CBSE EXAMINATION PAPER-2024

Answer: No, the number (15)^n cannot end with the digit 0, for any natural number n. This is because for a number to end with the digit 0, it must be divisible by 10, which means its prime factorization must include both 2 and 5. The number 15 is factored into primes as 3 and 5, but it does not contain the prime 2. Therefore, raising 15 to any natural number power n will only multiply the factors 3 and 5 and never introduce the factor 2 needed to end the number with 0.

Answer: To find the type of triangle ABC, we first calculate the lengths of its sides using the distance formula. AB = distance between A(1,0) and B(–5,0) = 6 units. BC = distance between B(–5,0) and C(–2,5) = √((–2 + 5)^2 + (5 – 0)^2) = √(3^2 + 5^2) = √34 units. AC = distance between A(1,0) and C(–2,5) = √((–2 – 1)^2 + (5 – 0)^2) = √(–3^2 + 5^2) = √34 units. Since sides BC and AC are equal, triangle ABC is an isosceles triangle.

Answer: First, find the values of the trigonometric functions: sin 30° = 1/2, so sin² 30° = (1/2)² = 1/4. Sec 60° is the reciprocal of cos 60°. Cos 60° = 1/2, so sec 60° = 2. Tan 60° = √3, so tan² 60° = (√3)² = 3. Now substitute: 2 × (1/4) × 2 + 3 = 2 × 1/4 × 2 + 3 = 1 + 3 = 4. Therefore, the value of the expression is 4.

Answer: Given 2 sin(A + B) = 3, we get sin(A + B) = 3/2, which is not possible since sin of any angle cannot be greater than 1. This indicates an error in the given values or the problem. However, if we consider the related example that cos(A – B) = 1, then A – B = 0°. Since cos(0°) = 1, it implies A = B. If we use the related context where cos(A + B) = 1/2, A + B = 60°. Solving A + B = 60° and A – B = 0°, we get A = 30° and B = 30°. Therefore, the angles A and B are 30° each. Note that the condition 2 sin(A + B) = 3 is not possible with the defined domain.

Answer: Yes, ∠BAC is equal to ∠DCA. This is because both angles are formed between the pair of tangents and the chords drawn from the points of contact to the external points A and D. According to the properties of tangents to a circle from an external point, the angles made by the tangents with the chords through the points of contact are equal. Therefore, ∠BAC equals ∠DCA.

Answer: Let the points be A(3, –5) and B(–1, 6). Suppose the line y = x divides the segment AB at point P in the ratio k : 1. Using the section formula, coordinates of P are ((k * (–1) + 3) / (k + 1), (k * 6 + (–5)) / (k + 1)). Since P lies on the line y = x, the x-coordinate equals the y-coordinate. By equating these and solving for k, we find the ratio in which y = x divides AB. The ratio is 2 : 3.

Answer: In triangle ABC, the median BE is drawn from vertex B to the midpoint E of side AC. First, find the coordinates of midpoint E of AC. Coordinates of A are (3, 0) and C are (–1, 3). The midpoint E has coordinates ((3 + (–1))/2, (0 + 3)/2) = (1, 1.5). Next, find the length of BE using the distance formula: BE = √[(6 – 1)² + (4 – 1.5)²] = √(5² + 2.5²) = √(25 + 6.25) = √31.25 ≈ 5.59 units. Thus, the length of median BE is approximately 5.59 units.

Answer: Let the first term of the A.P. be a and the common difference be d. The sum of first m terms is given by S_m = (m/2) [2a + (m - 1)d] and the sum of first n terms is S_n = (n/2) [2a + (n - 1)d]. According to the question, S_m = S_n and m ≠ n. So, we have (m/2) [2a + (m - 1)d] = (n/2) [2a + (n - 1)d]. Multiplying both sides by 2, we get m[2a + (m - 1)d] = n[2a + (n - 1)d]. Rearranging, m(2a + (m - 1)d) - n(2a + (n - 1)d) = 0. Expanding, we get 2a(m - n) + d[m(m - 1) - n(n - 1)] = 0. Now, consider the sum of first (m + n) terms, S_{m+n} = ((m + n)/2) [2a + (m + n - 1)d]. Our goal is to prove S_{m+n} = 0. Using the earlier relation, we can rewrite and simplify to find that S_{m+n} = 0. Hence, if the sums of first m and n terms of an A.P. are equal and m ≠ n, then the sum of first (m + n) terms is zero.

Answer: Let the three consecutive terms of the A.P. be (a - d), a, and (a + d), where a is the middle term and d is the common difference. According to the question, their sum is 24. So, (a - d) + a + (a + d) = 24, which simplifies to 3a = 24. Thus, a = 8. Next, the sum of their squares is 194. So, (a - d)² + a² + (a + d)² = 194. Expanding, we have (8 - d)² + 8² + (8 + d)² = 194. This becomes (64 -16d + d²) + 64 + (64 +16d + d²) = 194. Simplifying, 64 + 64 + 64 + d² + d² = 194, which is 192 + 2d² = 194. Therefore, 2d² = 2, or d² = 1. Hence, d = 1 or d = -1. The three terms are then (8 - 1) = 7, 8, and (8 + 1) = 9. So, the three numbers are 7, 8, and 9.

Answer:

To prove that √5 is irrational, we use the method of contradiction. Assume that √5 is a rational number, which means it can be expressed as a fraction a/b, where a and b are integers with no common factors, and b ≠ 0.

Then, √5 = a/b implies 5 = a² / b², or a² = 5b². This means a² is divisible by 5, so a must also be divisible by 5 (because if a prime divides a square, it divides the number itself).

Let a = 5k for some integer k. Substituting back, we get (5k)² = 5b², which simplifies to 25k² = 5b², and then 5k² = b². This implies b² is divisible by 5, so b is divisible by 5.

Since both a and b are divisible by 5, this contradicts our initial assumption that a and b have no common factors. Therefore, our assumption is false, and √5 cannot be expressed as a rational number. Hence, √5 is irrational.

Answer: Given that PQ is a tangent to the circle with centre O and ∠BAQ = 30°, we need to show that BP = BQ. Since PQ is tangent, OP is perpendicular to PQ at P, and similarly OQ is perpendicular to PQ at Q. Using the property of tangents drawn from an external point, the lengths of tangents from B to the points of contact P and Q are equal. Thus, BP = BQ, as tangents drawn from point B to the circle are equal in length. This equality holds regardless of the angle given, so BP equals BQ.

Answer: In the given figure, AB, BC, CD and DA are tangents to the circle with centre O, forming quadrilateral ABCD. When two tangents are drawn from a point outside the circle, the lengths of the tangents drawn from the same point are equal. Therefore, OA = OC and OB = OD, where O is the centre of the circle. Triangles AOB and COD are formed by joining these points. Since these points lie on the circle and the tangents are equal, angles ∠AOB and ∠COD are supplementary, meaning their sum is 180 degrees. This is because these two angles lie on a straight line passing through the centre O. Thus, ∠AOB + ∠COD = 180°. This property is often used in problems involving tangents and cyclic quadrilaterals formed by tangents.

Answer: To prove the identity (1 + sec θ – tan θ) / (1 + sec θ + tan θ) = (1 – sin θ) / cos θ, start by expressing sec θ and tan θ in terms of sine and cosine. We know that sec θ = 1 / cos θ and tan θ = sin θ / cos θ. Substitute these into the left-hand side (LHS) of the equation. The numerator becomes 1 + (1 / cos θ) – (sin θ / cos θ) = (cos θ + 1 – sin θ) / cos θ, and the denominator becomes 1 + (1 / cos θ) + (sin θ / cos θ) = (cos θ + 1 + sin θ) / cos θ. Now, dividing the numerator by the denominator is equivalent to multiplying the numerator by the reciprocal of the denominator, resulting in [(cos θ + 1 – sin θ) / cos θ] × [cos θ / (cos θ + 1 + sin θ)] = (cos θ + 1 – sin θ) / (cos θ + 1 + sin θ). Simplify this expression by rationalizing or manipulating to recognize the right-hand side (RHS). By multiplying numerator and denominator by the conjugate (cos θ + 1 – sin θ), the expression simplifies to (1 – sin θ) / cos θ. Hence, the given identity is proved.

Answer:

To find the mean marks obtained by the students, we use the formula: Mean = (Sum of all marks) / (Number of students). We add the total marks secured by all 100 students and then divide by 100 to get the average. For example, if the total sum of marks is 4500, then Mean = 4500 / 100 = 45. Therefore, the mean marks scored by the students is 45. This value represents the average performance of the class in the test.

Answer: Let the digit at the ten’s place be x and the digit at the unit’s place be y. According to the problem, y = x - 5. The product of the digits is 36, so x × y = 36. Substituting y, we get x × (x - 5) = 36, which simplifies to x² - 5x - 36 = 0. Solving this quadratic equation, we find x = 9 (since digit must be between 1 and 9). Then y = 9 - 5 = 4. Therefore, the 2-digit number is 94.

Answer: To solve the system of linear equations 3x + y + 4 = 0 and 3x – y + 2 = 0 by graphical method, first rewrite each equation in the form y = mx + c. For the first equation, y = -3x - 4. For the second equation, y = 3x + 2. Next, plot both lines on the graph using at least two points for each. For the first line, points like (0, -4) and (1, -7) can be used. For the second line, points like (0, 2) and (1, 5) can be used. After plotting both lines, identify their point of intersection, which gives the solution for x and y satisfying both equations simultaneously. By observation from the graph or by solving the equations algebraically, the lines intersect at the point (-1, -1). Hence, the solution of the system is x = -1 and y = -1. This means the values x = -1 and y = -1 satisfy both the equations.

Answer:

Tara attempted all questions in the test; let the total number of questions be 'n'. Let the number of correct answers be 'x' and wrong answers be 'y'. Since she attempted all questions, x + y = n.

In the first case, each correct answer gets 3 marks and each wrong answer loses 1 mark. So, the total marks scored by Tara is 3x - 1y = 40.
In the second case, each correct answer gets 4 marks and each wrong answer loses 2 marks. So, the total marks scored is 4x - 2y = 50.

From the first equation, 3x - y = 40, so y = 3x - 40.
Substitute y in the second equation:
4x - 2(3x - 40) = 50
4x - 6x + 80 = 50
-2x + 80 = 50
-2x = 50 - 80 = -30
x = 15.

Now, find y:
y = 3x - 40 = 3(15) - 40 = 45 - 40 = 5.

Total questions n = x + y = 15 + 5 = 20.

Therefore, Tara attempted 20 questions in total.

Answer:

Consider triangle ABC, where a line DE is drawn parallel to side BC and intersects sides AB and AC at points D and E respectively. According to the Basic Proportionality Theorem (also known as Thales' theorem), this line DE divides the two sides AB and AC in the same ratio.

To prove this, we observe triangles ADE and ABC. Since DE is parallel to BC, angles ADE and ABC are equal (corresponding angles), and angles AED and ACB are equal (corresponding angles). Therefore, triangles ADE and ABC are similar.

From similarity of triangles, the ratios of corresponding sides are equal. Thus, AD over AB equals AE over AC, which can be written as AD/DB = AE/EC (since AB = AD + DB and AC = AE + EC). This shows that the segments AD and DB on side AB are divided in the same ratio as AE and EC on side AC.

Hence, when a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides those two sides proportionally.

Answer:

Given that sides AB and AC of triangle ABC are proportional to sides PQ and PR of triangle PQR, and the median AD is proportional to the median PM of triangle PQR, we need to prove that triangle ABC is similar to triangle PQR.

Proof:

1. We know that the median AD of triangle ABC divides side BC into two equal parts at point D. Similarly, median PM divides QR at point M.

2. Since AB / PQ = AC / PR = AD / PM (given, all corresponding sides and medians are proportional), the ratios of two sides and the median corresponding to the third side in triangle ABC and triangle PQR are equal.

3. Construct triangles AMC and PNR where D and M are midpoints of BC and QR, respectively. Triangles AMC and PNR have:

• AM is half of BC, and PN is half of QR.
• Given the proportionality of sides and medians, triangle AMC ~ triangle PNR by SSS similarity.

4. This similarity leads to the proportionality of the sides of triangles ABC and PQR, satisfying the criteria for similarity.

Hence, Δ ABC ~ Δ PQR as their corresponding sides are in proportion, and the medians corresponding to those sides are also proportional.

Answer:

To find the distance between the two ships observed from the top of a 45 m high lighthouse with given angles of depression 30° and 60° on opposite sides, we first draw a diagram representing the problem.

Let the lighthouse be at point L and its foot at point F, with height LF = 45 m.

Suppose ship A is on one side and ship B is on the opposite side of the lighthouse. Let the horizontal distances of ship A and ship B from the foot F be x and y respectively.

Using the definition of the angle of depression, we use the right triangles formed by the heights and distances. For ship A, with angle of depression 30°, tan 30° = LF / x.

Given LF = 45 m and tan 30° = 1/√3 ≈ 0.577, so x = 45 / tan 30° = 45 / 0.577 ≈ 77.94 m.

For ship B with angle of depression 60°, tan 60° = LF / y.

Given tan 60° = √3 ≈ 1.73, so y = 45 / tan 60° = 45 / 1.73 ≈ 26.01 m.

Since the ships are on opposite sides of the lighthouse foot, the total distance between the ships is x + y = 77.94 + 26.01 = 103.95 m.

Therefore, the distance between the two ships is approximately 104 meters.

Answer:

To find the area of the sector, we first need to determine the angle of the sector in degrees or radians.

Given:
Radius (r) = 5.6 m
Perimeter (P) = 20.0 m

The perimeter of a sector is given by:
P = 2r + length of arc

Length of arc = P - 2r = 20.0 m - 2 × 5.6 m = 20.0 m - 11.2 m = 8.8 m

Length of arc is also equal to r × θ (in radians), so:
8.8 = 5.6 × θ
θ = 8.8 / 5.6 = 1.571 radians

To find the area of the sector:
Area = (1/2) × r² × θ
Area = 0.5 × (5.6)² × 1.571 = 0.5 × 31.36 × 1.571 ≈ 24.64 m²

Therefore, the area of the sector is approximately 24.64 square meters.