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Given:

Expression: x²ⁿ - y²ⁿ + 1

Divisor: xⁿ + yⁿ

Statement I: n is odd

Statement II: n is even

Formula Used:

(a² - b²) = (a - b)(a + b)

Remainder when P is divided by Q = P - (Q × Quotient)

Calculation:

Case 1: Using Statement I (n is odd)

⇒ x²ⁿ - y²ⁿ can be factored as (xⁿ - yⁿ)(xⁿ + yⁿ)

⇒ When divided by (xⁿ + yⁿ), remainder is (xⁿ - yⁿ) + 1

∴ Remainder can be determined using Statement I.

Case 2: Using Statement II (n is even)

⇒ x²ⁿ - y²ⁿ can be factored as (xⁿ + yⁿ)(xⁿ - yⁿ)

⇒ When divided by (xⁿ + yⁿ), remainder is (xⁿ - yⁿ) + 1

∴ Remainder can be determined using Statement II.

∴ The correct answer is Option 2.

Given:

Product of N and M = 252

N and M have the same digits in reverse order

Statement I: N + M = 33

Statement II: N > M

Formula Used:

Let N = 10a + b and M = 10b + a, where a and b are digits.

N × M = 252

N + M = 33

Calculation:

From N + M = 33:

⇒ (10a + b) + (10b + a) = 33

⇒ 11a + 11b = 33

⇒ a + b = 3

From N × M = 252:

⇒ (10a + b)(10b + a) = 252

⇒ 100ab + 10a² + 10b² + ab = 252

⇒ 101ab + 10a² + 10b² = 252

From a + b = 3, possible values are:

a = 1, b = 2 or a = 2, b = 1

Case 1: a = 1, b = 2:

⇒ N = 10(1) + 2 = 12

⇒ M = 10(2) + 1 = 21

⇒ N × M = 12 × 21 = 252 (valid)

Case 2: a = 2, b = 1:

⇒ N = 10(2) + 1 = 21

⇒ M = 10(1) + 2 = 12

⇒ N × M = 21 × 12 = 252 (valid)

From Statement II: N > M:

⇒ N = 21, M = 12 (valid)

∴ The correct answer is 21.

Given:

Number: (54D)¹⁰⁰

Last digit of expansion: 1

Statement I: D > 5

Statement II: D is a multiple of 3

Formula Used:

1. To find the last digit of a number raised to a power, consider the cyclicity of the last digit.
2. Cyclicity of a number is determined by the powers of its last digit.

Calculation:

⇒ The last digit of the number (54D) depends on D as the last digit of 54D is D.

⇒ For the expansion (54D)¹⁰⁰, the cyclicity of D determines the last digit.

⇒ To satisfy the condition that the last digit is 1, D must have a cyclicity where D⁴ gives 1 (as D¹⁰⁰ = (D⁴)²⁵).

⇒ D = 7 satisfies the condition as 7⁴ = 2401, and the last digit is 1.

Using Statement I:

⇒ D > 5 narrows down possibilities to D = 6, 7, 8, or 9.

⇒ From these, only D = 7 satisfies the cyclicity condition.

∴ Statement I alone is sufficient.

Using Statement II:

⇒ D is a multiple of 3 narrows down possibilities to D = 3, 6, or 9.

⇒ None of these satisfy the cyclicity condition for the last digit to be 1.

∴ Statement II alone is not sufficient.

∴ The correct answer is 1.

Given:

Statement-1: ABC is not an obtuse-angled triangle.

Statement-2: Angle C is acute.

Formula Used:

In a triangle, the sum of angles is 180°: ∠A + ∠B + ∠C = 180°.

Acute angle: An angle less than 90°.

Calculation:

We have,

∠A = ∠B - ∠C

⇒ ∠B = ∠A + ∠C

We know,

∠A + ∠B + ∠C = 180°

⇒ ∠A + (∠A + ∠C) + ∠C = 180°   (As  ∠B = ∠A + ∠C)

⇒ 2∠A + 2∠C = 180°

⇒ 2(∠A + ∠C) = 180° 

⇒ (∠A + ∠C) = 90°

Since  (∠A + ∠C) = 90°, it is mathematically certain that ∠A must be less than 90° (because ∠C must be greater than 0° for a triangle to exist).

So, we can answer this question even without using any of the Statements.

Hence, the correct answer is option (4).

Given:

Triangle ABC is right-angled at B.

AC = 20 cm.

Statement I: AB = 12 cm.

Statement-II: BC = 16 cm.

Formula Used:

Hypotenuse (AC)² = AB² + BC² (Pythagoras theorem)

Circum-radius = Hypotenuse / 2

Calculation:

Since Δ ABC is a right-angled triangle at B.

So, AC must be the Hypotenuse.

For the right-angled triangle:

Circum-radius = Hypotenuse / 2

So, Circum-radius of Δ ABC = AC/2 = 20/2 = 10 cm

Thus, the Question can be answered even without using any of the Statements 

∴ The correct answer is option (4).

Shortcut Trick

Area of parallelogram = side1 × side2 × sin(θ).

Given: Area = 7√3 and θ = 60° ⇒ AB × DA × sin(60°) = 7√3.

⇒ AB × DA × (√3/2) = 7√3 ⇒ AB × DA = 14.

For Statement-I: 14 = 2 × 7 (both prime). Perimeter = 2(2 + 7) = 18 units. Sufficient.

For Statement-II: 14 = 2 × 7 (both natural numbers > 1). Perimeter = 18 units. Sufficient.

∴ The correct answer is 18 units.

Alternate Method

Given: Parallelogram ABCD, ∠ABC = 60°, Area = 7√3 sq. units.

Formula Used: Area of parallelogram = a × b × sin(θ); Perimeter = 2(a + b).

⇒ Let sides AB = a and DA = b.

⇒ Area = a × b × sin(60°) = 7√3

⇒ a × b × (√3/2) = 7√3

⇒ a × b = 7 × 2 = 14

Case 1: Statement-I (a, b are prime numbers)

⇒ Factors of 14 are 1, 2, 7, 14.

⇒ Pairs (a, b) such that a × b = 14 are (1, 14) and (2, 7).

⇒ Since both must be prime, only (2, 7) is possible.

⇒ Perimeter = 2(2 + 7) = 18 units. (Sufficient)

Case 2: Statement-II (a, b are natural numbers > 1)

⇒ Factors of 14 are 1, 2, 7, 14.

⇒ Pairs (a, b) such that a × b = 14 and a, b > 1 is only (2, 7).

⇒ Perimeter = 2(2 + 7) = 18 units. (Sufficient)

∴ The correct answer is 18 units.

Additional Information

Area of Parallelogram

It can be calculated as base × height or as the product of adjacent sides and the sine of the included angle (ab sinθ).

Prime Numbers

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself (e.g., 2, 3, 5, 7).

Data Sufficiency

A question is answered by a statement if that statement leads to a unique value for the required quantity.

Given:

AP × PB = 48 square units

Statement-I: AP = 8 units

Statement-II: CP = 10 units

Concept Used:

In a circle, if two chords intersect (either inside or outside the circle), the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

Formula used:

If chords AB and CD intersect at P, then  AP × PB = CP × PD

Calculation:

So, to find CP × PD, we need AP × PB, which is already given in the question. 

So, we can answer this question without using any of the Statements.

Hence, the correct answer is option (3).

Given:

AB = 6 units

BC = 18 units

CD = 6 units

DA = 9 units

Formula Used:

1. For a quadrilateral, the diagonal can be determined using geometric properties and given lengths.
2. Additional conditions are provided in the statements to narrow down the possibilities.

Calculation:

Statement-I: The length of BD is an integer greater than 13.

⇒ Possible values for BD are 14, 15, 16, etc.

Statement-II: The length of BD is an even integer.

⇒ Possible values for BD are 14, 16, 18, etc.

Using both statements together:

⇒ From Statement-I, BD > 13 and integer. From Statement-II, BD is even.

⇒ The only value satisfying both is BD = 14.

∴ The correct answer is 3.

Shortcut Trick

For an isosceles triangle with sides 10, 10, and base x, area A = (x/4) × √(4×10² − x²) = 48.

By using Pythagorean triplets for height h and half-base (x/2): h² + (x/2)² = 10². Possible triplets (h, x/2) are (8, 6) or (6, 8).

Case 1: h = 8, x/2 = 6 ⇒ x = 12. Area = 1/2 × 12 × 8 = 48 units².

Case 2: h = 6, x/2 = 8 ⇒ x = 16. Area = 1/2 × 16 × 6 = 48 units².

Statement-I (x is even) applies to both 12 and 16. Statement-II (h > x/2) selects h=8, x=12. Thus, both are needed together.

∴ The correct answer is both the Statements together.

Alternate Method

Given: Isosceles triangle ABC, AB = AC = 10 units, Area = 48 square units.

Formula Used: Area = 1/2 × base × height; Pythagoras Theorem: h² + (b/2)² = a²

⇒ Let BC = b and height AD = h. In ΔABD, h² + (b/2)² = 10² = 100.

⇒ Area = 1/2 × b × h = 48 ⇒ b × h = 96.

⇒ Substitute h = 96/b into Pythagoras: (96/b)² + b²/4 = 100.

⇒ 9216/b² + b²/4 = 100 ⇒ 36864 + b⁴ = 400b² ⇒ b⁴ − 400b² + 36864 = 0.

⇒ Let y = b²: y² − 400y + 36864 = 0. Factors of 36864 that sum to 400 are 256 and 144.

⇒ b² = 256 or b² = 144 ⇒ b = 16 (then h = 6) or b = 12 (then h = 8).

⇒ Statement-I: b is an even integer. Both 12 and 16 are even. Not sufficient alone.

⇒ Statement-II: h > b/2. For b=16, h=6 (6 < 8 False). For b=12, h=8 (8 > 6 True). Not sufficient alone as we don't know if b must be an integer from this statement alone.

⇒ Using both: Statement-II narrows it to b = 12.

∴ The correct answer is using both the Statements together.

Additional Information

Area of Isosceles Triangle

Calculated as (b/4)√(4a² − b²) where a is the equal side and b is the base.

Data Sufficiency Logic

If multiple valid geometric configurations exist, supplementary statements are required to isolate a unique solution.

Altitude Property

In an isosceles triangle, the altitude to the base bisects the base and creates two congruent right-angled triangles.

Given:

Diagonals of the rhombus are in the ratio 5 : 12.

Sum of the diagonals = 34 cm (Statement-I).

Length of a side = 13 cm (Statement-II).

Formula Used:

Diagonal 1 = (5 / (5 + 12)) × Sum of diagonals.

Diagonal 2 = (12 / (5 + 12)) × Sum of diagonals.

Side of rhombus = √((Diagonal 1 / 2)² + (Diagonal 2 / 2)²).

Calculation:

Using Statement-I:

⇒ Sum of diagonals = 34.

⇒ Diagonal 1 = (5 / 17) × 34 = 10 cm.

⇒ Diagonal 2 = (12 / 17) × 34 = 24 cm.

Using formula for side:

⇒ Side = √((10 / 2)² + (24 / 2)²).

⇒ Side = √(5² + 12²).

⇒ Side = √(25 + 144) = √169 = 13 cm.

Using Statement-II:

Length of side is directly given as 13 cm.

From Statement-II, if side = 13 cm and diagonals are in ratio 5 : 12, we can calculate:

⇒ Side = √((Diagonal 1 / 2)² + (Diagonal 2 / 2)²).

⇒ Solving for diagonal lengths confirms the same result as from Statement-I.

∴ The correct answer is 2.