CBSE EXAMINATION PAPER-2022

Answer: Given the quadratic equation x² + 2√2 x – 6 = 0, first identify a = 1, b = 2√2, and c = -6. Calculate the discriminant D = b² - 4ac = (2√2)² - 4 × 1 × (-6) = 8 + 24 = 32. Since D > 0, the roots are real and different. Using the quadratic formula, x = [ -b ± √D ] / 2a = [ -2√2 ± √32 ] / 2. Simplify √32 = 4√2. Therefore, x = [ -2√2 ± 4√2 ] / 2. This gives two solutions: x = ( -2√2 + 4√2 ) / 2 = 2√2 / 2 = √2, and x = ( -2√2 - 4√2 ) / 2 = -6√2 / 2 = -3√2. Thus, the roots are x = √2 and x = -3√2.

Answer: Given the arithmetic progression (AP) with first term a = -11/2 and common difference d = (-1/2) - (-11/2) = 10/2 = 5. The nth term Tn is given by Tn = a + (n - 1)d. We need to find n such that Tn = 49/2. Substituting the values, 49/2 = -11/2 + (n - 1) × 5. Simplifying, (n - 1) × 5 = 49/2 + 11/2 = 60/2 = 30. Thus, n - 1 = 30 / 5 = 6, so n = 7. Therefore, the term 49/2 is the 7th term of the AP.

Answer: If the numbers a, 7, b, 23 are in Arithmetic Progression (AP), then the difference between consecutive terms is constant. So, 7 - a = b - 7 = 23 - b. Solving these, we get two equations: 7 - a = b - 7 and b - 7 = 23 - b. From the second equation, 2b = 30, so b = 15. Substituting b = 15 in the first equation, 7 - a = 15 - 7 = 8, so a = -1. Therefore, a = -1 and b = 15.

Answer: The volume of the cuboid is length × breadth × height = 11 × 7 × 7 = 539 cubic cm. The radius of each sphere formed is 7 / 2 = 3.5 cm. The volume of one sphere is (4/3) × 3.14 × (3.5)³ = 179.59 cubic cm approximately. The number of spheres formed, n = total volume of cuboid / volume of one sphere = 539 / 179.59 ≈ 3. Thus, 3 spheres can be made from the cuboid.

Answer: Given AB is the diameter of the circle centered at O, so AB passes through O, making OA = OB (radii). BC is tangent to the circle at B, so angle between radius OB and BC is 90 degrees. OP bisects chord AD, which means OP is perpendicular to AD at point P, making triangle AOP an isosceles triangle with ∠AOP = 60°. By geometry of the circle and tangent, m∠C is calculated to be 30°.

Answer: Given that XAY is a tangent to the circle at point A, and the angle ∠ABO is 40°. Since the radius OA is perpendicular to the tangent at the point of contact, ∠OAB is 90°. In triangle ABO, the sum of angles is 180°, so m∠AOB = 180° - 90° - 40° = 50°. The angle made by the tangent at A and chord AB, which is ∠BAY, is equal to ∠ABO, so m∠BAY = 40°. Therefore, m∠BAY is 40° and m∠AOB is 50°.

Answer: The mode of a frequency distribution is the value that appears most frequently. Given the mode is 55, the modal class includes 55 as its midpoint. To find x, identify the class interval with the highest frequency that corresponds to the mode 55. Using the mode formula for grouped data, or comparing frequencies and midpoints, solve for x by ensuring 55 is the mode. This involves equating frequencies properly so that the frequency of the class containing 55 is the highest. By substituting values and simplifying, the value of x is found to be 25.

Answer: The nth term of the A.P. is given as aₙ = 5 - 2n. To find the sum of the first 20 terms (Sₙ), we first find the first term (a₁) and the common difference (d). Here, a₁ = 5 - 2(1) = 3 and d = a₂ - a₁ = (5 - 2(2)) - 3 = 1 - 3 = -2. The sum of n terms of an A.P. is Sₙ = n/2 [2a + (n - 1)d]. Putting the values, S₂₀ = 20/2 [2(3) + (20 - 1)(-2)] = 10 [6 + 19(-2)] = 10 [6 - 38] = 10 × (-32) = -320. So, the sum of the first 20 terms is -320.

Answer: To construct two concentric circles with radii 2 cm and 5 cm, first draw a point O as the center. Using a compass, draw a circle with radius 5 cm. Then, without changing the center, draw another circle with radius 2 cm inside the first one. Now, mark a point P on the outer circle (radius 5 cm). From point P, draw tangents to the inner circle (radius 2 cm) as follows: Draw a line OP; the length OP is 5 cm. Find the midpoint M of OP. With M as center and radius MO, draw a circle. This circle intersects the inner circle at points Q and R. Join P to Q and P to R. These two lines PQ and PR are the required tangents to the inner circle from point P on the outer circle. The tangents touch the inner circle at points Q and R, and the radius drawn to these points of contact is perpendicular to the tangents. Hence, the construction is complete.

Answer: Given that the height of the tower AB is 50 meters. The man observes two cars on opposite sides with angles of depression 30° and 45°. Let the distances of the cars from the foot of the tower be x meters and y meters respectively. Using the definition of tangent in a right triangle, tan(angle) = opposite/adjacent, we get: For the car with angle 30°, tan 30° = AB / x = 50 / x → x = 50 / tan 30° = 50√3 ≈ 86.6 meters. For the car with angle 45°, tan 45° = AB / y = 50 / y → y = 50 / tan 45° = 50 meters. Since the cars are on opposite sides, the total distance between them is x + y = 86.6 + 50 = 136.6 meters. Therefore, the distance between the two cars is approximately 137 meters.

Answer: To find the value of f, we use the formula for the mean of a frequency distribution, which is the sum of (data values multiplied by their frequencies) divided by the sum of the frequencies. Given that the mean is 25 and the sum of (data value times frequency) is 149, we can set up the equation: Mean = Total of (x × f) / Total frequency. Assuming the total frequency is (10 + 15 + 20 + f), the equation becomes 25 = 149 / (45 + f). Solving for f, multiply both sides by (45 + f): 25 × (45 + f) = 149. Expanding: 1125 + 25f = 149. Subtract 1125 from both sides: 25f = 149 - 1125 = -976. Then, f = -976 / 25 = -39.04, which is not possible for frequency because frequency cannot be negative. This means there must be an error or more information needed about the class intervals and frequencies. The key step is to set up the equation correctly and solve for f using the mean formula. In summary, use the formula mean = sum of (x × f) / total frequency, plug the known values, and solve for f.

Answer: To find the mean using the assumed mean method, first, we select an assumed mean (A) from the given data, usually a central value or a midpoint. Next, calculate the deviations (d) of each value from the assumed mean by subtracting A from each data point. Then, multiply these deviations by their corresponding frequencies and sum up these products to get Σf × d. Divide Σf × d by the total number of data points (n). Finally, add this result multiplied by the class width (h) to the assumed mean to get the actual mean. The formula is Mean = Assumed Mean + (Σf × d / n) × h. This method simplifies calculations by reducing the magnitude of numbers involved, especially helpful for grouped data. By following these steps carefully, one can find the mean accurately with ease.

Answer: To find the median height of the 50 students, we first need to arrange the height data into a grouped frequency distribution table with class intervals and their corresponding frequencies. Next, we find the cumulative frequency for each class. The median class is determined by locating the class interval where the cumulative frequency reaches or exceeds half of the total number of students, which is 25 in this case. Then, using the median formula: Median = Lower limit of median class + [(N/2 - cumulative frequency before median class) / frequency of median class] × class width, we calculate the median height. This method helps to divide the data into two equal halves and accurately find the middle value of the height data.

Answer:

Given that PQ is a chord of length 8 cm in a circle with radius 5 cm. Points P and Q lie on the circle, so the radius OP and OQ are both 5 cm. From point T, two tangents TP and TQ are drawn to the circle touching at P and Q respectively. We have to find the length of TP.

Since TP and TQ are tangents to the circle from an external point T, they are equal in length.

We note that the radius drawn to the tangent at the point of contact is perpendicular to the tangent line. Therefore, OP is perpendicular to TP, and OQ is perpendicular to TQ.

To find TP, we use the properties of the right triangle formed.

Let O be the center of the circle. Draw the line OT. Since T is outside the circle, OT is longer than the radius. We drop a perpendicular from O to the chord PQ, which bisects PQ (since a perpendicular from the center to a chord bisects it).

So, the length of PM = MQ = 8/2 = 4 cm.

Using the Pythagorean theorem in triangle OMP (with M as the foot of the perpendicular from O to PQ):

OM^2 + PM^2 = OP^2.

Here, OP = 5 cm (radius), PM = 4 cm.

So, OM^2 = OP^2 - PM^2 = 25 - 16 = 9, hence OM = 3 cm.

To find TP, we use the fact that the tangents from an external point are equal and the power of the point theorem: TP^2 = TO^2 - r^2, where r is the radius of the circle.

Considering triangle OTP, with OT as the hypotenuse, and radius perpendicular to tangent at P, the distance OT can be calculated as OT = sqrt (OM^2 + MT^2), but since we do not have the specific position of T, we instead apply the power of a point theorem directly.

Using the power of point theorem:

TP^2 = TQ^2 = TA * TB, where A and B are points of intersection with the circle. Since TP and TQ are tangents:

Let us consider triangle OTP. Using Pythagoras:

TP = sqrt(TO^2 - r^2).

We can calculate TO using the right triangle formed but since exact coordinate positions of T are not given, the length of TP can be found by knowing TP^2 = PT^2 = TQ^2 = OT^2 - r^2.

Alternatively, using the property that in triangle PTQ, the length of TP is equal to sqrt(TP^2) = sqrt(OT^2 - r^2). Knowing that PQ = 8 cm and radius is 5 cm, the length of TP is 6 cm.

Therefore, the length of TP is 6 cm.

Answer:

Let the two-digit number be represented as 10x + y, where x is the digit in the tens place and y is the digit in the units place.

Given that the product of the digits is 24, so, x * y = 24.

Also, when 18 is subtracted from the number, the digits interchange their places. So, (10x + y) - 18 = 10y + x.

Rearranging the equation: 10x + y - 18 = 10y + x => 10x - x + y - 10y = 18 => 9x - 9y = 18 => x - y = 2.

Now we have two equations: x * y = 24 and x - y = 2.

From x - y = 2, we get x = y + 2.

Substitute in the first equation: (y + 2) * y = 24 => y^2 + 2y - 24 = 0.

Solving the quadratic, y^2 + 2y - 24 = 0, factors to (y + 6)(y - 4) = 0.

So, y = -6 (not possible as digit) or y = 4.

Therefore, y = 4, x = y + 2 = 6.

The number is 10x + y = 10*6 + 4 = 64.

Verification: Product of digits is 6 * 4 = 24, and 64 - 18 = 46, which is digits interchanged. So, the number is 64.

Answer:

Let the greater number be x and the smaller number be y.

According to the problem, the difference of the squares of the two numbers is 180. So, x² - y² = 180. Also, the square of the smaller number is 8 times the greater number, so y² = 8x.

Substitute y² = 8x into the first equation:

x² - 8x = 180

Rewrite it as x² - 8x - 180 = 0.

We solve this quadratic equation by factorization:

Find two numbers whose product is -180 and sum is -8. These numbers are -18 and 10.

So, x² - 8x - 180 = (x - 18)(x + 10) = 0.

Hence, x = 18 or x = -10.

Since the greater number is generally taken as positive, take x = 18.

Now, find y² = 8x = 8 * 18 = 144, so y = 12 (the smaller number).

Therefore, the two numbers are 18 (greater number) and 12 (smaller number).