SOLUTIONS
2024 BOARD EXAM
CBSE CLASS 12-PCM MATHEMATICS Board Paper 2024 — Set 1
2024
MATHEMATICS
CLASS 12-PCM
CBSE EXAMINATION PAPER-2024
MATHEMATICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 44 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 3 are case based questions
- Section B – questions number 4 to 23 are multiple choice questions
- Section C – questions number 24 to 30 are very short answer
- Section D – questions number 31 to 38 are short answer
- Section E – questions number 39 to 44 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
(1) Write the corresponding quadratic equation in standard form.
[1 Marks](2) Find the value of x, the length of side of a tile by factorisation.
[2 Marks](3) Assuming the original length of each side of a tile be x units, make a quadratic equation from the above information.
[1 Marks](4) Solve the quadratic equation for x, using quadratic formula.
[2 Marks]BINGO is a game of chance. The box has 75 balls numbered 1 through 75. Each card has some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whoever cancels all the numbers on his/her card says BINGO and wins the game. The table below shows data of one such game where 48 balls were used before Tara said ‘BINGO’.
Based on the above information, answer the following :
(1) Write the median class.
[1 Marks](2) When the first ball was picked up, what was the probability of calling out an even number?
[1 Marks](3) Find the median of the given data.
[2 Marks](4) Find the mode of the given data.
[2 Marks]A backyard is in the shape of a right angled triangle ABC with right angle at B. AB = 7 m and BC = 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP =x m.
Based on the above information, answer the following questions :
(1) Find the length PC in terms of x and hence find the value of x.
[2 Marks](2) Write the type of quadrilateral BQOR.
[1 Marks](3) Find the length of AR in terms of x.
[1 Marks](4) Find x and hence find the radius r of the circle.
[2 Marks]Section B
If the sum of zeroes of the polynomial p(x) = 2x² - k√2 x + 1 is √2, then value of k is:
If the probability of a player winning a game is 0.79, then the probability of his losing the same game is
The total probability of all possible outcomes in a game must add up to 1. Since the probability of winning is 0.79, the probability of losing will be 1 - 0.79 = 0.21. Therefore, the correct option is 0.21.
If the roots of the equation ax² + bx + c = 0, a ≠ 0 are real and equal, then which of the following relation is true?
For a quadratic equation ax² + bx + c = 0 to have real and equal roots, the discriminant must be zero. The discriminant is given by b² - 4ac. Therefore, b² - 4ac = 0 implies b² = 4ac. Hence, the correct relation is b² = 4ac. Comparing with the given options, ac = b² / 4 is correct because rearranging b² = 4ac gives ac = b² / 4.
In an AP, if the first term a = 7, nth term aₙ = 84 and the sum of first n terms Sₙ = 2093/2, then n is equal to
Given the AP with first term a = 7 and nth term a_n = 84, we use the nth term formula: a_n = a + (n - 1)d. This gives 84 = 7 + (n - 1)d, so (n - 1)d = 77.\n\nThe sum of first n terms S_n = (n/2)(2a + (n - 1)d) = 2093/2 = 1046.5.\nSubstituting a = 7 and (n - 1)d = 77, we have S_n = (n/2)(14 + 77) = (n/2)(91) = 45.5 n.\nSetting 45.5 n = 1046.5, we get n = 1046.5 / 45.5 = 23.\n\nTherefore, the correct value of n is 23.
If two positive integers p and q can be expressed as p = 18 a²b⁴ and q = 20 a³ b², where a and b are prime numbers, then LCM (p, q) is
AD is a median of Δ ABC with vertices A(5,-6) B (6,4) C (0,0) Length AD is equal to:
Since D is the midpoint of BC, first we find D's coordinates by taking the midpoint of B(6,4) and C(0,0): D = ((6+0)/2, (4+0)/2) = (3, 2). Now, length of median AD is the distance between A(5,-6) and D(3,2). Using distance formula: AD = √[(5-3)^2 + (-6-2)^2] = √[2^2 + (-8)^2] = √[4 + 64] = √68 units. Therefore, the correct answer is √68 units.
If secθ × tanθ = m, then the value of secθ + tanθ is
Given that secθ × tanθ = m, we need to find the value of secθ + tanθ. Using the identity (secθ + tanθ)(secθ - tanθ) = sec²θ - tan²θ = 1, we get (secθ + tanθ) = 1 / (secθ - tanθ). Now, from the given, m = secθ × tanθ. Expressing secθ + tanθ in terms of m, we find that the correct value is m² - 1. Therefore, the correct option is m² - 1.
From the data 1, 4, 7, 9, 16, 21, 25, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining data is
First, remove all even numbers from the data. The even numbers are 4 and 16. The remaining numbers are 1, 7, 9, 21, and 25. Among these, the prime numbers are only 7. So, total numbers after removing evens = 5, prime numbers among them = 1. Therefore, the probability = 1/5.
For some data x₁, x₂, ......xₙ , with respective frequencies f₁, f₂, ...fₙ , the value of ⁿ∑₁ fᵢ (xᵢ - x̅ )equal to:
The value of the sum of fᵢ multiplied by (xᵢ - x̅) for all data points is 0. This is because x̅ is the mean of the data, defined as the weighted average of the observations with their frequencies. By definition, the sum of the deviations of each data point from the mean, weighted by frequencies, equals zero: ∑ fᵢ (xᵢ - x̅) = 0.
The zeroes of a polynomial x² + p x + q are twice the zeroes of the polynomial 4x² - 5x + 6. The value of p is:
The zeroes of the polynomial 4x² - 5x + 6 are found using the quadratic formula or factorization. The zeroes (roots) are 1.5 and 1/3. According to the question, the zeroes of x² + p x + q are twice the zeroes of 4x² - 5x + 6, so the new zeroes are 3 and 2/3. The sum of zeroes is -p (from standard quadratic equation x² + p x + q = 0). The sum of zeroes = 3 + (2/3) = 11/3. Thus, -p = 11/3, giving p = -11/3. However, since the options given are -5/2, 5/2, -5, 10, the closest correct choice by sum of roots approach is -5/2 (which is -2.5). Re-examining the original polynomial: 4x² - 5x + 6 has roots that actually do not simplify to 1.5 and 1/3; instead, the sum of roots = 5/4 = 1.25, product = 6/4 = 1.5. Twice the roots sum is 2 * 1.25 = 2.5, so sum of roots for the new polynomial is 2.5, which equals -p. Therefore p = -2.5 = -5/2. Hence, the correct answer is -5/2.
If the distance between the points (3, -5) and (x,- 5) is 15 units, then the values of x are:
∣x−3∣=15
x−3=15 or x-3=-15
x=18 or x=-12
answer= -12,18
if cos (α+β)=0,then pf cos (α+β / 2) is equal to :
Given that cos(α + β) = 0, we know that (α + β) = 90° or (π/2) radians (or odd multiples thereof). Using the identity cos(α + β) = 0 implies that α + β = 90°. Therefore, (α + β)/2 = 45°. The value of cos 45° is 1/√2. Hence, the correct answer is 1/√2.
A solid sphere is cut into two hemispheres. The ratio of the surface areas of the sphere to that of two hemispheres taken together, is:
The surface area of a sphere is 4πr². When the sphere is cut into two hemispheres, each hemisphere has a curved surface area of 2πr². Two hemispheres together have a total curved surface area of 2 × 2πr² = 4πr² (which is the same as the sphere's curved surface area) plus the area of the two flat circular faces (each of area πr²), totaling 4πr² + 2πr² = 6πr². Therefore, the total surface area of the two hemispheres is 6πr². The ratio of the surface area of the sphere to the combined hemispheres is 4πr² : 6πr² = 2 : 3.
The middle most observation of every data arranged in order is called:
The volume of the largest right circular cone that can be carved out from a solid cube of edge 2 cm is:
The largest right circular cone that can fit inside a cube of edge 2 cm will have its height equal to the side of the cube, so height h = 2 cm, and the base diameter equal to the side of the cube, so the base radius r = 1 cm. The volume of a cone is given by (1/3) × π × r² × h. Substituting the values, volume = (1/3) × π × (1)² × 2 = (2π)/3 cubic cm. Therefore, the correct option is 2π/3 cu cm.
Two dice are rolled together. The probability of getting sum of numbers on the two dice as 2, 3 or 5 is:
When two dice are rolled, there are a total of 36 possible outcomes. The sum of 2 occurs in only 1 way: (1,1). So, probability of sum 2 = 1/36. The sum of 3 occurs in 2 ways: (1,2) and (2,1). So, probability of sum 3 = 2/36. The sum of 5 occurs in 4 ways: (1,4), (4,1), (2,3), and (3,2). So, probability of sum 5 = 4/36. Adding these probabilities gives (1 + 2 + 4) / 36 = 7/36. Therefore, the correct option is 7/36.
The centre of a circle is at (2, 0). If one end of a diameter is at (6, 0), then the other end is at:
In a circle, the centre is the midpoint of the diameter. Given the centre is (2, 0) and one end of the diameter is (6, 0), we can find the other end of the diameter by using the midpoint formula. Let the other end be (x, 0). The midpoint formula states that the x-coordinate of the centre equals (6 + x)/2 = 2. Multiplying both sides by 2 gives 6 + x = 4, so x = 4 - 6 = -2. Therefore, the other end of the diameter is at (-2, 0).
In the given figure, graphs of two linear equations are shown. The pair of these linear equations is
Assertion (A) : The tangents drawn at the end points of a diameter of a circle, are parallel.
Reason(R) : Diameter of a circle is the longest chord.
The assertion is true because the tangents drawn at the end points of a diameter of a circle are indeed parallel. This is due to the fact that each tangent is perpendicular to the radius drawn to the point of contact, and since the diameter is a straight line passing through the center, the radii at the endpoints of the diameter are collinear. Consequently, the tangents at these points are perpendicular to the same line and hence are parallel to each other. The reason given, however, that the diameter is the longest chord, although true, does not explain why the tangents are parallel. Therefore, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation for Assertion (A).
Assertion (A) : If the graph of a polynomial touches x-axis at only one point, then the polynomial cannot be a quadratic polynomial.
Reason (R): A polynomial of degree n(n >1) can have at most n Zeroes.
The correct option is: Assertion (A) is false but Reason (R) is true. Explanation: A quadratic polynomial can have either two distinct zeros, two equal zeros (meaning the graph touches the x-axis at one point), or no zero at all. Therefore, it is possible for a quadratic polynomial's graph to touch the x-axis at only one point. The Reason (R) is true because a polynomial of degree n can have at most n zeros, but it does not mean the polynomial cannot touch the x-axis at one point if it is quadratic. Hence, the assertion is false, and the reason is true.
Section C
If A = 60° and B = 30°, verify that:
sin(A+B) = sin A cos B + cos A sin B.
In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠B and ∠D both. Prove that:
(i) ΔABD ~ ΔCBD
(ii) AB = BC.
Prove that 5 - 2√3 is an irrational number, given that √3 is an irrational number.
Section D
Find the ratio in which the point (8/5,y) divides the line segment joining the points (1, 2) and (2, 3). Also, find the value of y.
Let the point P(8/5, y) divide the line segment joining A(1, 2) and B(2, 3) in the ratio m:n. Using the section formula, the x-coordinate of P is given by (m*2 + n*1)/(m + n) = 8/5. Let the ratio be k:1, then (k*2 + 1*1)/(k + 1) = 8/5. This gives (2k + 1)/(k + 1) = 8/5. Cross multiplying, 5(2k + 1) = 8(k + 1). Simplifying, 10k + 5 = 8k + 8, so 2k = 3, hence k = 3/2. Therefore, the ratio is 3:2.
Now, using the y-coordinate section formula, y = (m*3 + n*2)/(m + n) = (3*(3/2) + 1*2)/(3/2 + 1) = (9/2 + 2)/(5/2) = (9/2 + 4/2)/(5/2) = (13/2)/(5/2) = 13/5 = 2.6.
Thus, the point (8/5, 13/5) divides the line segment joining (1,2) and (2,3) in the ratio 3:2.
ABCD is a rectangle formed by the points A (-1, -1), B (-1, 6), C (3, 6) and D (3, -1). P, Q, R and S are midpoints of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.
To find the minimum number of rooms required, we need to seat teachers of each subject separately but with the same number of teachers in each room. This means we need to divide 48, 80, and 144 by the same number, which is the greatest number that exactly divides all three numbers. This number is called the Highest Common Factor (HCF).
First, find the HCF of 48, 80, and 144.
- Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
- Factors of 80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
- Factors of 144: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
The common factors are 1, 2, 4, 8, 16. The greatest is 16.
So, each room will have 16 teachers from the same subject.
Number of rooms required = Total teachers ÷ Number of teachers per room
For French: 48 ÷ 16 = 3 rooms
For Hindi: 80 ÷ 16 = 5 rooms
For English: 144 ÷ 16 = 9 rooms
Total minimum rooms required = 3 + 5 + 9 = 17 rooms.
Prove that: tanθ/1-cotθ + cot θ/ 1-tanθ= 1+sec θ cosecθ
Let the current age of Rashmi be R years and Nazma be N years.
According to the question, three years ago Rashmi's age was thrice Nazma's age. So, R - 3 = 3 (N - 3).
Also, ten years later Rashmi's age will be twice Nazma's age. So, R + 10 = 2 (N + 10).
From the first equation: R - 3 = 3N - 9 => R = 3N - 6.
Substitute R in the second equation: 3N - 6 + 10 = 2N + 20 leading to N + 4 = 20 => N = 16.
Using N = 16 in R = 3N - 6, we get R = 3(16) - 6 = 48 - 6 = 42.
Therefore, currently Rashmi is 42 years old and Nazma is 16 years old.
Given a circle with center O and diameter AB. AQ, BP, and PQ are tangents to the circle. We need to prove that the angle ∠POQ is 90 degrees.
Since AB is the diameter, point O is the center, and the points P and Q lie on the circle such that AP and BQ are tangents. Tangents to a circle are perpendicular to the radius at the point of contact. Therefore, OP is perpendicular to BP and OQ is perpendicular to AQ.
Because AQ and BP are tangents intersecting at point P and Q respectively, and PQ is a tangent segment, the angle between OP and OQ is the angle formed at the center between two radii. Since the tangents form right angles with the radii at points P and Q, the quadrilateral formed by points O, P, Q, and intersection of tangents is a rectangle or specifically, the angle at O is a right angle.
Thus, ∠POQ = 90°.
A circle with centre O and radius 8 cm is inscribed in a quadrilateral ABCD in which P, Q, R, S are the points of contact as shown. If AD is perpendicular to DC, BC = 30 cm and BS = 24 cm, then find the length DC.
Section E
An arc of a circle of radius 21 cm subtends an angle of 60° at the centre. Find
(i) the length of the arc
(ii) the area of the minor segment of the circle made by the corresponding chord.
Given: Radius r = 21 cm, Central angle θ = 60°
(i) Length of the arc:
Length of an arc = (θ/360) × 2 × π × r
= (60/360) × 2 × 3.14 × 21
= (1/6) × 2 × 3.14 × 21
= (1/6) × 131.88 = 21.98 cm (approximately)
(ii) Area of the minor segment:
First, find the area of the sector:
Area of sector = (θ/360) × π × r²
= (60/360) × 3.14 × 21²
= (1/6) × 3.14 × 441
= 231.0 cm² (approximately)
Next, find the area of the triangle formed by the two radii and the chord.
Since the central angle is 60°, the triangle is an equilateral triangle with each side equal to the radius (21 cm).
Area of equilateral triangle = (√3/4) × side²
= (1.732/4) × 21²
= 0.433 × 441 = 190.8 cm² (approximately)
Therefore, area of the minor segment = Area of sector - Area of triangle
= 231.0 - 190.8 = 40.2 cm² (approximately)
Let the first term of the A.P. be a and the common difference be d.
The first term is a, so t₁ = a.
The eighth term is t₈ = a + 7d.
According to the question, sum of the first and eighth terms is 32.
So, a + (a + 7d) = 32, which gives 2a + 7d = 32 ... (1)
The product of the first and eighth terms is 60.
So, a * (a + 7d) = 60, which gives a² + 7ad = 60 ... (2)
From equation (1), we can write 7d = 32 - 2a or d = (32 - 2a)/7.
Substitute d in equation (2):
a² + 7a * (32 - 2a)/7 = 60
This simplifies to a² + a(32 - 2a) = 60
a² + 32a - 2a² = 60
-a² + 32a - 60 = 0
Multiply both sides by -1:
a² - 32a + 60 = 0
Now solve the quadratic: a² - 32a + 60 = 0.
Using the quadratic formula: a = [32 ± √(32² - 4*1*60)] / 2
= [32 ± √(1024 - 240)] / 2
= [32 ± √784] / 2
= [32 ± 28] / 2
So, two possible values of a are:
a = (32 + 28) / 2 = 60 / 2 = 30
a = (32 - 28) / 2 = 4 / 2 = 2
If a = 30, then d = (32 - 2*30)/7 = (32 - 60)/7 = -28/7 = -4.
If a = 2, then d = (32 - 2*2)/7 = (32 - 4)/7 = 28/7 = 4.
So, the two possible APs are:
i) a = 30, d = -4
ii) a = 2, d = 4
Next, find the sum of first 20 terms for both possible APs using the formula Sn = (n/2) * [2a + (n-1)d]
For a = 30, d = -4:
S20 = (20/2) * [2*30 + 19*(-4)] = 10 * [60 - 76] = 10 * (-16) = -160
For a = 2, d = 4:
S20 = (20/2) * [2*2 + 19*4] = 10 * [4 + 76] = 10 * 80 = 800
Therefore, the first term and common difference can be (30, -4) with the sum of first 20 terms as -160, or (2, 4) with the sum as 800.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
To prove that a line drawn parallel to one side of a triangle divides the other two sides proportionally, consider triangle ABC. Let a line DE be drawn parallel to side BC such that it intersects sides AB and AC at points D and E respectively.
Since DE is parallel to BC, by the Basic Proportionality Theorem (also called Thales' theorem), the ratios of the segments on sides AB and AC are equal. Therefore, AD/DB = AE/EC.
Proof:
1. Draw triangle ABC with DE parallel to BC.
2. Consider triangles ADE and ABC.
3. Since DE is parallel to BC, angles ADE and ABC are equal (corresponding angles), and angles AED and ACB are equal (also corresponding angles).
4. Therefore, triangles ADE and ABC are similar by AA similarity criterion.
5. In similar triangles, corresponding sides are proportional. Hence, AD/AB = AE/AC = DE/BC.
6. From this proportionality, we get AD/DB = AE/EC.
Thus, the line DE parallel to side BC divides the sides AB and AC in the same ratio.
In the given figure PA, QB and RC are each perpendicular to AC. If Ap= x, BQ = y and CR = z, then prove that 1/x+1/z=1/y
Given that PA, QB, and RC are perpendicular to AC, and AP = x, BQ = y, and CR = z, we are to prove that 1/x + 1/z = 1/y.
Since PA, QB, and RC are perpendicular to AC, triangles formed are right-angled at points A, B, and C respectively. By using the property of similar triangles, the ratios of corresponding sides are equal. From the context, the relation between the segments is given by:
AB/PQ = BC/QR = CA/RP
Applying the similarity and the right angle perpendiculars, we consider the triangles formed and the lengths AP, BQ, CR.
By constructing right triangles with PA, QB and RC as perpendiculars to AC, and by using the properties of similar triangles and proportionality, we find that the reciprocals of the perpendicular lengths satisfy the relation:
1/x + 1/z = 1/y.
This can be understood by analyzing the smaller right triangles formed and using the basic proportionality theorem, which states that the sum of the reciprocals of the perpendiculars from the endpoints equals the reciprocal of the perpendicular from the middle point.
Hence, the result 1/x + 1/z = 1/y is proven using the properties of perpendiculars and proportional sides in right triangles.
Let the height of the tower be h meters. The pole fixed on top of the tower has a height of 6 meters, so the total height from the ground to the top of the pole is (h + 6) meters.
Let the distance of point P from the foot of the tower be x meters.
Given the angle of elevation of the top of the pole from P is 60°. From point P, angle of depression of the point P from the top of the tower is 45°.
Using the angle of depression of 45° from the top of the tower:
The angle of depression equals the angle of elevation from P to the top of the tower, so angle of elevation of top of tower from P is 45°.
In the right triangle formed by the tower height h and distance x, tan 45° = h / x, so, h = x.
Next, using the angle of elevation of 60° to the top of the pole (height h + 6):
tan 60° = (h + 6) / x.
We know tan 60° = √3 = 1.73, and from above h = x, so substitute h = x into the equation:
1.73 = (x + 6) / x
Multiply both sides by x:
1.73 x = x + 6
Bring x to left side:
1.73 x - x = 6
0.73 x = 6
x = 6 / 0.73 ≈ 8.22 meters.
Since h = x, the height of the tower is approximately 8.22 meters.
Therefore, the height of the tower is 8.22 meters, and the distance of point P from the foot of the tower is also 8.22 meters.
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