CBSE Class 12 PCM Mathematics Board Paper 2022 — Set 1 Solutions

Answer: The given AP is -30, -24, -18, ...... Here, the first term a = -30 and the common difference d = -24 - (-30) = 6. The formula for the sum of first n terms of an AP is Sₙ = n/2 × [2a + (n-1)d]. For n = 30, S₃₀ = 30/2 × [2 × (-30) + (30-1) × 6] = 15 × [-60 + 174] = 15 × 114 = 1710. Therefore, the sum of the first 30 terms is 1710.

Answer: Given the sum of the first n terms of an AP, Sₙ = n (4n + 1). To find the AP, first find the first term a by putting n = 1: a = S₁ = 1 × (4 × 1 + 1) = 5. Now, find the second term T₂ = S₂ - S₁ = 2 × (8 + 1) - 5 = 18 - 5 = 13. The common difference d = T₂ - a = 13 - 5 = 8. Therefore, the AP is 5, 13, 21, 29, ... with first term 5 and common difference 8.

Answer: First, calculate the volume of the original solid sphere. The volume of a sphere is given by (4/3) × π × r³. Here, r = 10.5 cm, so volume of sphere = (4/3) × π × (10.5)³. Next, calculate the volume of one small cone using the formula (1/3) × π × r² × h. Here, r = 3.5 cm and h = 3 cm. Then, divide the volume of the sphere by the volume of one cone to find the number of cones formed.

Answer: To find the value of m for which the given quadratic equation has two real and equal roots, we use the condition for equal roots: the discriminant (b² - 4ac) must be zero. Here, a = (m - 1), b = 2(m - 1), and c = 1. Calculating the discriminant, we get [2(m - 1)]² - 4 * (m - 1) * 1 = 0. This simplifies to 4(m - 1)² - 4(m - 1) = 0, or 4(m - 1)[(m - 1) - 1] = 0. Therefore, either (m - 1) = 0, giving m = 1, or (m - 1) - 1 = 0, giving m = 2. Hence, the quadratic has two equal roots when m = 1 or m = 2.

Answer: Given the quadratic equation √3x² + 10x + 7√3 = 0, we solve it by factorisation. First, rewrite the equation as (√3x - √2)(√3x - √2) = 0. This implies √3x - √2 = 0. Solving for x gives x = √2 / √3, which can be simplified to x = √6 / 3. Hence, the root of the equation is x = √6 / 3.

Answer: The mode of a frequency distribution is the value or class interval that appears most frequently in the data set. To find the mode, first identify the class interval with the highest frequency, called the modal class. For example, if the class 40-55 has the highest frequency, then the mode lies in this class. Mode indicates the value that occurs most frequently and helps us understand the most common data point in the set.

Answer: Let Rehan's present age be x years. Five years ago, his age was (x - 5). Seven years from now, his age will be (x + 7). According to the problem, the product of these two ages is one more than twice his present age. So, (x - 5) multiplied by (x + 7) equals 2x + 1. Expanding the product: x squared plus 7x minus 5x minus 35 equals 2x + 1, which simplifies to x squared plus 2x minus 35 equals 2x plus 1. Subtract 2x from both sides: x squared minus 35 equals 1. Then subtract 1 from both sides: x squared minus 36 equals 0. This factors into (x - 6)(x + 6) equals 0. So, x equals 6 or x equals -6. Age cannot be negative, so Rehan’s present age is 6 years.

Answer: Given two concentric circles with radii 4 cm (larger circle) and 3 cm (smaller circle), a chord of the larger circle that touches the smaller circle is a tangent to the smaller circle. The distance from the center to the chord equals the radius of the smaller circle (3 cm). Using the Pythagoras theorem in the right triangle formed by the radius of the larger circle (4 cm), the distance to the chord (3 cm), and half the chord length, we find half the chord length = √(4² - 3²) = √(16 - 9) = √7 cm. Therefore, the length of the chord = 2 × √7 cm.

Answer: To find the value of x for which the median of the given frequency distribution is 34.5, first calculate the total frequency by adding all the frequencies including x. Next, find N/2 where N is the total frequency. Then, find the cumulative frequencies until the class interval containing the median position is identified. The median class is the one where the cumulative frequency just exceeds N/2. Using the formula for median in grouped data: Median = L + [(N/2 - F) / f] × h, where L is the lower boundary of median class, F is cumulative frequency before the median class, f is frequency of the median class, and h is class width. Substitute the median value 34.5 into the formula and solve for x. This will give the required value of x that makes the median 34.5.

Answer: To draw the circle and construct tangents from points P and Q, follow these steps: First, draw a circle with center O and radius 3 cm using a compass. Next, draw a diameter and extend it on both sides. Mark points P and Q on this extended diameter such that both are 7 cm away from the center O. Now, to construct tangents from P to the circle, draw a circle centered at P with radius equal to the distance from P to the point of contact on the circle (this can be found using the perpendicular bisector method or by constructing right triangles). Repeat a similar procedure for point Q. The lines drawn from P and Q to the points where the smaller circles touch the original circle are the tangent lines. These tangents touch the circle exactly at one point each, fulfilling the definition of tangent lines from an external point.

Answer: Let the height of the building be h meters and the distance between the building and the tower be d meters. Given the height of the tower is 50 meters. From the foot of the tower, the angle of elevation to the top of the building is 30°, so using the tangent function, tan 30° = h / d. From the foot of the building, the angle of elevation to the top of the tower is 60°, so tan 60° = 50 / d. We know tan 30° is 1/√3 and tan 60° is √3. From the second equation, d = 50 / √3. Substitute this d value into the first equation: (1/√3) = h / (50 / √3). Cross-multiplying, h = (1/√3) * (50 / √3) = 50 / 3 ≈ 16.67 meters. Therefore, the height of the building is approximately 16.67 meters.

Answer: Let the width of the river be AB. Let P be the point on the bridge which is 3 meters above the river banks. The angles of depression from P to the banks A and B are 30° and 45° respectively. Draw perpendiculars from P to the banks at points A and B, and let the foot of the perpendiculars be D on the line AB. Since the angles of depression are given, angle PAD is 30° and angle PBD is 45°. Using right triangle properties, For the bank with 45° angle, Height PD = 3 m, Let distance BD = x, Tan 45° = PD / BD = 3 / x, So, x = 3 m. For the bank with 30° angle, Height PD = 3 m, Let distance AD = y, Tan 30° = PD / AD = 3 / y, So, y = 3 / tan 30° = 3 / (1/√3) = 3√3 m. Therefore, the width of the river AB = AD + BD = 3√3 + 3 = 3 (√3 + 1) meters.

Answer:

To find the mean daily expenditure of the employees, we first organize the data of daily expenditures along with the number of employees corresponding to each expenditure. Using the method of the assumed mean or direct method, we multiply each expenditure by the number of employees to get the total expenditure for that group. Then, we add all these totals to get the sum of all daily expenditures of 30 employees. Finally, we divide the total expenditure by 30, which is the total number of employees. This gives the mean daily expenditure per employee. The mean is a measure of central tendency and shows the average amount spent daily on lunch by each employee.

Answer:

Given a solid cylinder with height 30 cm and radius 7 cm, and a conical cavity with the same radius 7 cm and height 24 cm is hollowed out from it. We need to find the total surface area of the remaining solid.

First, calculate the slant height (l) of the cone using Pythagoras theorem: l = sqrt(radius² + height²) = sqrt(7² + 24²) = sqrt(49 + 576) = sqrt(625) = 25 cm.

The curved surface area of the original cylinder is 2 × π × radius × height = 2 × 3.14 × 7 × 30 = 1319.6 cm².

The base area of the cylinder is π × radius² = 3.14 × 7 × 7 = 153.86 cm².

The curved surface area of the conical cavity is π × radius × slant height = 3.14 × 7 × 25 = 549.5 cm².

After hollowing out the cone, the inside curved surface of the conical cavity becomes a new surface of the remaining solid.

The total surface area of the remaining solid = curved surface area of the cylinder + base area of the cylinder + curved surface area of the conical cavity.

Therefore, total surface area = 1319.6 + 153.86 + 549.5 = 2023 cm² (approx).

Answer: Given triangle ABC with angle B equal to 90 degrees, AB is taken as the diameter of a circle. According to the property of the circle, if AB is the diameter, then any point P on the circle lies such that angle APB is a right angle. Since point P lies on the circle and also on AC, we consider the tangent drawn at point P to the circle. We need to prove that this tangent bisects the side BC. To prove this, we use the property that the tangent at point P is perpendicular to the radius OP, where O is the centre of the circle with diameter AB. Since AB is the diameter, center O is the midpoint of AB; thus, OA = OB. Using the geometry of the figure, we observe triangles formed by the radius OP, the sides BC and the tangent at P. By showing congruence of the triangles formed by this tangent with segments of BC, we prove that the tangent passes through the midpoint of BC. Therefore, the tangent drawn at point P bisects BC, as required.

Answer:

First, calculate the volume of water flowing through the canal in one hour. The cross-sectional area of the canal is width × depth = 8 m × 6 m = 48 m². The speed of water flow is 12 km/hour, which converts to 12000 m/hour. So, the volume of water flowing in one hour = cross-sectional area × speed = 48 m² × 12000 m = 576,000 m³.

Next, to find the area that can be irrigated using this volume of water, knowing 0.05 m of standing water is required, use the formula: Volume = Area × Depth. Thus, the area irrigated = Volume / Depth = 576,000 m³ / 0.05 m = 11,520,000 m².

Therefore, the canal will irrigate an area of 11,520,000 square meters in one hour when 0.05 meters of standing water is required.