CDS-II 2025 (Elementary Mathematics) Official Paper (Held On: 14 Sept, 2025) (Previous Year Paper)

Given:

p(x) leaves a remainder of 2 when divided by (x - 1).

p(x) leaves a remainder of 1 when divided by (x - 2).

We need the remainder when p(x) is divided by (x - 1)(x - 2).

Formula used:

The remainder when p(x) is divided by (x - 1)(x - 2) will be a linear polynomial of the form ax + b.

From the given conditions: p(1) = 2 and p(2) = 1.

Calculations:

Assume remainder = ax + b.

From p(1) = 2:

⇒ a(1) + b = 2

⇒ a + b = 2

From p(2) = 1:

⇒ a(2) + b = 1

⇒ 2a + b = 1

Solving the two equations:

1. a + b = 2
2. 2a + b = 1

⇒ Subtract equation 1 from equation 2:

⇒ (2a + b) - (a + b) = 1 - 2

⇒ a = -1

Substitute a = -1 in equation 1:

⇒ -1 + b = 2

⇒ b = 3

Final Remainder:

Remainder = ax + b = -x + 3 = 3 - x

∴ The correct answer is option (3).

Given:

I. x + (1/x) > 1

II. (x + (1/x))² > 2

III. (x + (1/x))⁴ > 9

Formula used:

For positive real numbers, x + (1/x) ≥ 2 when x > 0.

Calculation:

Check I: x + (1/x) > 1

⇒ x > 0 ensures x + (1/x) ≥ 2 > 1

⇒ I is correct.

Check II: (x + (1/x))² > 2

⇒ x + (1/x) ≥ 2

⇒ (x + (1/x))² ≥ 2² = 4

⇒ II is correct.

Check III: (x + (1/x))⁴ > 9

⇒ (x + (1/x)) ≥ 2

⇒ (x + (1/x))⁴ ≥ 2⁴ = 16 > 9

⇒ III is correct.

∴ The correct answer is option (4).

Given:

Let q² - 5p - 4 be negative, where q > p and p, q are natural numbers.

Formula used:

For q2 - 5p - 4 < 0 , solve for p, considering q > p.

Calculation:

To find the largest p:

Let q = p + 1 (since q > p ).

⇒ (p + 1)² - 5p - 4 < 0

⇒ p² + 2p + 1 - 5p - 4 < 0

⇒ p² - 3p - 3 < 0

Factorize or test values for p:

For p = 3:

⇒ 3² - 3×3 - 3 = 9 - 9 - 3 = -3 (negative).

For p = 4:

⇒ 4² - 3×4 - 3 = 16 - 12 - 3 = 1 (positive).

Largest p is 3 where q² - 5p - 4 < 0 .

∴ The correct answer is option (1).

Given:

x, y are natural numbers < 20

x, y, (x + y), (x − y) are prime numbers

Formula used:

Prime rule: Except 2, all primes are odd

Calculation:

x + y is prime

⇒ Sum of two odd primes is even (not prime)

⇒ One number must be 2

Since (x − y) is prime

⇒ x > y

⇒ y = 2

Now x is prime < 20 and x > 2

Check values:

x = 3 ⇒ x − y = 1 (not prime) 

x = 5 ⇒ x − y = 3, x + y = 7 

x = 7 ⇒ x + y = 9 

x = 11 ⇒ x − y = 9 

x = 13 ⇒ x + y = 15 

x = 17 ⇒ x − y = 15 

x = 19 ⇒ x + y = 21 

Valid set:

(x, y, x + y, x − y) = (5, 2, 7, 3)

∴ Number of possible combinations = One (Option 1).

Given:

(x + 1)(x + p)(x² + p²) = x⁴ − 1

Formula used:

x⁴ − 1 = (x − 1)(x + 1)(x² + 1)

Calculation:

x⁴ − 1 = (x − 1)(x + 1)(x² + 1)

Given expression:

(x + 1)(x + p)(x² + p²)

Common factor: (x + 1)

⇒ Remaining factors must match

⇒ (x + p) = (x − 1)

⇒ p = −1

Check:

p² = 1

⇒ (x² + p²) = (x² + 1)

Hence LHS = (x + 1)(x − 1)(x² + 1)

= x⁴ − 1

∴ The value of p is −1.

Given:

(2 + √3)^x + (2 - √3)^x = 2

Formula used:

Let a = 2 + √3, b = 2 - √3. Then a × b = 1, so b^x = a^-x.

Calculations:

Let y = (2 + √3)^x. Then given ⇒ y + 1/y = 2.

⇒ y + 1/y = 2

⇒ y² + 1 = 2y

⇒ y² - 2y + 1 = 0

⇒ (y - 1)² = 0

⇒ y = 1 ⇒ (2 + √3)^x = 1 ⇒ x = 0

Therefore, (2 + √3)^x - (2 - √3)^x = 1 - 1 = 0

∴ The required value is 0.

Given:

$\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{5}{6}$

$\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{13}{36}$

Formula used:

$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2 = \dfrac{1}{a^2}+\dfrac{1}{b^2}+2\cdot\dfrac{1}{ab}$

$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3 = \dfrac{1}{a^3}+\dfrac{1}{b^3}+3\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\cdot\dfrac{1}{ab}$

Calculation:

⇒ S = $\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{5}{6}$

⇒ S² = $\dfrac{25}{36}$

⇒ Use S² = (1/a²+1/b²) + 2×(1/ab)

⇒ $\dfrac{25}{36} = \dfrac{13}{36} + 2\times\dfrac{1}{ab}$

⇒ 2×(1/ab) = $\dfrac{12}{36}=\dfrac{1}{3}$

⇒ $\dfrac{1}{ab}=\dfrac{1}{6}$

⇒ Now S³ = $\dfrac{125}{216}$

⇒ Use S³ = (1/a³+1/b³) + 3×S×(1/ab)

⇒ $\dfrac{125}{216} = \left(\dfrac{1}{a^3}+\dfrac{1}{b^3}\right) + 3\times\dfrac{5}{6}\times\dfrac{1}{6}$

⇒ 3×S×(1/ab) = 3×(5/6)×(1/6) = $\dfrac{15}{36}=\dfrac{5}{12}=\dfrac{90}{216}$

⇒ $\dfrac{1}{a^3}+\dfrac{1}{b^3} = \dfrac{125}{216}-\dfrac{90}{216}=\dfrac{35}{216}$

∴ The correct answer is $\dfrac{35}{216}$.

Shortcut Trick

To find the remainder when a polynomial is divided by (x² + 1), substitute x² = −1 into the polynomial.

Given polynomial: x⁶

Rewrite as: (x²)³

Substitute x² = −1 ⇒ (−1)³ = −1

∴ The correct answer is −1.

Alternate Method

Given: Dividend = x⁶, Divisor = x² + 1

Formula Used: Remainder Theorem for polynomials.

⇒ Let f(x) = x⁶

⇒ We use the identity: a³ + b³ = (a + b)(a² − ab + b²)

⇒ Consider x⁶ + 1 = (x²)³ + 1³

⇒ x⁶ + 1 = (x² + 1)(x⁴ − x² + 1)

⇒ x⁶ = (x² + 1)(x⁴ − x² + 1) − 1

⇒ This is in the form: Dividend = (Divisor × Quotient) + Remainder

⇒ Comparing the terms, Remainder = −1

∴ The correct answer is −1.

Additional Information

Remainder Theorem

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

Imaginary Unit Substitution

When dividing by (x² + 1), the roots are x = ±i. Substituting x² = −1 is a direct application of this property.

Degree of Remainder

The degree of the remainder is always strictly less than the degree of the divisor. For a quadratic divisor, the remainder is at most linear (ax + b).

Given:

(x + 2) is a factor of which one of the following polynomials:

Formula used:

If (x + c) is a factor of a polynomial f(x), then f(-c) = 0.

Calculation:

Here, (x + 2) is the factor, so c = -2. Substitute x = -2 into the given polynomials:

Option 1:

f(x) = x⁵ - 4x⁴ - 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ - 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 - 64 + 24 + 32 + 28 + 12

= 0

Since f(-2) = 0, (x + 2) is a factor of Option 1.

Option 2:

f(x) = x⁵ + 4x⁴ - 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ + 4(-2)⁴ - 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 + 64 + 24 + 32 + 28 + 12

= 128 (Not 0)

Option 3:

f(x) = x⁵ - 4x⁴ + 3x³ + 8x² - 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ + 3(-2)³ + 8(-2)² - 14(-2) + 12

= -32 - 64 - 24 + 32 + 28 + 12

= -48 (Not 0)

Option 4:

f(x) = x⁵ - 4x⁴ - 3x³ + 8x² + 14x + 12

f(-2) = (-2)⁵ - 4(-2)⁴ - 3(-2)³ + 8(-2)² + 14(-2) + 12

= -32 - 64 + 24 + 32 - 28 + 12

= -56 (Not 0)

∴ The correct answer is Option 1.

Given:

log₁₀2 = 0.301

log₁₀3 = 0.477

Find number of digits in 60⁶⁰

Formula used:

The number of digits in any number ‘n’ is equal to + 1, where X is a floor function and it gives the largest integer less than or equal to x.

Calculations:

log₁₀60 = log₁₀(6 × 10) = log₁₀6 + 1

log₁₀6 = log₁₀2 + log₁₀3 = 0.301 + 0.477 = 0.778

⇒ log₁₀60 = 0.778 + 1 = 1.778

⇒ log₁₀60⁶⁰ = 60 × 1.778 = 106.68

⇒ No. of digits in (60)⁶⁰ = 106.68 + 1 = 106 + 1 = 107

∴ Number of digits = 107.

Given:

(17²⁵ + 19²⁵) ÷ 18

Formula used:

a mod m → reduce base first

Calculation:

⇒ 17 mod 18 = 17

⇒ 19 mod 18 = 1

⇒ 17²⁵ mod 18 = 17²⁵ (17 ≡ −1 mod 18)

⇒ (−1)²⁵ = −1 ⇒ remainder = 17

⇒ 19²⁵ mod 18 = 1²⁵ = 1

⇒ Total remainder = 17 + 1 = 18

⇒ 18 mod 18 = 0

∴ Remainder = 0.

Given:

HCF of x and y = H

p = (x³ + y³) ÷ (x² − xy + y²)

q = (x³ − y³) ÷ (x² + xy + y²)

Formula used:

x³ + y³ = (x + y)(x² − xy + y²)

x³ − y³ = (x − y)(x² + xy + y²)

Calculation:

p = (x + y)(x² − xy + y²) ÷ (x² − xy + y²)

⇒ p = x + y

q = (x − y)(x² + xy + y²) ÷ (x² + xy + y²)

⇒ q = x − y

HCF(p, q) = HCF(x + y, x − y)

Let x = Ha, y = Hb where HCF(a, b) = 1

⇒ p = H(a + b)

⇒ q = H(a − b)

HCF(p, q) = H × HCF(a + b, a − b)

If a and b both odd ⇒ a + b and a − b are even

⇒ HCF(a + b, a − b) = 2

⇒ HCF(p, q) = 2H

If one of a, b is even ⇒ HCF(a + b, a − b) = 1

⇒ HCF(p, q) = H

∴ Both statements I and II are correct.

Given:

x⁴ = x² + 1 , x > 0

Formula used:

Put y = x² ⇒ x⁴ = y²

Calculation:

⇒ y² = y + 1

⇒ y² − y − 1 = 0

⇒ y = (1 + √5) / 2 (positive root)

⇒ x² = (1 + √5)/2

⇒ x⁴ = y² = ((1 + √5)/2)²

⇒ x⁴ = (6 + 2√5)/4 = (3 + √5)/2

⇒ 2x⁴ = 3 + √5

∴ The correct answer is 3 + √5.

Given:

$\dfrac{p+q}{q+r}=\dfrac{r+s}{s+p}$

Formula used:

Cross multiplication: (p+q)(s+p) = (r+s)(q+r)

(A−B)(A+B) = A² − B²

Calculation:

(p + q)(s + p) = (r + s)(q + r)

⇒ ps + p² + qs + pq = rq + r² + sq + sr

⇒ (p² − r²) + (ps − sr) + (pq − rq) = 0

⇒ (p − r)(p + r) + s(p − r) + q(p − r) = 0

⇒ (p − r)(p + r + s + q) = 0

⇒ (p − r) = 0 or (p + r + s + q) = 0

⇒ p = r or (p + r + s + q) = 0.

∴ The correct answer is: Either p + q + r + s = 0 or p = r.

Given:

n ∈ N, n < 7

Find number of n for which gcd(12n + 2, 8n + 1) = 1

Formula used:

Use Euclid: gcd(a, b) = gcd(b, a − q×b)

Calculations:

⇒ gcd(12n + 2, 8n + 1) = gcd(8n + 1, (12n + 2) − (8n + 1))

⇒ gcd(8n + 1, 4n + 1)

⇒ gcd(4n + 1, (8n + 1) − 2×(4n + 1))

⇒ gcd(4n + 1, −1) = 1

⇒ gcd(12n + 2, 8n + 1) = 1 for every n

n values = 1, 2, 3, 4, 5, 6 ⇒ count = 6

∴ Number of values = 6.

Given:

f = x³ + y³ + 3xy − 1

g = (x − y)⁴ − 1

Formula used:

HCF(f, g) = greatest common divisor of polynomials. Factor both and compare common factors.

Calculation:

⇒ Factor g: (x + y)⁴ − 1 = (x + y − 1)(x + y + 1)(x² + 2xy + y² + 1)

⇒ Factor f: x³ + y³ + 3xy − 1 = (x + y − 1)(x² − x y + x + y² + y + 1)

⇒ Common factor: (x + y - 1)

∴ HCF = (x + y - 1).

Given:

x = n(n + 1)(n + 2)

n is an even natural number

Formula used:

Three consecutive numbers always include:

• one multiple of 3

• at least two even numbers if n is even

Calculation:

Let n = 2k

⇒ x = 2k(2k + 1)(2k + 2)

⇒ x = 2k(2k + 1)2(k + 1)

⇒ x = 4k(k + 1)(2k + 1)

From k and (k + 1), one is even 

⇒ x has factor 8

Among k, (k + 1), (2k + 1) one is divisible by 3 

⇒ x is always divisible by 24

Check Statement I:

48 = 16 × 3

x has only 8 × 3

Example: n = 2 ⇒ x = 2 × 3 × 4 = 24

⇒ Not divisible by 48 

Statement I is false

Check Statement II:

x is divisible by 24

⇒ x² is divisible by 24² = 576

144 = 16 × 9 ⊂ 576

⇒ x² is always divisible by 144 

∴ Only Statement II is correct.

Correct option: II only.

Given:

Find LCM of A = x⁴ + x²y² + y⁴, B = x³y + y⁴, C = x⁴y² − x³y³

Formula used:

LCM = product of distinct irreducible factors with highest powers present in A, B, C

Calculations:

⇒ A = x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² − xy + y²)

⇒ B = x³y + y⁴ = y(x³ + y³) = y(x + y)(x² − xy + y²)

⇒ C = x⁴y² − x³y³ = x³y²(x − y)

Highest powers needed: x³ (from C), y² (from C), factors (x − y), (x + y), (x² + xy + y²), (x² − xy + y²)

⇒ Combine: LCM = x³ y² (x − y)(x + y)(x² + xy + y²)(x² − xy + y²)

⇒ (x − y)(x + y)(x² + xy + y²)(x² − xy + y²) = x⁶ − y⁶

⇒ LCM = x³ y² (x⁶ − y⁶)

The correct answer is Option (2).

Given:

Let XYZ be a 3-digit number. Let D be the difference between XYZ and ZYX.

We need to find the remainder when D is divided by 99.

Calculation:

Let XYZ = 100X + 10Y + Z and ZYX = 100Z + 10Y + X

D = (100X + 10Y + Z) - (100Z + 10Y + X)

⇒ D = 100X - X + 10Y - 10Y + Z - 100Z

⇒ D = 99X - 99Z

⇒ D = 99(X - Z)

When D is divided by 99:

⇒ Remainder = 0

∴ The correct answer is option (1).

Given:

p, q are natural numbers

(p + q)^p+q is divisible by 512

Formula used:

512 = 2⁹

If a = 2^k × odd, then aⁿ contains 2^k×n

Calculation:

Let s = p + q

Then (p + q)^p+q = s^s

Check options:

s = 4

4 = 2²

⇒ s^s = 2^2×4 = 2⁸ < 2⁹ 

s = 6

6 = 2 × 3

⇒ s^s contains 2⁶ < 2⁹ 

s = 8

8 = 2³

⇒ s^s = 2^3×8 = 2²⁴ ≥ 2⁹ 

s = 12

12 > 8 but not least

∴ The least value of (p + q) is 8.

Shortcut Trick

In cyclic or symmetric algebraic expressions, substitute specific values for variables.

Let a = 1, b = 0, and c = 2 (Satisfies a ≠ b, b ≠ c, c ≠ a).

Term 1 = 1² ÷ [(2 − 1)(2 + 1 + 0)] = 1 ÷ 3 = 1/3.

Term 2 = [(1 + 0) × 2] ÷ [2² − 0 − 1² − 0] = 2 ÷ 3 = 2/3.

Term 3 = (1 + 0 + 2) ÷ [2(2 − 1)] = 3 ÷ 2 = 1.5.

Expression = 1/3 + 2/3 − 1.5 = 1 − 1.5 = −0.5 = −1/2.

∴ The correct answer is −1/2.

Alternate Method

Given: a ≠ b, b ≠ c, c ≠ a

Formula Used: Rational Expression Simplification and Substitution.

Term 1 1/3 + Term 2 2/3 − Term 3 1.5 Result: 1/3 + 2/3 − 1.5 = −0.5

⇒ Let a = 1, b = 0, c = 2.

⇒ Term 1 = (1 + 0)² ÷ [(2 − 1)(2 + 1 + 0)] = 1 ÷ [1 × 3] = 1/3.

⇒ Term 2 = (1 + 0) × 2 ÷ [2² − (0 × 2) − 1² − (1 × 0)] = 2 ÷ [4 − 1] = 2/3.

⇒ Term 3 = (1 + 2 × 0 + 2) ÷ [2(2 − 1)] = 3 ÷ 2 = 1.5.

⇒ Expression = 1/3 + 2/3 − 3/2 = 1 − 1.5 = −0.5.

∴ The correct answer is −1/2.

Additional Information

Substitution Method in Algebra

Complex algebraic expressions that are independent of variables can be solved by substituting convenient integers that satisfy the given constraints.

Cyclic and Symmetric Expressions

Expressions involving terms like (a−b), (b−c), and (c−a) are cyclic. If the degree of the numerator and denominator match across terms, the result is often a constant.

Rationalization Property

Always check for common factors in denominators, such as (c² − a²) = (c − a)(c + a), to simplify sums of fractions.

Given:

a^b = b^a

Find: $\dfrac{a \times \left(\frac{a}{b}\right)^{\frac{a}{b}}}{a^{\frac{a}{b}}}$

Formula used:

Properties of exponents: (u^p)/(v^p) = (u/v)^p and from a^b = b^a ⇒ a^1/a = b^1/b

Calculations:

Let E = $\dfrac{a \times \left(\frac{a}{b}\right)^{\frac{a}{b}}}{a^{\frac{a}{b}}}$

⇒ E = a × $\dfrac{\left(\frac{a}{b}\right)^{\frac{a}{b}}}{a^{\frac{a}{b}}}$

⇒ E = a × $\left(\dfrac{a/b}{a}\right)^{\frac{a}{b}}$ = a × $\left(\dfrac{1}{b}\right)^{\frac{a}{b}}$

⇒ E = a × b^-a/b

From a^b = b^a ⇒ a^1/a = b^1/b

⇒ raise both sides to power a: b^a/b = a

⇒ b^-a/b = 1/a

⇒ E = a × (1/a) = 1

∴ Correct answer = 1.

Given:

x = 2 + 2^1/2 + 2^3/2

2^1/2 = √2 and 2^3/2 = 2√2

⇒ x = 2 + √2 + 2√2 = 2 + 3√2

Formula used:

(a + b)² = a² + 2ab + b²

Calculation:

x = 2 + 3√2

x² = (2 + 3√2)²

⇒ x² = 4 + 12√2 + 18 = 22 + 12√2

Now compute x² − 4x − 10

⇒ (22 + 12√2) − 4(2 + 3√2) − 10

⇒ 22 + 12√2 − 8 − 12√2 − 10

⇒ 4

∴ Correct answer = 4.

Given:

$\dfrac{\sqrt{p+x} + \sqrt{p-x}}{\sqrt{p+x} - \sqrt{p-x}} = p$

Formula used:

Cross multiplication and algebraic manipulation.

Calculation:

$\dfrac{\sqrt{p+x} + \sqrt{p-x}}{\sqrt{p+x} - \sqrt{p-x}} = p$

⇒ $\sqrt{p+x} + \sqrt{p-x} = p(\sqrt{p+x} - \sqrt{p-x})$

⇒ $\sqrt{p+x} + \sqrt{p-x} = p\sqrt{p+x} - p\sqrt{p-x}$

⇒ $\sqrt{p+x} - p\sqrt{p+x} = -\sqrt{p-x} - p\sqrt{p-x}$

⇒ $(1-p)\sqrt{p+x} = -(1+p)\sqrt{p-x}$

⇒ Squaring both sides:

$(1-p)^2(p+x) = (1+p)^2(p-x)$

⇒ $(1-p)^2p + (1-p)^2x = (1+p)^2p - (1+p)^2x$

⇒ $(1-p)^2x + (1+p)^2x = (1+p)^2p - (1-p)^2p$

⇒ $x[(1-p)^2 + (1+p)^2] = p[(1+p)^2 - (1-p)^2]$

⇒ $x = \dfrac{p[(1+p)^2 - (1-p)^2]}{(1-p)^2 + (1+p)^2}$

⇒ $x = \dfrac{p[1 + 2p + p^2 - (1 - 2p + p^2)]}{1 - 2p + p^2 + 1 + 2p + p^2}$

⇒ $x = \dfrac{p[4p]}{2(1 + p^2)}$

⇒ $x = \dfrac{2p^2}{p^2+1}$

∴ The correct answer is option (4).

Given:

$\left(\frac{a-b}{2}\right)x^2-\left(\frac{a+b}{2}\right)x+b=0$

Formula used:

If x = r is a root, then substituting x = r makes LHS = 0

Calculation:

Multiply whole equation by 2 to simplify

⇒ (a − b)x² − (a + b)x + 2b = 0

Check x = 1

⇒ (a − b) − (a + b) + 2b

⇒ a − b − a − b + 2b

⇒ 0 

So one root is x = 1

Sum of roots = (a + b)/(a − b)

Let second root = r

⇒ 1 + r = (a + b)/(a − b)

⇒ r = (a + b − a + b)/(a − b)

⇒ r = 2b/(a − b)

∴ The roots are 1 and 2b/(a − b).

Given:

$\rm x-\frac{1}{x}=2$

Formula used:

(x - 1/x)² = x² + 1/x² - 2

Calculation:

⇒ (x − 1/x)² = 2² = 4

⇒ x² + 1/x² − 2 = 4

⇒ x² + 1/x² = 6

We need x² − 1/x²

⇒ (x − 1/x)(x + 1/x) = x² − 1/x²

⇒ x + 1/x = √(x² + 1/x² + 2)

⇒ x + 1/x = √(6 + 2) = √8 = 2√2

⇒ x² − 1/x² = 2 × 2√2 = 4√2

∴ x² − 1/x² = 4√2.

Given:

(a − b)² + (b − c)² + (c − a)² = 6

a² + b² + c² = 29

Formula used:

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

Calculation:

6 = 2(a² + b² + c² − ab − bc − ca)

⇒ 3 = a² + b² + c² − (ab + bc + ca)

Given a² + b² + c² = 29

⇒ 3 = 29 − (ab + bc + ca)

⇒ ab + bc + ca = 26

Now,

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

⇒ (a + b + c)² = 29 + 52 = 81

⇒ a + b + c = ±9

∴ The correct answer is Option (1).

Given:

$\rm p=\frac{\sqrt5-2}{\sqrt5+2}$ and $\rm q=\frac{\sqrt5+2}{\sqrt5-2}$

Formula used:

$\dfrac{p}{q}+\dfrac{q}{p} = \dfrac{p^2+q^2}{pq}$

Calculation:

$\dfrac{p}{q} = \left(\dfrac{\sqrt5-2}{\sqrt5+2}\right)^2$

⇒ $\dfrac{( \sqrt5-2 )^2}{( \sqrt5+2 )^2}$

⇒ $\dfrac{5+4-4\sqrt5}{5+4+4\sqrt5}$

⇒ $\dfrac{9-4\sqrt5}{9+4\sqrt5}$

Similarly, $\dfrac{q}{p} = \dfrac{9+4\sqrt5}{9-4\sqrt5}$

⇒ Required value = $\dfrac{9-4\sqrt5}{9+4\sqrt5}+\dfrac{9+4\sqrt5}{9-4\sqrt5}$

⇒ $\dfrac{(9-4\sqrt5)^2+(9+4\sqrt5)^2}{9^2-(4\sqrt5)^2}$

⇒ $\dfrac{(81+80)+(81+80)}{81-80}$

⇒  $\dfrac{322}{1}$

⇒ 322

∴ The value of (p/q + q/p) = 322.

Given:

Find digit at hundreds place of 25¹⁰

Formula used:

Last 3 digits of N determine hundreds digit; 25ⁿ for n ≥ 2 ends with 625.

Calculations:

25² = 625

⇒ 25ⁿ ≡ 625 (mod 1000) for all n ≥ 2

⇒ 25¹⁰ ≡ 625 (mod 1000)

Last 3 digits = 625 ⇒ hundreds digit = 6

∴ Required digit = 6.

Given:

A number N leaves remainder 3 when divided by 4, 6, 7 and 9

Formula used:

If a number leaves same remainder r when divided by several numbers, then

N − r is divisible by LCM of those numbers

Calculation:

LCM of 4, 6, 7, 9

LCM = 2² × 3² × 7 = 252

N − 3 = multiple of 252

⇒ N = 252k + 3

Smallest 4-digit N:

252 × 3 + 3 = 759 (not 4-digit)

252 × 4 + 3 = 1008 + 3 = 1011 

∴ The smallest 4-digit number is 1011.

Given:

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+....}}}}=cosecθ$

Formula used:

For infinite nested radicals:

$\sqrt{k+\sqrt{k+\sqrt{k+....}}} = x \text{ where } x^2 = k+x$

Also, $cosec \theta = \frac{1}{\sin \theta}$

Calculation:

$\sqrt{2+\sqrt{2+\sqrt{2+....}}} = x$

⇒ $cosec \theta = x$

⇒ $cosec \theta = x \text{ and } x^2 = 2+x$

⇒$cosec \theta = x$, and solving $x^2 - x - 2 = 0$

⇒ \((x-2)(x+1) = 0 \Rightarrow x = 2 \text{ (taking positive root as \(cosec \theta > 0\))}\)

⇒ $cosec \theta = 2$

⇒ $\sin \theta = \frac{1}{cosec \theta} = \frac{1}{2}$

∴ The correct answer is option (4).

Given:

8 sinθ − cosθ = 4, 0 < θ < π/2

Formula used:

cosθ = 8 sinθ − 4; sin²θ + cos²θ = 1; solve quadratic in sinθ

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

Calculations:

⇒ cosθ = 8 sinθ − 4

⇒ Since sin²θ + cos²θ = 1 

⇒ sin²θ + (8 sinθ − 4)² = 1

⇒ sin²θ + 64 sin²θ − 64 sinθ + 16 = 1

⇒ 65 sin²θ − 64 sinθ + 15 = 0

⇒ Discriminant = 64² − 4 × 65 × 15 = 196, √196 = 14

⇒ sinθ = (64 ± 14)/130

⇒ sinθ = 78/130 = 3/5 (valid)

⇒ sinθ = 50/130 = 5/13 (invalid, cosθ negative)

⇒ Valid sinθ = 3/5 ⇒ cosecθ = 1 / sinθ = 5/3

∴ cosecθ = 5/3.

Given:

2 tanθ = sec² θ - 2 , 0 < θ < π/2

Formula used:

sec² θ = 1 + tan² θ

cotθ = 1 / tanθ

Calculations:

⇒ 2 tanθ = sec² θ - 2

⇒ 2 tanθ = (1 + tan² θ) - 2

⇒ 2 tanθ = tan² θ - 1

⇒ tan² θ - 2 tanθ - 1 = 0

Let t = tanθ

⇒ t² - 2t - 1 = 0

⇒ t = (2 ± √(4 + 4)) / 2 = (2 ± √8)/2 = 1 ± √2

In 0 < θ < π/2, tanθ > 0 ⇒ t = 1 + √2

⇒ cotθ = 1 / (1 + √2)

⇒ cotθ = √2 - 1 (after rationalisation)

∴ Correct option: √2 - 1.

Given:

$(\sec \theta - \tan \theta) - \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}}$

Formula used:

$\sec^2 \theta - \tan^2 \theta = 1$

 $\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}} = \sec \theta - \tan \theta$

Calculation:

$(\sec \theta - \tan \theta) - \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}}$

⇒ $(\sec \theta - \tan \theta) - (\sec \theta - \tan \theta)$

⇒ 0

∴ The correct answer is option (1).

Given:

cot θ = √7

Formula used:

cosec²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

tan θ = 1 / cot θ

Calculation:

⇒ tan θ = 1 / √7

⇒ sec²θ = 1 + 1/7 = 8/7

⇒ cosec²θ = 1 + 7 = 8

⇒ Numerator = 8 − 8/7 = (56/7 − 8/7) = 48/7

⇒ Denominator = 8 + 8/7 = (56/7 + 8/7) = 64/7

⇒ Value = (48/7) ÷ (64/7)

⇒ 48/64 = 3/4

∴ The correct answer is 3/4.

Given:

The difference between the two acute angles in a right-angled triangle is $\dfrac{\pi}{12}$ radian.

One of the acute angles of the triangle is to be determined.

Formula used:

In a right-angled triangle:

$\text{Sum of acute angles} = \dfrac{\pi}{2}$

Let one angle be $\theta$ and the other be $\theta - \dfrac{\pi}{12}$.

$\theta + \left(\theta - \dfrac{\pi}{12}\right) = \dfrac{\pi}{2}$

Calculations:

$\theta + \theta - \dfrac{\pi}{12} = \dfrac{\pi}{2}$

⇒ $\theta = \dfrac{\pi}{4} + \dfrac{\pi}{24}$

⇒ $\theta = \dfrac{6\pi + \pi}{24}$

⇒ $\theta = \dfrac{7\pi}{24}$

Converting $\theta$ to degrees:

$\theta = \dfrac{7 \times 180}{24} = 52.5^\circ$

∴ The correct answer is option (3): 52.5º.

Given:

$\log_{10}\big[998+\sqrt{x^2-18x+76}\big]=3$

Formula used:

If log₁₀(A) = 3 ⇒ A = 10³

Calculations:

⇒ 998 + √(x² - 18x + 76) = 10³ = 1000

⇒ √(x² - 18x + 76) = 1000 - 998 = 2

⇒ x² - 18x + 76 = 2² = 4

⇒ x² - 18x + 72 = 0

⇒ x2 - 12x - 6x + 72 = 0

⇒ (x - 12) (x - 6) = 0

⇒ x = 12 or x = 6

⇒ Let α = 12, β = 6

⇒ (α - β)² = (12 - 6)² = 6² = 36

∴ (α - β)² = 36.

Given:

x⁴ + y⁴ = 14x²y²

Formula used:

x⁴ + y⁴ = (x² + y²)² − 2x²y²

log a + log b = log(ab)

Calculation:

x⁴ + y⁴ = (x² + y²)² − 2x²y²

⇒ (x² + y²)² − 2x²y² = 14x²y²

⇒ (x² + y²)² = 16x²y²

⇒ x² + y² = 4xy

Check Statement I:

log(x² + y²) = log(4xy)

⇒ = log4 + logx + logy

⇒ = 2log2 + logx + logy 

Statement I is correct

Check Statement II:

x² − y² = (x − y)(x + y)

From x² + y² = 4xy

⇒ (x − y)² = 2xy

⇒ x − y = √(2xy)

x + y = √(x² + y² + 2xy)

⇒ x + y = √(6xy)

⇒ x² − y² = √(12x²y²)

⇒ = 2√3 xy

log(x² − y²) = log(2√3 xy)

⇒ logx + logy + log2 + 0.5log3 

Statement II is correct

∴ Both I and II are correct.

Given:

(3x + y)² + (3x + y)(x + 5y) − 20(x + 5y)²

Formula used:

a² + ab − 20b² = (a + 5b)(a − 4b)

Calculation:

Let a = (3x + y) and b = (x + 5y)

⇒ Given expression = a² + ab − 20b²

⇒ (a + 5b)(a − 4b)

⇒ [(3x + y) + 5(x + 5y)] × [(3x + y) − 4(x + 5y)]

⇒ (3x + y + 5x + 25y) × (3x + y − 4x − 20y)

⇒ (8x + 26y) × (−x − 19y)

⇒ 2(4x + 13y) × (−1)(x + 19y)

⇒ Factors are (4x + 13y) and (x + 19y)

∴ The correct answer is Both I and II.

Given:

$\dfrac{\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}}{\dfrac{x+y}{x-y}+\dfrac{y+z}{y-z}+\dfrac{z+x}{z-x}+3}$ = ?

Formula used:

y = x - (x - y) ⇒ y/(x - y) = x/(x - y) - 1 

Calculations:

Let A = $\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}$

Then sum $\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x}$ =

⇒ using y/(x-y) = x/(x-y) - 1:

⇒ $\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x} = \dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x} - 3$

So denominator D = $\dfrac{x+y}{x-y}+\dfrac{y+z}{y-z}+\dfrac{z+x}{z-x}+3$ =

⇒ $\Big(\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}\Big) + \Big(\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x}\Big) + 3$

⇒ substitute previous result:

⇒ D = A + (A - 3) + 3 = 2A

Therefore the whole expression =

⇒ $\dfrac{A}{2A} = \dfrac{1}{2}$

∴ The value of the given expression is 1/2.

Given:

p sin² α + q cos² α = m

q sin² β + p cos² β = n

p ≠ m, n and q ≠ m, n

Formula used:

sin² θ + cos² θ = 1

tan² θ = sin² θ ÷ cos² θ

Calculation:

p sin² α + q cos² α = m

⇒ p sin² α + q(1 − sin² α) = m

⇒ (p − q) sin² α = m − q

⇒ sin² α = (m − q) ÷ (p − q)

⇒ cos² α = (p − m) ÷ (p − q)

⇒ tan² α = (m − q) ÷ (p − m)

q sin² β + p cos² β = n

⇒ q sin² β + p(1 − sin² β) = n

⇒ (q − p) sin² β = n − p

⇒ sin² β = (n − p) ÷ (q − p)

⇒ cos² β = (q − n) ÷ (q − p)

⇒ tan² β = (n − p) ÷ (q − n)

⇒ (tan α ÷ tan β)² = [(m − q) ÷ (p − m)] × [(q − n) ÷ (n − p)]

⇒ (tan α ÷ tan β)² = (m − q)(n − p) ÷ (m − p)(n − q)

∴ The correct answer is option (4).

Given:

p sin² α + q cos² α = m

q sin² β + p cos² β = n

α and β are complementary ⇒ α + β = 90°

Find correct relation among m and n

Formula used:

If α + β = 90°

sin β = cos α

cos β = sin α

Calculation:

Given:

m = p sin² α + q cos² α

n = q sin² β + p cos² β

Using complementary property:

⇒ sin² β = cos² α

⇒ cos² β = sin² α

Substitute in n:

⇒ n = q cos² α + p sin² α

⇒ n = p sin² α + q cos² α

⇒ n = m

⇒ m − n = 0

∴ The correct answer is m − n = 0.

Given:

cosec θ − sin θ = p

sec θ − cos θ = q

Find p sin θ + q cos θ = ?

Formula used:

cosec θ = 1/sin θ

sec θ = 1/cos θ

Calculation:

p sin θ + q cos θ

= (cosec θ − sin θ) sin θ + (sec θ − cos θ) cos θ

= cosec θ × sin θ − sin² θ + sec θ × cos θ − cos² θ

⇒ 1 − sin² θ + 1 − cos² θ

⇒ 2 − (sin² θ + cos² θ)

⇒ 2 − 1

⇒ 1

∴ The required value is 1.

Given:

p = cosec θ − sin θ

q = sec θ − cos θ

Find p²q²(p² + q² + 3) = ?

Formula used:

sin² θ + cos² θ = 1

Calculation:

Assume θ = 45°

sin θ = cos θ = 1/√2

cosec θ = sec θ = √2

⇒ p = √2 − 1/√2 = 1/√2

⇒ q = √2 − 1/√2 = 1/√2

⇒ p² = 1/2 , q² = 1/2

⇒ p²q² = 1/4

⇒ p² + q² + 3 = 1/2 + 1/2 + 3 = 4

⇒ p²q²(p² + q² + 3) = 1/4 × 4

⇒ 1

∴ The required value is 1.

Given:

$\frac{\sin \alpha}{\sin \beta}=\frac{4\sqrt2}{3}$

$\frac{\cos \alpha}{\cos \beta}=\frac{2\sqrt3}{9}$

Find tan² α = ?

Formula used:

sin²θ + cos²θ = 1

tan θ = sin θ / cos θ

Calculation:

Let sin α = 4√2 × k , sin β = 3k

Let cos α = 2√3 × m , cos β = 9m

⇒ For α :

(4√2k)² + (2√3m)² = 1

⇒ 32k² + 12m² = 1 ……(1)

⇒ For β :

(3k)² + (9m)² = 1

⇒ 9k² + 81m² = 1 ……(2)

⇒ Solving (1) and (2)

⇒ m² = 1/108 , k² = 1/36

⇒ sin α = 4√2 × 1/6 = 2√2/3

⇒ cos α = 2√3 × 1/√108 = 1/3

⇒ tan α = (2√2/3) ÷ (1/3)

⇒ tan α = 2√2

⇒ tan² α = (2√2)²

⇒ tan² α = 8

∴ The required value is 8.

Given:

$\frac{\sin \alpha}{\sin \beta}=\frac{4\sqrt2}{3}$

$\frac{\cos \alpha}{\cos \beta}=\frac{2\sqrt3}{9}$

Find tan² β = ?

Formula used:

sin²θ + cos²θ = 1

tan θ = sin θ / cos θ

Calculation:

From previous calculation:

k² = 1/36 , m² = 1/108

⇒ sin β = 3k = 3 × 1/6 = 1/2

⇒ cos β = 9m = 9 × 1/√108

⇒ √108 = 6√3

⇒ cos β = 9/(6√3) = 3/(2√3) = √3/2

⇒ tan β = (1/2) ÷ (√3/2)

⇒ tan β = 1/√3

⇒ tan² β = (1/√3)²

⇒ tan² β = 1/3

∴ The required value is 1/3.

Given:

$\frac{1+\sin \theta}{\cos \theta}=p+\sqrt{p^2+1}$

Find sec θ = ?

Formula used:

(1 + sin θ)/cos θ = sec θ + tan θ

sec² θ − tan² θ = 1

Calculation:

(1 + sin θ)/cos θ = sec θ + tan θ

⇒ sec θ + tan θ = p + √(p² + 1)

Using identity:

(sec θ + tan θ)(sec θ − tan θ) = 1

⇒ sec θ − tan θ = 1/(p + √(p² + 1))

Rationalising:

⇒ sec θ − tan θ = √(p² + 1) − p

Adding both equations:

⇒ 2 sec θ = (p + √(p² + 1)) + (√(p² + 1) − p)

⇒ 2 sec θ = 2√(p² + 1)

⇒ sec θ = √(p² + 1)

∴ The correct answer is √(p² + 1).

Given:

$\frac{1+\sin \theta}{\cos \theta}=p+\sqrt{p^2+1}$

Find tan θ = ?

Formula used:

(1 + sin θ)/cos θ = sec θ + tan θ

(sec θ + tan θ)(sec θ − tan θ) = 1

Calculation:

sec θ + tan θ = p + √(p² + 1)

⇒ sec θ − tan θ = 1/(p + √(p² + 1))

⇒ sec θ − tan θ = √(p² + 1) − p

Subtracting:

⇒ 2 tan θ = (p + √(p² + 1)) − (√(p² + 1) − p)

⇒ 2 tan θ = 2p

⇒ tan θ = p

∴ The correct answer is p.

Given:

sin θ + cos θ = p , p ≠ 1

sec θ + cosec θ = q

Find relation between p and q

Formula used:

(sin θ + cos θ)² = 1 + 2 sin θ cos θ

sec θ + cosec θ = (sin θ + cos θ)/(sin θ cos θ)

Calculation:

(sin θ + cos θ)² = p²

⇒ 1 + 2 sin θ cos θ = p²

⇒ 2 sin θ cos θ = p² − 1

⇒ sin θ cos θ = (p² − 1)/2

Now,

q = sec θ + cosec θ

⇒ q = (sin θ + cos θ)/(sin θ cos θ)

⇒ q = p ÷ [(p² − 1)/2]

⇒ q = 2p/(p² − 1)

⇒ 2p = q(p² − 1)

∴ The correct relation is 2p = q(p² − 1).

Given:

sin θ + cos θ = p

sec θ + cosec θ = q

Find tan θ + cot θ = ?

Formula used:

tan θ + cot θ = (sin² θ + cos² θ)/(sin θ × cos θ)

sin² θ + cos² θ = 1

Calculation:

sec θ + cosec θ = 1/cos θ + 1/sin θ

⇒ q = (sin θ + cos θ)/(sin θ × cos θ)

⇒ q = p/(sin θ × cos θ)

⇒ sin θ × cos θ = p/q

tan θ + cot θ = 1/(sin θ × cos θ)

⇒ tan θ + cot θ = 1 ÷ (p/q)

⇒ tan θ + cot θ = q/p

∴ The correct answer is q/p.

Given:

Speed of Train X = 100 km/hr

Speed of Train Y = 60 km/hr

Relative speed of Train X with respect to Train Y = (100 - 60) km/hr = 40 km/hr

Time taken to cross = 9 seconds

Formula used:

Length of Train X = Relative Speed × Time

Note: To convert speed from km/hr to m/s, multiply by (5/18)

Calculations:

Relative Speed = 40 km/hr = 40 × (5/18) m/s = 100/9 m/s

⇒ Length of Train X = (Relative Speed) × Time

⇒ Length of Train X = (100/9) × 9

⇒ Length of Train X = 100 m

∴ The correct answer is option (2).

Given:

Speed of X = 4.8 km/hr

Speed of Y = 6 km/hr

Time difference = 10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours

Formula used:

Distance = Speed × Time

Relative Speed (when Y catches up with X) = Speed of Y - Speed of X

Calculation:

Relative Speed = 6 - 4.8 = 1.2 km/hr

Distance covered by X in $\frac{1}{6}$ hours:

Distance = Speed × Time

⇒ Distance = 4.8 × $\frac{1}{6}$ = 0.8 km

Now, time taken by Y to catch up with X:

Time = Distance / Relative Speed

⇒ Time = 0.8 / 1.2 = $\frac{2}{3}$ hours

Distance covered by X in 2/3 hour = 2/3 × 4.8 = 3.2 km

So, distance from P when both will meet = 0.8 + 3.2 = 4 km

∴ The correct answer is option (3).

Given:

X's salary is increased by 12% and then decreased by 10%.

Y's salary is increased by 10% and then decreased by 12%.

X and Y's current salaries are equal.

Formula used:

Net salary after increase and decrease = Initial salary × (1 + increase%/100) × (1 - decrease%/100)

Ratio of initial salaries = Final salary of X / Final salary of Y

Calculations:

Net salary of X: $\text{Initial salary of X} \times (1+\frac{12}{100}) \times (1-\frac{10}{100})$

⇒ Net salary of X = Initial salary of X × 1.12 × 0.9

⇒ Net salary of X = Initial salary of X × 1.008

Net salary of Y: $\text{Initial salary of Y} \times (1+\frac{10}{100}) \times (1-\frac{12}{100})$

⇒ Net salary of Y = Initial salary of Y × 1.1 × 0.88

⇒ Net salary of Y = Initial salary of Y × 0.968

Since the final salaries are equal:

⇒ Initial salary of X × 1.008 = Initial salary of Y × 0.968

⇒ Ratio of initial salaries = $\frac{\text{Initial salary of X}}{\text{Initial salary of Y}} = \frac{0.968}{1.008}$

⇒ Ratio = $\frac{968}{1008}$

⇒ Simplify the ratio:

⇒ Ratio = $\frac{121}{126}$

∴ The correct answer is option (3): 121 : 126.

Given:

Village population (p) = total number of persons

x% = percentage of total persons who can read and write

y% = percentage of males who can read and write

z% = percentage of females who can read and write

x > z and y > z

Formula used:

Number of males = $\dfrac{p(x - z)}{y - z}$

Calculation:

Given formula is:

⇒ Number of males = $\dfrac{p(x - z)}{y - z}$

∴ The correct answer is option (1).

Given:

Alloy X: Cu : Zn = 5 : 4

Alloy Y: Cu : Zn = 5 : 13

Equal quantities of X and Y are mixed to form alloy Z.

Formula used:

In equal quantities mixing, the average ratio of Cu and Zn is taken.

Cu in Z = (Cu in X + Cu in Y) ÷ 2

Zn in Z = (Zn in X + Zn in Y) ÷ 2

Calculation:

Let the quantity of each alloy (X and Y) be 9 units (LCM of 5+4 and 5+13).

For Alloy X:

Cu = (5/9) × 9 = 5 units

Zn = (4/9) × 9 = 4 units

For Alloy Y:

Cu = (5/18) × 9 = 2.5 units

Zn = (13/18) × 9 = 6.5 units

In Alloy Z:

Total Cu = 5 + 2.5 = 7.5 units

Total Zn = 4 + 6.5 = 10.5 units

Cu : Zn in Z = 7.5 : 10.5

Divide by 1.5:

⇒ Cu : Zn = 5 : 7

∴ The correct answer is option (2).

Given:

Score of X = Score of Y + 20

Score of X = 75% of (Score of X + Score of Y)

Formula used:

Let Score of Y = y

Score of X = y + 20

Score of X = 75% × (Score of X + Score of Y)

Calculations:

Let Score of Y = y

⇒ y + 20 = 75% × (y + 20 + y)

⇒ y + 20 = 75% × (2y + 20)

⇒ y + 20 = (75/100) × (2y + 20)

⇒ y + 20 = (3/4) × (2y + 20)

⇒ 4 × (y + 20) = 3 × (2y + 20)

⇒ 4y + 80 = 6y + 60

⇒ 80 - 60 = 6y - 4y

⇒ 20 = 2y

⇒ y = 10

Score of Y = 10

Score of X = y + 20 = 10 + 20 = 30

Ratio of Score of X to Score of Y = X : Y

⇒ 30 : 10

⇒ 3 : 1

∴ The correct answer is option (3).

Given:

Equation: 2x² − 5p x + 2p² = 0

One root exceeds the other by 4

Formula used:

Difference between roots = √(Discriminant) ÷ a

D = b² − 4ac

Calculations:

a = 2, b = −5p, c = 2p²

Difference of roots = 4

⇒ √D / a = 4

⇒ √D / 2 = 4

⇒ √D = 8

⇒ D = 64

Now compute Discriminant:

D = b² − 4ac

= (−5p)² − 4 × 2 × 2p²

= 25p² − 16p²

= 9p²

So, 9p² = 64

⇒ p² = 64/9

⇒ p = ±8/3

∴ The value of p = 8/3.

Given:

Principal (P) = ₹10,000

Rate of interest (r) = 10% per annum

Time (t) = 3 years

Installments = 3 equal annual payments

Formula used:

Installment Amount = $\dfrac{P \times r}{1-(1+r)^{-n}}$

Where, n = number of installments

Calculation:

Installment Amount = $\dfrac{10000 \times 0.10}{1-(1+0.10)^{-3}}$

⇒ Installment Amount = $\dfrac{1000}{1-(1.10)^{-3}}$

⇒ Installment Amount = $\dfrac{1000}{1 - 0.7513}$

⇒ Installment Amount = $\dfrac{1000}{0.2487}$

⇒ Installment Amount ≈ ₹4021

∴ The correct answer is option (3).