)/3 cm.

Additional Information

Apollonius' Theorem

Relates the lengths of the sides of a triangle to the length of its median: AB² + AC² = 2(AM² + BM²), where AM is the median to side BC.

Centroid Property

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

Sum of Squares of Medians

3 × (Sum of squares of sides) = 4 × (Sum of squares of medians). For ΔABC: 3(AB² + BC² + CA²) = 4(m_a² + m_b² + m_c²).

Shortcut Trick

In triangle ABC, since PQ is parallel to BC, the segments on the sides are proportional (BPT).

Ratio of segments on side AC → AQ : QC = 4√2 : 6√2 = 2 : 3.

By property, the same ratio applies to side AB → AP : PB = 2 : 3.

Total parts for AB = 2 + 3 = 5 parts.

Given AB = 20 cm, so 5 parts = 20 cm ⇒ 1 part = 4 cm.

PB = 3 parts = 3 × 4 = 12 cm.

∴ The correct answer is 12 cm.

Alternate Method

Given: AQ = 4√2 cm, QC = 6√2 cm, AB = 20 cm, PQ || BC.

Formula Used: Basic Proportionality Theorem (BPT) → AP/PB = AQ/QC

Let PB = x cm. Then AP = AB − PB = (20 − x) cm.

According to Basic Proportionality Theorem:

⇒ AP / PB = AQ / QC

⇒ (20 − x) / x = (4√2) / (6√2)

⇒ (20 − x) / x = 4 / 6

⇒ (20 − x) / x = 2 / 3

⇒ 3 × (20 − x) = 2x

⇒ 60 − 3x = 2x

⇒ 5x = 60

⇒ x = 60 ÷ 5

⇒ x = 12 cm.

∴ The correct answer is 12 cm.

Additional Information

Basic Proportionality Theorem (Thales's Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Similar Triangles Property (AA Criterion)

When PQ || BC, ΔAPQ ∼ ΔABC. This implies AP/AB = AQ/AC = PQ/BC.

Ratio of Areas

Area(ΔAPQ) / Area(ΔABC) = (AP/AB)² = (AQ/AC)² = (PQ/BC)².

Shortcut Trick

In the given triangle, ∠AED = ∠ABC and ∠A is common, so ▵AED ∼ ▵ABC.

By similarity property, the ratio of corresponding sides is equal: AE ÷ AB = AD ÷ AC.

8 ÷ (12 + BD) = 12 ÷ (8 + 14)

8 ÷ (12 + BD) = 12 ÷ 22 ⇒ 2 ÷ (12 + BD) = 3 ÷ 22

12 + BD = 44 ÷ 3 ⇒ BD = 44/3 − 12 = 8/3 cm.

∴ The correct answer is 8/3 cm.

Alternate Method

Given: AD = 12 cm, AE = 8 cm, EC = 14 cm.

Formula: If two angles of one triangle are equal to two angles of another, the triangles are similar (AA similarity).

⇒ Total length AC = AE + EC = 8 + 14 = 22 cm

⇒ In ▵AED and ▵ABC:

⇒ ∠A = ∠A (Common)

⇒ ∠AED = ∠ABC (Given)

⇒ ∴ ▵AED ∼ ▵ABC (AA Similarity)

⇒ AE ÷ AB = AD ÷ AC

⇒ 8 ÷ (AD + BD) = 12 ÷ 22

⇒ 8 ÷ (12 + BD) = 6 ÷ 11

⇒ 6 × (12 + BD) = 8 × 11

⇒ 72 + 6BD = 88

⇒ 6BD = 88 − 72 = 16

⇒ BD = 16 ÷ 6 = 8/3 cm

∴ The correct answer is 8/3 cm.

Additional Information

AA Similarity Criterion

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar and their corresponding sides are in the same ratio.

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(▵1) ÷ Area(▵2) = (s₁ ÷ s₂)².

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Shortcut Trick

Check for Pythagorean triplet: 9² + 12² = 81 + 144 = 225 = 15².

Since the radii and the distance between centers form a right-angled triangle, the altitude to the hypotenuse is exactly half of the common chord.

Altitude (h) = (Radius₁ × Radius₂) ÷ Distance between centres

Length of common chord = 2 × (9 × 12 ÷ 15) = 2 × 108 ÷ 15 = 2 × 7.2 = 14.4 cm.

∴ The correct answer is 14.4 cm.

Alternate Method

Given: Radii r₁ = 9 cm, r₂ = 12 cm, Distance (d) = 15 cm.

In ΔC₁PC₂, where P is the intersection point:

⇒ Sides are 9 cm, 12 cm, and 15 cm.

⇒ Since 9² + 12² = 15², ΔC₁PC₂ is a right-angled triangle with ∠C₁PC₂ = 90°.

⇒ Area of ΔC₁PC₂ = (1 ÷ 2) × base × height = (1 ÷ 2) × 9 × 12 = 54 cm².

⇒ Let h be the perpendicular from P to C₁C₂. This h is half the common chord.

⇒ Area = (1 ÷ 2) × C₁C₂ × h

⇒ 54 = (1 ÷ 2) × 15 × h

⇒ 54 = 7.5 × h

⇒ h = 54 ÷ 7.5 = 7.2 cm.

⇒ Length of common chord PQ = 2 × h = 2 × 7.2 = 14.4 cm.

∴ The correct answer is 14.4 cm.

Additional Information

Common Chord Property

The line joining the centres of two intersecting circles is the perpendicular bisector of their common chord.

General Formula

Length of common chord = (2 × Area of triangle formed by radii and distance between centres) ÷ Distance between centres.

Condition for Orthogonal Circles

Two circles are said to be orthogonal if the square of the distance between their centres is equal to the sum of the squares of their radii (d² = r₁² + r₂²).

Shortcut Trick

Calculate distance from larger circle center (O₁) to P: O₁P = √(15² + 20²) = 25 cm.

Distance from point X (where circles touch) to P: XP = O₁P − R = 25 − 15 = 10 cm.

Using similarity ratio for radii: r / R = (XP − r) / O₁P

r / 15 = (10 − r) / 25 ⇒ 5r = 3(10 − r) ⇒ 8r = 30

r = 3.75 cm.

∴ The correct answer is 3.75 cm.

Alternate Method

Given: Radius of larger circle (R) = 15 cm, Tangent length (CP) = 20 cm. Point X is the point of contact between circles.

Formula: In a right-angled triangle, Hypotenuse² = Base² + Perpendicular². Similar triangles property: Ratio of corresponding sides is equal.

⇒ In right ▵O₁CP, O₁C = 15 cm and CP = 20 cm.

⇒ O₁P = √(O₁C² + CP²) = √(15² + 20²) = √(225 + 400) = √625 = 25 cm.

⇒ The distance from the center O₁ to the point of contact X is the radius R = 15 cm.

⇒ The distance XP = O₁P − O₁X = 25 − 15 = 10 cm.

⇒ Let the radius of the smaller circle be r. Then O₂X = r and O₂P = XP − O₂X = 10 − r.

⇒ By similarity of ▵O₂DP and ▵O₁CP (where D is contact point on smaller circle):

⇒ r / R = O₂P / O₁P

⇒ r / 15 = (10 − r) / 25

⇒ 25r = 15(10 − r) ⇒ 25r = 150 − 15r

⇒ 40r = 150 ⇒ r = 150 ÷ 40 = 3.75 cm.

∴ The correct answer is 3.75 cm.

Additional Information

Direct Common Tangents (DCT)

Length of DCT = √(d² − (R − r)²), where d is the distance between centers.

Transverse Common Tangents (TCT)

Length of TCT = √(d² − (R + r)²). TCTs exist only if d > R + r.

Circles Touching Externally

When two circles touch externally, the distance between their centers d = R + r and they have exactly 3 common tangents.

Shortcut Trick

When two circles touch each other externally, the common internal tangent bisects the segment of the common external tangent.

The point of intersection A of the tangents lies such that AY = AX and AZ = AX.

Therefore, AX is exactly half of the common tangent YZ.

Formula: YZ = 2 × AX

YZ = 2 × 16 = 32 cm.

∴ The correct answer is 32 cm.

Alternate Method

Given:

Length of the internal tangent segment from intersection point A to contact point X, AX = 16 cm.

Formula Used:

The lengths of tangents drawn from an external point to a circle are equal.

Calculations:

⇒ For the first circle, A is an external point. Tangents are AY and AX.

⇒ ∴ AY = AX = 16 cm

⇒ For the second circle, A is an external point. Tangents are AZ and AX.

⇒ ∴ AZ = AX = 16 cm

⇒ The total length of the common tangent YZ = AY + AZ

⇒ YZ = 16 + 16

⇒ YZ = 32 cm

∴ The correct answer is 32 cm.

Additional Information

Length of Direct Common Tangent (DCT)

For two circles with radii r₁ and r₂ and distance between centers d, DCT = √(d² − (r₁ − r₂)²).

Length of Transverse Common Tangent (TCT)

The length of the transverse common tangent is given by TCT = √(d² − (r₁ + r₂)²).

Angle Subtended at Point of Contact

When two circles touch externally, the angle subtended by the common tangent segment YZ at the point of contact X is always 90° (∠YXZ = 90°).

Shortcut Trick

In a circle with 8 equidistant points, each arc segment is 360° ÷ 8 = 45°.

The arc FGH spans 2 segments (FG and GH), so Arc FGH = 2 × 45° = 90°.

The inscribed angle ∠FDH subtends arc FGH.

By the inscribed angle theorem, ∠FDH = ½ × Arc FGH = ½ × 90° = 45°.

∴ The correct answer is 45°.

Alternate Method

Given:

Points A, B, C, D, E, F, G, H are 8 equidistant points on a circle.

Formula:

Angle subtended by an arc at the circumference = ½ × Angle subtended by the same arc at the center.

Calculations:

⇒ Total points = 8

⇒ Central angle of each arc segment (AB, BC, ...) = 360° ÷ 8 = 45°

⇒ Arc FGH = Arc FG + Arc GH

⇒ Arc FGH = 45° + 45° = 90°

⇒ Since ∠FDH is an inscribed angle subtended by Arc FGH:

⇒ ∠FDH = ½ × (Measure of Arc FGH)

⇒ ∠FDH = ½ × 90° = 45°

∴ The correct answer is 45°.

Additional Information

Inscribed Angle Theorem

An angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle.

Angles Subtended by Equal Arcs

Equal arcs (or chords) of a circle subtend equal angles at the center and equal angles at any point on the remaining part of the circumference.

Angle in a Semicircle

An angle inscribed in a semicircle is always a right angle (90°), as it subtends an arc of 180°.

Shortcut Trick

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

Therefore, Angle at Circumference = Angle at Centre ÷ 2

∠RPQ = ∠QOR ÷ 2

∠RPQ = 50° ÷ 2 = 25°

∴ The correct answer is 25°.

Alternate Method

Given: O is the centre of the circle and ∠QOR = 50°.

Formula Used: 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.

Calculations:

⇒ ∠QOR = 2 × ∠RPQ

⇒ 50° = 2 × ∠RPQ

⇒ ∠RPQ = 50° ÷ 2

⇒ ∠RPQ = 25°

∴ The correct answer is 25°.

Additional Information

Central Angle Theorem

The angle subtended by an arc at the center is exactly twice the angle it subtends at the circumference (∠Center = 2 × ∠Circumference).

Angles in the Same Segment

All angles subtended by the same arc at any point on the circumference are equal (∠P₁ = ∠P₂).

Angle in a Semicircle

An angle subtended by the diameter at the circumference is always a right angle (90°).

Cyclic Quadrilateral

In a cyclic quadrilateral, the sum of opposite internal angles is always 180° (∠A + ∠C = 180°).

Shortcut Trick

When circles are placed such that they touch two converging lines and also touch each other externally, their radii (r₁, r₂, r₃, ...) form a Geometric Progression (G.P.).

For three such circles, the common ratio is constant: r₂/r₁ = r₃/r₂.

Applying the G.P. property: r₂² = r₁ × r₃.

Taking square root on both sides: r₂ = √(r₁r₃).

∴ The correct answer is √(r₁r₃).

Alternate Method

Given: Three circles C₁, C₂, and C₃ with radii r₁, r₂, and r₃ touching two lines and each other.

Formula Used: Similarity of triangles and trigonometric ratios in right-angled triangles.

⇒ Let the angle between the two lines be 2θ. The centers of the circles lie on the angle bisector.

⇒ In the right triangle formed by the vertex O and center of circle C_n: sin θ = r_n / d_n, where d_n is the distance from O to the center.

⇒ d₁ = r₁ / sin θ and d₂ = r₂ / sin θ.

⇒ Distance between centers C₁ and C₂ is r₁ + r₂.

⇒ d₂ − d₁ = r₁ + r₂

⇒ (r₂ / sin θ) − (r₁ / sin θ) = r₁ + r₂

⇒ (r₂ − r₁) / sin θ = r₁ + r₂ ⇒ sin θ = (r₂ − r₁) / (r₂ + r₁)

⇒ Similarly for circles C₂ and C₃: sin θ = (r₃ − r₂) / (r₃ + r₂)

⇒ (r₂ − r₁) / (r₂ + r₁) = (r₃ − r₂) / (r₃ + r₂)

⇒ (r₂ − r₁)(r₃ + r₂) = (r₃ − r₂)(r₂ + r₁)

⇒ r₂r₃ + r₂² − r₁r₃ − r₁r₂ = r₃r₂ + r₁r₃ − r₂² − r₁r₂

⇒ 2r₂² = 2r₁r₃ ⇒ r₂² = r₁r₃

⇒ r₂ = √(r₁r₃)

∴ The correct answer is √(r₁r₃).

Additional Information

Geometric Progression of Radii

If 'n' circles are inscribed in an angle touching each other and the sides, their radii r₁, r₂, ..., r_n are in G.P. with common ratio k = (1 + sin θ) / (1 − sin θ).

Direct Common Tangent

The length of the direct common tangent between two circles with radii r₁ and r₂ touching each other externally is 2√(r₁r₂).

Angle Bisector Property

The centers of all circles that touch two intersecting lines must lie on the angle bisector of those lines.

Shortcut Trick

An equilateral triangle of side 3 units can be divided into 9 smaller equilateral triangles of side 1 unit.

Cutting the three vertices (each an equilateral triangle of side 1 unit) leaves a regular hexagon.

Area of Hexagon = Area of Triangle − 3 × (Area of small triangle at vertex).

Area of Hexagon = 9 units − 3 units = 6 units.

Required Percentage = (6 ÷ 9) × 100 = 66.66%.

∴ The correct answer is 66.66%.

Alternate Method

Given:

Area of original equilateral triangle = 300 cm²

Side of the regular hexagon = 1/3 × Side of the original triangle

Formula:

Area of Equilateral Triangle = (√3 ÷ 4) × (side)²

Area of Regular Hexagon = 6 × [(√3 ÷ 4) × (side of hexagon)²]

Calculations:

⇒ Let the side of the original equilateral triangle be a.

⇒ Side of the regular hexagon = a/3.

⇒ Area of original triangle (T) = (√3 ÷ 4) × a² = 300 cm².

⇒ The hexagon is made of 6 small equilateral triangles, each with side a/3.

⇒ Area of one small triangle (t) = (√3 ÷ 4) × (a/3)² = [(√3 ÷ 4) × a²] ÷ 9.

⇒ Area of one small triangle (t) = 300 ÷ 9 = 33.33 cm².

⇒ Area of Hexagon (H) = 6 × Area of small triangle (t).

⇒ Area of Hexagon (H) = 6 × 33.33 = 200 cm².

⇒ Required Percentage = (Area of Hexagon ÷ Area of Triangle) × 100.

⇒ Percentage = (200 ÷ 300) × 100 = 66.66%.

∴ The correct answer is 66.66%.

Additional Information

Regular Hexagon Formation

A regular hexagon can be formed by cutting three equilateral triangles of side a/3 from the corners of a larger equilateral triangle of side a.

Area Ratio of Similar Triangles

If the ratio of sides of two equilateral triangles is x:y, then the ratio of their areas is x²:y².

Regular Hexagon Area

The area of a regular hexagon with side s is given by 3√3/2 × s², which is equivalent to 6 times the area of an equilateral triangle with the same side.

Shortcut Trick

Area of equilateral ΔPQR = (√3 / 4) × side² = (√3 / 4) × 12 × 12 = 36√3 cm².

Since S and T are midpoints, lines QS and RT are medians intersecting at centroid O.

The centroid divides an equilateral triangle into 3 triangles of equal area.

Shaded Area (ΔOQR) = (1/3) × Total Area = (1/3) × 36√3 = 12√3 cm².

∴ The correct answer is 12√3 cm².

Alternate Method

Given: Side of equilateral triangle (a) = 12 cm. S and T are midpoints of PQ and PR.

Formula Used: Area of equilateral triangle = (√3/4)a²; Centroid Property: Area(OQR) = Area(PQR)/3.

⇒ Total Area of ΔPQR = (√3 / 4) × (12)²

⇒ Total Area = (√3 / 4) × 144 = 36√3 cm²

⇒ S and T are midpoints, so QS and RT are medians.

⇒ Let O be the point of intersection of medians (Centroid).

⇒ In an equilateral triangle, the centroid O divides it into 3 equal areas: ΔPOQ, ΔPOR, and ΔOQR.

⇒ Shaded Area (ΔOQR) = Total Area ÷ 3

⇒ Shaded Area = 36√3 ÷ 3 = 12√3 cm²

∴ The correct answer is 12√3 cm².

Additional Information

Area of Equilateral Triangle

The area is given by (√3/4) × a² and the height is (√3/2) × a, where 'a' is the side length.

Centroid and Area

The centroid divides the triangle into three smaller triangles of equal area and each median into a 2:1 ratio.

Midpoint Theorem

The segment joining the midpoints of two sides of a triangle (ST) is parallel to the third side (QR) and is half its length.

Shortcut Trick

For any point P on or inside a rectangle ABCD, the relation holds: AP² + CP² = BP² + DP².

Substitute the given values: AP² + 10² = 6² + 13²

⇒ AP² + 100 = 36 + 169

⇒ AP² = 205 − 100 = 105

⇒ AP = √105

∴ The correct answer is √105.

Alternate Method

Given:

Rectangle ABCD where DP = 13, CP = 10, BP = 6 and P lies on side AB.

Formula Used:

Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular²

Calculations:

In rectangle ABCD, let the height AD = BC = h.

Since ABCD is a rectangle, ∠PBC = 90° and ∠PAD = 90°.

In ▵PBC:

⇒ BC² + BP² = CP²

⇒ h² + 6² = 10²

⇒ h² + 36 = 100

⇒ h² = 64

In ▵PAD:

⇒ AD² + AP² = DP²

⇒ h² + AP² = 13²

⇒ 64 + AP² = 169

⇒ AP² = 169 − 64

⇒ AP² = 105

⇒ AP = √105

∴ The correct answer is √105.

Additional Information

British Flag Theorem

For any point P in the plane of a rectangle ABCD, the sum of the squares of distances to opposite vertices is equal: PA² + PC² = PB² + PD².

Rectangle Properties

Opposite sides are equal and parallel (AD = BC, AB = CD), and all interior angles are 90°.

Pythagorean Triplets

Common sets of integers satisfying a² + b² = c² include (3, 4, 5), (6, 8, 10), (5, 12, 13), and (8, 15, 17).

Shortcut Trick

The area of a triangle formed by joining alternate vertices of a regular hexagon is always exactly half of the hexagon's total area.

Area of Regular Hexagon = (3√3 ÷ 2) × (side)²

Area of Hexagon = (3√3 ÷ 2) × 12² = (3√3 ÷ 2) × 144 = 216√3 cm².

Area of ΔSQU = 216√3 ÷ 2 = 108√3 cm².

∴ The correct answer is 108√3 cm².

Alternate Method

Given: Side of the regular hexagon (a) = 12 cm.

Formula: Area of an equilateral triangle = (√3 ÷ 4) × (side)².

⇒ In a regular hexagon, the distance between alternate vertices (longer diagonal) s = a√3.

⇒ Side of equilateral ΔSQU (s) = 12 × √3 = 12√3 cm.

⇒ Area of ΔSQU = (√3 ÷ 4) × (12√3)²

⇒ Area = (√3 ÷ 4) × (144 × 3)

⇒ Area = √3 × 36 × 3

⇒ Area = 108√3 cm².

∴ The correct answer is 108√3 cm².

Additional Information

Properties of Regular Hexagon

A regular hexagon has 6 equal sides and 6 equal interior angles of 120° each. Area = (3√3 ÷ 2) × a².

Hexagon Diagonals

Short diagonal (joining alternate vertices) = a√3. Long diagonal (joining opposite vertices) = 2a.

Internal Triangles

Joining alternate vertices forms one large equilateral triangle. Joining all opposite vertices forms 6 small equilateral triangles.

Shortcut Trick

Internal angle of Equilateral Triangle ∠ABE = 60°.

Internal angle of Square ∠ABC = 90°.

Internal angle of Regular Pentagon ∠CBZ = [(5 − 2) × 180°] ÷ 5 = 108°.

Sum of all angles around point B = 360°.

∠EBZ = 360° − (∠ABE + ∠ABC + ∠CBZ) = 360° − (60° + 90° + 108°) = 102°.

∴ The correct answer is 102°.

Alternate Method

Given:

ABCD is a square ⇒ ∠ABC = 90°.

BCXYZ is a regular pentagon ⇒ Interior angle = [(n − 2) × 180°] ÷ n.

ABE is an equilateral triangle ⇒ ∠ABE = 60°.

Formula for interior angle of a regular polygon with n sides = [(n − 2) × 180°] ÷ n

⇒ Interior angle of regular pentagon (n=5) = [(5 − 2) × 180°] ÷ 5

⇒ ∠CBZ = (3 × 180°) ÷ 5 = 540° ÷ 5 = 108°

⇒ For square ABCD, ∠ABC = 90°

⇒ For equilateral triangle ABE, ∠ABE = 60°

⇒ Since angles ∠ABE, ∠ABC, ∠CBZ, and ∠EBZ meet at a single point B:

⇒ ∠EBZ + ∠ABE + ∠ABC + ∠CBZ = 360°

⇒ ∠EBZ + 60° + 90° + 108° = 360°

⇒ ∠EBZ + 258° = 360°

⇒ ∠EBZ = 360° − 258° = 102°

∴ The correct answer is 102°.

Additional Information

Sum of Interior Angles

The sum of the interior angles of any polygon with n sides is given by (n − 2) × 180°.

Exterior Angle of Regular Polygon

Each exterior angle of a regular polygon is 360° ÷ n. The sum of interior and exterior angles at a vertex is always 180°.

Common Polygon Angles

Regular Hexagon (n=6) interior angle = 120°; Regular Octagon (n=8) interior angle = 135°.

Complete Angle

The total angle at a point or around a vertex in a plane is always equal to 360°.

Shortcut Trick

Check for Pythagorean triplet: 21, 28, 35 → (3, 4, 5) × 7. Since 21² + 28² = 35², ΔABC is right-angled at B.

Property: For a right-angled triangle, the area of the lunes (shaded parts) formed by semicircles on the legs is equal to the area of the triangle itself.

Area of shaded part = Area of ΔABC = 1/2 × Base × Height

Area = 1/2 × 21 × 28 = 21 × 14 = 294 cm².

∴ The correct answer is 294 cm².

Alternate Method

Given:

Sides of the triangle: AB = 21 cm, BC = 28 cm, AC = 35 cm.

Formula Used:

Area of triangle = 1/2 × Base × Height

Area of semicircle = 1/2 × πr²

Checking the triangle type:

⇒ AB² + BC² = 21² + 28² = 441 + 784 = 1225

⇒ AC² = 35² = 1225

⇒ Since AB² + BC² = AC², ΔABC is a right-angled triangle at B.

Calculating the areas:

⇒ Area of semicircle on AB (S₁) = 1/2 × π × (21/2)²

⇒ Area of semicircle on BC (S₂) = 1/2 × π × (28/2)²

⇒ Area of semicircle on AC (S₃) = 1/2 × π × (35/2)²

⇒ Area of shaded part = (S₁ + S₂ + Area of ΔABC) − S₃

⇒ Since S₁ + S₂ = S₃ (by Pythagoras theorem), the S₁, S₂ and S₃ cancel out.

⇒ Area of shaded part = Area of ΔABC

⇒ Area = 1/2 × 21 × 28 = 294 cm².

∴ The correct answer is 294 cm².

Additional Information

Lunes of Hippocrates

This specific geometry problem is based on the Lunes of Hippocrates, which states that the sum of the areas of the two lunes formed on the legs of a right triangle is equal to the area of the triangle.

Pythagorean Triplets

Common triplets like (3, 4, 5), (5, 12, 13), and (7, 24, 25) are essential for identifying right-angled triangles. In this question, (21, 28, 35) is a multiple of (3, 4, 5).

Area of Semicircle

The area of a semicircle is given by A = (πr²)/2 or (πd²)/8, where r is the radius and d is the diameter.

Shortcut Trick

Area difference = π(R² − r²) = π(R + r)(R − r) = 2002.

Given R + r = 91, so (22/7) × 91 × (R − r) = 2002.

22 × 13 × (R − r) = 2002 ⇒ 286 × (R − r) = 2002 ⇒ R − r = 7.

Larger radius R = (Sum + Difference) ÷ 2 = (91 + 7) ÷ 2 = 98 ÷ 2 = 49 cm.

∴ The correct answer is 49 cm.

Alternate Method

Given:

Sum of radii (R + r) = 91 cm

Difference of areas (πR² − πr²) = 2002 cm²

Formula Used:

Area of circle = πr² | Algebraic identity: a² − b² = (a + b)(a − b)

⇒ πR² − πr² = 2002

⇒ π(R² − r²) = 2002

⇒ (22/7) × (R + r)(R − r) = 2002

⇒ (22/7) × 91 × (R − r) = 2002

⇒ 22 × 13 × (R − r) = 2002

⇒ 286 × (R − r) = 2002

⇒ R − r = 2002 ÷ 286

⇒ R − r = 7 --- (Eq. 1)

⇒ R + r = 91 --- (Eq. 2)

Adding Eq. 1 and Eq. 2:

⇒ 2R = 98

⇒ R = 49 cm

∴ The correct answer is 49 cm.

Additional Information

Area of a Circle

The total space occupied by the circle is given by A = πr², where π ≈ 22/7 or 3.14.

Area of a Ring (Annulus)

The area between two concentric circles is π(R² − r²) = π(R + r)(R − r).

Circumference of a Circle

The boundary length of a circle is calculated as C = 2πr or C = πd, where d is the diameter.

Diameter-Radius Relation

The diameter is the longest chord and is exactly twice the radius: d = 2r.

Shortcut Trick

Volume of a prism = Base Area × Height

Base is an equilateral triangle with side a = 15 cm.

Base Area = (√3/4) × a² = (√3/4) × 15² = 225√3/4 cm²

Volume = (225√3/4) × 20√3 = 225 × 5 × (√3 × √3) = 225 × 5 × 3

Volume = 225 × 15 = 3375 cm³

∴ The correct answer is 3375 cm³.

Alternate Method

Given:

Side of equilateral triangular base (a) = 15 cm

Height of the prism (h) = 20√3 cm

Formula Used:

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

Volume of Prism = Base Area × Height

Calculations:

⇒ Area of the base = (√3/4) × (15)²

⇒ Area of the base = (√3/4) × 225 cm²

⇒ Volume of the prism = [(√3/4) × 225] × 20√3

⇒ Volume = (225 × 20 × √3 × √3) ÷ 4

⇒ Volume = (225 × 20 × 3) ÷ 4

⇒ Volume = 225 × 5 × 3

⇒ Volume = 225 × 15

⇒ Volume = 3375 cm³

∴ The correct answer is 3375 cm³.

Additional Information

Lateral Surface Area (LSA) of Prism

LSA = Perimeter of base × Height of the prism. For this prism, LSA = (3 × 15) × 20√3 = 900√3 cm².

Total Surface Area (TSA) of Prism

TSA = Lateral Surface Area + 2 × Area of base. TSA = (Perimeter × h) + 2 × [(√3/4) × a²].

Volume of a Pyramid

The volume of a pyramid is exactly 1/3 the volume of a prism with the same base and height: Volume = (1/3) × Base Area × Height.

Shortcut Trick

Height of original cone (H) = 45 cm; Height of smaller cone (h) = 45 − 15 = 30 cm.

Ratio of heights (Small : Large) = 30 : 45 = 2 : 3.

Ratio of volumes = (Height ratio)³ = 2³ : 3³ = 8 : 27.

Given, 8 units = 18480 cm³ ⇒ 1 unit = 2310 cm³.

Volume of original cone = 27 units = 27 × 2310 = 62370 cm³.

∴ The correct answer is 62370 cm³.

Alternate Method

Given:

Total Height (H) = 45 cm

Cutting Height from base = 15 cm

Volume of smaller cone (v) = 18480 cm³

Formula Used:

Ratio of volumes of two similar cones = (Ratio of their heights)³

Calculations:

⇒ Height of smaller cone (h) = 45 − 15 = 30 cm

⇒ Ratio of heights (k) = h ÷ H = 30 ÷ 45 = 2/3

⇒ Volume of small cone ÷ Volume of large cone = (h/H)³

⇒ 18480 ÷ V = (2/3)³

⇒ 18480 ÷ V = 8/27

⇒ V = (18480 × 27) ÷ 8

⇒ V = 2310 × 27

⇒ V = 62370 cm³

∴ The correct answer is 62370 cm³.

Additional Information

Volume of Frustum

The volume of the bottom part remaining after the cut (frustum) is V_frustum = (1/3)πh(R² + r² + Rr), or simply V_large − V_small.

Similarity Property of Solids

For any two similar 3D figures, the ratio of their volumes is the cube of the ratio of any corresponding linear dimensions (height, radius, or slant height).

Surface Area Ratio

The ratio of the curved surface areas of the smaller cone to the original cone is (h/H)². In this case, (2/3)² = 4/9.

Shortcut Trick

Ratio of CSA : TSA = h : (h + r) = 2 : 5 → h = 2 units, r = 3 units.

Since TSA is 5 units = 3080 cm², then CSA (2πrh) = 2 units = 1232 cm².

Area of two bases (2πr²) = TSA − CSA = 3080 − 1232 = 1848 cm².

πr² = 924 ⇒ r² = 294 ⇒ r = 7√6 cm and h = (2/3)r = 14√6 / 3 cm.

Volume = πr²h = 924 × (14√6 / 3) = 4312√6 cm³.

∴ The correct answer is 4312√6 cm³.

Alternate Method

Given:

Total Surface Area (TSA) of cylinder = 3080 cm²

Ratio of Curved Surface Area (CSA) to TSA = 2 : 5

Formula Used:

CSA = 2πrh

TSA = 2πr(h + r)

Volume (V) = πr²h

Calculations:

⇒ CSA / TSA = (2πrh) / [2πr(h + r)] = 2 / 5

⇒ h / (h + r) = 2 / 5

⇒ 5h = 2h + 2r

⇒ 3h = 2r ⇒ h = 2r / 3

⇒ CSA = (2/5) × TSA = (2/5) × 3080 = 1232 cm²

⇒ TSA − CSA = Area of 2 circular bases = 3080 − 1232 = 1848 cm²

⇒ 2πr² = 1848

⇒ 2 × (22/7) × r² = 1848

⇒ r² = (1848 × 7) / 44 = 42 × 7 = 294

⇒ r = √294 = 7√6 cm

⇒ h = (2/3) × 7√6 = 14√6 / 3 cm

⇒ Volume = πr²h = (22/7) × 294 × (14√6 / 3)

⇒ Volume = 924 × (14√6 / 3)

⇒ Volume = 308 × 14√6 = 4312√6 cm³

∴ The correct answer is 4312√6 cm³.

Additional Information

Cylinder Surface Areas

The lateral surface area is 2πrh, and the total surface area includes the top and bottom circles: 2πrh + 2πr².

Volume Comparison

A cylinder has 3 times the volume of a cone with the same radius and height: V_cylinder = 3 × V_cone = πr²h.

Hollow Cylinder

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

Shortcut Trick

Volume of cylinder depends on r² and h (V ∝ r × r × h).

Successive % increase of 2% and 2% = 2 + 2 + (2 × 2)/100 = 4.04%.

Successive % increase of 4.04% and 2% = 4.04 + 2 + (4.04 × 2)/100 = 6.04 + 0.0808 = 6.1208%.

Approximate increase ≈ 6.12%.

∴ The correct answer is 6.12%.

Alternate Method

Given: Increase in radius (r) = 2%, Increase in height (h) = 2%.

Formula Used: Volume of cylinder (V) = πr²h

⇒ Let original radius (r) = 100 and original height (h) = 100.

⇒ Original Volume (V₁) = π × (100)² × 100 = 1,000,000π.

⇒ New radius (r') = 100 + 2% of 100 = 102.

⇒ New height (h') = 100 + 2% of 100 = 102.

⇒ New Volume (V₂) = π × (102)² × 102 = π × 10404 × 102 = 1,061,208π.

⇒ Increase in Volume = V₂ − V₁ = 1,061,208π − 1,000,000π = 61,208π.

⇒ Percentage Increase = (61,208π / 1,000,000π) × 100 = 6.1208%.

⇒ Approximate Percentage Increase ≈ 6.12%.

∴ The correct answer is 6.12%.

Additional Information

Successive Percentage for 3 Variables

For three successive changes a%, b%, and c%, the net change is given by [a + b + c + (ab + bc + ca)/100 + (abc)/10000]%.

Curved Surface Area (CSA) Change

CSA of cylinder = 2πrh. If r and h both increase by x%, the net increase in CSA is (x + x + x²/100)%.

Small Change Approximation

If a quantity depends on x^ay^b and variables change by very small %, the net change is approximately (a × %change in x + b × %change in y).

Shortcut Trick

Each octant (1/8th part) has 1/8th of the sphere's curved area and 3 flat quadrant faces.

TSA of 1 part = (1/8 × 4πr²) + (3 × 1/4πr²) = 1.25 × πr².

Direct Calculation: 1.25 × (22/7) × 21 × 21.

TSA = 1.25 × 1386 = 1732.5 cm².

∴ The correct answer is 1732.5 cm².

Alternate Method

Given:

Radius of sphere (r) = 21 cm

Sphere is cut into 8 identical parts by 3 perpendicular cuts passing through the center.

Formula Used:

Curved Surface Area (CSA) of sphere = 4πr²

Total Surface Area (TSA) of each part = (1/8 × CSA of sphere) + Area of 3 flat surfaces

Area of each flat surface = Quarter of a circle = (1/4)πr²

Calculations:

⇒ Area of 1/8th curved surface = (1/8) × 4 × (22/7) × 21 × 21

⇒ Area of 1/8th curved surface = (1/2) × 22 × 3 × 21 = 11 × 63 = 693 cm²

⇒ Area of 3 flat faces = 3 × (1/4) × (22/7) × 21 × 21

⇒ Area of 3 flat faces = (3/4) × 22 × 3 × 21 = (3/2) × 11 × 63 = 2079 / 2 = 1039.5 cm²

⇒ Total Surface Area (TSA) = 693 + 1039.5

⇒ TSA = 1732.5 cm²

∴ The correct answer is 1732.5 cm².

Additional Information

Surface Area of Sphere and Parts

Total Surface Area of a sphere is 4πr². When cut into two hemispheres, the TSA of one hemisphere becomes 3πr² (curved surface 2πr² + flat base πr²).

Effect of Cuts through Center

Each full circular cut through the center of a sphere adds an additional surface area equal to 2 × (πr²). Three perpendicular cuts create 3 full circular cross-sections, adding 6πr² to the original 4πr².

Volume of Sphere Parts

Volume of a full sphere is (4/3)πr³. When cut into 8 identical parts, the volume of each part is simply 1/8 of the total: V = (1/6)πr³.

Shortcut Trick

Side of cube (a) = 14 cm

Radius of maximum hemisphere (r) = a ÷ 2 = 14 ÷ 2 = 7 cm

Volume of 2 hemispheres = Volume of 1 full sphere = (4/3) × π × r³

Remaining Volume = a³ − (4/3) × (22/7) × 7³ = 2744 − 1437.33 = 1306.67 cm³

∴ The correct answer is 1306.67 cm³.

Alternate Method

Given:

Side of the solid cube (a) = 14 cm

Number of hemispheres = 2 (cut from opposite faces)

Formula Used:

Volume of Cube = a³

Volume of Hemisphere = (2/3) × π × r³

Calculations:

⇒ For maximum size, the diameter of each hemisphere must be equal to the side of the cube.

⇒ Diameter = 14 cm ⇒ Radius (r) = 14 ÷ 2 = 7 cm

⇒ Volume of the cube = a³ = 14 × 14 × 14 = 2744 cm³

⇒ Volume of one hemisphere = (2/3) × π × r³

⇒ Volume of two hemispheres = 2 × (2/3) × (22/7) × 7 × 7 × 7

⇒ Volume of two hemispheres = (4/3) × 22 × 49 = 4312 ÷ 3 ≈ 1437.33 cm³

⇒ Remaining volume = Volume of cube − Volume of two hemispheres

⇒ Remaining volume = 2744 − 1437.33 = 1306.67 cm³

∴ The correct answer is 1306.67 cm³.

Additional Information

Volume and Surface Area of Hemisphere

Volume = (2/3)πr³, Curved Surface Area = 2πr², and Total Surface Area = 3πr².

Maximum Sphere in a Cube

The radius of the largest sphere that can be carved out of a cube of side 'a' is a/2.

Cube Mensuration

Volume = a³, Total Surface Area = 6a², and Diagonal = a√3.

Shortcut Trick

Side of the largest possible cube (a) = HCF of 65, 26, and 3.9 = 1.3 cm.

Number of cubes (n) = (65 ÷ 1.3) × (26 ÷ 1.3) × (3.9 ÷ 1.3) = 50 × 20 × 3 = 3000.

Total Surface Area (TSA) of one cube = 6a² = 6 × (1.3)² = 6 × 1.69 = 10.14 cm².

Total TSA of all cubes = n × TSA of one cube = 3000 × 10.14 = 30420 cm².

∴ The correct answer is 30420 cm².

Alternate Method

Given:

Dimensions of cuboid (L, B, H) = 65 cm, 26 cm, 3.9 cm.

Formula Used:

1. Side of largest cube (a) = HCF (L, B, H)
2. Number of cubes (n) = (Volume of cuboid) ÷ (Volume of one cube)
3. Total Surface Area of n cubes = n × 6a²

Calculations:

⇒ To find HCF(65, 26, 3.9), convert all to mm: 650 mm, 260 mm, 39 mm.

⇒ 650 = 13 × 50; 260 = 13 × 20; 39 = 13 × 3.

⇒ HCF(650, 260, 39) = 13 mm = 1.3 cm.

⇒ Side of the cube (a) = 1.3 cm.

⇒ Number of cubes (n) = (65 × 26 × 3.9) ÷ (1.3 × 1.3 × 1.3)

⇒ n = (65 ÷ 1.3) × (26 ÷ 1.3) × (3.9 ÷ 1.3) = 50 × 20 × 3 = 3000.

⇒ Total Surface Area = n × 6a² = 3000 × 6 × (1.3)²

⇒ Total Surface Area = 18000 × 1.69 = 180 × 169

⇒ Total Surface Area = 30420 cm².

∴ The correct answer is 30420 cm².

Additional Information

HCF of Decimals

To find the HCF of decimals, multiply them by a power of 10 to make them integers, find the HCF of those integers, and then divide by the same power of 10.

Total Surface Area vs Volume

When a solid is cut into smaller pieces, the total volume remains constant, but the total surface area increases due to the creation of new exposed surfaces.

Mensuration Formulas

Volume of cuboid = L × B × H | Volume of cube = a³ | Surface area of cube = 6a².

Shortcut Trick

Ratio of heights of three similar pyramids formed from the top = 1 : 2 : 3

Ratio of volumes of these nested pyramids = 1³ : 2³ : 3³ = 1 : 8 : 27

V₁ (Top part) = 1 unit

V₂ (Middle part) = 8 − 1 = 7 units

V₃ (Bottom part) = 27 − 8 = 19 units

∴ The correct answer is 1 : 7 : 19.

Alternate Method

Given:

Altitude of the pyramid is trisected by 2 planes parallel to the base.

Let the total height of the pyramid be H.

Heights of the three resulting pyramids (from the apex) are h, 2h, and 3h where h = H/3.

Formula Used:

For similar solids, the ratio of their volumes is the ratio of the cubes of their corresponding heights (V₁/V₂ = (h₁/h₂)³).

Calculations:

⇒ Volume of first pyramid (Top) = v₁ ∝ h³

⇒ Volume of second pyramid (Top + Middle) = v₂ ∝ (2h)³ = 8h³

⇒ Volume of third pyramid (Total) = v₃ ∝ (3h)³ = 27h³

⇒ V₁ = v₁ = 1 unit (assuming h³ = 1)

⇒ V₂ = v₂ − v₁ = 8 − 1 = 7 units

⇒ V₃ = v₃ − v₂ = 27 − 8 = 19 units

⇒ Ratio V₁ : V₂ : V₃ = 1 : 7 : 19

∴ The correct answer is 1 : 7 : 19.

Additional Information

Property of Similar Solids

When a solid is cut by planes parallel to the base, the smaller solids formed are similar to the original. Ratios of side/height = k, Ratios of areas = k², Ratios of volumes = k³.

Volume of a Pyramid

The volume of any pyramid is given by V = (1/3) × Base Area × Height.

Frustum of a Pyramid

The middle and bottom parts in this question are called frustums. The volume of a frustum with base areas A₁ and A₂ and height h is (1/3)h(A₁ + A₂ + √(A₁ × A₂)).

Shortcut Trick

Apply the Sum-to-Product identities for the numerator and denominator.

Numerator: cos 7A + cos 5A = 2 cos((7A+5A)/2) cos((7A−5A)/2) = 2 cos 6A cos A.

Denominator: sin 7A − sin 5A = 2 cos((7A+5A)/2) sin((7A−5A)/2) = 2 cos 6A sin A.

Dividing Numerator by Denominator: (2 cos 6A cos A) ÷ (2 cos 6A sin A) = cos A / sin A = cot A.

∴ The correct answer is cot A.

Alternate Method

Given: Expression = [(cos 7A + cos 5A) / (sin 7A − sin 5A)]

Formula Used:

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

Calculations:

⇒ Let C = 7A and D = 5A.

⇒ Numerator = 2 cos((7A + 5A)/2) cos((7A − 5A)/2)

⇒ Numerator = 2 cos(12A/2) cos(2A/2) = 2 cos 6A cos A

⇒ Denominator = 2 cos((7A + 5A)/2) sin((7A − 5A)/2)

⇒ Denominator = 2 cos(12A/2) sin(2A/2) = 2 cos 6A sin A

⇒ Value = [2 cos 6A cos A] / [2 cos 6A sin A]

⇒ Cancel out common term 2 cos 6A: Value = cos A / sin A

⇒ Value = cot A

∴ The correct answer is cot A.

Additional Information

Sum-to-Product Identities

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

Basic Trigonometric Ratios

tan θ = sin θ / cos θ and cot θ = cos θ / sin θ.

Double Angle Formulas

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

Shortcut Trick

Use value substitution: Let A = 30°.

Expression = [1 − sin(90° − 60°)] ÷ [1 + sin(90° + 60°)]

⇒ [1 − sin 30°] ÷ [1 + sin 150°] = [1 − 0.5] ÷ [1 + 0.5]

⇒ 0.5 ÷ 1.5 = 1/3.

Check options: tan²(30°) = (1/√3)² = 1/3.

∴ The correct answer is tan²A.

Alternate Method

Given: [1 − sin(90 − 2A)] ÷ [1 + sin(90 + 2A)]

Formula Used:

1. sin(90 − θ) = cos θ
2. sin(90 + θ) = cos θ
3. 1 − cos 2θ = 2 sin²θ
4. 1 + cos 2θ = 2 cos²θ

⇒ Numerator: 1 − sin(90 − 2A) = 1 − cos 2A

⇒ Denominator: 1 + sin(90 + 2A) = 1 + cos 2A

⇒ Expression = (1 − cos 2A) ÷ (1 + cos 2A)

⇒ Using identities: 1 − cos 2A = 2 sin²A and 1 + cos 2A = 2 cos²A

⇒ Value = (2 sin²A) ÷ (2 cos²A)

⇒ Value = sin²A ÷ cos²A = tan²A

∴ The correct answer is tan²A.

Additional Information

Trigonometric Quadrant Rules

In the first quadrant (90 − θ), all ratios are positive. In the second quadrant (90 + θ), only sine and cosecant are positive. Here, sin(90 + θ) changes to cos θ and remains positive.

Double Angle Identities

cos 2A = cos²A − sin²A = 1 − 2 sin²A = 2 cos²A − 1. These are essential for converting linear trig terms into squared terms.

Complementary Angle Formulas

sin(90 − θ) = cos θ, cos(90 − θ) = sin θ, tan(90 − θ) = cot θ.

Shortcut Trick

Apply the Sum-to-Product formula: sin C + sin D = 2 sin((C + D) ÷ 2) cos((C − D) ÷ 2).

Here, C = 75° and D = 15°.

Sum = 2 sin((75° + 15°) ÷ 2) cos((75° − 15°) ÷ 2) = 2 sin 45° cos 30°.

Value = 2 × (1 ÷ √2) × (√3 ÷ 2) = √3 ÷ √2 = √(3/2).

∴ The correct answer is √(3/2).

Alternate Method

Given: sin 75° + sin 15°

Formula Used: sin(A ± B) = sin A cos B ± cos A sin B

⇒ sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

⇒ sin 75° = (1/√2) × (√3/2) + (1/√2) × (1/2) = (√3 + 1) / (2√2)

⇒ sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30°

⇒ sin 15° = (1/√2) × (√3/2) − (1/√2) × (1/2) = (√3 − 1) / (2√2)

⇒ Sum = [(√3 + 1) / (2√2)] + [(√3 − 1) / (2√2)]

⇒ Sum = (√3 + 1 + √3 − 1) / (2√2) = (2√3) / (2√2) = √3 / √2

⇒ Sum = √(3/2)

∴ The correct answer is √(3/2).

Additional Information

Sum-to-Product Identities

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

Complementary Angles

sin θ = cos(90° − θ). Thus, sin 75° = cos 15° and sin 15° = cos 75°.

Values of 15° and 75°

sin 15° = (√3−1)/(2√2) and sin 75° = (√3+1)/(2√2). These are frequently used in competitive exams.

Shortcut Trick

Put θ = 0° in the given expression.

Value = [(cos 0° + 2 cos 0° + cos 0°) ÷ (cos 0° + 2 cos 0° + cos 0°)] + sin 0°.tan 0°

⇒ [(1 + 2(1) + 1) ÷ (1 + 2(1) + 1)] + 0 × 0 = [4 ÷ 4] + 0 = 1

Now check options for θ = 0°: (1) cos(2 × 0°) = 1, (2) sin 0° = 0, (3) tan 0° = 0.

Option 1 matches the value.

∴ The correct answer is cos2θ.

Alternate Method

Given Expression: [(cos3θ + 2cos5θ + cos7θ) ÷ (cosθ + 2cos3θ + cos5θ)] + sin2θ.tan3θ

Formula used: cos C + cos D = 2 cos[(C + D)/2] cos[(C − D)/2]

⇒ Numerator = (cos 7θ + cos 3θ) + 2 cos 5θ = 2 cos 5θ cos 2θ + 2 cos 5θ = 2 cos 5θ (cos 2θ + 1)

⇒ Denominator = (cos 5θ + cos θ) + 2 cos 3θ = 2 cos 3θ cos 2θ + 2 cos 3θ = 2 cos 3θ (cos 2θ + 1)

⇒ Fraction = [2 cos 5θ (cos 2θ + 1)] ÷ [2 cos 3θ (cos 2θ + 1)] = cos 5θ ÷ cos 3θ

⇒ Total = (cos 5θ ÷ cos 3θ) + (sin 2θ × sin 3θ ÷ cos 3θ)

⇒ Total = (cos 5θ + sin 2θ sin 3θ) ÷ cos 3θ

⇒ Total = [cos(3θ + 2θ) + sin 2θ sin 3θ] ÷ cos 3θ

⇒ Total = [cos 3θ cos 2θ − sin 3θ sin 2θ + sin 2θ sin 3θ] ÷ cos 3θ

⇒ Total = (cos 3θ cos 2θ) ÷ cos 3θ = cos 2θ

∴ The correct answer is cos2θ.

Additional Information

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).

Compound Angle Identities

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

Double Angle Formulas

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

Shortcut Trick

Use the substitution method by assuming θ = 0°.

Numerator = 2 sin(45° + 0°) sin(45° − 0°) = 2 sin 45° sin 45° = 2 × (1/2) = 1.

Denominator = cos(2 × 0°) = cos 0° = 1.

Expression Value = 1 ÷ 1 = 1.

∴ The correct answer is 1.

Alternate Method

Given: [2sin(45° + θ) sin(45° − θ)] ÷ cos 2θ

Formula Used: 2 sin A sin B = cos(A − B) − cos(A + B)

⇒ Let A = 45° + θ and B = 45° − θ

⇒ A − B = (45° + θ) − (45° − θ) = 2θ

⇒ A + B = (45° + θ) + (45° − θ) = 90°

⇒ Numerator = cos(2θ) − cos(90°)

⇒ Numerator = cos 2θ − 0 = cos 2θ

⇒ Expression = cos 2θ ÷ cos 2θ = 1

∴ The correct answer is 1.

Additional Information

Product to Sum Identities

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

Double Angle Formulas

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

Complementary Angles

For any angle θ, sin(90° − θ) = cos θ and cos(90° − θ) = sin θ.

Shortcut Trick

Substitute A = 0° in the expression: sin(90° + 0)[4 − cos²(90° − 0)] = 1 × [4 − 0] = 4.

Check Option (3): 4(cos⁶ 0 + sin⁶ 0) = 4(1 + 0) = 4.

Check Option (4): 4(cos⁶ 0 − sin⁶ 0) = 4(1 − 0) = 4.

Substitute A = 30°: Expression = cos 60°[4 − sin² 60°] = 0.5 × (4 − 0.75) = 1.625.

Check Option (4) for A = 30°: 4[(√3/2)⁶ − (1/2)⁶] = 4[27/64 − 1/64] = 26/16 = 1.625.

∴ The correct answer is 4(cos⁶A − sin⁶A).

Alternate Method

Given: sin(90° + 2A)[4 − cos²(90° − 2A)]

Formula Used: sin(90° + θ) = cos θ; cos(90° − θ) = sin θ; cos 2A = cos² A − sin² A; sin 2A = 2 sin A cos A; a³ − b³ = (a − b)(a² + ab + b²)

⇒ sin(90° + 2A) = cos 2A

⇒ cos(90° − 2A) = sin 2A

⇒ Expression = cos 2A [4 − sin² 2A]

⇒ Substitute sin 2A = 2 sin A cos A: cos 2A [4 − (2 sin A cos A)²]

⇒ cos 2A [4 − 4 sin² A cos² A] = 4 cos 2A [1 − sin² A cos² A]

⇒ Substitute cos 2A = cos² A − sin² A and 1 = (cos² A + sin² A)²

⇒ 4 (cos² A − sin² A) [(cos² A + sin² A)² − sin² A cos² A]

⇒ 4 (cos² A − sin² A) [cos⁴ A + sin⁴ A + 2 sin² A cos² A − sin² A cos² A]

⇒ 4 (cos² A − sin² A) [cos⁴ A + sin⁴ A + sin² A cos² A]

⇒ Since (x − y)(x² + xy + y²) = x³ − y³, let x = cos² A and y = sin² A

⇒ Expression = 4 [(cos² A)³ − (sin² A)³] = 4 (cos⁶ A − sin⁶ A)

∴ The correct answer is 4(cos⁶A − sin⁶A).

Additional Information

Allied Angle Relations

Functions change at 90° and 270° (sin ↔ cos, tan ↔ cot) and signs depend on the quadrant.

Double Angle Identities

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

Sum and Difference of Cubes

x³ − y³ = (x − y)(x² + xy + y²) and x³ + y³ = (x + y)(x² − xy + y²).

Shortcut Trick

Apply ASTC (All-Silver-Tea-Cups) quadrant rules for each term.

cos(90° + A) = −sin A and sec(270° + A) → Here expression has sec(270° − A) = −cosec A.

sin(270° + A) = −cos A and cosec(630° − A) = cosec(360° + 270° − A) = cosec(270° − A) = −sec A.

Expression = [−sin A ÷ (−cosec A)] + [−cos A ÷ (−sec A)] = sin²A + cos²A = 1.

∴ The correct answer is 1.

Alternate Method

Given: [cos(90° + A) ÷ sec(270° − A)] + [sin(270° + A) ÷ cosec(630° − A)]

Formula Used: Quadrant transformations and sin²θ + cos²θ = 1.

⇒ Term 1: cos(90° + A) is in the 2nd quadrant. cos changes to sin and is negative.

⇒ cos(90° + A) = −sin A

⇒ Term 2: sec(270° − A) is in the 3rd quadrant. sec changes to cosec and is negative.

⇒ sec(270° − A) = −cosec A

⇒ Term 3: sin(270° + A) is in the 4th quadrant. sin changes to cos and is negative.

⇒ sin(270° + A) = −cos A

⇒ Term 4: cosec(630° − A) = cosec(360° + 270° − A) = cosec(270° − A)

⇒ 270° − A is in the 3rd quadrant. cosec changes to sec and is negative.

⇒ cosec(630° − A) = −sec A

⇒ Expression = [−sin A ÷ (−cosec A)] + [−cos A ÷ (−sec A)]

⇒ Expression = [sin A × sin A] + [cos A × cos A]

⇒ Expression = sin²A + cos²A

⇒ Expression = 1

∴ The correct answer is 1.

Additional Information

ASTC Rule

In the Cartesian plane, Trigonometric ratios are positive as follows: Q-I (All), Q-II (Sine/Cosec), Q-III (Tan/Cot), Q-IV (Cos/Sec).

Transformation at (n × 90° ± θ)

If n is odd, the ratio changes (sin↔cos, tan↔cot, sec↔cosec). If n is even, the ratio remains the same.

Trigonometric Identities

sin²θ + cos²θ = 1 | 1 + tan²θ = sec²θ | 1 + cot²θ = cosec²θ.

Shortcut Trick

Use the direct distance formula: d = h(cot θ₁ − cot θ₂), where d is distance moved.

Here, d = 100 m, θ₁ = 45° and θ₂ = 60°.

100 = h(cot 45° − cot 60°) ⇒ 100 = h(1 − 1/√3).

100 = h(√3 − 1)/√3 ⇒ h = 100√3/(√3 − 1).

Rationalizing: h = [100√3 × (√3 + 1)] / 2 = 50(3 + √3).

∴ The correct answer is 50(3 + √3) m.

Alternate Method

Given: Initial angle θ₁ = 45°, Final angle θ₂ = 60°, Distance moved CD = 100 m.

Formula: tan θ = Perpendicular / Base

⇒ Let the height of the building AB be h and the initial distance BC be x.

⇒ In right-angled ΔABC, tan 60° = AB / BC

⇒ √3 = h / x ⇒ x = h / √3 --- (Equation 1)

⇒ In right-angled ΔABD, tan 45° = AB / BD

⇒ 1 = h / (x + 100) ⇒ h = x + 100 --- (Equation 2)

⇒ Substituting x from Eq 1 into Eq 2: h = (h / √3) + 100

⇒ h − h / √3 = 100 ⇒ h(1 − 1/√3) = 100

⇒ h(√3 − 1) / √3 = 100 ⇒ h = 100√3 / (√3 − 1)

⇒ Rationalizing the denominator: h = [100√3 × (√3 + 1)] / [(√3 − 1) × (√3 + 1)]

⇒ h = [100(3 + √3)] / (3 − 1) ⇒ h = 100(3 + √3) / 2

⇒ h = 50(3 + √3) m.

∴ The correct answer is 50(3 + √3) m.

Additional Information

Distance-Height Relation

When moving towards an object, the relationship is h = d / (cot θ₁ − cot θ₂). If moving away, h = d / (cot θ₂ − cot θ₁).

Complementary Angles

If the angles of elevation from two points at distances 'a' and 'b' from the base are complementary, the height h = √(ab).

Standard Trig Ratios

tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. Note that cot θ = 1/tan θ.

Shortcut Trick

For a broken tree forming angle θ with ground at distance d from the root:

Total Height = d × (tan θ + sec θ)

Given: d = 25 m and θ = 60°

Total Height = 25 × (tan 60° + sec 60°) = 25 × (√3 + 2)

Total Height = 25 × (1.732 + 2) = 25 × 3.732 = 93.3 m

∴ The correct answer is 93.3 m.

Alternate Method

Given:

Distance from root to top on ground (BC) = 25 m

Angle of elevation (θ) = 60°

Formula Used:

tan θ = Perpendicular / Base

cos θ = Base / Hypotenuse

Total height of tree = Vertical part (AB) + Broken part (AC)

In right-angled ΔABC:

⇒ tan 60° = AB / BC

⇒ √3 = AB / 25

⇒ AB = 25 × 1.732 = 43.3 m

⇒ cos 60° = BC / AC

⇒ 1/2 = 25 / AC

⇒ AC = 25 × 2 = 50 m

⇒ Total Height = AB + AC = 43.3 + 50 = 93.3 m

∴ The correct answer is 93.3 m.

Additional Information

Trigonometric Ratios in Broken Tree Problems

The height of the vertical standing part is found using tan θ = h/d, and the length of the fallen part (hypotenuse) is found using cos θ = d/l or sec θ = l/d.

Standard Values for √3 and √2

Commonly used approximations in competitive exams: √3 ≈ 1.732 and √2 ≈ 1.414.

Pythagoras Theorem Relation

For any such problem, the relation (Broken Part)² = (Remaining Height)² + (Distance from Root)² always holds true.

Shortcut Trick

Height of taller tower (H) = 300 m

Horizontal distance (d) = 120 m

Difference in height (Δh) = d × tan(60°) = 120 × √3

Δh = 120 × 1.732 = 207.84 m

Height of smaller tower (h) = H − Δh = 300 − 207.84 = 92.16 m

∴ The correct answer is 92.15 m.

Alternate Method

Given:

Height of taller tower (AB) = 300 m

Angle of depression = 60°

Horizontal distance (BD) = 120 m

Let the height of the smaller tower be CD = h.

From the diagram, the distance EC = BD = 120 m.

In ΔAEC, ∠ACE = 60° (Alternate interior angle to the angle of depression).

The height AE = AB − EB = 300 − h.

Using Trigonometric ratio in ΔAEC:

⇒ tan(60°) = AE / EC

⇒ √3 = (300 − h) / 120

⇒ 1.732 × 120 = 300 − h

⇒ 207.84 = 300 − h

⇒ h = 300 − 207.84

⇒ h = 92.16 m

Rounding to the nearest option provided:

∴ The correct answer is 92.15 m.

Additional Information

Angle of Depression

It is the angle formed between the horizontal line of sight and the line of sight to an object below the observer's eye level.

Trigonometric Values

Important values for Height & Distance: tan(30°) = 1/√3, tan(45°) = 1, and tan(60°) = √3.

Complementary Angles Rule

If the angle of elevation from the top of one tower to another is θ₁ and the angle of depression is θ₂, then the height difference is calculated using the respective tangents of these angles.

Number of employees of department A = 10% of 80000 = 8000

⇒ Number of employees from city Z = (8000 × 1) / (1 + 9) = 800

Number of employees of department C = 12% of 80000 = 9600

⇒ Number of employees from city Z = (9600 × 5) / (5 + 1) = 8000

∴ Total number of employees from city Z from departments A and C = 800 + 8000 = 8800

Number of employees in department E = 36% of 80000 = 28800

⇒ Number of males = (28800 × 8) / (8 + 1) = 25600

Number of employees of department A = 10% of 80000 = 8000

⇒ Number of employees from city Z = (8000 × 1) / (1 + 9) = 800

∴ Required percentage = (25600/800) × 100 = 3200%

Total number of employees in departments A, B, D and E are 10% of 80000, 22% of 80000, 20% of 80000 and 36% of 80000 i.e. 8000, 17600, 16000, 28800 respectively

⇒ Female population of A = (3/10) × 8000 = 2400

⇒ Female population of department E = 28800 × 1/9 = 3200

⇒ Female population of A + E = 5600

⇒ Male population of B = (13/22) × 17600 = 10400

⇒ Male population of D = (3/5) × 16000 = 9600

⇒ Total male population of B + D = 20000

∴ Required ratio = 20000 : 5600 = 25 : 7

Number of residents of city Y = [Σ {Percentage of employees in that department × Y / (Y + Z)}] / 100 × Total number of employees

= {10 × 9/10 + 22 × 19/22 + 12 × 1/6 + 20 × 3/4 + 36 × 13/18} /100 × 80000

= {9 + 19 + 2 + 15 + 26} × 800 = 56800

Average = 56800/5 = 11360

⇒ Total percentage of employees in department A + E = 10 + 36 = 46%

Total number of employees = 80000

⇒ Total number of employees in department A + E = 46% of 80000 = 36800

Shortcut Trick

Apply the Alligation cross method:

Tea 1 (Dearer) = 300, Tea 2 (Cheaper) = 200, Mean Price = 225

Difference 1 (225 − 200) = 25

Difference 2 (300 − 225) = 75

Required Ratio = 25 : 75 = 1 : 3

∴ The correct answer is 1 : 3.

Alternate Method

Given:

Cost of Tea 1 = ₹300/kg

Cost of Tea 2 = ₹200/kg

Target Mixture Cost = ₹225/kg

Calculations:

⇒ Let the quantity of Tea 1 be x and Tea 2 be y.

⇒ According to Alligation rule, Ratio (x : y) = (Mean Price − Cost 2) : (Cost 1 − Mean Price)

⇒ Ratio = (225 − 200) : (300 − 225)

⇒ Ratio = 25 : 75

⇒ Ratio = 1 : 3

∴ The correct answer is 1 : 3.

Additional Information

Rule of Alligation

It is used to find the ratio in which two or more ingredients at their respective prices are mixed to produce a mixture of a specified price. Ratio = (Dearer Price − Mean Price) / (Mean Price − Cheaper Price).

Mean Price

The mean price is the cost price of a unit quantity of the mixture. It always lies between the cost price of the cheaper and dearer ingredients.

Replacement Formula

If x liters are removed from a container of C liters of pure liquid and replaced by water, after n operations, Quantity of pure liquid = C × (1 − x/C)ⁿ.

Shortcut Trick

Profit Ratio A : B : C = (Investment × Time) for each partner.

Investment Ratio → A : B : C = 5 : 6 : (2/3 × 6) = 5 : 6 : 4.

Profit Ratio → (5 × 12 months) : (6 × 12 months) : (4 × 6 months) = 60 : 72 : 24 = 5 : 6 : 2.

C profit = 2 units = ₹21,600 ⇒ 1 unit = ₹10,800.

Total Profit = (5 + 6 + 2) = 13 units = 13 × 10,800 = ₹140,400.

∴ The correct answer is ₹140,400.

Alternate Method

Given:

Investment ratio of A and B = 5 : 6

C investment = 2/3^rd of B investment

Time period for A and B = 12 months (end of year)

Time period for C = 6 months (joined after 6 months)

C profit share = ₹21,600

Calculations:

Let the investments of A and B be 5x and 6x respectively.

⇒ C investment = (2 ÷ 3) × 6x = 4x

⇒ Profit share of A = 5x × 12 = 60x

⇒ Profit share of B = 6x × 12 = 72x

⇒ Profit share of C = 4x × 6 = 24x

⇒ Ratio of Profit A : B : C = 60x : 72x : 24x

⇒ Simplified Ratio = 5 : 6 : 2

⇒ Share of C in profit = 2 ÷ (5 + 6 + 2) = 2/13

⇒ 2/13 × Total Profit = 21,600

⇒ Total Profit = (21,600 × 13) ÷ 2

⇒ Total Profit = 10,800 × 13

⇒ Total Profit = 140,400

∴ The correct answer is ₹140,400.

Additional Information

Basic Partnership Law

Profit is always distributed in the ratio of the product of Capital (Investment) and Time: P₁ : P₂ = (C₁ × T₁) : (C₂ × T₂).

Total Profit Calculation

If the share of one partner is given as 's' out of 'n' total ratio parts, then Total Profit = (Given Share Value ÷ s) × n.

Sleeping Partner Concept

A partner who only invests money but does not participate in management is a sleeping partner. Active partners may sometimes take a salary before the profit is divided.

Shortcut Trick

Investment ratio A : B = 2 : 5 → Total distributed units = 7

A's share = 2 units = 3.6 lakhs

⇒ 1 unit = 1.8 lakhs

Distributed profit (50% of total) = 7 units = 7 × 1.8 = 12.6 lakhs

Total profit (100%) = 12.6 × 2 = 25.2 lakhs

∴ The correct answer is 25.2 lakhs.

Alternate Method

Given:

Investment ratio of A and B = 2 : 5

Charity deduction = 50% of total profit

A's share in remaining profit = ₹3.6 lakhs

Formula Used:

Individual Profit Share = (Individual Investment ÷ Total Investment) × Distributed Profit

Distributed Profit = Total Profit − Charity

Calculations:

⇒ Let the total profit be P lakhs.

⇒ Amount given to charity = 50% of P = 0.5P

⇒ Remaining profit for distribution = P − 0.5P = 0.5P

⇒ Total parts in investment ratio = 2 + 5 = 7 units

⇒ A's share = (2 ÷ 7) × 0.5P

⇒ 3.6 = (2 ÷ 7) × (1 ÷ 2) × P

⇒ 3.6 = (1 ÷ 7) × P

⇒ P = 3.6 × 7

⇒ P = 25.2 lakhs

∴ The correct answer is 25.2 lakhs.

Additional Information

Partnership Investment Rule

Profit is always shared in the ratio of the product of Capital and Time (P₁:P₂ = C₁×T₁ : C₂×T₂).

Net Distributed Profit

When a portion of profit is donated or kept as a reserve, only the remaining amount is divided among partners according to their investment ratio.

Active Partner Salary

If an active partner receives a salary, it is subtracted from the total profit before the rest is shared in the capital ratio.

Shortcut Trick

Efficiency ratio A : C = 3 : 1 → Combined efficiency (A+C) = 4 units/day.

Total Work = (A+C) efficiency × Days = 4 × 22.5 = 90 units.

B efficiency = 90 ÷ 15 = 6 units/day.

Remaining work after (A+C) work for 15 days = 90 − (4 × 15) = 30 units.

Days taken by (A+B+C) to finish = Remaining Work ÷ Total Efficiency = 30 ÷ (4 + 6) = 3 days.

∴ The correct answer is 3 days.

Alternate Method

Given: Efficiency of A = 3 × Efficiency of C. (A+C) time = 22.5 days. B's alone time = 15 days.

Formula Used: Total Work = Efficiency × Time; Remaining Work = Total Work − Work Done.

⇒ Total efficiency of (A + C) = 4 units/day.

⇒ Total work = 4 × 22.5 = 90 units.

⇒ Efficiency of B = 90 ÷ 15 = 6 units/day.

⇒ Work done by A and C in 15 days = 4 × 15 = 60 units.

⇒ Remaining work = 90 − 60 = 30 units.

⇒ Combined efficiency when B joins (A + B + C) = 4 + 6 = 10 units/day.

⇒ Time taken to finish remaining work = 30 ÷ 10 = 3 days.

∴ The correct answer is 3 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a task. If efficiency ratio is x : y, then time ratio is y : x.

Work and Wages

Wages are always distributed in the ratio of the actual work done by each person (Efficiency × Number of days worked).

Pipes and Cisterns

The same logic applies where efficiency is the rate of filling or emptying a tank per unit of time.

Shortcut Trick

Total Work = LCM(12, 18, 36) = 36 units.

Efficiency: A = 3, B = 2, C = 1 (Combined = 6 units/day).

Work completed in 2 days = 6 × 2 = 12 units.

Remaining work = 36 − 12 = 24 units.

Time required for A and C = 24 ÷ (3 + 1) = 6 days.

∴ The correct answer is 6 days.

Alternate Method

Given:

Time taken by A = 12 days

Time taken by B = 18 days

Time taken by C = 36 days

A, B, and C work together for 2 days, then B quits.

Method Used:

LCM Method for Time & Work

Calculations:

⇒ Total efficiency of A, B, and C = 3 + 2 + 1 = 6 units/day

⇒ Total work done by all three in 2 days = 6 × 2 = 12 units

⇒ Remaining work = Total work − Work done = 36 − 12 = 24 units

⇒ Efficiency of A and C together = 3 + 1 = 4 units/day

⇒ Time taken by A and C to finish remaining work = Remaining work ÷ Combined efficiency

⇒ Time taken = 24 ÷ 4 = 6 days

∴ The correct answer is 6 days.

Additional Information

Pipes & Cisterns

Similar to the Time & Work concept, inlet pipes have positive efficiency while outlet pipes (leaks) have negative efficiency. Total capacity is taken as the LCM of individual fill/empty times.

Work & Wages

Wages are always distributed in the ratio of the total work done by each individual. If workers work for the same number of days, the wages are distributed directly in the ratio of their efficiencies.

Alternate Day Work

Calculate the combined work done in a complete cycle (e.g., 2 days for 2 people). Multiply the cycle to get close to the total work, then assign the remaining work sequentially to find the final time.

Shortcut Trick

Assume Total Work = LCM of (4, 4.5, 12) = 36 units.

Efficiency (A+B+C) = 36 ÷ 4 = 9 | Efficiency (A+C) = 36 ÷ 4.5 = 8 | Efficiency (B+C) = 36 ÷ 12 = 3.

Efficiency of B = Eff(A+B+C) − Eff(A+C) = 9 − 8 = 1.

Efficiency of C = Eff(B+C) − Eff(B) = 3 − 1 = 2.

Time taken by C alone = 36 ÷ 2 = 18 days.

∴ The correct answer is 18 days.

Alternate Method

Given:

Time (A+B+C) = 4 days

Time (A+C) = 4.5 days = 9/2 days

Time (B+C) = 12 days

Formula used: Efficiency = Total Work ÷ Time

⇒ Efficiency of (A+B+C) = 9 units/day

⇒ Efficiency of (A+C) = 8 units/day

⇒ Efficiency of B = (A+B+C) − (A+C) = 9 − 8 = 1 unit/day

⇒ Efficiency of (B+C) = 3 units/day

⇒ Efficiency of C = (B+C) − B = 3 − 1 = 2 units/day

⇒ Days taken by C alone = Total Work ÷ Efficiency of C

⇒ Days = 36 ÷ 2 = 18 days

∴ The correct answer is 18 days.

Additional Information

Work and Efficiency Relation

Efficiency is inversely proportional to time taken. If time ratio is x:y, the efficiency ratio is y:x.

LCM Method for Fractions

LCM of fractions (a/b, c/d) = LCM(a, c) ÷ HCF(b, d). This is used to find total work when days are in fractions.

Combined Work Formula

If A does a job in x days and B in y days, together they do it in (x × y) ÷ (x + y) days.

Shortcut Trick

When the time taken by person A and the time taken by (A + B) together are given, time for B is calculated as (Product of days) ÷ (Difference of days).

Time taken by B = (40 × 8) ÷ (40 − 8)

Time taken by B = 320 ÷ 32 = 10 days

∴ The correct answer is 10 days.

Alternate Method

Given: A alone = 40 days, A + B together = 8 days.

Formula: Efficiency = Total Work ÷ Time

⇒ Total Work = LCM(40, 8) = 40 units

⇒ Efficiency of A = 40 ÷ 40 = 1 unit/day

⇒ Combined efficiency (A + B) = 40 ÷ 8 = 5 units/day

⇒ Efficiency of B = (Combined efficiency) − (Efficiency of A)

⇒ Efficiency of B = 5 − 1 = 4 units/day

⇒ Time taken by B = Total Work ÷ Efficiency of B = 40 ÷ 4 = 10 days

∴ The correct answer is 10 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken. If the ratio of efficiency is x:y, then the ratio of time taken is y:x.

Work and Wages

Wages are always distributed in the ratio of the work done by the individuals, which is same as the ratio of their efficiencies when they work for the same duration.

Combined Work for Three People

If three people A, B, and C can do a job in x, y, and z days respectively, then together they take (x × y × z) ÷ (xy + yz + zx) days.

Shortcut Trick

Total Marked Price (MP) for 4 bottles = 240 × 4 = ₹960.

Selling Price (SP) of the pack = ₹768.

Discount = MP − SP = 960 − 768 = ₹192.

Discount % = (Discount ÷ MP) × 100 = (192 ÷ 960) × 100 = 20%.

∴ The correct answer is 20%.

Alternate Method

Given:

Cost of 1 bottle (Marked Price) = ₹240

Number of bottles in a pack = 4

Pack Selling Price = ₹768

Formula Used: Discount % = [(Marked Price − Selling Price) ÷ Marked Price] × 100

⇒ Total Marked Price of 4 bottles = 4 × 240 = ₹960 ⇒ Selling Price of the pack = ₹768 ⇒ Discount Amount = 960 − 768 = ₹192 ⇒ Discount % = (192 ÷ 960) × 100 ⇒ Discount % = (1 ÷ 5) × 100 ⇒ Discount % = 20%

∴ The correct answer is 20%.

Additional Information

Successive Discounts

When multiple discounts are applied one after another, the equivalent discount is given by: a + b − (a × b ÷ 100).

Relation between CP, MP, and SP

Selling Price = MP × (100 − Discount%) ÷ 100 and CP = SP × 100 ÷ (100 + Profit%).

Buy X Get Y Free

In such offers, the discount percentage is calculated as: [Free Items (Y) ÷ Total Items (X + Y)] × 100.

Shortcut Trick

Markup = 100% → Marked Price (MP) = 200% of CP.

Discount = 20% → Selling Price (SP) = 80% of MP.

Effective SP = 80% of 200% of CP = 160% of CP.

Given SP = 1200 ⇒ 160% = 1200.

Value of x (100%) = (1200 ÷ 160) × 100 = 7.5 × 100 = 750.

∴ The correct answer is ₹750.

Alternate Method

Given:

Cost Price (CP) = ₹x

Markup % = 100%

Discount % = 20%

Selling Price (SP) = ₹1200

Formula Used:

MP = CP × (100 + Markup%)/100

SP = MP × (100 − Discount%)/100

Calculations:

⇒ Marked Price (MP) = x + (100% of x) = 2x

⇒ Selling Price (SP) = MP − (20% of MP)

⇒ 1200 = 2x − (20 ÷ 100 × 2x)

⇒ 1200 = 2x − 0.4x

⇒ 1200 = 1.6x

⇒ x = 1200 ÷ 1.6

⇒ x = 750

∴ The correct answer is ₹750.

Additional Information

Markup and Discount Relation

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

Effective Profit Percentage

Net Profit% = Markup% − Discount% − (Markup% × Discount% ÷ 100). In this case: 100 − 20 − (100 × 20 ÷ 100) = 60% profit.

Successive Percentage Rule

If a value is increased by a% and then decreased by b%, the net change is [a − b − (ab/100)]%.

Shortcut Trick

Total MP of Cookies = 2 × 1000 = ₹2000

Total MP of Chocolates = 3 × 400 = ₹1200

Ratio of Total MP = 2000 : 1200 = 5 : 3

Effective Discount % (Weighted Average) = (5 × 10% + 3 × 8%) ÷ (5 + 3)

Effective Discount % = (50 + 24) ÷ 8 = 74 ÷ 8 = 9.25%

∴ The correct answer is 9.25%.

Alternate Method

Given:

MP of 1 cookie box = ₹1000, Discount = 10%

MP of 1 chocolate bar = ₹400, Discount = 8%

Quantity: 2 cookie boxes and 3 chocolate bars

⇒ Total MP = (2 × 1000) + (3 × 400) = 2000 + 1200 = ₹3200

⇒ Discount on Cookies = 10% of 2000 = (10/100) × 2000 = ₹200

⇒ Discount on Chocolates = 8% of 1200 = (8/100) × 1200 = ₹96

⇒ Total Discount = 200 + 96 = ₹296

⇒ Effective Discount % = (Total Discount ÷ Total MP) × 100

⇒ Effective Discount % = (296 ÷ 3200) × 100 = 296 ÷ 32 = 9.25%

∴ The correct answer is 9.25%.

Additional Information

Effective Discount Formula

When buying multiple items with different discounts, the effective discount is (Total Discount ÷ Total Marked Price) × 100.

Successive Discounts

If two discounts d₁% and d₂% are given successively on a single item, the equivalent single discount is (d₁ + d₂ − d₁×d₂/100)%.

Relationship between MP, CP, and Profit

MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%). This is useful for finding the markup percentage.