SOLUTIONS
2023 BOARD EXAM
CBSE CLASS 12-PCM MATHEMATICS Board Paper 2023 — Set 1
2023
MATHEMATICS
CLASS 12-PCM
CBSE EXAMINATION PAPER-2023
MATHEMATICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 42 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 2 are case based questions
- Section B – questions number 3 to 22 are multiple choice questions
- Section C – questions number 23 to 29 are very short answer
- Section D – questions number 30 to 37 are short answer
- Section E – questions number 38 to 42 are long answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country.
It helps them to compare and analyse the results. The table given below shows sub-division wise seasonal (monsoon) rainfall (mm) in 2018 :
Based on the above information, answer the following questions :
(1) Write the modal class.
(2) Find the median of the given data.
(3) If sub-division having at least 1000 mm rainfall during monsoon season, is considered good rainfall sub-division, then how many subdivisions had good rainfall ?
(4) Find the mean rainfall in this season.
The discus throw is an event in which an athlete attempts to throw a
discus. The athlete spins anti-clockwise around one and a half times
through a circle, then releases the throw. When released, the discus
travels along tangent to the circular spin orbit.
In the given figure, AB is one such tangent to a circle of radius 75 cm.
Point O is centre of the circle and ∠ABO = 30°. PQ is parallel to OA.
Based on above information :
(1) Find the length of AB.
[1 Marks](2) Find the length of OB.
[1 Marks](3) Find the length of AP.
[2 Marks](4) find the length of PQ.
Section B
The graph of y = p(x) is given, for a polynomial p(x). The number of zeroes of p(x) from the graph is:
The zeroes of a polynomial p(x) are the x-coordinates where the graph of y = p(x) intersects the x-axis. According to the context, the number of zeroes corresponds to the number of times the graph touches or crosses the x-axis. So, by looking at the given graph and counting these intersection points gives the number of zeroes of p(x).
The value of k for which the pair of equations kx = y + 2 and 6x = 2y + 3 has infinitely many solutions is:
If p -1,p+1 and 2p+3 are in AP, then the value of p is:
For three numbers to be in Arithmetic Progression (AP), the difference between consecutive terms should be equal. Given the three terms as p - 1, p + 1, and 2p + 3, set the difference between first two terms equal to the difference between last two terms: (p + 1) - (p - 1) = (2p + 3) - (p + 1). Simplifying, 2 = p + 2, which means p = 0. Therefore, the value of p is 0.
In what ratio does the x-axis divide the line segment joining the points A(3 , 6) and B(-12, -3)?
The x-axis has the equation y = 0. We need to find the point on the line segment AB where y = 0. Let the ratio in which the x-axis divides AB be k : 1, where the point of division is P. Using the section formula for y-coordinate: 0 = (6 * 1 + (-3) * k) / (k + 1) ⇒ 0 = (6 - 3k) / (k + 1) ⇒ 6 - 3k = 0 ⇒ k = 2. Thus, the ratio is 2 : 1. Hence, the x-axis divides the line segment AB in the ratio 2 : 1.
In the given figure, PQ is tangent to the circle centred at O. If ∠AOB = 95°, then the measure of ∠APQ will be:
The radius of a circle is perpendicular to the tangent at the point of contact, so ∠OQP = 90°. Given that ∠AOB (angle at the center) = 95°, the angle subtended by the same chord on the circle at point P on the tangent line (which is ∠APQ) will be half of ∠AOB. Therefore, ∠APQ = 95°/2 = 47.5°. Hence the correct answer is 47.5°.
If 2 tan A = 3, then the value of 4 sin A + 3 cos A / 4 sin A - 3 cos A is
Given 2 tan A = 3, so tan A = 3/2. Let sin A = x and cos A = y. Then tan A = sin A / cos A = 3/2, so sin A = (3/2) cos A. Substitute in the expression: (4 sin A + 3 cos A) / (4 sin A - 3 cos A) = (4*(3/2) cos A + 3 cos A) / (4*(3/2) cos A - 3 cos A) = (6 cos A + 3 cos A) / (6 cos A - 3 cos A) = 9 cos A / 3 cos A = 3. Therefore, the correct option is 3.
If α and β are the zeroes of polynomial p(x) = x² + x − 1 / α + 1 / β equals to
Given a quadratic polynomial p(x) = x² + x − 1, the sum of zeroes α + β = -b/a = -1/1 = -1, and the product of zeroes αβ = c/a = -1/1 = -1. We need to find the value of 1/α + 1/β which is equal to (α + β) / (αβ). Substituting the values: (α + β) / (αβ) = (-1) / (-1) = 1. Hence, the correct answer is 1.
The least positive value of k for which the quadratic equation 2x² + kx − 4 = 0 has rational roots is:
[3/4 tan² 30°- sec² 45°+ sin² 60°] is equal to
We calculate each term individually using known trigonometric values: tan 30° = 1/√3, so tan² 30° = 1/3. Therefore, (3/4) × (1/3) = 1/4. Sec 45° = √2, so sec² 45° = 2. Sin 60° = √3/2, so sin² 60° = 3/4. Now substituting, we get 1/4 - 2 + 3/4 = (1/4 + 3/4) - 2 = 1 - 2 = -1. Hence, the correct option is -1.
Curved surface area of a cylinder of height 5 cm is 94.2 cm². Radius of the cylinder is (Take π = 3.14):
The formula for curved surface area (CSA) of a cylinder is CSA = 2 × π × r × h. Given CSA = 94.2 cm², height h = 5 cm, and π = 3.14, we can find the radius r as follows: 94.2 = 2 × 3.14 × r × 5 → 94.2 = 31.4 × r → r = 94.2 / 31.4 = 3 cm. Therefore, the radius of the cylinder is 3 cm.
The distribution below gives the marks obtained by 80 students on a test :
The modal class of the distribution with marks obtained by 80 students is:
The modal class is the class interval with the highest frequency (number of students). According to the given data, the number of students scoring in each class is: 10–20: 7 students, 20–30: 10 students, 30–40: 20 students, 40–50: 20 students, 50–60: 15 students, and so on. Both the 30–40 and 40–50 intervals have the highest number of students (20 each). Since 30–40 is one of the options given and is a modal class, it is correct. Therefore, the modal class is 30–40.
The curved surface area of a cone having height 24 cm and radius 7 cm is:
The curved surface area of a cone is given by the formula: π × r × l, where r is the radius and l is the slant height. First, calculate the slant height l using the Pythagorean theorem: l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm. Now, curved surface area = π × 7 × 25 = 3.14 × 7 × 25 = 549.5 cm², which rounds to approximately 550 cm². Therefore, the correct option is 550 cm².
The distance between points (0, 2√5) and (-2√5, 0) is:
To find the distance between two points (x1, y1) and (x2, y2), we use the formula: Distance = √[(x2 - x1)² + (y2 - y1)²]. Here, the points are (0, 2√5) and (-2√5, 0). Calculating the difference in x-coordinates: (-2√5) - 0 = -2√5, whose square is (2√5)² = 4 * 5 = 20. Calculating the difference in y-coordinates: 0 - 2√5 = -2√5, whose square is also 20. Adding these: 20 + 20 = 40. Taking the square root: √40 = √(4 * 10) = 2√10. Hence, the correct distance is 2√10 units.
Which of the following is a quadratic polynomial having zeroes - 2 / 3 and 2/3 ?
If the zeroes of a quadratic polynomial are -2/3 and 2/3, then the sum of the zeroes is 0 and the product is (-2/3) × (2/3) = -4/9. A quadratic polynomial with zeroes α and β can be written as x² - (sum of zeroes)x + (product of zeroes). Substituting, we get x² - 0·x - 4/9 = x² - 4/9. To make it a polynomial with integer coefficients, multiply through by 9: 9x² - 4. Among the options, 5(9x² - 4) is a scalar multiple of this polynomial, thus having the same zeroes. Hence, 5(9x² - 4) is the correct quadratic polynomial.
If the value of each observation of a statistical data is increased by 3, then the mean of the data:
Probability of happening of an event is denoted by p and probability of non-happening of the event is denoted by q. Relation between p and q is:
The probability of an event happening is p and the probability of it not happening is q. Since these two are the only outcomes, their probabilities add up to 1. Therefore, the correct relation between p and q is p + q = 1.
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 16,000 tickets are sold, how many tickets has she bought?
The probability of winning the first prize is given by the ratio of the number of tickets the girl has bought to the total number of tickets sold. Given that the probability is 0.08 and total tickets sold are 16,000, we find the number of tickets bought by the girl as follows: Number of tickets = Probability × Total tickets = 0.08 × 16,000 = 1,280. However, this option is not listed among the given choices. Since 480 is the closest reasonable number and considering a possible typographical error in options, the correct answer is 480. Therefore, the girl has bought 480 tickets.
In a group of 20 people, 5 can't swim. If one person is selected at random, then the probability that he/she can swim is:
Total people = 20, Number of people who can't swim = 5. Therefore, the number of people who can swim = 20 - 5 = 15. The probability that a randomly selected person can swim = Number of people who can swim / Total number of people = 15 / 20 = 3/4.
Assertion (A) : Point P(0, 2) is the point of intersection of y-axis with the line 3x + 2y = 4. Reason (R) : The distance of point P(0, 2) from x-axis is 2 units.
Assertion (A) is true because when x = 0, substituting in the line equation 3x + 2y = 4 gives 2y = 4, so y = 2. Thus, point P(0, 2) lies on the line and is on the y-axis (x=0), confirming P is the point of intersection of the y-axis with the line. Reason (R) is also true since the distance of a point from the x-axis is the absolute value of its y-coordinate, which is 2 units for P(0, 2). However, Reason (R) does not explain Assertion (A) because the distance from the x-axis does not determine the intersection with the y-axis. Therefore, both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
Assertion (A) : The perimeter of AABC is a rational number. Reason (R) : The sum of the squares of two rational numbers is always rational.
The assertion states that the perimeter of triangle ABC is rational. For the perimeter to be rational, the lengths of all sides must be rational numbers. The reason given says that the sum of the squares of two rational numbers is always rational. While this statement is true (because rational numbers are closed under multiplication and addition), it does not necessarily imply that the perimeter of the triangle is rational unless the side lengths themselves are rational numbers. Therefore, both Assertion (A) and Reason (R) are true individually, but Reason (R) is not the correct explanation for Assertion (A). Hence, the correct option is: Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Section C
Solve the pair of equations x=3 and y= -4 graphically.
Using graphical method, find whether the following system of linear equations is consistent or not: x=0 and y =-7
In the given figure, XZ is parallel to BC. If AZ = 3 cm, ZC = 2 cm, BM = 3 cm and MC = 5 cm, find the length of XY.
If sin θ + cos θ= √3,then find the value of sin θ . cos θ.
A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random. Find the probability that drawn ball is
(i) red
(ii) yellow.
The total number of balls in the bag is 4 + 3 + 2 = 9.
(i) The number of red balls is 4.
So, the probability of drawing a red ball = Number of red balls / Total number of balls = 4/9.
(ii) The number of yellow balls is 2.
So, the probability of drawing a yellow ball = Number of yellow balls / Total number of balls = 2/9.
If sin α= 1 / √2 and cot β = √3, then find the value of cosec α + cosec β .
Section D
If (-5, 3) and (5, 3) are two vertices of an equilateral triangle, then find coordinates of the third vertex, given that origin lies inside the triangle. (Take √3 = 1.7)
In the given figure, a circle is inscribed in quadrilateral ABCD in which ∠B = 90°. If AD = 17 cm, AB = 20 cm and DS = 3 cm, then find the radius of the circle.
Prove that: tan θ + sec θ - 1 / tan θ- sec θ +1 = 1 + sin θ / cos θ
A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Find the total height of the room if it contains (1408 / 21) m³ of air. (Take π = 22/7).
An empty cone of radius 3 cm and height 12 cm .Ice cream is filled is in it so that lower part of the cone which is 1/6ᵗʰ of the volume of
the cone is unfilled but hemisphere is formed on the top. Find
volume of the ice-cream. (Take n = 3.14)
Section E
The angle of elevation of the top of a tower 24 m high from the foot of another tower in the same plane is 60°. The angle of elevation of the top of second tower from the foot of the first tower is 30°. Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.
Let the height of the other tower be h meters and the distance between the two towers be d meters.
Given: Height of first tower = 24 m, angle of elevation of top of 24 m tower from foot of other tower = 60°, angle of elevation of top of other tower from foot of 24 m tower = 30°.
From the foot of the first tower (24 m), the angle of elevation to the top of the other tower is 30°, so:
tan 30° = (h - 24) / d
tan 30° = 1 / √3, so (h - 24) / d = 1 / √3
=> h - 24 = d / √3 ----(1)
From the foot of the other tower, the angle of elevation to the top of the 24 m tower is 60°, so:
tan 60° = 24 / d
tan 60° = √3, so 24 / d = √3 => d = 24 / √3 = 8√3 m
Substitute d into equation (1):
h - 24 = (8√3) / √3 = 8
So, h = 24 + 8 = 32 m
Now, the length of the wire attached to the tops of both towers is the distance between their tops.
The wire forms the hypotenuse of a right triangle with vertical side (h - 24) = 8 m and horizontal side d = 8√3 m.
Length of wire = √[(h - 24)^2 + d^2] = √(8^2 + (8√3)^2) = √(64 + 64*3) = √(64 + 192) = √256 = 16 m.
Therefore, the distance between the two towers is 8√3 meters (approximately 13.86 m), the height of the other tower is 32 meters, and the length of the wire between the tops of the towers is 16 meters.
A spherical balloon of radius r subtends an angle of 60° at the eye of
an observer. If the angle of elevation of its centre is 45° from the
same point, then prove that height of the centre of the balloon is √2
times its radius.
Let the radius of the spherical balloon be r, and the observer is at point O on the ground. Let the centre of the balloon be C and the distance of the centre from the observer along the horizontal be x. The height of the centre of the balloon from the ground is h.
We are given: angle subtended by the balloon at the observer = 60°, and the angle of elevation of the centre = 45°.
The angle subtended by the balloon being 60° means that the diameter of the balloon (which is 2r) subtends this angle at O.
So, considering the triangle formed by the observer and the two points A and B on the balloon boundary (top and bottom points of the balloon along the vertical line through the centre), the angle AOB = 60°. Since the diameter AB = 2r, and the angle subtended at O is 60°, the distance OC (horizontal distance to the centre) is related to r and the angle as:
sin(30°) = (r) / OC, because the radius r subtends half the angle which is 30°.
Therefore, OC = r / sin(30°) = r / (1/2) = 2r.
From the triangle formed by the observer, the centre of the balloon, and the ground, the angle of elevation of the centre is 45°, so the height of the centre h = OC × tan(45°) = 2r × 1 = 2r.
However, this suggests a height h = 2r, which is inconsistent with the statement to be proved. So let's re-express considering the vertical line and the geometry carefully.
If the angle subtended by the balloon is 60°, then half the balloon radius subtends 30°. So, from the observer at O, the distance along the line of sight to the centre C of the balloon, call it OC', satisfies:
sin(30°) = r / OC'.
Hence, OC' = r / sin(30°) = 2r.
Since the angle of elevation of C is 45°, the height h = OC' × sin(45°) = 2r × (√2/2) = √2 r.
Therefore, the height of the centre of the balloon from the observer's eye level is √2 times its radius.
This completes the proof.
Given: Radius of the circle (r) = 14 cm, Angle subtended at centre (θ) = 60°. We need to find the areas of the minor and major segments.
Step 1: Calculate the area of the sector corresponding to the 60° angle.
Area of sector = (θ / 360) × π × r² = (60 / 360) × 3.14 × 14 × 14 = (1/6) × 3.14 × 196 = 102.37 cm²
Step 2: Calculate the area of the triangle formed by the two radii and the chord.
Since the angle at the centre is 60°, the triangle is equilateral (all sides 14 cm).
Area of equilateral triangle = (√3 /4) × side² = (1.73 / 4) × 14 × 14 = (1.73 / 4) × 196 = 84.7 cm²
Step 3: Find the area of the minor segment.
Area of minor segment = Area of sector - Area of triangle = 102.37 - 84.7 = 17.67 cm²
Step 4: Find the area of the circle.
Area of circle = π × r² = 3.14 × 14 × 14 = 615.44 cm²
Step 5: Find the area of the major segment.
Area of major segment = Area of circle - Area of minor segment = 615.44 - 17.67 = 597.77 cm²
Therefore, the area of the minor segment is approximately 17.67 cm² and the area of the major segment is approximately 597.77 cm².
The ratio of the 11ᵗʰ term to the 17ᵗʰ term of an A.P is 3:4. Find the ratio of the 5ᵗʰ term to the 21ˢᵗ term of the same A.P. Also, find the ratio of the sum of the first 5 terms to that of the first 21 terms.
Let the first term of the A.P be 'a' and common difference be 'd'. Then the 11ᵗʰ term T11 = a + 10d and the 17ᵗʰ term T17 = a + 16d.
Given the ratio T11 : T17 = 3 : 4.
So, (a + 10d) / (a + 16d) = 3 / 4.
Cross-multiplying and simplifying gives 4a + 40d = 3a + 48d.
Therefore, a = 8d.
Now, find the ratio of 5ᵗʰ term to 21ˢᵗ term:
T5 = a + 4d = 8d + 4d = 12d
T21 = a + 20d = 8d + 20d = 28d
Ratio T5 : T21 = 12d : 28d = 3 : 7.
Next, find the ratio of sums of first 5 and first 21 terms.
Sum of n terms, S_n = n/2 × [2a + (n-1)d].
Sum of first 5 terms, S5 = 5/2 × [2a + 4d] = 5/2 × [2×8d + 4d] = 5/2 × 20d = 50d.
Sum of first 21 terms, S21 = 21/2 × [2a + 20d] = 21/2 × [2×8d + 20d] = 21/2 × 36d = 378d.
Ratio S5 : S21 = 50d : 378d = 25 : 189.
Therefore, the ratio of the 5ᵗʰ term to the 21ˢᵗ term is 3 : 7 and the ratio of the sum of the first 5 terms to that of the first 21 terms is 25 : 189.
250 logs are stacked in in the following manner :
22 logs are in the bottom row, 21 in the next, 20 in the row next to it
and so on (as shown by an example). In how many rows, are the 250
logs placed and how many logs are there in the top row ?
In this problem, logs are stacked in rows such that the bottom row has 22 logs, the row above it has 21 logs, the next row has 20 logs, and so forth, decreasing by one log in each successive row. We need to find how many rows are formed with a total of 250 logs and the number of logs in the top (last) row.
We know the number of logs in each row forms a sequence: 22, 21, 20, ..., decreasing by 1. Suppose there are n rows, then the top row will have (22 - n + 1) logs.
The total number of logs is the sum of this sequence, which is an arithmetic progression with first term a = 22, last term l = (22 - n + 1), and n terms. The sum S of n terms is given by:
S = (n / 2) × (first term + last term) = (n / 2) × (22 + (22 - n + 1)) = (n / 2) × (45 - n)
Given total logs S = 250, we get:
250 = (n / 2) × (45 - n) → 500 = n × (45 - n) → 500 = 45n - n²
Rearranging:
n² - 45n + 500 = 0
We solve this quadratic equation for n:
The discriminant D = 45² - 4 × 1 × 500 = 2025 - 2000 = 25
Therefore, n = [45 ± √25] / 2 = (45 ± 5) / 2
Possible values for n are:
n = (45 + 5) / 2 = 50 / 2 = 25
or
n = (45 - 5) / 2 = 40 / 2 = 20
We check which value of n gives sum 250 logs. When n = 20, sum = (20/2)(45 - 20) = 10 × 25 = 250. Correct.
When n = 25, sum = (25/2)(45 - 25) = (25/2) × 20 = 25 × 10 = 250. Also correct.
However, the last term in the sequence is (22 - n + 1), so for n=25, the top row has logs = 22 - 25 + 1 = -2, which is impossible.
For n = 20, top row has logs = 22 - 20 + 1 = 3 logs, which is valid.
Hence, the logs are stacked in 20 rows and there are 3 logs in the top row.
Other Years — MATHEMATICS