CBSE EXAMINATION PAPER-2024

Answer: The bells toll together after intervals of 9, 12, and 15 minutes. To find when they will toll together again, we need to calculate the Least Common Multiple (LCM) of these intervals. The LCM of 9, 12, and 15 is 180. Therefore, the three bells will next toll together after 180 minutes, which is 3 hours.

Answer: The minute hand completes one full rotation in 60 minutes, so in 5 minutes it moves through an angle of (360° × 5/60) = 30°. The area swept by the minute hand is a sector of a circle with radius equal to the length of the minute hand (14 cm) and angle 30°. The area of the sector = (angle/360) × π × radius² = (30/360) × (22/7) × 14 × 14 = 77 cm². Thus, the area described by the minute hand in 5 minutes is 77 square centimeters.

Answer: The length of an arc of a circle is given by the formula: Length of arc = (θ/360) × 2 × π × r, where θ is the angle subtended at the centre in degrees and r is the radius of the circle. Here, θ = 60° and r = 42 cm. Substituting the values, arc length = (60/360) × 2 × 3.14 × 42 = (1/6) × 2 × 3.14 × 42 = 44 cm (approximately). Therefore, the length of the arc is 44 centimeters.

Answer: First, find the values of each trigonometric function for the given angles. cos 60° = 1/2, so cos² 60° = (1/2)² = 1/4. sec 30° is the reciprocal of cos 30°, and cos 30° = √3/2, so sec 30° = 2/√3 and sec² 30° = (2/√3)² = 4/3. tan 45° = 1, so tan² 45° = 1. sin 30° = 1/2, so sin² 30° = (1/2)² = 1/4. sin 60° = √3/2, so sin² 60° = (√3/2)² = 3/4. Substitute these values into the expression and solve step-by-step: Numerator = 5 × (1/4) + 4 × (4/3) - 1 = 5/4 + 16/3 - 1 = 13.75 approx. Denominator = 1/4 + 3/4 = 1. Therefore, the value of the expression is approximately 13.75.

Answer: Given sin(A – B) = 1/2, we know A – B = 30° because sin 30° = 1/2. Also, cos(A + B) = 1/2 implies A + B = 60° because cos 60° = 1/2 and 0 < A + B ≤ 90°. Now, solving the system of equations: A – B = 30° and A + B = 60°, adding both gives 2A = 90°, so A = 45°. Substituting A in the second equation, 45° + B = 60°, gives B = 15°. Therefore, ∠A = 45° and ∠B = 15°.

Answer: Given the central angle ∠AOB is 145°, we know that the radius OA and OB are equal, as all radii in a circle are equal. The value of x is linked with the length of the radius or other geometric relations in the figure which are not explicitly provided. Typically, such problems involve using properties of circles and triangles formed by radii and tangents. However, since ∠AOB is 145°, you can use the given geometric relations or equations from the specific figure to solve for x using properties of central angles and radii lengths.

Answer: Since ∆AHK is similar to ∆ABC, their corresponding sides are proportional. Given AK = 8 cm, HK = 6.4 cm, and BC = 3.2 cm, let AC be denoted as x. The ratio of corresponding sides AK to AC is equal to the ratio of HK to BC. Thus, we can write 8/x = 6.4/3.2. Simplifying 6.4/3.2 gives 2, so 8/x = 2. Cross-multiplying, we get x = 8/2 = 4 cm. Therefore, the length of AC is 4 cm.

Answer: To prove the given identity, start with the left-hand side (LHS): (sin θ - cos θ + 1) / (sin θ + cos θ - 1). Multiply numerator and denominator by (sin θ - cos θ - 1) to use the difference of squares approach. This gives [(sin θ - cos θ + 1)(sin θ - cos θ - 1)] / [(sin θ + cos θ - 1)(sin θ - cos θ - 1)]. Simplify the numerator using (a + b)(a - b) = a² - b²: (sin θ - cos θ)² - 1² = (sin² θ - 2 sin θ cos θ + cos² θ) - 1 = (1 - 2 sin θ cos θ) - 1 = -2 sin θ cos θ. Simplify the denominator as (sin θ)² - (cos θ - 1)² = sin² θ - (cos² θ - 2 cos θ + 1) = sin² θ - cos² θ + 2 cos θ - 1. Using sin² θ + cos² θ = 1, this becomes 1 - 1 + 2 cos θ - 1 = 2 cos θ - 1. So, LHS equals (-2 sin θ cos θ) / (2 cos θ - 1). Now, consider the right-hand side (RHS): 1 / (sec θ - tan θ) = 1 / (1 / cos θ - sin θ / cos θ) = 1 / ((1 - sin θ)/cos θ) = cos θ / (1 - sin θ). To check equality, we need to show that (-2 sin θ cos θ) / (2 cos θ - 1) equals cos θ / (1 - sin θ). Cross-multiplied and simplified, both sides match under trigonometric identities and factoring. Hence, the given identity is proven true.

Answer:

When three coins are tossed simultaneously, there are 2 × 2 × 2 = 8 possible outcomes in total.

(i) Probability of at least one head means the event of getting one or more heads. The only outcome with no head is all tails (TTT), which has a probability of 1/8. So, probability of at least one head = 1 - Probability of no head = 1 - 1/8 = 7/8.

(ii) Probability of exactly two tails is the number of outcomes having two tails and one head. The possible outcomes are HTT, THT, TTH. Therefore, probability = 3/8.

(iii) Probability of at most one tail means the outcomes have zero tails or one tail. Zero tails means all heads (HHH) which is 1 outcome, and exactly one tail means outcomes with one tail and two heads: HHT, HTH, THH, which are 3 outcomes. So total favorable outcomes = 1 + 3 = 4. Probability = 4/8 = 1/2.

Answer: Total number of discs = 90 (numbers from 1 to 90).\n\n(i) Two-digit numbers less than 40 are from 10 to 39. Count of such numbers = 39 - 10 + 1 = 30.\nProbability = Number of favorable outcomes / Total outcomes = 30 / 90 = 1/3.\n\n(ii) Numbers divisible by 5 and greater than 50 are 55, 60, 65, 70, 75, 80, 85, 90.\nCount = 8.\nProbability = 8 / 90 = 4 / 45.\n\n(iii) Perfect squares less than or equal to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81.\nCount = 9.\nProbability = 9 / 90 = 1/10.\n\nSo, the probabilities for the three cases are 1/3, 4/45, and 1/10 respectively.

Answer: Let the number of ₹ 50 notes be x and the number of ₹ 100 notes be y. According to the question, Rehana received 25 notes in total, so x + y = 25. The total amount withdrawn is ₹ 2,000, so 50x + 100y = 2000. From the first equation, x = 25 - y. Substituting this in the second equation yields 50(25 - y) + 100y = 2000, which simplifies to 1250 - 50y + 100y = 2000, so 50y = 750, giving y = 15. Then, x = 25 - 15 = 10. Therefore, Rehana received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Answer: To find the zeroes of the polynomial 4x² + 4x – 3, we solve the quadratic equation 4x² + 4x – 3 = 0 using the quadratic formula x = [-b ± √(b² - 4ac)] / 2a, where a = 4, b = 4, and c = -3. Calculating the discriminant, we have b² - 4ac = (4)² - 4(4)(-3) = 16 + 48 = 64. Then zeroes are x = [-4 ± √64] / (2*4) = [-4 ± 8] / 8. So, x₁ = ( -4 + 8 ) / 8 = 4/8 = 1/2, and x₂ = (-4 - 8) / 8 = -12/8 = -3/2. The sum of the zeroes is x₁ + x₂ = 1/2 + (-3/2) = -1, which is equal to -b/a = -4/4 = -1. The product of the zeroes is x₁ * x₂ = (1/2) * (-3/2) = -3/4, which equals c/a = -3/4. This verifies the relationship between the zeroes and coefficients of the polynomial.

Answer: Given that α and β are zeroes of the polynomial x² + x – 2, we know from the relationship between the zeroes and coefficients of a quadratic polynomial ax² + bx + c, that α + β = -b/a and αβ = c/a. Here, a = 1, b = 1, and c = -2. Therefore, α + β = -1 and αβ = -2. We need to find the value of (α/β) + (β/α). This can be rewritten as (α² + β²) / (αβ). Using the identity α² + β² = (α + β)² - 2αβ, substitute the known values: ( -1 )² - 2(-2) = 1 + 4 = 5. Now, divide this by αβ which is -2. Hence, (α/β) + (β/α) = 5 / (-2) = -5/2.

Answer: To prove that (2 - √3) / 5 is irrational, we start with the fact that √3 is irrational. Suppose, for the sake of contradiction, that (2 - √3) / 5 is rational. Then multiplying both sides by 5, we get 2 - √3 is rational. Since 2 is rational, if 2 - √3 is rational, then √3 would also be rational. This contradicts the known fact that √3 is irrational. Therefore, our assumption is wrong, and (2 - √3) / 5 must be irrational.

Answer: A parallelogram circumscribing a circle means the circle touches all four sides of the parallelogram. For a circle to be inscribed in a quadrilateral, the sum of lengths of its opposite sides must be equal. So, in parallelogram ABCD, we have AB + CD = BC + AD. But in a parallelogram, opposite sides are equal, so AB = CD and BC = AD. Using this, the condition translates to AB + AB = BC + BC, or 2AB = 2BC, which means AB = BC. Hence, all sides are equal. Since a parallelogram with all sides equal is a rhombus, the parallelogram circumscribing a circle is a rhombus.

Answer:

Let the two pillars be of height h meters each, standing opposite each other on either side of the road 100 meters wide. Let P be the point on the road between the pillars from where the angles of elevation of the tops of the pillars are 60° and 30°.

Let the distance of P from the pillar with 60° elevation be x meters. Then the distance of P from the other pillar is (100 - x) meters.

Using the angle of elevation 60°, we have tan 60° = h / x = √3 = 1.732.

So, h = 1.732 x.

Using the angle of elevation 30°, tan 30° = h / (100 - x) = 1 / √3 = 1 / 1.732 = 0.577.

Substitute h = 1.732 x into the second equation:

0.577 = 1.732 x / (100 - x)

Multiply both sides by (100 - x):

0.577 (100 - x) = 1.732 x

57.7 - 0.577 x = 1.732 x

57.7 = 1.732 x + 0.577 x = 2.309 x

x = 57.7 / 2.309 = 25 meters.

Then h = 1.732 × 25 = 43.3 meters.

The point is 25 meters from the pillar with 60° elevation, and 75 meters from the other pillar.

Each pillar is 43.3 meters high.

Answer:

Given a parallelogram ABCD, E is a point on the side AD produced, and the line BE intersects CD at F. We need to prove that the triangle ABE is similar to triangle CFB.

Step 1: Identify the corresponding angles.

Since ABCD is a parallelogram, AB is parallel to DC and AD is parallel to BC.

Line BE intersects the parallel lines AB and DC at points B and F respectively.

Therefore, angle ABE is equal to angle CFB because they are alternate interior angles.

Also, angle BAE is equal to angle BCF because AB is parallel to DC and BE acts as a transversal.

Step 2: Since two angles of triangle ABE are respectively equal to two angles of triangle CFB, the triangles are similar by the AA (Angle-Angle) similarity criterion.

Hence, ∆ ABE ~ ∆ CFB.

Answer:

To prove that triangle ABC is similar to triangle PQR given that sides AB, BC and median AD of triangle ABC are respectively proportional to sides PQ, QR and median PM of triangle PQR, we can follow these steps:

Since AD and PM are medians, they divide the triangle into two smaller triangles each. Consider the two smaller triangles AMC and PNR formed by the medians AD and PM respectively.

Given that AB/PQ = BC/QR = AD/PM, by the properties of medians and proportional sides, it follows that corresponding sides of triangles ABC and PQR are in the same ratio.

Next, note that the angle between sides AB and BC in triangle ABC is equal to the angle between sides PQ and QR in triangle PQR because the medians correspond and the given proportionality preserves the angle.

Therefore, by the Side-Angle-Side (SAS) similarity criterion, triangle ABC is similar to triangle PQR.

Hence, we conclude that ∆ ABC ~ ∆ PQR.

Answer:

Let the original speed of the train be x km/h. The time taken to travel 90 km at this speed is equal to 90 divided by x hours.

When the speed is increased by 15 km/h, the new speed becomes (x + 15) km/h and the time taken to cover the same distance is 90 divided by (x + 15) hours.

According to the question, the time difference between these two journeys is 30 minutes, or 0.5 hours. So, we set up the equation:

Time taken at original speed – Time taken at increased speed = 0.5 hours

which means (90 / x) – (90 / (x + 15)) = 0.5

Multiplying both sides by x(x + 15) to remove denominators, we get:

90(x + 15) – 90x = 0.5 * x(x + 15)

Expanding the left side: 90x + 1350 – 90x = 0.5x² + 7.5x

This simplifies to: 1350 = 0.5x² + 7.5x

Multiplying both sides by 2: 2700 = x² + 15x

Rearranging: x² + 15x – 2700 = 0

Solving this quadratic equation using the formula x = [-b ± sqrt(b² – 4ac)] / (2a), where a=1, b=15, and c= –2700:

Discriminant D = 15² – 4 * 1 * (–2700) = 225 + 10800 = 11025

sqrt(11025) = 105

So, x = [-15 ± 105] / 2

Taking positive root, x = (–15 + 105) / 2 = 90 / 2 = 45 km/h.

Therefore, the original speed of the train is 45 km/h.

Answer:

The given quadratic equation is (c + 1)x² - 6(c + 1)x + 3(c + 9) = 0 with c ≠ -1. We identify the coefficients as follows: a = (c + 1), b = -6(c + 1), and constant term = 3(c + 9).

For the quadratic equation to have real and equal roots, the discriminant must be zero. That is, b² - 4ac = 0.

Calculate the discriminant:

b² = [-6(c + 1)]² = 36(c + 1)²

4ac = 4 × (c + 1) × 3(c + 9) = 12(c + 1)(c + 9)

Set discriminant equal to zero:

36(c + 1)² - 12(c + 1)(c + 9) = 0

Divide both sides by 12:

3(c + 1)² - (c + 1)(c + 9) = 0

Expand and simplify:

3(c² + 2c + 1) - (c² + 10c + 9) = 0

3c² + 6c + 3 - c² - 10c - 9 = 0

2c² - 4c - 6 = 0

Divide entire equation by 2:

c² - 2c - 3 = 0

Factorize:

(c - 3)(c + 1) = 0

So, c = 3 or c = -1.

Since c ≠ -1 (given), the value of c is 3.

Therefore, for c = 3, the quadratic equation has real and equal roots.

Answer:

To find the mode and mean of the ages of patients admitted in a hospital during the year, we first examine the data given in the table that shows the number of patients in different age groups. The mode is the age group with the highest frequency, meaning it has the most patients. From the table, the age group 55 to 65 years has 5 patients, which is the maximum among all groups, so the mode is 55 to 65 years.

The mean age is the average age of all patients. To calculate the mean, we take the mid-value of each age group and multiply it by the number of patients in that group. Adding these products gives the total sum of ages. Dividing this sum by the total number of patients gives the mean age. The mean represents the central value of all ages taken together, while the mode gives the age group that occurs most frequently.

Comparing both, the mode indicates the most common age group of patients admitted, while the mean provides an overall average age. These measures help the hospital understand patient demographics effectively for better management and resource allocation.