SSC CGL Tier 2 Quant Previous Year Paper (Held On: 20 Feb 2018)

Given: I. 333 > 333 II. 333 > (33)3 Formula used: (am)n = am × n Calculations: For Statement I: 333 = 33 × 33 × 33 = 35937 333 = (33)11 = 2711 ⇒ 35937 < 2711 ⇒ Statement I is FALSE For Statement II: (33)3 = 33 × 3 = 39 ⇒ 39 = 19683 ⇒ 333 < 19683 ⇒ Statement II is FALSE ∴ The correct answer is option (4).

Given:

P = 2² + 6² + 10² + 14² + ... + 94²

Q = 1² + 5² + 9² + ... + 81²

Formula used:

n = $\frac{l - a}{d}$ + 1

a² - b² = (a - b)(a + b)

S_n = $\frac{n}{2}$ × (a + l)

Calculation:

n for P = $\frac{94 - 2}{4}$ + 1 = 24

n for Q = $\frac{81 - 1}{4}$ + 1 = 21

⇒ P - Q = (2² - 1²) + (6² - 5²) + ... + (82² - 81²) + (86² + 90² + 94²)

⇒ P - Q = (2-1)(2+1) + (6-5)(6+5) + ... + (82-81)(82+81) + 7396 + 8100 + 8836

⇒ P - Q = (3 + 11 + 19 + ... + 163) + 24332

⇒ In AP: a = 3, l = 163, n = 21

⇒ Sum = $\frac{21}{2}$ × (3 + 163) = 1743

⇒ P - Q = 1743 + 24332 = 26075

∴ The correct answer is option (2).

Given:

A = $\dfrac{1}{0.4} + \dfrac{1}{0.04} + \dfrac{1}{0.004} + \dots$ upto 8 terms

Formula used:

Sum of G.P. = \(\dfrac{a(rⁿ - 1)}{r - 1}\)

Calculation:

A = 2.5 + 25 + 250 + $\dots$ (8 terms)

⇒ a = 2.5, r = 10, n = 8

⇒ A = \(\dfrac{2.5 \times (10⁸ - 1)}{10 - 1}\)

⇒ A = $\dfrac{2.5 \times 99999999}{9}$

⇒ A = 2.5 × 11111111

⇒ A = 27777777.5

∴ The correct answer is option (3).

Given:

M = 0.1 + (0.1)² + (0.01)²

N = 0.3 + (0.03)² + (0.003)²

Formula used:

Decimal powers and addition

Calculation:

⇒ M = 0.1 + 0.01 + 0.0001

⇒ M = 0.1101

⇒ N = 0.3 + 0.0009 + 0.000009

⇒ N = 0.300909

⇒ M + N = 0.1101 + 0.300909

⇒ M + N = 0.411009

∴ The correct answer is option (1).

Given: P = 96 / (95 × 97) Q = 97 / (96 × 98) R = 1 / 97 Formula used: (a - 1)(a + 1) = a2 - 1 Calculation: P = $\frac{96}{96^2 - 1}$ Q = $\frac{97}{97^2 - 1}$ R = $\frac{1}{97} = \frac{97}{97^2}$ ⇒ 972 - 1 < 972 ⇒ $\frac{97}{97^2 - 1} > \frac{97}{97^2}$ ⇒ Q > R ⇒ P = $\frac{96}{9215}$ ⇒ Q = $\frac{97}{9408}$ ⇒ 96 × 9408 > 97 × 9215 ⇒ 903168 > 893855 ⇒ P > Q ⇒ R < Q < P ∴ The correct answer is option (2).

Given:

I. $11\frac{1}{2} + 17\frac{3}{4} - 5\frac{1}{5} - 2\frac{1}{10} = \frac{439}{20}$

II. $\frac{9}{1078} > \frac{11}{1127} > \frac{12}{1219}$

III. $\frac{149}{151} > \frac{153}{155} > \frac{157}{159}$

Formula used:

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

Fraction Rule: For $\frac{x}{x+k}$ where k > 0, the value increases as x increases.

Calculations:

Statement I:

⇒ $(11 + 17 - 5 - 2) + (\frac{1}{2} + \frac{3}{4} - \frac{1}{5} - \frac{1}{10})$

⇒ $21 + (\frac{10 + 15 - 4 - 2}{20})$

⇒ $21 + \frac{19}{20} = \frac{420 + 19}{20} = \frac{439}{20}$ (True)

Statement II:

⇒ $\frac{9}{1078} \approx 0.0083$

⇒ $\frac{11}{1127} \approx 0.0097$

⇒ $\frac{12}{1219} \approx 0.0098$

⇒ $0.0083 < 0.0097 < 0.0098$ (False)

Statement III:

⇒ $\frac{149}{149+2} , \frac{153}{153+2} , \frac{157}{157+2}$

⇒ $149 < 153 < 157$

⇒ $\frac{149}{151} < \frac{153}{155} < \frac{157}{159}$ (False)

∴ The correct answer is option (1).

Given:

I. $\frac{2}{3\sqrt{5}} < \frac{3}{2\sqrt{5}} < \frac{5}{4\sqrt{3}}$

II. $\frac{3}{2\sqrt{5}} < \frac{2}{3\sqrt{3}} < \frac{7}{4\sqrt{5}}$

Formula Used:

Comparison of positive terms using squaring: x < y ⇒ x² < y²

Calculation:

For statement I:

⇒ $(\frac{2}{3\sqrt{5}})^2 < (\frac{3}{2\sqrt{5}})^2 < (\frac{5}{4\sqrt{3}})^2$

⇒ $\frac{4}{9 \times 5} < \frac{9}{4 \times 5} < \frac{25}{16 \times 3}$

⇒ $\frac{4}{45} < \frac{9}{20} < \frac{25}{48}$

⇒ 0.088 < 0.45 < 0.52 (True)

For statement II:

⇒ $\frac{3}{2\sqrt{5}} < \frac{2}{3\sqrt{3}}$

Squaring both sides:

⇒ $\frac{3^2}{(2\sqrt{5})^2} < \frac{2^2}{(3\sqrt{3})^2}$

⇒ $\frac{9}{20} < \frac{4}{27}$

⇒ 9 × 27 < 4 × 20

⇒ 243 < 80 (False)

∴ The correct answer is option (1).

Shortcut Trick

I. 72 = 2³ × 3² → Factors count = (3+1) × (2+1) = 4 × 3 = 12.

II. Sum of first n odd numbers = n² → 20² = 400.

III. Primes below 100: 89, 97. Clearly 97 is the largest two-digit prime.

All three statements are verified to be correct.

∴ The correct answer is All are true.

Alternate Method

Statement I: Total number of positive factors of 72.

⇒ Prime factorization: 72 = 8 × 9 = 2³ × 3²

⇒ Formula: If N = p^a × q^b, number of factors = (a + 1)(b + 1)

⇒ Factors = (3 + 1) × (2 + 1) = 4 × 3 = 12. (Statement I is TRUE)

Statement II: Sum of first 20 odd numbers.

⇒ The series is 1, 3, 5, ..., up to 20 terms.

⇒ Sum of first n odd numbers = n²

⇒ Sum = 20² = 400. (Statement II is TRUE)

Statement III: Largest two-digit prime number.

⇒ Two-digit numbers end at 99. 99 is 9 × 11 (Composite).

⇒ 98 is even (Composite).

⇒ 97 is not divisible by 2, 3, 5, or 7. (Statement III is TRUE)

∴ The correct answer is All are true.

Additional Information

Number of Factors Formula

For any number N = p^a × q^b × r^c (where p, q, r are primes), the total number of factors is (a+1)(b+1)(c+1).

Arithmetic Progression Sums

Sum of first n even numbers = n(n + 1). Sum of first n natural numbers = n(n + 1) ÷ 2.

Prime Numbers under 100

There are exactly 25 prime numbers between 1 and 100. The smallest is 2 and the largest is 97.

Sum of Factors

Sum of factors for N = p^a × q^b is [(p^a+1 − 1)÷(p − 1)] × [(q^b+1 − 1)÷(q − 1)].

Given:

M = (3/7) ÷ (6/5) × (2/3) + (1/5) × (3/2)

N = (2/5) × (5/6) ÷ (1/3) + (3/5) × (2/3) ÷ (3/5)

Formula used:

BODMAS rule

Calculations:

M = $\frac{3}{7} \times \frac{5}{6} \times \frac{2}{3} + \frac{1}{5} \times \frac{3}{2}$

⇒ M = $\frac{5}{21} + \frac{3}{10}$

⇒ M = $\frac{50+63}{210} = \frac{113}{210}$

N = $\frac{2}{5} \times \frac{5}{6} \times \frac{3}{1} + \frac{3}{5} \times \frac{2}{3} \times \frac{5}{3}$

⇒ N = $1 + \frac{2}{3} = \frac{5}{3}$

Value of M/N:

⇒ $\frac{M}{N} = \frac{113}{210} \div \frac{5}{3}$

⇒ $\frac{M}{N} = \frac{113}{210} \times \frac{3}{5}$

⇒ $\frac{M}{N} = \frac{113}{70 \times 5} = \frac{113}{350}$

∴ The correct answer is option (3).

Shortcut Trick

Common Difference (k) = Divisor − Remainder = 4 − 2 = 5 − 3 = 6 − 4 = 7 − 5 = 2.

LCM(4, 5, 6, 7) = 420.

Largest 4-digit multiple of 420: 9999 ÷ 420 ≈ 23.8 → 420 × 23 = 9660.

M = 9660 − 2 = 9658.

Remainder of 9658 ÷ 9: Sum of digits = 9 + 6 + 5 + 8 = 28 → 28 ÷ 9 leaves remainder 1.

∴ The correct answer is 1.

Alternate Method

Given:

Divisors: 4, 5, 6, 7

Corresponding Remainders: 2, 3, 4, 5

Formula Used:

Number (M) = LCM(Divisors) × n − k (where k is the constant difference between divisor and remainder)

Calculations:

⇒ Common difference (k) = 4 − 2 = 2, 5 − 3 = 2, 6 − 4 = 2, 7 − 5 = 2.

⇒ LCM(4, 5, 6, 7):

⇒ 4 = 2²

⇒ 5 = 5

⇒ 6 = 2 × 3

⇒ 7 = 7

⇒ LCM = 2² × 3 × 5 × 7 = 420.

⇒ To find the largest 4-digit number, divide 9999 by 420:

⇒ 9999 ÷ 420 = 23 with remainder 339.

⇒ Largest multiple of 420 below 9999 = 9999 − 339 = 9660.

⇒ Therefore, M = 9660 − 2 = 9658.

⇒ Remainder when M is divided by 9:

⇒ 9658 ÷ 9

⇒ Remainder = 1.

∴ The correct answer is 1.

Additional Information

LCM Remainder Theorem (Constant Difference)

If a number leaves remainders r₁, r₂... when divided by d₁, d₂... and (d₁ − r₁) = (d₂ − r₂) = k, the number is of the form [LCM(d₁, d₂...) × n] − k.

Divisibility Rule for 9

A number is divisible by 9 if the sum of its digits is divisible by 9. The remainder when a number is divided by 9 is the same as the remainder when the sum of its digits is divided by 9.

LCM Calculation Tip

To find LCM of small numbers, take the largest number and check its multiples for divisibility by the other numbers: 7, 14, 21... until 420 is reached.

Shortcut Trick

If (a + b) = (c + d), then (√a + √b) > (√c + √d) if the product (a × b) > (c × d).

Statement I: 11 + 7 = 18 and 10 + 8 = 18. Product 11 × 7 = 77 and 10 × 8 = 80. Since 77 < 80, √11 + √7 < √10 + √8 is True.

Statement II: 17 + 11 = 28 and 15 + 13 = 28. Product 17 × 11 = 187 and 15 × 13 = 195. Since 187 < 195, √17 + √11 < √15 + √13. Thus, II is False.

∴ The correct answer is Only I.

Alternate Method

Given Statements:

I. √11 + √7 < √10 + √8

II. √17 + √11 > √15 + √13

Method: Square both sides of the inequality to compare values.

Evaluating Statement I:

⇒ L.H.S = (√11 + √7)² = 11 + 7 + 2√(11 × 7) = 18 + 2√77

⇒ R.H.S = (√10 + √8)² = 10 + 8 + 2√(10 × 8) = 18 + 2√80

⇒ Since 77 < 80, 18 + 2√77 < 18 + 2√80. Statement I is True.

Evaluating Statement II:

⇒ L.H.S = (√17 + √11)² = 17 + 11 + 2√(17 × 11) = 28 + 2√187

⇒ R.H.S = (√15 + √13)² = 15 + 13 + 2√(15 × 13) = 28 + 2√195

⇒ Since 187 < 195, 28 + 2√187 < 28 + 2√195. Statement II is False.

∴ The correct answer is Only I.

Additional Information

Comparison Trick for Constant Sum

When (a + b) = (c + d), the sum of roots (√a + √b) is larger when the difference |a − b| is smaller.

Comparison Trick for Constant Difference

When (a − b) = (c − d), the difference of roots (√a − √b) is smaller when the values of a and b are larger.

Square Root Approximation

For quick estimation, use the formula √x ≈ √y + (x − y) ÷ (2√y), where y is the nearest perfect square to x.

Given: I. √12 > 3√16 > 4√24 II. 3√25 > 4√32 > 6√48 III. 4√9 > 3√15 > 6√24 Formula used: To compare surds, make the indices of the roots equal using the LCM of the root values. Calculation: Statement I: √12, 3√16, 4√24 LCM of indices (2, 3, 4) = 12 ⇒ $(12^{1/2})^{12}$ = 126 = 2,985,984 ⇒ $(16^{1/3})^{12}$ = 164 = 65,536 ⇒ $(24^{1/4})^{12}$ = 243 = 13,824 Since 2,985,984 > 65,536 > 13,824 ⇒ Statement I is TRUE. Statement II: 3√25, 4√32, 6√48 LCM of indices (3, 4, 6) = 12 ⇒ $(25^{1/3})^{12}$ = 254 = 390,625 ⇒ $(32^{1/4})^{12}$ = 323 = 32,768 ⇒ $(48^{1/6})^{12}$ = 482 = 2,304 Since 390,625 > 32,768 > 2,304 ⇒ Statement II is TRUE. Statement III: 4√9, 3√15, 6√24 LCM of indices (4, 3, 6) = 12 ⇒ $(9^{1/4})^{12}$ = 93 = 729 ⇒ $(15^{1/3})^{12}$ = 154 = 50,625 Since 729 < 50,625, the order √49 > √315 is FALSE. ⇒ Statement III is FALSE. ∴ The correct answer is option (1).

Shortcut Trick

Use the direct identity: (x − y)² + (y − z)² + (z − x)² = 2[(x + y + z)² − 3(xy + yz + zx)]

Given: x + y + z = 22 and xy + yz + zx = 35

Substitute values: 2[(22)² − 3(35)]

⇒ 2[484 − 105] = 2 × 379 = 758

∴ The correct answer is 758.

Alternate Method

Given: x + y + z = 22, xy + yz + zx = 35

Formula Used: (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

⇒ First, find the sum of squares: x² + y² + z²

⇒ (22)² = x² + y² + z² + 2(35)

⇒ 484 = x² + y² + z² + 70

⇒ x² + y² + z² = 484 − 70 = 414

⇒ Now, expand the target expression:

⇒ (x − y)² + (y − z)² + (z − x)² = (x² − 2xy + y²) + (y² − 2yz + z²) + (z² − 2zx + x²)

⇒ Expression = 2(x² + y² + z²) − 2(xy + yz + zx)

⇒ Expression = 2(414) − 2(35)

⇒ Expression = 828 − 70 = 758

∴ The correct answer is 758.

Additional Information

Algebraic Identity for Sum of Squares

The relation between the square of a trinomial and its components is: (x+y+z)² = x²+y²+z² + 2(xy+yz+zx).

Half-Sum of Squares Difference

The expression x²+y²+z² − xy − yz − zx is exactly equal to ½[(x−y)² + (y−z)² + (z−x)²].

Cubic Expansion Identity

x³+y³+z³ − 3xyz = (x+y+z)(x²+y²+z² − xy − yz − zx), which can also be written using the differences of terms.

Shortcut Trick

Pick simple values satisfying (x + y) / z = 2. Avoid values that make denominators zero.

Let z = 2, x = 3, and y = 1. Then (3 + 1) / 2 = 4 / 2 = 2. The condition is satisfied.

Now, substitute these values into the expression [y / (y − z)] + [x / (x − z)]:

⇒ [1 / (1 − 2)] + [3 / (3 − 2)]

⇒ [1 / (−1)] + [3 / 1] = −1 + 3 = 2.

∴ The correct answer is 2.

Alternate Method

Given: (x + y) / z = 2

Formula used: Algebraic simplification and term rearrangement.

⇒ (x + y) / z = 2

⇒ x + y = 2z

⇒ x + y = z + z

⇒ x − z = z − y

⇒ Target Expression = [y / (y − z)] + [x / (x − z)]

⇒ Substitute (x − z) = (z − y) into the second term:

⇒ [y / (y − z)] + [x / (z − y)]

⇒ Since (z − y) = −(y − z):

⇒ [y / (y − z)] − [x / (y − z)]

⇒ (y − x) / (y − z)

⇒ From x + y = 2z, we can also write y − z = z − x, so y = 2z − x

⇒ Numerator: (y − x) = (2z − x − x) = 2z − 2x = 2(z − x)

⇒ Denominator: (y − z) = (2z − x − z) = z − x

⇒ Expression = 2(z − x) / (z − x)

⇒ Expression = 2

∴ The correct answer is 2.

Additional Information

Algebraic Substitution Rule

If an expression must hold true for all variables satisfying a condition, picking specific numerical values (avoiding division by zero) will always yield the correct constant result.

Symmetric Expressions

In many competitive exam questions, if the variables x, y, and z are balanced in the condition (like x + y = 2z), the result often simplifies to a small integer like 0, 1, or 2.

Linear Equation in Three Variables

A single equation with three variables like x + y − 2z = 0 defines a plane in 3D space, allowing two degrees of freedom for value selection.

Shortcut Trick

For the equation x² − 2x + 4 = 0, the sum α + β = 2 and product αβ = 4.

Multiplying by (α + 2), we get α³ + 2³ = 0 ⇒ α³ = −8. Similarly, β³ = −8.

New root 1: α³/β² = −8/β². Since β = 4/α, Root 1 = −8 ÷ (16/α²) = −α²/2.

From the original equation: α² = 2α − 4 ⇒ −α²/2 = −(2α − 4)/2 = 2 − α. Since α + β = 2, then 2 − α = β.

The new roots are simply β and α, making the new equation identical to the original.

∴ The correct answer is x² − 2x + 4 = 0.

Alternate Method

Given equation: x² − 2x + 4 = 0

Sum of roots (α + β) = 2

Product of roots (αβ) = 4

To find the equation with roots p = α³/β² and q = β³/α²:

⇒ Sum (S) = p + q = α³/β² + β³/α² = (α⁵ + β⁵) / (αβ)²

⇒ α² + β² = (α + β)² − 2αβ = 2² − 2(4) = 4 − 8 = −4

⇒ α³ + β³ = (α + β)(α² + β² − αβ) = 2(−4 − 4) = −16

⇒ α⁵ + β⁵ = (α² + β²)(α³ + β³) − α²β²(α + β)

⇒ α⁵ + β⁵ = (−4)(−16) − (4²)(2) = 64 − 32 = 32

⇒ S = 32 / 4² = 32 / 16 = 2

⇒ Product (P) = p × q = (α³/β²) × (β³/α²) = αβ = 4

⇒ Required equation: x² − Sx + P = 0

⇒ x² − 2x + 4 = 0

∴ The correct answer is x² − 2x + 4 = 0.

Additional Information

Relationship between Roots and Coefficients

For a quadratic equation ax² + bx + c = 0, the sum of roots is −b/a and the product of roots is c/a.

Formation of Quadratic Equation

If the roots are p and q, the quadratic equation is given by x² − (p + q)x + pq = 0.

Complex Roots of Unity

Equations of the form x² ± ax + a² = 0 have roots related to the cube roots of −a³ or a³, involving terms like aω and aω².

Newton's Sums for Powers

Higher powers of roots can be found using the relation S_n = αⁿ + βⁿ, where aS_n + bS_n−1 + cS_n−2 = 0.

Shortcut Trick If roots of Ax2 + Bx + C = 0 are in ratio m : n, then (m + n)2 / (m × n) = B2 / AC. Given ratio (m : n) = 2.5 : 1 → 5 : 2. Substitute m = 5 and n = 2 in the formula: (5 + 2)2 / (5 × 2) = B2 / AC 72 / 10 = B2 / AC ⇒ 49 / 10 = B2 / AC ∴ The correct answer is 10B2 = 49CA. Alternate Method Given Equation: Ax2 + Bx + C = 0 Let the roots be α and β. Given: α = 2.5β = (5/2)β Sum of roots (α + β) = −B/A ⇒ (5/2)β + β = −B/A ⇒ (7/2)β = −B/A ⇒ β = −2B / 7A Product of roots (α × β) = C/A ⇒ (5/2)β × β = C/A ⇒ (5/2)β2 = C/A Substitute the value of β from above: ⇒ (5/2) × (−2B / 7A)2 = C/A ⇒ (5/2) × (4B2 / 49A2) = C/A ⇒ (10B2 / 49A2) = C/A ⇒ 10B2 = 49A2 × (C/A) ⇒ 10B2 = 49AC ∴ The correct answer is 10B2 = 49CA. Additional Information Nature of Roots The nature depends on the discriminant D = B2 − 4AC. If D > 0 roots are real and distinct; if D = 0 roots are real and equal; if D < 0 roots are imaginary. Roots Ratio Relation If one root is n times the other root, the condition is nB2 = (n + 1)2AC. Sum and Product of Roots For a quadratic equation ax2 + bx + c = 0, sum = −b/a and product = c/a.

Shortcut Trick Shift the variable: Let (x − 4) = k, which means x = k + 4. Equation becomes (k + 4)2 − 12(k + 4) + 33 = 0. On simplifying: k2 + 8k + 16 − 12k − 48 + 33 = 0 ⇒ k2 − 4k + 1 = 0. Dividing by k gives k + 1/k = 4. Required value k2 + 1/k2 = 42 − 2 = 14. ∴ The correct answer is 14. Alternate Method Given: x2 − 12x + 33 = 0 Formula used: (a − b)2 = a2 − 2ab + b2 and if k + 1/k = n, then k2 + 1/k2 = n2 − 2 ⇒ x2 − 12x + 33 = 0 ⇒ (x2 − 8x + 16) − 4x + 17 = 0 ⇒ (x − 4)2 − 4x + 16 + 1 = 0 ⇒ (x − 4)2 − 4(x − 4) + 1 = 0 ⇒ Let t = (x − 4) ⇒ t2 − 4t + 1 = 0 ⇒ t − 4 + 1/t = 0 ⇒ t + 1/t = 4 ⇒ (t + 1/t)2 = 42 ⇒ t2 + 1/t2 + 2 = 16 ⇒ t2 + 1/t2 = 16 − 2 ⇒ (x − 4)2 + 1/(x − 4)2 = 14 ∴ The correct answer is 14. Additional Information Algebraic Identity for Squares If x + 1/x = k, then x2 + 1/x2 = k2 − 2. This is derived by squaring both sides of the equation. Algebraic Identity for Cubes If x + 1/x = k, then x3 + 1/x3 = k3 − 3k. This is derived using the formula (a + b)3 = a3 + b3 + 3ab(a + b). Discriminant of Quadratic Equation For ax2 + bx + c = 0, the nature of roots is determined by D = b2 − 4ac. If D > 0, roots are real and distinct.

Shortcut Trick

Directly substitute the given value b = 1 into the equation.

a⁴ + 1 = a²(4(1) − 1 − 1) ⇒ a⁴ + 1 = 2a².

Rearrange the terms: a⁴ − 2a² + 1 = 0.

This is a perfect square: (a² − 1)² = 0, which implies a² = 1.

Thus, a⁴ = 1. Therefore, a⁴ + b⁴ = 1 + 1⁴ = 2.

∴ The correct answer is 2.

Alternate Method

Given equation: a⁴ + 1 = [a²/b²](4b² − b⁴ − 1)

Given value: b = 1

Substitute b = 1 into the equation:

⇒ a⁴ + 1 = [a²/1²] × (4(1)² − 1⁴ − 1)

⇒ a⁴ + 1 = a² × (4 − 1 − 1)

⇒ a⁴ + 1 = a² × 2

⇒ a⁴ − 2a² + 1 = 0

Let a² = x. Then the equation becomes:

⇒ x² − 2x + 1 = 0

⇒ (x − 1)² = 0

⇒ x = 1

⇒ a² = 1

Now, calculate the value of a⁴ + b⁴:

⇒ a⁴ = (a²)² = 1² = 1

⇒ b⁴ = 1⁴ = 1

⇒ a⁴ + b⁴ = 1 + 1 = 2

∴ The correct answer is 2.

Additional Information

Perfect Square Formula

The expression (x − y)² = x² − 2xy + y² is a fundamental identity used to factorize quadratic equations into perfect squares.

Algebraic Substitution

Substituting known numerical values for variables early in a calculation simplifies complex algebraic expressions and reduces the chance of error.

Quadratic Form

Equations involving higher powers like a⁴ can often be solved by treating them as quadratics in terms of a squared variable (a²).

Shortcut Trick

Let x = √((1 − a)/a). The equation simplifies to 3x + 9 = 19 − 3/x.

Rearranging gives 3x + 3/x = 10, or x + 1/x = 10/3.

Since 10/3 = 3 + 1/3, it is clear by inspection that x = 3 or x = 1/3.

If √((1 − a)/a) = 3, then (1 − a)/a = 9 ⇒ 1 − a = 9a ⇒ a = 1/10.

If √((1 − a)/a) = 1/3, then (1 − a)/a = 1/9 ⇒ 9 − 9a = a ⇒ a = 9/10.

∴ The correct answer is 1/10, 9/10.

Alternate Method

Given: 3√((1 − a)/a) + 9 = 19 − 3√(a/(1 − a))

Formula: Substitution method for radical equations.

⇒ 3√((1 − a)/a) + 3√(a/(1 − a)) = 19 − 9

⇒ 3√((1 − a)/a) + 3√(a/(1 − a)) = 10

⇒ Let t = √((1 − a)/a). Then √(a/(1 − a)) = 1/t.

⇒ 3t + 3/t = 10

⇒ 3t² + 3 = 10t

⇒ 3t² − 10t + 3 = 0

⇒ 3t² − 9t − t + 3 = 0

⇒ 3t(t − 3) − 1(t − 3) = 0

⇒ (3t − 1)(t − 3) = 0

⇒ t = 3 or t = 1/3

⇒ Case 1: √((1 − a)/a) = 3 → (1 − a)/a = 9 → 1 − a = 9a → 10a = 1 → a = 1/10

⇒ Case 2: √((1 − a)/a) = 1/3 → (1 − a)/a = 1/9 → 9 − 9a = a → 10a = 9 → a = 9/10

∴ The correct answer is 1/10, 9/10.

Additional Information

Substitution Method

Equations involving complex radical terms can be simplified by replacing the radical expression with a single variable, transforming the equation into a solvable polynomial.

Reciprocal Equations

An equation of the form x + 1/x = k + 1/k generally yields roots x = k and x = 1/k. This property is useful for identifying roots by inspection.

Quadratic Roots

The roots of ax² + bx + c = 0 can be found using the quadratic formula: x = [−b ± √(b² − 4ac)] / 2a.

Shortcut Trick

Squaring the equation: a/b − 13 = −b/a − 11

⇒ a/b + b/a = 13 − 11 = 2

If x + 1/x = 2, then x = 1 ⇒ a/b = 1 → a = b

Since a + b = 10 and a = b, then a = 5 and b = 5

Target: 3(25) + 4(25) + 5(25) = (3 + 4 + 5) × 25 = 12 × 25 = 300

∴ The correct answer is 300.

Alternate Method

Given:

a + b = 10

√[(a/b) − 13] = √[(−b/a) − 11]

Formula Used:

If x + 1/x = 2, then x = 1

(a − b)² = a² + b² − 2ab

Calculations:

⇒ Square both sides of the equation: √[(a/b) − 13] = √[(−b/a) − 11]

⇒ a/b − 13 = −b/a − 11

⇒ a/b + b/a = 13 − 11

⇒ a/b + b/a = 2

⇒ (a² + b²) / ab = 2

⇒ a² + b² = 2ab

⇒ a² + b² − 2ab = 0

⇒ (a − b)² = 0 ⇒ a = b

Substitute a = b in a + b = 10:

⇒ b + b = 10 ⇒ 2b = 10 ⇒ b = 5

⇒ a = 5

Now, calculate the value of 3ab + 4a² + 5b²:

⇒ 3(5 × 5) + 4(5²) + 5(5²)

⇒ 3(25) + 4(25) + 5(25)

⇒ 75 + 100 + 125

⇒ 300

∴ The correct answer is 300.

Additional Information

Reciprocal Identity

If a real number x satisfies x + 1/x = 2, then x must be 1. Similarly, if x + 1/x = −2, then x = −1.

Elimination of Radicals

Equations involving square roots (√f(x) = √g(x)) are solved by squaring both sides to get f(x) = g(x).

Algebraic Expansion

The square of a binomial difference is given by (a − b)² = a² − 2ab + b². If this equals 0, then a = b.

Shortcut Trick

Rearrange equations to isolate variables: 3x + 4y − 2z = 8, 7x + 2y + 11z = 15, and 5x + 9y + 6z = 22.

Observe the sum of coefficients: (3 + 7 + 5) = 15, (4 + 2 + 9) = 15, and (−2 + 11 + 6) = 15.

Summing all equations: 15x + 15y + 15z = 8 + 15 + 22 ⇒ 15(x + y + z) = 45.

So, x + y + z = 3. Therefore, x + y + z − 34 = 3 − 34 = −31.

∴ The correct answer is −31.

Alternate Method

Given Equations:

Eq(1): 3x + 4y − 2z + 9 = 17 ⇒ 3x + 4y − 2z = 8

Eq(2): 7x + 2y + 11z + 8 = 23 ⇒ 7x + 2y + 11z = 15

Eq(3): 5x + 9y + 6z − 4 = 18 ⇒ 5x + 9y + 6z = 22

Calculations:

⇒ Adding Eq(1), Eq(2), and Eq(3):

⇒ (3x + 7x + 5x) + (4y + 2y + 9y) + (−2z + 11z + 6z) = 8 + 15 + 22

⇒ 15x + 15y + 15z = 45

⇒ 15(x + y + z) = 45

⇒ x + y + z = 45 ÷ 15 = 3

⇒ Required value = (x + y + z) − 34

⇒ Required value = 3 − 34 = −31

∴ The correct answer is −31.

Additional Information

Symmetric Coefficients

When the sum of coefficients for each variable is identical across a set of equations, the sum of the variables can be found by adding all equations and dividing by that common sum.

Linear Independence

A system of n linear equations with n variables has a unique solution if the equations are linearly independent (one cannot be formed by a linear combination of others).

Direct Substitution

In competitive exams, always check if summing or subtracting the equations yields the desired expression (like x+y+z) before solving for individual variables x, y, or z.

Shortcut Trick

Simplify the equations to linear form: 2x + 6y − z = 12, 3x + 4y + 6z = 99, and x + y + 15z = 63.

Eliminate z using pairs: 6 × Eq(1) + Eq(2) → 15x + 40y = 171.

Eliminate z using pairs: 15 × Eq(1) + Eq(3) → 31x + 91y = 243.

Sum the results: (15x + 31x) + (40y + 91y) = 171 + 243 = 414.

∴ The correct answer is 414.

Alternate Method

Given:

1. (x + 3y − 2z/4) = 6 ⇒ x + 3y − z/2 = 6

2. x + (2/3)(2y + 3z) = 33

3. (1/7)(x + y + z) + 2z = 9

Calculations:

⇒ From (1): 2x + 6y − z = 12 ... (Eq. i)

⇒ From (2): 3x + 4y + 6z = 99 ... (Eq. ii)

⇒ From (3): x + y + z + 14z = 63 ⇒ x + y + 15z = 63 ... (Eq. iii)

⇒ Multiply (Eq. i) by 6 and add to (Eq. ii):

⇒ (12x + 36y − 6z) + (3x + 4y + 6z) = 72 + 99

⇒ 15x + 40y = 171 ... (Eq. iv)

⇒ Multiply (Eq. i) by 15 and add to (Eq. iii):

⇒ (30x + 90y − 15z) + (x + y + 15z) = 180 + 63

⇒ 31x + 91y = 243 ... (Eq. v)

⇒ Adding (Eq. iv) and (Eq. v):

⇒ (15x + 31x) + (40y + 91y) = 171 + 243

⇒ 46x + 131y = 414

∴ The correct answer is 414.

Additional Information

System of Linear Equations

A system of three linear equations in three variables requires at least three independent equations to find unique values for x, y, and z.

Elimination Method

This involves multiplying equations by suitable constants so that adding or subtracting them eliminates one variable at a time, reducing the system complexity.

Cramer's Rule

For a system AX = B, the solution is given by x = D_x/D, y = D_y/D, and z = D_z/D, where D is the determinant of the coefficient matrix.

Shortcut Trick

In a triangle, the length of median XM = 1/2 √(2(XY² + XZ²) − YZ²).

Given XY = 11, XZ = 7, YZ = 14.

XM = 1/2 √(2(11² + 7²) − 14²) = 1/2 √(2(121 + 49) − 196) = 1/2 √(340 − 196).

XM = 1/2 √144 = (1/2) × 12 = 6 cm.

Centroid G divides median XM in ratio XG : GM = 2 : 1, so GM = (1/3) × XM.

GM = (1/3) × 6 = 2 cm.

∴ The correct answer is 2 cm.

Alternate Method

Given: Triangle XYZ with sides XY = 11 cm, YZ = 14 cm, and XZ = 7 cm. G is the centroid and XM is the median to side YZ.

Formula Used: Apollonius Theorem ⇒ 2(XM² + YM²) = XY² + XZ², where M is the midpoint of YZ.

⇒ M is the midpoint of YZ, so YM = YZ ÷ 2 = 14 ÷ 2 = 7 cm.

⇒ By Apollonius Theorem: 2(XM² + 7²) = 11² + 7²

⇒ 2(XM² + 49) = 121 + 49

⇒ 2(XM² + 49) = 170

⇒ XM² + 49 = 85

⇒ XM² = 85 − 49 = 36

⇒ XM = √36 = 6 cm

⇒ Centroid G divides median XM such that XG : GM = 2 : 1

⇒ GM = 1/(2 + 1) × XM = 1/3 × 6 = 2 cm

∴ The correct answer is 2 cm.

Additional Information

Centroid Property

The centroid of a triangle is the intersection point of its three medians and divides each median in the ratio 2 : 1 from the vertex.

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).

Median Area Property

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

Shortcut Trick

Side of the inscribed square (a) = Radius × √2 = 4 × √2 = 4√2 cm.

Since PQ is produced to Y and QR = QY = a, then PY = PQ + QY = a + a = 2a.

In right-angled triangle SPY, ∠SPY = 90° (as PS ⊥ PQ).

Using Pythagoras theorem: SY = √(SP² + PY²) = √(a² + (2a)²) = a√5.

Substituting a = 4√2: SY = 4√2 × √5 = 4√10 cm.

∴ The correct answer is 4√10 cm.

Alternate Method

Given:

Radius of circle (r) = 4 cm

Diagonal of square PQRS = Diameter of circle = 2 × 4 = 8 cm

QR = QY

Formula Used:

Side of square (a) = Diagonal / √2

Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular²

Calculations:

⇒ Side of square (PQ = QR = RS = SP) = 8 / √2 = 4√2 cm

⇒ Given QR = QY, so QY = 4√2 cm

⇒ Length PY = PQ + QY = 4√2 + 4√2 = 8√2 cm

⇒ In ▵SPY, ∠SPY = 90° (Since PQRS is a square and PQ is produced to Y)

⇒ SY² = SP² + PY²

⇒ SY² = (4√2)² + (8√2)²

⇒ SY² = 32 + 128 = 160

⇒ SY = √160 = √(16 × 10) = 4√10 cm

∴ The correct answer is 4√10 cm.

Additional Information

Inscribed Square Property

When a square is inscribed in a circle, its diagonal coincides with the diameter of the circle. Side a = Diameter / √2 = r√2.

Tangent Property

A tangent at any point on a circle is perpendicular to the radius through the point of contact. Here, radius OR ⊥ tangent RY.

Pythagorean Triplet for a and 2a

In a right triangle with legs in ratio 1:2, the hypotenuse is always √5 × (shorter leg). SY = SP × √5.

Shortcut Trick

The ratio of the lengths of the parallel sides of a trapezium is equal to the ratio in which the diagonals divide each other.

Ratio of segments = 2 : 9

Ratio of parallel sides = 2 : 9

Given: Larger side (9 units) = 45 cm

⇒ 1 unit = 45 ÷ 9 = 5 cm

⇒ Smaller side (2 units) = 2 × 5 = 10 cm

∴ The correct answer is 10 cm.

Alternate Method

Given:

Ratio of diagonal segments = 2 : 9

Larger parallel side = 45 cm

Let the trapezium be ABCD with AB || CD and diagonals AC and BD intersecting at O.

⇒ In ΔAOB and ΔCOD:

⇒ ∠OAB = ∠OCD (Alternate interior angles as AB || CD)

⇒ ∠OBA = ∠ODC (Alternate interior angles as AB || CD)

⇒ ΔAOB ≈ ΔCOD (By AA similarity criterion)

⇒ Corresponding sides are proportional:

⇒ AB ÷ CD = OB ÷ OD = OA ÷ OC

⇒ Given the diagonal ratio is 2 : 9, then AB ÷ CD = 2 ÷ 9

⇒ AB ÷ 45 = 2 ÷ 9

⇒ AB = (2 × 45) ÷ 9

⇒ AB = 2 × 5

⇒ AB = 10 cm

∴ The correct answer is 10 cm.

Additional Information

Similarity Property of Diagonals

In any trapezium, the triangles formed by the diagonals and the parallel sides are always similar. The ratio of their areas is the square of the ratio of their parallel sides: Area(ΔAOB) ÷ Area(ΔCOD) = (AB ÷ CD)².

Area of a Trapezium

The area is calculated using the formula: Area = 1 ÷ 2 × (Sum of parallel sides) × Height.

Median of a Trapezium

The line segment joining the midpoints of the non-parallel sides is called the median. Its length is the average of the two parallel sides: Median = (a + b) ÷ 2.

Shortcut Trick

Total diameter AB = CD = 2 × 5 = 10 cm.

Point P divides AB in 2:3 ratio → PB = (2/5) × 10 = 4 cm. Distance OP = 5 − 4 = 1 cm.

Point N divides CD in 2:3 ratio → CN = (2/5) × 10 = 4 cm. Distance ON = 1 cm.

M is the intersection of perpendiculars, so M is 1 cm from both diameters.

S lies on the circle at a vertical distance of 1 cm from AB → SR = √(5² − 1²) = √24 = 2√6 cm.

SM = SR − RM = 2√6 − 1 cm.

∴ The correct answer is (2√6) − 1 cm.

Alternate Method

Given: Radius (r) = 5 cm. Diameters AB and CD are perpendicular.

Let O be the origin (0,0). Then A = (−5,0), B = (5,0), C = (0,5), D = (0,−5).

Ratio PB : PA = 2 : 3. Since length AB = 10 cm:

⇒ PB = (2 / 5) × 10 = 4 cm

⇒ P is at (5 − 4, 0) = (1, 0). So, OP = 1 cm.

Ratio CN : ND = 2 : 3. Since length CD = 10 cm:

⇒ CN = (2 / 5) × 10 = 4 cm

⇒ N is at (0, 5 − 4) = (0, 1). So, ON = 1 cm.

M is the intersection of the perpendicular from P and the perpendicular from N.

⇒ Coordinates of M = (1, 1).

S lies on the circle x² + y² = 5² and is on the horizontal line SR (y = 1) passing through N and M.

⇒ x² + 1² = 25

⇒ x² = 24

⇒ x = √24 = 2√6 cm.

The point R is the intersection of the horizontal line with diameter CD, so R = (0, 1).

⇒ SR = 2√6 cm and RM = OP = 1 cm.

⇒ SM = SR − RM = 2√6 − 1 cm.

∴ The correct answer is (2√6) − 1 cm.

Additional Information

Equation of a Circle

For a circle with center (h, k) and radius r, the equation is (x − h)² + (y − k)² = r². In this problem, center is (0,0).

Pythagoras Theorem in Circles

Any chord perpendicular to a radius forms a right-angled triangle where r² = (distance from center)² + (half chord length)².

Section Formula

A point dividing a segment with endpoints (x₁, y₁) and (x₂, y₂) in ratio m:n has coordinates ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).

Shortcut Trick

Side of square PQRS = 20 cm → Center O₁ is at (10, 10).

In right ▵QRT, QR = 20, QT = 25 ⇒ RT = √(25² − 20²) = 15 cm.

Inradius r₂ of ▵QRT = (Base + Height − Hypotenuse) ÷ 2 = (20 + 15 − 25) ÷ 2 = 5 cm.

Position of O₂ relative to O₁: Δx = 10 + 5 = 15 cm, Δy = 10 − 5 = 5 cm.

Distance O₁O₂ = √(15² + 5²) = √(225 + 25) = √250 = 5√10 cm.

∴ The correct answer is 5√10 cm.

Alternate Method

Given:

Side of square (a) = 20 cm

Hypotenuse of ▵QRT (QT) = 25 cm

Formula Used:

Inradius of a right-angled triangle = (Leg₁ + Leg₂ − Hypotenuse) ÷ 2

Distance between two points (x₁, y₁) and (x₂, y₂) = √((x₂ − x₁)² + (y₂ − y₁)²)

Calculations:

Let S be the origin (0, 0).

⇒ Coordinates: S(0, 0), R(20, 0), Q(20, 20), P(0, 20).

⇒ In ▵QRT, QR = 20 cm and QT = 25 cm.

⇒ RT = √(QT² − QR²) = √(25² − 20²) = √(625 − 400) = 15 cm.

⇒ T is on extended SR → T = (20 + 15, 0) = (35, 0).

⇒ O₁ is the center of the square incircle → O₁ = (20 ÷ 2, 20 ÷ 2) = (10, 10).

⇒ Inradius of ▵QRT (r₂) = (20 + 15 − 25) ÷ 2 = 5 cm.

⇒ O₂ is shifted from R(20, 0) by r₂ along x-axis and r₂ along y-axis.

⇒ O₂ = (20 + 5, 0 + 5) = (25, 5).

⇒ Distance O₁O₂ = √((25 − 10)² + (5 − 10)²)

⇒ O₁O₂ = √(15² + (−5)²) = √(225 + 25) = √250

⇒ O₁O₂ = 5√10 cm.

∴ The correct answer is 5√10 cm.

Additional Information

Inradius of a Right-Angled Triangle

The inradius (r) can be calculated using the formula r = (P + B − H) ÷ 2, where P is perpendicular, B is base, and H is hypotenuse.

Distance Formula in Coordinate Geometry

The distance between two points A(x₁, y₁) and B(x₂, y₂) is given by AB = √((x₂ − x₁)² + (y₂ − y₁)²).

Center of a Square

The center of a square is the point of intersection of its diagonals and also the center of its inscribed and circumscribed circles.

Shortcut Trick

For a square of side 'a', the radius 'r' of a circle passing through two vertices and tangent to the opposite side is given by r = 5a/8.

Given: Side of square (a) = 6 cm.

Applying the shortcut formula:

r = (5 × 6) ÷ 8 = 30 ÷ 8 = 3.75 cm.

∴ The correct answer is 3.75 cm.

Alternate Method

Given: PQSR is a square with side length a = 6 cm. Circle passes through P and R, and is tangent to side QS at point M.

Formula used: Pythagoras theorem in a circle, r² = (a − r)² + (a/2)².

⇒ Let the radius of the circle be r cm.

⇒ Let the center of the circle be O. Since the circle passes through P and R, O lies on the perpendicular bisector of PR.

⇒ Distance from O to side PR = Side − Radius = 6 − r.

⇒ In the right-angled triangle formed by the center O, the midpoint of PR, and vertex R:

⇒ Hypotenuse = r, Base = 6 − r, Perpendicular = 6/2 = 3.

⇒ r² = (6 − r)² + 3²

⇒ r² = 36 + r² − 12r + 9

⇒ 12r = 45

⇒ r = 45 ÷ 12 = 3.75 cm.

∴ The correct answer is 3.75 cm.

Additional Information

Tangent-Radius Property

The radius of a circle is always perpendicular to the tangent at the point of contact.

Perpendicular Bisector of a Chord

The perpendicular bisector of any chord (like PR in this case) always passes through the center of the circle.

Circle in a Square

If a circle is inscribed in a square of side 'a', its radius is a/2. If it passes through vertices, the radius depends on the specific points of contact or intersection.

Shortcut Trick

In a right ΔPQR, altitude QS to hypotenuse PR follows: QS² = PS × SR.

First, find PS using Pythagoras in ΔPQS: PS = √(35² − 28²) = √(1225 − 784) = √441 = 21 cm.

Apply geometric mean theorem: SR = QS² ÷ PS.

SR = 28² ÷ 21 = 784 ÷ 21 = 37.33 cm.

∴ The correct answer is 37.33 cm.

Alternate Method

Given: PQ = 35 cm, QS = 28 cm, ∠PQR = 90°. Semicircle passes through Q, R, and S.

Formula: In right ΔPQR, if QS ⊥ PR, then PS² + QS² = PQ² and QS² = PS × SR.

⇒ In ΔPQS, applying Pythagoras theorem:

⇒ PS = √(PQ² − QS²)

⇒ PS = √(35² − 28²)

⇒ PS = √(1225 − 784) = √441 = 21 cm

⇒ For a semicircle to pass through Q, R, S where S lies on the hypotenuse, ∠QSR must be 90° (angle in a semicircle).

⇒ Using property of altitude to hypotenuse: QS² = PS × SR

⇒ 28² = 21 × SR

⇒ 784 = 21 × SR

⇒ SR = 784 ÷ 21 = 37.333...

∴ The correct answer is 37.33 cm.

Additional Information

Altitude to Hypotenuse Properties

In a right triangle with altitude h to hypotenuse dividing it into segments x and y, h² = x × y. Also, leg² = adjacent segment × hypotenuse.

Circle Properties in Triangles

The angle subtended by a diameter at any point on the circumference is 90°. If ∠QSR = 90°, then QR is the diameter of the circle passing through Q, S, and R.

Pythagorean Triplets

Common triplets like 3, 4, 5 can be scaled. Here, 21, 28, 35 is a 3:4:5 triplet multiplied by 7, which helps in quick calculation of PS.

Shortcut Trick

Let ∠RSP = x. Since QS = PR and PR is the radius, QS = Radius = PQ.

In isosceles ΔPQS (PQ = QS), ∠QPS = ∠QSP = x.

Exterior ∠PQR of ΔPQS = ∠QPS + ∠QSP = x + x = 2x.

In isosceles ΔPQR (PQ = PR = radii), ∠PRQ = ∠PQR = 2x.

Exterior ∠TPR of ΔPSR = ∠RSP + ∠PRQ = x + 2x = 3x.

∴ The correct answer is 1 ∶ 3.

Alternate Method

Given:

P is the centre of the circle. PR is the radius. QS = PR. Points S, Q, R are collinear.

Formula Used:

1. Angles opposite to equal sides in a triangle are equal.
2. Exterior angle of a triangle is equal to the sum of two opposite interior angles.

Calculations:

⇒ Let ∠RSP = x

⇒ Join PQ. Since PQ and PR are radii, PQ = PR.

⇒ Given QS = PR, ∴ QS = PQ = PR = Radius.

⇒ In ΔPQS, side PQ = side QS ⇒ ∠QPS = ∠QSP = x

⇒ In ΔPQS, exterior ∠PQR = ∠QPS + ∠QSP = x + x = 2x

⇒ In ΔPQR, side PQ = side PR (radii) ⇒ ∠PRQ = ∠PQR = 2x

⇒ In ΔPSR, ∠TPR is the exterior angle at vertex P.

⇒ ∠TPR = ∠RSP + ∠PRQ = x + 2x = 3x

⇒ Ratio = ∠RSP ∶ ∠TPR = x ∶ 3x = 1 ∶ 3

∴ The correct answer is 1 ∶ 3.

Additional Information

Exterior Angle Theorem

The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.

Isosceles Triangle Theorem

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

Circle Secant Property

In this specific configuration where the external segment of a secant equals the radius, the intercepted central angle is always three times the external angle.

Shortcut Trick

Length of Direct Common Tangent (DCT) = √(d² − (r₁ − r₂)²)

Here, d = 60 cm, r₁ = 35 cm, r₂ = 24 cm

Difference in radii (r₁ − r₂) = 35 − 24 = 11 cm

DCT = √(60² − 11²) = √(3600 − 121) = √3479 cm

∴ The correct answer is √3479 cm.

Alternate Method

Given:

Distance between centers (d) = 60 cm

Radius of first circle (r₁) = 35 cm

Radius of second circle (r₂) = 24 cm

Formula Used:

Length of Direct Common Tangent = √[(Distance between centers)² − (r₁ − r₂)²]

Calculations:

⇒ Difference in radii = 35 − 24 = 11 cm

⇒ DCT = √[60² − (11)²]

⇒ DCT = √[3600 − 121]

⇒ DCT = √3479 cm

∴ The correct answer is √3479 cm.

Additional Information

Transverse Common Tangent (TCT)

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

Circles touching externally

When two circles touch externally, d = r₁ + r₂. In this case, DCT = 2√(r₁ × r₂) and TCT = 0.

Circles touching internally

When two circles touch internally, d = |r₁ − r₂|. In this case, DCT = 0 and TCT is not defined.

Shortcut Trick

Extend PQ and RS to meet at a point T to form an equilateral triangle TQR.

Area of ΔTQR (side = 18 cm) = (√3/4) × 18² = 81√3 cm².

The smaller triangle ΔTPS is a right-angled triangle with ∠S = 90°, ∠T = 60°, and PS = 9 cm.

Area of ΔTPS = 1/2 × PS × (PS / tan 60°) = 1/2 × 9 × (9/√3) = 13.5√3 cm².

Area of quadrilateral PQRS = Area(ΔTQR) − Area(ΔTPS) = 81√3 − 13.5√3 = 67.5√3 cm².

∴ The correct answer is (135√3)/2 cm².

Alternate Method

Given: QR = 18 cm, PS = 9 cm. In quadrilateral PQRS, ∠Q = 60°, ∠R = 60°, and ∠S = 90°.

Formula Used: Area of equilateral Δ = (√3/4) × side² | Area of right Δ = 1/2 × base × height.

⇒ Extend PQ and RS to meet at point T. Since ∠Q = 60° and ∠R = 60°, ΔTQR is an equilateral triangle.

⇒ Area of ΔTQR = (√3/4) × (18)² = (√3/4) × 324 = 81√3 cm².

⇒ In right-angled triangle TPS, ∠S = 90°, ∠T = 60°, and PS = 9 cm (given).

⇒ TS = PS / tan 60° = 9 / √3 = 3√3 cm.

⇒ Area of ΔTPS = 1/2 × PS × TS = 1/2 × 9 × 3√3 = 13.5√3 cm².

⇒ Area of quadrilateral PQRS = Area of ΔTQR − Area of ΔTPS

⇒ Area PQRS = 81√3 − 13.5√3 = 67.5√3 cm².

⇒ Area PQRS = (135√3)/2 cm².

∴ The correct answer is (135√3)/2 cm².

Additional Information

Area of Equilateral Triangle

The area of an equilateral triangle with side 'a' is given by the formula A = (√3/4) × a².

Trigonometric Ratios in Right Triangles

In a right-angled triangle, tan θ = Perpendicular / Base. For 60°, tan 60° = √3 and sin 60° = √3/2.

Sum of Angles in a Quadrilateral

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

Shortcut Trick

Angle bisector PS divides the base QR in the ratio of sides PQ and PR.

Ratio of Area(ΔPQS) : Area(ΔPSR) = PQ : PR = 2 : 3.

Total ratio parts = 2 + 3 = 5 units.

5 units = 180 cm² → 1 unit = 36 cm².

Area of ΔPSR = 3 units = 3 × 36 = 108 cm².

∴ The correct answer is 108 cm².

Alternate Method

Given:

Area of ΔPQR = 180 cm²

PQ : PR = 2 : 3

PS is the internal angle bisector of ∠QPR.

Formula Used:

Angle Bisector Theorem: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. ⇒ QS/SR = PQ/PR

Area of a triangle with constant height is proportional to its base.

Calculations:

According to the Angle Bisector Theorem:

⇒ QS / SR = PQ / PR = 2 / 3

Let QS = 2x and SR = 3x

Triangles ΔPQS and ΔPSR share the same vertex P and their bases QS and SR lie on the same line QR.

Thus, they have the same altitude (h) from P to QR.

⇒ Area(ΔPQS) = (1/2) × QS × h = (1/2) × 2x × h

⇒ Area(ΔPSR) = (1/2) × SR × h = (1/2) × 3x × h

⇒ Area(ΔPQS) : Area(ΔPSR) = 2x : 3x = 2 : 3

Total Area(ΔPQR) = Area(ΔPQS) + Area(ΔPSR)

⇒ 180 = 2k + 3k (where k is a constant)

⇒ 5k = 180

⇒ k = 180 ÷ 5 = 36

Area of ΔPSR = 3k

⇒ Area(ΔPSR) = 3 × 36

⇒ Area(ΔPSR) = 108 cm²

∴ The correct answer is 108 cm².

Additional Information

Internal Angle Bisector Theorem

In any triangle, the bisector of an interior angle divides the opposite side into segments whose lengths are in the same ratio as the lengths of the other two sides.

Area of Triangles with Same Altitude

If two triangles have the same height, the ratio of their areas is equal to the ratio of their corresponding bases. Area₁/Area₂ = Base₁/Base₂.

External Angle Bisector Theorem

For an external angle bisector at P intersecting extended QR at S, the ratio remains the same: QS/SR = PQ/PR.

Shortcut Trick

Each square formed by joining midpoints has half the area of the outer square.

Area of square LMNO = 1/2 × Area of square EFGH = 1/4 × Area of square ABCD.

Area of innermost square LMNO = (4/π) × Area of inscribed circle.

Area of LMNO = (4 × 38.5) ÷ (22/7) = (4 × 38.5 × 7) ÷ 22 = 49 cm².

Area of ABCD = 4 × Area of LMNO = 4 × 49 = 196 cm².

∴ The correct answer is 196 cm².

Alternate Method

Given:

Area of inscribed circle in LMNO = 38.5 cm²

ABCD, EFGH, and LMNO are nested squares formed by midpoints.

Calculations:

⇒ Area of circle = πr² = 38.5

⇒ (22/7) × r² = 38.5

⇒ r² = (38.5 × 7) ÷ 22 = 12.25

⇒ r = √12.25 = 3.5 cm

⇒ Side of square LMNO = Diameter of circle = 2r = 2 × 3.5 = 7 cm

⇒ Area of square LMNO = 7 × 7 = 49 cm²

⇒ Area of square EFGH = 2 × Area of LMNO = 2 × 49 = 98 cm²

⇒ Area of square ABCD = 2 × Area of EFGH = 2 × 98 = 196 cm²

∴ The correct answer is 196 cm².

Additional Information

Midpoint Square Area Property

When the midpoints of the sides of a square are joined, a new square is formed whose area is exactly half (1/2) the area of the original square.

Inscribed Circle Relationship

For a circle inscribed in a square of side 's', the radius 'r' of the circle is s/2, meaning the side of the square is equal to the diameter of the circle.

Area Comparison

The ratio of the area of a square to the area of its inscribed circle is always 4 : π, which is approximately 14 : 11.

Shortcut Trick

Area of a triangle formed by two adjacent sides and a diagonal (like ΔECD) in a regular hexagon is equal to the area of one equilateral triangle of the same side length.

Area of equilateral triangle = (√3 / 4) × (side)²

Area of ΔECD = (√3 / 4) × 12²

Area of ΔECD = (√3 / 4) × 144 = 36√3 cm².

∴ The correct answer is 36√3 cm².

Alternate Method

Given:

Side of regular hexagon (s) = 12 cm

Formula Used:

Interior angle of a regular hexagon = (n − 2) × 180° / n

Area of triangle = (1/2) × a × b × sin(θ)

Calculations:

⇒ In regular hexagon ABCDEF, internal angle ∠CDE = (6 − 2) × 180° / 6 = 120°

⇒ In ΔECD, CD = 12 cm and DE = 12 cm (sides of regular hexagon)

⇒ Area of ΔECD = (1/2) × CD × DE × sin(∠CDE)

⇒ Area = (1/2) × 12 × 12 × sin(120°)

⇒ sin(120°) = sin(180° − 60°) = sin(60°) = √3 / 2

⇒ Area = (1/2) × 144 × (√3 / 2)

⇒ Area = 72 × (√3 / 2) = 36√3 cm²

∴ The correct answer is 36√3 cm².

Additional Information

Area of Regular Hexagon

Total area of a regular hexagon with side 's' = 6 × (√3 / 4) × s² = (3√3 / 2) × s².

Internal Angle and Diagonals

Each internal angle is 120°. A regular hexagon has 9 diagonals. The longest diagonal (e.g., AD) is 2 × side.

Area Partitioning

A regular hexagon is composed of 6 congruent equilateral triangles meeting at the center.

Shortcut Trick

For a square of side 'a', the side of the largest regular octagon 's' is given by the formula: s = a / (√2 + 1).

By rationalizing the denominator, s = a × (√2 − 1).

Given side of square a = 16 cm.

s = 16 × (√2 − 1) = 16√2 − 16 cm.

∴ The correct answer is 16√2 − 16 cm.

Alternate Method

Given: Side of square (a) = 16 cm. Octagon is regular.

Let the length cut from each corner of the square be 'x'.

The hypotenuse of the corner triangle (octagon side) is s = √(x² + x²) = x√2.

The remaining part on the square edge (octagon side) is s = 16 − 2x.

⇒ x√2 = 16 − 2x

⇒ x√2 + 2x = 16

⇒ x(√2 + 2) = 16

⇒ x = 16 / (2 + √2)

Substitute x in s = x√2:

⇒ s = (16 × √2) / (2 + √2)

⇒ s = (16√2) / [√2(√2 + 1)]

⇒ s = 16 / (√2 + 1)

⇒ s = 16(√2 − 1) / [(√2 + 1)(√2 − 1)]

⇒ s = 16(√2 − 1) / (2 − 1)

⇒ s = 16√2 − 16 cm

∴ The correct answer is 16√2 − 16 cm.

Additional Information

Area of Regular Octagon

The area of a regular octagon with side 's' is given by A = 2(1 + √2)s².

Interior and Exterior Angles

For a regular octagon, each exterior angle is 360° ÷ 8 = 45° and each interior angle is 180° − 45° = 135°.

Apothem of Regular Octagon

The distance from the center to any side (inradius) is r = (s/2) × (1 + √2).

Shortcut Trick

Radius of the semicircle (R) = side QR = 7 cm.

The small circle is tangent to the corner of a square of side R and the semicircle.

Shortcut formula for radius (r) in this configuration: r = R × (3 − 2√2).

Substituting R = 7: r = 7 × (3 − 2√2) = 21 − 14√2 cm.

∴ The correct answer is 21 − 14√2 cm.

Alternate Method

Given: QR = 7 cm, Semicircle diameter = SR. PQRS is a rectangle.

Formula: Distance between centers of tangent circles d = R₁ + R₂.

⇒ Let radius of semicircle be R = 7 cm and radius of small circle be r.

⇒ Let center of semicircle be O(7, 0) and center of small circle be C(14 − r, 7 − r).

⇒ Small circle is tangent to semicircle externally, so OC = R + r = 7 + r.

⇒ Using distance formula: OC² = (14 − r − 7)² + (7 − r − 0)²

⇒ (7 + r)² = (7 − r)² + (7 − r)²

⇒ 49 + 14r + r² = 2(49 − 14r + r²)

⇒ 49 + 14r + r² = 98 − 28r + 2r²

⇒ r² − 42r + 49 = 0

⇒ Using quadratic formula: r = [42 − √(42² − 4 × 1 × 49)] / 2

⇒ r = [42 − √(1764 − 196)] / 2 = [42 − √1568] / 2

⇒ r = [42 − 28√2] / 2 = 21 − 14√2

∴ The correct answer is 21 − 14√2 cm.

Additional Information

Distance Between Centers of Tangent Circles

For two circles with radii R and r touching externally, distance d = R + r. For internal touch, d = R − r.

Circle in a Corner

The radius r of a circle in the corner of a square of side R, tangent to an inscribed arc of radius R, is given by r = R(3 − 2√2).

Quadratic Formula

For ax² + bx + c = 0, the roots are x = [−b ± √(b² − 4ac)] / 2a.

Shortcut Trick

Radius of a circle (r) inscribed in a quadrant of radius R is: r = R × (√2 − 1).

Given R = 7 cm ⇒ r = 7(√2 − 1) cm.

Area of circle = πr² = (22/7) × [7(√2 − 1)]².

Area = (22/7) × 49 × (2 + 1 − 2√2) = 154 × (3 − 2√2).

Area = 462 − 308√2 cm².

∴ The correct answer is 462 − 308√2 cm².

Alternate Method

Given: Radius of the quadrant (R) = 7 cm.

Formula Used: Radius of inscribed circle r = R / (√2 + 1); Area of circle = πr².

In the quadrant PQR, let O be the center and r be the radius of the inscribed circle.

⇒ PO = Diagonal of a square of side r = r√2.

⇒ Total radius of quadrant (R) = PO + Radius of circle = r√2 + r.

⇒ 7 = r(√2 + 1).

⇒ r = 7 / (√2 + 1).

Rationalizing the denominator:

⇒ r = [7 × (√2 − 1)] / [(√2 + 1) × (√2 − 1)] = 7(√2 − 1) cm.

⇒ Area of circle = πr² = (22/7) × [7(√2 − 1)]².

⇒ Area = (22/7) × 49 × (2 + 1 − 2√2).

⇒ Area = 154 × (3 − 2√2).

⇒ Area = 462 − 308√2 cm².

∴ The correct answer is 462 − 308√2 cm².

Additional Information

Radius of Inscribed Circle in a Quadrant

For a quadrant of radius R, the radius r of the largest inscribed circle is given by r = R/(√2 + 1) or r = R(√2 − 1).

Area of Circle and Quadrant

Area of a full circle = πr², while the area of a quadrant with radius R is (1/4)πR².

Diagonal of a Square

In geometry, the distance from the vertex of a quadrant to the center of the inscribed circle forms the diagonal of a square of side r, which is always r√2.

Shortcut Trick

Base Area of regular hexagon = (3√3 / 2) × side² = (3√3 / 2) × 6² = 54√3 cm².

Total Surface Area (TSA) = 2 × Base Area + Perimeter × Height.

216√3 = 2 × (54√3) + (6 × 6) × h.

216√3 = 108√3 + 36h ⇒ 36h = 108√3.

h = 108√3 ÷ 36 = 3√3 cm.

∴ The correct answer is 3√3 cm.

Alternate Method

Given:

Base side (s) = 6 cm

Total Surface Area (TSA) = 216√3 cm²

Formula Used:

Area of regular hexagon = (3√3 / 2) × s²

Perimeter of hexagon = 6 × s

TSA of Prism = 2 × Base Area + Lateral Surface Area (Perimeter × h)

Calculations:

⇒ Base Area = (3√3 / 2) × 6²

⇒ Base Area = (3√3 / 2) × 36 = 54√3 cm²

⇒ Perimeter of Base = 6 × 6 = 36 cm

⇒ TSA = 2 × (54√3) + 36 × h

⇒ 216√3 = 108√3 + 36h

⇒ 216√3 − 108√3 = 36h

⇒ 108√3 = 36h

⇒ h = 108√3 ÷ 36

⇒ h = 3√3 cm

∴ The correct answer is 3√3 cm.

Additional Information

Volume of Hexagonal Prism

Volume = Base Area × Height = (3√3 / 2) × s² × h.

Lateral Surface Area (LSA)

LSA of any prism = Perimeter of base × height of prism.

Regular Hexagon Properties

A regular hexagon can be divided into 6 equilateral triangles of side 's'. Area = 6 × (√3 / 4)s² = (3√3 / 2)s².

Shortcut Trick

Heights of the three cones from the vertex are h₁ = 7 cm, h₂ = 14 cm, and h₃ = 21 cm.

Ratio of heights (h₁ : h₂ : h₃) = 7 : 14 : 21 = 1 : 2 : 3.

Since CSA ∝ (height)² for similar cones, ratio of CSAs of cones = 1² : 2² : 3² = 1 : 4 : 9.

Ratio of CSA of parts (top : middle : bottom) = 1 : (4 − 1) : (9 − 4) = 1 : 3 : 5.

∴ The correct answer is 1 : 3 : 5.

Alternate Method

Given:

Total Height (H) = 21 cm, Base Radius (R) = 9 cm.

Cuts are at 7 cm and 14 cm from the base.

Formula:

Curved Surface Area (CSA) of cone = πrl

For similar cones formed by parallel cuts, CSA₁ : CSA₂ : CSA₃ = h₁² : h₂² : h₃²

Calculations:

⇒ Height of small cone (top) from vertex = 21 − 14 = 7 cm

⇒ Height of medium cone (top + middle) from vertex = 21 − 7 = 14 cm

⇒ Height of original cone (total) = 21 cm

⇒ Ratio of heights = 7 : 14 : 21 = 1 : 2 : 3

⇒ Ratio of CSA of three cones = 1² : 2² : 3² = 1 : 4 : 9

⇒ CSA of Top part = 1 unit

⇒ CSA of Middle part = CSA of medium cone − CSA of top cone = 4 − 1 = 3 units

⇒ CSA of Bottom part = CSA of total cone − CSA of medium cone = 9 − 4 = 5 units

⇒ Required Ratio = 1 : 3 : 5

∴ The correct answer is 1 : 3 : 5.

Additional Information

Volume Ratio of Similar Cones

If heights are in ratio a : b : c, then the volumes are in ratio a³ : b³ : c³.

CSA of Frustum

CSA of a frustum = π(r₁ + r₂)l, where r₁, r₂ are radii of bases and l is the slant height of the frustum.

Slant Height Relationship

For a cone, l = √(r² + h²). In similar cones, the ratio of slant heights is equal to the ratio of their vertical heights.

Shortcut Trick

Each cut parallel to the base creates 2 new circular surfaces (2 × πr²).

For 2 cuts, total added area = 4 × πr².

Percentage increase = [Total added area ÷ Initial Total Surface Area] × 100.

Shortcut formula: % Increase = [(n × r) ÷ (h + r)] × 100, where n is the number of cuts.

% Increase = [(2 × 7) ÷ (18 + 7)] × 100 = (14 ÷ 25) × 100 = 56%.

∴ The correct answer is 56%.

Alternate Method

Given: Radius (r) = 7 cm, Height (h) = 18 cm. Number of cuts = 2.

Formula: Total Surface Area (TSA) of cylinder = 2πr(h + r). Area of circular base = πr².

⇒ Initial TSA = 2 × π × 7 × (18 + 7)

⇒ Initial TSA = 2 × π × 7 × 25 = 350π cm²

⇒ Each cut parallel to the base exposes 2 new circular surfaces of radius r.

⇒ Total additional area for 2 cuts = 2 × (2 × πr²) = 4πr²

⇒ Total additional area = 4 × π × 7² = 4 × 49 × π = 196π cm²

⇒ Percentage increase = (Additional Area ÷ Initial TSA) × 100

⇒ Percentage increase = (196π ÷ 350π) × 100

⇒ Percentage increase = (14 ÷ 25) × 100 = 56%

∴ The correct answer is 56%.

Additional Information

TSA and CSA of Cylinder

Curved Surface Area (CSA) = 2πrh. Total Surface Area (TSA) = 2πrh + 2πr². The CSA remains unchanged when a cylinder is cut parallel to its base.

Cutting Logic

n cuts parallel to the base divide the cylinder into (n + 1) smaller cylinders and create 2n new circular faces, each with area πr².

Volume Invariance

The sum of the volumes of the individual parts always equals the volume of the original cylinder (πr²h), regardless of how it is cut.

Shortcut Trick

Ratio of Curved Surface Area (CSA) to Volume (V) = 2πrh / πr²h = 2/r

Given 2/r = 1/7 ⇒ r = 14 units

Ratio of Total Surface Area (TSA) to Volume (V) = 2πr(r + h) / πr²h = 2(r + h) / rh

Substitute r = 14: 2(14 + h) / 14h = 187 / 770 ⇒ (14 + h) / h = (187 × 7) / 770 = 1.7

1 + 14/h = 1.7 ⇒ 14/h = 0.7 ⇒ h = 20 units

∴ The correct answer is 7 ∶ 10.

Alternate Method

Given:

Ratio of CSA to Volume = 1 ∶ 7

Ratio of TSA to Volume = 187 ∶ 770

Formula Used:

Curved Surface Area (CSA) = 2πrh

Total Surface Area (TSA) = 2πr(r + h)

Volume (V) = πr²h

Calculations:

⇒ CSA / V = 2πrh / πr²h = 2/r

⇒ 2/r = 1/7

⇒ r = 14

⇒ TSA / V = [2πr(r + h)] / πr²h = 2(r + h) / rh

⇒ 2(14 + h) / (14 × h) = 187 / 770

⇒ (14 + h) / 7h = 187 / 770

⇒ 14 + h = (187 × 7h) / 770

⇒ 14 + h = (187 × h) / 110

⇒ 14 + h = 1.7h

⇒ 1.7h − h = 14

⇒ 0.7h = 14

⇒ h = 14 / 0.7 = 20

⇒ Required Ratio = r ∶ h = 14 ∶ 20 = 7 ∶ 10

∴ The correct answer is 7 ∶ 10.

Additional Information

Surface Area of Cylinder

Curved Surface Area is the area of the side wall, 2πrh. Total Surface Area adds the two circular bases, 2πrh + 2πr².

Volume of Cylinder

The volume represents the space inside the cylinder and is calculated as Base Area × Height = πr²h.

Ratio Shortcut for Cylinder

The ratio of TSA to CSA is always (r + h) / h. The ratio of V to CSA is always r / 2.

Shortcut Trick

For any sphere, the ratio of Surface Area to Volume is 3/R.

Given TSA/Volume = 1/7 ⇒ 3/R = 1/7 ⇒ R = 21 cm.

Radius of small sphere r = R/6 = 21/6 = 3.5 cm.

Number of small spheres (n) = (R/r)³ = 6³ = 216.

Sum of CSAs = n × 4πr² = 216 × 154 = 33264 cm².

∴ The correct answer is 33264 cm².

Alternate Method

Given: Ratio of Total Surface Area (TSA) to Volume (V) of large sphere = 1 ∶ 7.

Formula: TSA = 4πR², Volume = (4/3)πR³, Area of small sphere = 4πr².

⇒ (4πR²) ÷ (4/3 πR³) = 1/7

⇒ 3/R = 1/7

⇒ R = 21 cm

⇒ Radius of small sphere (r) = R/6 = 21/6 = 3.5 cm

⇒ Total volume remains same: (4/3)πR³ = n × (4/3)πr³

⇒ n = (R/r)³ = (21 / 3.5)³ = 6³ = 216

⇒ CSA of one small sphere = 4 × (22/7) × 3.5 × 3.5 = 154 cm²

⇒ Sum of all CSAs = 216 × 154 = 33264 cm²

∴ The correct answer is 33264 cm².

Additional Information

Sphere Properties

The Total Surface Area (TSA) and Curved Surface Area (CSA) of a sphere are identical, given by the formula 4πr². The volume is (4/3)πr³.

Melting and Recasting

When a solid object is melted and recast into several smaller objects, the total volume of all new objects is equal to the volume of the original object.

Ratio Shortcut for Sphere

For any sphere, the ratio of Surface Area to Volume is always 3/r. This is useful for quickly finding the radius when the ratio is given.

Shortcut Trick

Ratio of CSA_Hemisphere : TSA_Cube = (2 × 22/7 × r²) : (6 × a²) = 11 : 42

⇒ (44/7 × r²) / 6a² = 11/42 ⇒ 44r² / 42a² = 11/42 ⇒ 4r² = a² ⇒ a = 2r

Total Height = a + r = 2r + r = 3r = 21 cm ⇒ r = 7 cm and a = 14 cm

Total Volume = a³ + (2/3)πr³ = 14³ + (2/3) × (22/7) × 7³ = 2744 + 718.67 = 3462.67 cm³

∴ The correct answer is 3462.67 cm³.

Alternate Method

Given: Total height = 21 cm, Ratio of CSA_hemisphere : TSA_cube = 11 : 42

Formula Used: CSA of hemisphere = 2πr², TSA of cube = 6a², Volume of cube = a³, Volume of hemisphere = (2/3)πr³

⇒ CSA_hemisphere / TSA_cube = 11 / 42

⇒ (2 × 22/7 × r²) / (6 × a²) = 11 / 42

⇒ (44r² / 7) / 6a² = 11 / 42

⇒ 44r² / 42a² = 11 / 42

⇒ 44r² = 11a²

⇒ 4r² = a² ⇒ a = 2r

⇒ Total height = side of cube + radius of hemisphere = a + r = 21

⇒ 2r + r = 21 ⇒ 3r = 21 ⇒ r = 7 cm

⇒ a = 2 × 7 = 14 cm

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

⇒ Volume of hemisphere = (2/3) × (22/7) × 7 × 7 × 7 = 2156 / 3 = 718.666... cm³

⇒ Total Volume = 2744 + 718.67 = 3462.67 cm³

∴ The correct answer is 3462.67 cm³.

Additional Information

Surface Area and Volume of Sphere

Total surface area of sphere = 4πr² and Volume = (4/3)πr³. For a hemisphere, CSA = 2πr² and TSA = 3πr².

Cube Properties

Lateral surface area of cube = 4a², Total surface area = 6a², Volume = a³, and Face Diagonal = a√2.

Composite Solids

When one solid is placed on another, the total volume is the sum of individual volumes, while the total surface area may exclude the contact area.

Shortcut Trick

Each of the 4 parts is a triangular prism with a right-isosceles triangle base.

Area of Top + Bottom faces per part = 2 × (Area of square face ÷ 4) = 2 × (8 × 8 ÷ 4) = 32 cm².

Area of one vertical outside face = 8 × 8 = 64 cm².

Area of two internal cut faces = 2 × [Height × (Diagonal ÷ 2)] = 2 × [8 × (8√2 ÷ 2)] = 64√2 cm².

Total Surface Area = 32 + 64 + 64√2 = 96 + 64√2 cm².

∴ The correct answer is 96 + 64√2 cm².

Alternate Method

Given:

Side of cube (a) = 8 cm.

Number of equal parts after diagonal cuts = 4.

Formula Used:

Total Surface Area (TSA) of a part = Sum of areas of all individual faces.

Calculations:

⇒ Area of Top face of one part = (1/4) × Area of square top = (1/4) × 8² = 16 cm².

⇒ Area of Bottom face of one part = 16 cm² (identical to top).

⇒ Area of the vertical outer square face = 8 × 8 = 64 cm².

⇒ Diagonal of the top face = a√2 = 8√2 cm.

⇒ The cut faces are rectangles with height = 8 cm and width = half of diagonal = (8√2)/2 = 4√2 cm.

⇒ Area of one internal cut face = 8 × 4√2 = 32√2 cm².

⇒ Since each part has 2 such internal faces, Area = 2 × 32√2 = 64√2 cm².

⇒ Total Surface Area = (Top + Bottom) + Outer Face + Internal Faces

⇒ Total Surface Area = (16 + 16) + 64 + 64√2

⇒ Total Surface Area = 32 + 64 + 64√2 = 96 + 64√2 cm².

∴ The correct answer is 96 + 64√2 cm².

Additional Information

Surface Area of Cube

For a cube of side 'a', the Total Surface Area = 6a² and Lateral Surface Area = 4a².

Cube Diagonals

Face diagonal of a cube = a√2 and Space diagonal = a√3.

Effect of Cutting

Whenever a solid is cut, the total surface area increases by twice the area of the cutting plane because two new surfaces are created.

Shortcut Trick

Volume of Square Pyramid = 1/3 × (Side)² × Height = 1/3 × 14 × 14 × 22

Volume of Sphere = 4/3 × (22/7) × r³

Equating: 1/3 × 196 × 22 = 4/3 × (22/7) × r³

r³ = (196 × 22 × 7) ÷ (4 × 22) = 196 × 7 ÷ 4 = 49 × 7 = 343

r = ∛343 = 7 cm

∴ The correct answer is 7 cm.

Alternate Method

Given:

Height of square pyramid (h) = 22 cm

Base side of pyramid (a) = 14 cm

Volume of Pyramid = Volume of Sphere

Formula Used:

Volume of Square Pyramid = (1/3) × a² × h

Volume of Sphere = (4/3) × π × r³

Calculations:

⇒ Volume of Pyramid = (1/3) × 14² × 22

⇒ Volume of Pyramid = (1/3) × 196 × 22 cm³

⇒ Volume of Sphere = (4/3) × (22/7) × r³

⇒ According to the question: (1/3) × 196 × 22 = (4/3) × (22/7) × r³

⇒ 196 × 22 = 4 × (22/7) × r³

⇒ 196 = (4/7) × r³

⇒ r³ = (196 × 7) ÷ 4

⇒ r³ = 49 × 7

⇒ r³ = 343

⇒ r = ∛343 = 7 cm

∴ The correct answer is 7 cm.

Additional Information

Volume of a Cone

The volume of a circular cone is given by V = (1/3) × π × r² × h, which is exactly one-third of a cylinder with the same base and height.

Surface Area of Sphere

The total surface area of a sphere is 4 π r², while for a hemisphere, it is 3 π r² (including the base).

Regular Pyramid Properties

For a regular pyramid, the slant height (l) is related to vertical height (h) and side (a) by l = √(h² + (a/2)²).

Shortcut Trick

Use the substitution method by setting the variable z = 0.

Numerator = sin(y − 0) + sin(y + 0) + 2siny = siny + siny + 2siny = 4siny.

Denominator = sin(x − 0) + sin(x + 0) + 2sinx = sinx + sinx + 2sinx = 4sinx.

Expression = 4siny ÷ 4sinx = siny/sinx.

∴ The correct answer is siny/sinx.

Alternate Method

Given: [sin(y − z) + sin(y + z) + 2siny] ÷ [sin(x − z) + sin(x + z) + 2sinx]

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

Applying the formula to the numerator:

⇒ sin(y − z) + sin(y + z) = 2 sin y cos z

⇒ Numerator = 2 sin y cos z + 2 sin y

⇒ Numerator = 2 sin y (cos z + 1)

Applying the formula to the denominator:

⇒ sin(x − z) + sin(x + z) = 2 sin x cos z

⇒ Denominator = 2 sin x cos z + 2 sin x

⇒ Denominator = 2 sin x (cos z + 1)

Now, dividing the numerator by the denominator:

⇒ Value = [2 sin y (cos z + 1)] ÷ [2 sin x (cos z + 1)]

⇒ Value = sin y ÷ sin x

∴ The correct answer is siny/sinx.

Additional Information

Sum to Product Identities

These identities convert sums of sine or cosine functions into products: sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2).

Compound Angle Formulas

Essential for expansion: sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B.

Pythagorean Identity

The fundamental relation between sine and cosine is sin²θ + cos²θ = 1.

Shortcut Trick

Simplify first part: Use sin(A+B) + sin(A-B) = 2sinA cosB. Result is [2sinx(cosy − 1)] / [2cosx(cosy − 1)] = tanx.

Simplify second part: Apply (sinC − sinD) / (cosC + cosD) = tan[(C−D)/2]. Result is tan[(10x − 8x)/2] = tanx.

Multiply the results: tanx × tanx = tan²x.

∴ The correct answer is tan²x.

Alternate Method

Given: [sin(x + y) − 2sinx + sin(x − y)] / [cos(x − y) + cos(x + y) − 2cosx] × [(sin 10x − sin 8x) / (cos 10x + cos 8x)]

Formula Used:

1. sin(A + B) + sin(A − B) = 2sinA cosB
2. cos(A + B) + cos(A − B) = 2cosA cosB
3. sinC − sinD = 2cos[(C+D)/2] sin[(C−D)/2]
4. cosC + cosD = 2cos[(C+D)/2] cos[(C−D)/2]

⇒ Term 1 = [2sinx cos y − 2sinx] / [2cosx cos y − 2cosx]

⇒ Term 1 = [2sinx(cos y − 1)] / [2cosx(cos y − 1)] = sinx / cosx = tanx

⇒ Term 2 = [2cos((10x+8x)/2) sin((10x−8x)/2)] / [2cos((10x+8x)/2) cos((10x−8x)/2)]

⇒ Term 2 = [2cos 9x sin x] / [2cos 9x cos x] = sinx / cosx = tanx

⇒ Expression = tanx × tanx = tan²x

∴ The correct answer is tan²x.

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).

Compound Angle Identities

sin(A ± B) = sinA cosB ± cosA sinB and cos(A ± B) = cosA cosB ∓ sinA sinB.

Quotient Identity

The expression (sin C − sin D) / (cos C + cos D) always simplifies to tan((C−D)/2).

Shortcut Trick

Reduce using quadrant rules: sin(90° − 10θ) = cos 10θ and cos(π − 6θ) = −cos 6θ.

Denominator: cos(π/2 − 10θ) = sin 10θ and sin(π − 6θ) = sin 6θ.

Expression becomes (cos 10θ + cos 6θ) ÷ (sin 10θ − sin 6θ).

Apply CD formulas: [2 cos 8θ cos 2θ] ÷ [2 cos 8θ sin 2θ].

Cancel 2 cos 8θ to get cos 2θ ÷ sin 2θ = cot 2θ.

∴ The correct answer is cot 2θ.

Alternate Method

Given: [sin(90° − 10θ) − cos(π − 6θ)] ÷ [cos(π/2 − 10θ) − sin(π − 6θ)]

Formula Used:

1. sin(90° − A) = cos A | 2. cos(π − A) = −cos A | 3. cos(π/2 − A) = sin A | 4. sin(π − A) = sin A
2. cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2)
3. sin C − sin D = 2 cos((C+D)/2) sin((C−D)/2)

⇒ Numerator = sin(90° − 10θ) − cos(π − 6θ)

⇒ Numerator = cos 10θ − (−cos 6θ) = cos 10θ + cos 6θ

⇒ Denominator = cos(π/2 − 10θ) − sin(π − 6θ)

⇒ Denominator = sin 10θ − sin 6θ

⇒ Expression = [2 cos((10θ + 6θ)/2) cos((10θ − 6θ)/2)] ÷ [2 cos((10θ + 6θ)/2) sin((10θ − 6θ)/2)]

⇒ Expression = [2 cos 8θ cos 2θ] ÷ [2 cos 8θ sin 2θ]

⇒ Expression = cos 2θ ÷ sin 2θ = cot 2θ

∴ The correct answer is cot 2θ.

Additional Information

CD Formulas (Sum to Product)

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).

Quadrant Rules (ASTC Rule)

Trigonometric ratios change (sin → cos, tan → cot) at 90° (π/2) and 270° (3π/2) but remain same at 180° (π) and 360° (2π).

Complementary and Supplementary Angles

sin(90°−θ) = cosθ, cos(90°−θ) = sinθ, sin(180°−θ) = sinθ, and cos(180°−θ) = −cosθ.

Shortcut Trick

Expand the given equation: sec θ(cos θ + sin θ) = √2 → 1 + tan θ = √2.

Solving for tan θ, we get tan θ = √2 − 1.

Divide the required expression by cos θ: 2 sin θ / (cos θ − sin θ) = 2 tan θ / (1 − tan θ).

Substitute tan θ = √2 − 1 → 2(√2 − 1) / (1 − (√2 − 1)) = 2(√2 − 1) / (2 − √2) = √2.

∴ The correct answer is √2.

Alternate Method

Given: sec θ(cos θ + sin θ) = √2

Formula Used: sec θ = 1/cos θ | tan θ = sin θ/cos θ

⇒ sec θ cos θ + sec θ sin θ = √2

⇒ (1/cos θ) × cos θ + (1/cos θ) × sin θ = √2

⇒ 1 + tan θ = √2

⇒ tan θ = √2 − 1

⇒ Now, consider the expression: 2sin θ / (cos θ − sin θ)

⇒ Dividing numerator and denominator by cos θ:

⇒ (2sin θ / cos θ) / (cos θ/cos θ − sin θ/cos θ)

⇒ = 2tan θ / (1 − tan θ)

⇒ = 2(√2 − 1) / [1 − (√2 − 1)]

⇒ = 2(√2 − 1) / [2 − √2]

⇒ = 2(√2 − 1) / [√2(√2 − 1)]

⇒ = 2 / √2 = √2

∴ The correct answer is √2.

Additional Information

Trigonometric Definitions

sec θ = 1/cos θ and tan θ = sin θ/cos θ. These identities are used to transform expressions into a single trigonometric ratio.

Rationalization

For any positive number a, a/√a = √a. This is used to simplify fractions with square roots in the denominator.

Special Values

The value tan 22.5° = √2 − 1. Recognizing this can sometimes help in identifying the angle θ directly.

Shortcut Trick

Substitute θ = 45° into the expression.

sin(90° − 45°) = sin 45° = 1/√2 and cos(90° − 45°) = cos 45° = 1/√2.

Expression = 1/(1/√2)⁴ + 1/[(1/√2)² − 1] = 4 + 1/(−1/2) = 4 − 2 = 2.

Check Option 1: tan²45°sec²45° = (1)² × (√2)² = 1 × 2 = 2.

The value matches the result obtained from the shortcut substitution.

∴ The correct answer is tan²θsec²θ.

Alternate Method

Given: 1/sin⁴(90° − θ) + 1/[cos²(90° − θ) − 1]

Formula Used: sin(90° − θ) = cosθ, cos(90° − θ) = sinθ, sin²θ + cos²θ = 1, and tanθ = sinθ/cosθ.

⇒ Substitute complementary ratios: sin(90° − θ) = cosθ and cos(90° − θ) = sinθ.

⇒ Expression = 1/cos⁴θ + 1/(sin²θ − 1)

⇒ Using the identity sin²θ + cos²θ = 1 → sin²θ − 1 = −cos²θ.

⇒ Expression = 1/cos⁴θ − 1/cos²θ

⇒ Taking 1/cos⁴θ as common: (1 − cos²θ) / cos⁴θ

⇒ Expression = sin²θ / cos⁴θ

⇒ Expression = (sin²θ / cos²θ) × (1 / cos²θ)

⇒ Expression = tan²θ × sec²θ

∴ The correct answer is tan²θsec²θ.

Additional Information

Complementary Angle Identities

For any angle θ, sin(90°−θ) = cosθ, cos(90°−θ) = sinθ, tan(90°−θ) = cotθ, and sec(90°−θ) = cosecθ.

Pythagorean Identities

The three main identities are sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Relations

Secant and Cosine are reciprocals (secθ = 1/cosθ), as are Cosecant/Sine (cosecθ = 1/sinθ) and Cotangent/Tangent (cotθ = 1/tanθ).

Shortcut Trick

Pick a specific value for angle A to simplify the expression quickly.

Let A = 22.5°, then 2A = 45° and 90 − A = 67.5°.

Numerator = [tan 67.5° + cot 67.5°]² = (2 cosec 45°)² = (2√2)² = 8.

Denominator = 2 sec²(90 − 45°) = 2 sec² 45° = 2 × (√2)² = 4.

Final Value = 8 ÷ 4 = 2.

∴ The correct answer is 2.

Alternate Method

Given: [tan(90 − A) + cot(90 − A)]² ÷ [2sec²(90 − 2A)]

Formula Used: tan(90 − θ) = cot θ; cot(90 − θ) = tan θ; sec(90 − θ) = cosec θ; tan θ + cot θ = 2 cosec 2θ

⇒ Numerator = [tan(90 − A) + cot(90 − A)]²

⇒ Numerator = [cot A + tan A]²

⇒ cot A + tan A = (cos A / sin A) + (sin A / cos A) = (cos² A + sin² A) / (sin A cos A) = 1 / (sin A cos A)

⇒ 1 / (sin A cos A) = 2 / (2 sin A cos A) = 2 / sin 2A = 2 cosec 2A

⇒ Numerator = (2 cosec 2A)² = 4 cosec² 2A

⇒ Denominator = 2 sec²(90 − 2A) = 2 cosec² 2A

⇒ Value = 4 cosec² 2A ÷ 2 cosec² 2A = 2

∴ The correct answer is 2.

Additional Information

Complementary Angle Relations

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

Sum of Tan and Cot

The identity tan θ + cot θ = 2 cosec 2θ is derived from 1/(sin θ cos θ) using the double angle formula sin 2θ = 2 sin θ cos θ.

Pythagorean Identity

The fundamental identity sin² θ + cos² θ = 1 is often used to simplify fractional sums of trigonometric ratios.

Shortcut Trick

Pick specific values for x and y to simplify the expression quickly.

Let x = 90° and y = 0°.

Expression = {sin(90° − 90°) × cos(π − (90° − 0°))} + {cos(90° − 90°) × sin(π − (0° − 90°))}

= {sin 0° × cos(π/2)} + {cos 0° × sin(3π/2)} = {0 × 0} + {1 × (−1)} = −1.

Evaluating options for y = 0°: Option (1) −cos 0° = −1. This matches our result.

∴ The correct answer is -cos y.

Alternate Method

Given: {sin (90 − x) × cos[π − (x − y)]} + {cos(90 − x) × sin [π − (y − x)]}

Formula Used: Allied angles and compound angle identities.

Applying Allied Angle identities:

⇒ sin(90 − x) = cos x

⇒ cos[π − (x − y)] = −cos(x − y)

⇒ cos(90 − x) = sin x

⇒ sin[π − (y − x)] = sin(y − x)

Substitute these into the expression:

⇒ [cos x × (−cos(x − y))] + [sin x × sin(y − x)]

⇒ −cos x × cos(x − y) + sin x × (−sin(x − y)) [Since sin(y − x) = −sin(x − y)]

⇒ −[cos x × cos(x − y) + sin x × sin(x − y)]

Using the identity cos(A − B) = cos A cos B + sin A sin B, where A = x and B = x − y:

⇒ −[cos(x − (x − y))]

⇒ −[cos(x − x + y)]

⇒ −cos y

∴ The correct answer is -cos y.

Additional Information

Allied Angle Rules (ASTC Rule)

In the 2^nd quadrant (π − θ), only sine and cosecant are positive. In the 1^st quadrant (90 − θ), all ratios are positive but trig functions change to co-functions (sin → cos, tan → cot, sec → cosec).

Compound Angle Formulas

cos(A − B) = cos A cos B + sin A sin B and sin(A − B) = sin A cos B − cos A sin B. These are fundamental for simplifying multi-variable trig expressions.

Negative Angle Identity

sin(−θ) = −sin θ and cos(−θ) = cos θ. This explains why sin(y − x) = −sin(x − y) while cos(y − x) = cos(x − y).

Shortcut Trick

Distance (D) covered by an object moving at constant height (h) is: D = h(cot θ₂ − cot θ₁)

Here, h = 4500 m, θ₁ = 60°, θ₂ = 30°

D = 4500 × (cot 30° − cot 60°) = 4500 × (√3 − 1/√3) = 4500 × (2/√3) = 3000√3 m

Speed = Distance ÷ Time = 3000√3 ÷ 30 = 100√3 m/s

∴ The correct answer is 100√3 m/s.

Alternate Method

Given:

Height of aeroplane (h) = 4500 m

Initial angle of elevation (θ₁) = 60°

Final angle of elevation (θ₂) = 30°

Time taken (t) = 30 seconds

Formula Used:

tan θ = Perpendicular ÷ Base

Speed = Distance ÷ Time

Calculations:

In the first position (60°):

⇒ tan 60° = 4500 ÷ x

⇒ √3 = 4500 ÷ x

⇒ x = 4500 / √3 = 1500√3 m

In the second position (30°):

⇒ tan 30° = 4500 ÷ (x + D)

⇒ 1/√3 = 4500 ÷ (1500√3 + D)

⇒ 1500√3 + D = 4500√3

⇒ D = 4500√3 − 1500√3 = 3000√3 m

Now, finding the speed:

⇒ Speed = 3000√3 ÷ 30

⇒ Speed = 100√3 m/s

∴ The correct answer is 100√3 m/s.

Additional Information

Angle of Elevation

The angle formed by the line of sight with the horizontal when the object is above the horizontal level.

Trigonometric Ratios for Standard Angles

tan 30° = 1/√3, tan 60° = √3, cot 30° = √3, cot 60° = 1/√3.

Speed-Distance Relation

Speed = Distance ÷ Time. To convert km/h to m/s, multiply by 5/18.

Shortcut Trick

In a 30°−60°−90° right-angled triangle, the ratio of sides is 1 : √3 : 2.

The side opposite to 30° (Height) is always half of the hypotenuse (String length).

Given Hypotenuse (String) = 420 m.

Height = 420 ÷ 2 = 210 m.

∴ The correct answer is 210 m.

Alternate Method

Given:

Length of the string (AC) = 420 m

Angle of elevation (θ) = 30°

Formula Used:

sin θ = Perpendicular ÷ Hypotenuse

Calculations:

In right-angled triangle ABC:

⇒ sin 30° = AB ÷ AC

⇒ 1 ÷ 2 = AB ÷ 420

⇒ AB = 420 × (1 ÷ 2)

⇒ AB = 210 m

∴ The correct answer is 210 m.

Additional Information

Trigonometric Ratios of Standard Angles

Standard values: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2. These are essential for height and distance problems.

30°-60°-90° Triangle Rule

In such a triangle, the sides are in the ratio 1 : √3 : 2. Side opposite to 30° is x, side opposite to 60° is x√3, and hypotenuse is 2x.

Angle of Elevation

The angle formed by the line of sight with the horizontal when looking up at an object. It is equal to the angle of depression when viewed from the object looking down.

Shortcut Trick

In a 30°−60°−90° triangle, the sides are in the ratio 1 : √3 : 2.

Side opposite to 30° (Base) = 450√3 m.

Side opposite to 60° (Height) = Base × √3 = 450√3 × √3 = 1350 m.

Total time = 6 minutes = 360 seconds.

Speed = Height ÷ Time = 1350 ÷ 360 = 3.75 m/s.

∴ The correct answer is 3.75 m/s.

Alternate Method

Given:

Distance of observer from point P (Base) = 450√3 m

Angle of elevation (θ) = 60°

Time taken (t) = 6 minutes = 6 × 60 seconds = 360 s

Formula Used:

tan θ = Perpendicular ÷ Base

Speed = Distance ÷ Time

⇒ In right-angled triangle OPB, tan 60° = PB ÷ OP

⇒ √3 = PB ÷ 450√3

⇒ PB = 450√3 × √3

⇒ PB = 450 × 3 = 1350 m

⇒ This height (1350 m) is the distance traveled by the balloon in 360 seconds.

⇒ Speed = 1350 ÷ 360

⇒ Speed = 135 ÷ 36 = 15 ÷ 4

⇒ Speed = 3.75 m/s

∴ The correct answer is 3.75 m/s.

Additional Information

Angle of Elevation

It is the angle formed between the horizontal line and the line of sight when an observer looks upward at an object.

Speed Conversion

To convert speed from km/h to m/s, multiply by 5/18. To convert minutes to seconds, multiply by 60.

Trigonometric Values

Standard values for height and distance: tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3.

Total manufactured bats by company T = 550000

Since plastic bats manufactured by T are 45%, wooden bats will be 55%,

⇒ Total no. of wooden bats manufactured by company T = 550000 × 0.55 = 302500

Since the ratio of wooden bats of Brand A and B manufactured by T = 6 ∶ 19

∴ Total number of wooden bats of brand A manufactured by company T = 6/25 × 302500 = 72600

N = Wooden bats of Brand B manufactured by U = 14/55 × 0.25 × 550000 = 35000

M = Total wooden bats manufactured by R and W together = (0.45 + 0.6) × 550000 = 577500

∴ N/M = 35000/577500 = 0.061

P = Sum of wooden bats of Brand B manufactured by S and wooden bats of Brand A manufactured by W,

⇒ P = 7/15 × 0.3 × 550000 + 5/11 × 0.60 × 550000

⇒ P = 77000 + 150000

⇒ P = 227000

Q = Difference of Brand B wooden bats and Brand A wooden bats manufactured by U

⇒ Q = 27/55 × 0.25 × 550000

⇒ Q = 67500

∴ P - Q = 227000 - 67500 = 159500

Taking all 6 companies together, wooden bats of Brand A produced,

⇒ 550000 × (21/25 × 0.45 + 8/15 × 0.3 + 6/25 × 0.55 + 41/55 × 0.25 + 7/22 × 0.4 + 5/11 × 0.6)

⇒ 550000 × (0.378 + 0.16 + 0.132 + 0.186 + 0.127 + 0.273)

⇒ 550000 × (1.256)

⇒ 690800 which is approximately 691000

X = Average of plastic bats manufactured by V, U and T,

⇒ X = 1/3 × 550000 × (0.6 + 0.75 + 0.45)

⇒ X = 1/3 × 550000 × 1.8

⇒ X = 330000

Y = Wooden bats of Brand A manufactured by V,

⇒ Y = 7/22 × 550000 × 0.4

⇒ Y = 70000

∴ X - Y = 330000 - 70000 = 260000

Shortcut Trick

Remaining Ethanol = Initial Quantity × [1 − (Replaced Quantity / Initial Quantity)]ⁿ

Remaining Ethanol = 80 × [1 − 20/80]² = 80 × (3/4)²

Remaining Ethanol = 80 × 9/16 = 45 litres.

Water present = Total Mixture − Remaining Ethanol = 80 − 45 = 35 litres.

∴ The correct answer is 35 litres.

Alternate Method

Given: Initial Ethanol = 80 litres, Volume replaced each time = 20 litres, Operations = 2.

⇒ Total volume of drum = 80 litres.

⇒ In the first operation, 20 litres of ethanol is replaced by 20 litres of water.

⇒ Ethanol remaining = 80 − 20 = 60 litres | Water = 20 litres.

⇒ Ratio of Ethanol : Water in the mixture = 60 : 20 = 3 : 1.

⇒ In the second operation, 20 litres of mixture is removed.

⇒ Ethanol removed = (3 / 4) × 20 = 15 litres.

⇒ Final amount of Ethanol = 60 − 15 = 45 litres.

⇒ Total volume remains 80 litres (since 20 litres of water was added again).

⇒ Final amount of Water = 80 − 45 = 35 litres.

∴ The correct answer is 35 litres.

Additional Information

Repeated Replacement Formula

The final quantity of the original liquid after n replacements is given by: Q_n = Q₀ × (1 − x/V)ⁿ, where Q₀ is initial quantity, x is volume replaced, and V is total volume.

Mixture Consistency

In replacement problems where the replaced quantity equals the added quantity, the total volume of the mixture remains constant throughout the process.

Ratio Method for Two Components

If the original liquid is reduced by a fraction f in each step, the remaining quantity after n steps is Initial × (1 − f)ⁿ. Here f = 20/80 = 1/4.

Shortcut Trick

Weighted average cost per kg = [(Ratio A × Cost A) + (Ratio B × Cost B)] ÷ (Total Ratio parts)

Cost per kg = (3 × 2000 + 1 × 400) ÷ (3 + 1) = 6400 ÷ 4 = ₹1600/kg

Total cost for 8 kg alloy = 1600 × 8 = ₹12800

∴ The correct answer is ₹12800.

Alternate Method

Given: Metal A = ₹2000/kg, Metal B = ₹400/kg, Ratio A : B = 3 : 1

Formula: Total Cost = (Quantity_A × Price_A) + (Quantity_B × Price_B)

⇒ Total ratio parts = 3 + 1 = 4 units ⇒ Weight of 4 units = 8 kg ⇒ Weight of 1 unit = 8 ÷ 4 = 2 kg ⇒ Weight of Metal A = 3 × 2 = 6 kg ⇒ Weight of Metal B = 1 × 2 = 2 kg ⇒ Cost of Metal A = 6 × 2000 = ₹12000 ⇒ Cost of Metal B = 2 × 400 = ₹800 ⇒ Total Cost of 8 kg alloy = 12000 + 800 = ₹12800

∴ The correct answer is ₹12800.

Additional Information

Weighted Average Formula

The average price of a mixture is calculated as (n₁c₁ + n₂c₂) ÷ (n₁ + n₂), where n is quantity and c is cost per unit.

Rule of Alligation

It relates the ratio of quantities to the differences between the mean price and individual prices: (Qty A) ÷ (Qty B) = (Cost B − Mean) ÷ (Mean − Cost A).

Ratio Distribution

To find individual quantities from a total, multiply the total by (Individual Ratio ÷ Sum of Ratios).

Shortcut Trick

Let A invest x. Then B = x + 50,000 and C = (x + 50,000) − 25,000 = x + 25,000.

Total Investment: 3x + 75,000 = 3,00,000 ⇒ 3x = 2,25,000 ⇒ x = 75,000.

Ratio A : B : C = 75,000 : 1,25,000 : 1,00,000 = 3 : 5 : 4.

C's share = (4 / (3 + 5 + 4)) × 14,400 = (1 / 3) × 14,400 = 4,800.

∴ The correct answer is ₹4800.

Alternate Method

Given:

Total Investment = ₹3,00,000

Total Profit = ₹14,400

B = A + 50,000

C = B − 25,000

Formula Used:

Profit Share ∝ Investment × Time (Since time is constant, Profit Ratio = Investment Ratio)

Calculations:

⇒ Let A's investment = x

⇒ B's investment = x + 50,000

⇒ C's investment = (x + 50,000) − 25,000 = x + 25,000

⇒ Total = x + x + 50,000 + x + 25,000 = 3,00,000

⇒ 3x + 75,000 = 3,00,000

⇒ 3x = 2,25,000 ⇒ x = 75,000

⇒ A = ₹75,000; B = ₹1,25,000; C = ₹1,00,000

⇒ Investment Ratio (A : B : C) = 75,000 : 1,25,000 : 1,00,000 = 3 : 5 : 4

⇒ Sum of ratio parts = 3 + 5 + 4 = 12

⇒ C's profit share = (4 / 12) × 14,400

⇒ C's profit share = 1/3 × 14,400 = 4,800

∴ The correct answer is ₹4800.

Additional Information

Basic Partnership Rule

Profit is distributed in the ratio of the product of capital and time: P1 : P2 = C1 × T1 : C2 × T2.

Active vs Sleeping Partner

An active partner manages the business and may receive a salary or fixed percentage of profit before the remaining profit is divided.

Investment Calculation

If profit and time are known, the ratio of investments can be calculated as C1 : C2 = (P1 ÷ T1) : (P2 ÷ T2).

Shortcut Trick

Ratio of Profit A : B = 5 : 8

A's share (5 units) = ₹87,500 → 1 unit = ₹17,500

Total distributed profit (13 units) = 13 × 17,500 = ₹2,27,500

This distributed amount is 70% of total profit (100% − 30% reinvested)

Total Profit = ₹2,27,500 × (100 / 70) = ₹3,25,000

∴ The correct answer is ₹325000.

Alternate Method

Given:

Investment ratio A : B = 5 : 8

Reinvestment percentage = 30%

A's share of distributed profit = ₹87,500

Calculations:

⇒ Let the total profit be P.

⇒ Reinvested amount = 30% of P

⇒ Distributed profit = 100% − 30% = 70% of P = 0.7P

⇒ This 0.7P is shared in the ratio 5 : 8.

⇒ A's share = (5 / (5 + 8)) × 0.7P

⇒ 87,500 = (5 / 13) × 0.7P

⇒ 87,500 × 13 = 3.5P

⇒ 11,37,500 = 3.5P

⇒ P = 11,37,500 ÷ 3.5

⇒ P = ₹3,25,000

∴ The correct answer is ₹325000.