SOLUTIONS
2025 BOARD EXAM
CBSE Class 12 PCM Mathematics Board Paper 2025 — Set 1 Solutions
2025
Mathematics
Class 12 PCM
Practice version of the official CBSE paper
Official CBSE paper80 marks
This practice set88 marksall “OR” options included
Optional (“OR”) questions are added here as separate questions, so the maximum marks are higher than in the official CBSE paper.
Why are the marks different?
Some questions in the official CBSE paper offer a choice (“OR”), and students attempt only one option. On Prepzy’s practice PYQs, we treat every option as a separate question so you can practise all of them. As a result, the total number of questions and the maximum marks shown here may be higher than in the original CBSE paper. For the official question paper, marking scheme and total marks, please refer to CBSE’s official website.
CBSE EXAMINATION PAPER-2025
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
A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.
Based on the information given above, answer the following questions :
(1) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed.
The nth term of an arithmetic progression is given by: a_n = a + (n - 1)d.
Therefore,
4th term = 300 + (4 - 1) × 50 = 300 + 150 = 450 metres,
5th term = 300 + (5 - 1) × 50 = 300 + 200 = 500 metres,
6th term = 300 + (6 - 1) × 50 = 300 + 250 = 550 metres.
(2) Determine the distance of the 8ᵗʰ round.
(3) Find the total distance run after completing all 10 rounds.
The distances for 10 rounds are: 300, 350, 400, 450, 500, 550, 600, 650, 700, and 750 metres.
To find the total distance, add these distances:
Total distance = 300 + 350 + 400 + 450 + 500 + 550 + 600 + 650 + 700 + 750
This is an arithmetic series with first term 300, common difference 50, and number of terms 10.
Sum of arithmetic series = (number of terms / 2) × (first term + last term)
= (10 / 2) × (300 + 750)
= 5 × 1050
= 5250 metres.
Therefore, the total distance run after completing all 10 rounds is 5250 metres.
(4) If a runner completes only the first 6 rounds, what is the total distance run by the runner?
A brooch is a decorative piece often worn on clothing like jackets, blouses or dresses to add elegance. Made from precious metals and decorated with gemstones, brooches come in many shapes and designs. One such brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure.
Based on the above given information, answer the following questions :
(1) Find the central angle of each sector.
[1 Mark](2) Find the length of the arc ACB.
[1 Mark](3) Find the area of each sector of the brooch.
[2 Marks](4) Find the total length of the silver wire used.
[2 Marks]Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be 60°. Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be 45°.
Based on the above given information, answer the following questions :
(1) IF CD is h meter, find the distance BD in term of 'h'
(2) Find distance BC in term of ' h'
(3) Find the height CE of the lighthouse. (Use √3 = 1.73)
[2 Marks](4) Find distance AE, if AC = 100 m.
[2 Marks]Section B
If α and β are the zeroes of polynomial 3x² + 6x + k such that α + β + αβ =-2/3, then the value of k is:
For the quadratic polynomial 3x² + 6x + k, the sum of zeroes α + β = -b/a = -6/3 = -2, and the product αβ = c/a = k/3. According to the given condition, α + β + αβ = -2/3, substituting sum and product we get: -2 + (k/3) = -2/3. Solving for k: k/3 = -2/3 + 2 = 4/3, so k = 4. Hence, the correct value of k is 4.
If x = 1 and y = 2 is a solution of the pair of linear equations 2x - 3y + a = 0 and 2x + 3y - b = 0, then:
For the first equation: 2(1) - 3(2) + a = 0 ⇒ 2 - 6 + a = 0 ⇒ a - 4 = 0 ⇒ a = 4.
For the second equation: 2(1) + 3(2) - b = 0 ⇒ 2 + 6 - b = 0 ⇒ 8 - b = 0 ⇒ b = 8.
Therefore, a = 4 and b = 8, which shows a = 2b. Thus, the correct option is 'a = 2b'. This solution is found by direct substitution from the context of solving linear equations.
The mid-point of the line segment joining the points P(-4, 5) and Q(4, 6) lies on:
The mid-point of a line segment joining two points P(x1, y1) and Q(x2, y2) is calculated as ((x1 + x2)/2, (y1 + y2)/2). Here, P is (-4, 5) and Q is (4, 6). Calculating the midpoint gives ((-4 + 4)/2, (5 + 6)/2) = (0, 11/2) = (0, 5.5). Since the x-coordinate of the midpoint is 0 and the y-coordinate is positive, the midpoint lies on the y-axis. Therefore, the correct option is 'y-axis'.
If θ is an acute angle and 7+ 4 sin θ = 9, then the value of θ is:
The value of tan²θ - (1/cosθ x secθ) is:
We know sec θ is the reciprocal of cos θ, so sec θ = 1/cos θ. Therefore, 1/cos θ × sec θ = 1/cos θ × 1/cos θ = 1/cos² θ, which is equal to sec² θ. Using the Pythagorean identity: tan² θ + 1 = sec² θ, we can rewrite tan² θ - (1/cos θ × sec θ) as tan² θ - sec² θ = tan² θ - (tan² θ + 1) = -1. Hence, the correct value is -1.
If HCF(98, 28) = m and LCM(98, 28) = n, then the value of n -7 m is:
The tangents drawn at the extremities of the diameter of a circle are always:
In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE. Then, the two triangles are:
For two triangles to be similar, their corresponding angles must be equal and their corresponding sides must be proportional. Here, although ∠B = ∠E and ∠C = ∠F, the side AB is equal to 3 times DE, which means the sides are not proportional. Without proportional sides, the triangles cannot be similar. Also, for congruency, all corresponding sides and angles must be equal, which is not the case here. Therefore, the triangles are neither congruent nor similar.
If (−1)ⁿ + (−1)⁸ = 0, then n is:
Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is:
Mode and Mean of a data are 15x and 18x, respectively. Then the median of the data is:
According to the relation between mean, median, and mode in statistics: Mode = 3 × Median - 2 × Mean. Given Mode = 15x and Mean = 18x, substituting gives 15x = 3 × Median - 2 × 18x. Simplifying, 15x = 3 × Median - 36x; thus, 3 × Median = 51x; hence, Median = 17x. Therefore, the median of the data is 17x.
A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is:
There are 52 cards in total, with 26 red cards (13 diamonds and 13 hearts). The face cards in each suit are Jack, Queen, and King, so there are 3 face cards per suit. For red cards, the total number of red face cards is 3 (diamonds) + 3 (hearts) = 6. Therefore, the probability of selecting a red face card = number of favorable outcomes / total number of cards = 6 / 52 = 3 / 26. Hence, the correct option is 3/26.
Which of the following is a rational number between √3 and √5?
Given that √3 ≈ 1.732 and √5 ≈ 2.236, we need to find a rational number between these two values. The options are: 1.4142387954012... (which is approx √2 and irrational), π (irrational), 1.857142 (which is 13/7, a rational number), and 2.326̅ (which is a repeating decimal, hence rational). Among these, 1.857142 lies between 1.732 and 2.236 and is rational. Therefore, the correct answer is 1.857142.
If a sector of a circle has an area of 40π sq. units and a central angle of 72°, the radius of the circle is:
The area of a sector of a circle is given by the formula: (θ / 360) × π × r², where θ is the central angle and r is the radius. Given the sector area is 40π and θ = 72°, we have: (72 / 360) × π × r² = 40π. Simplifying, (1/5) × π × r² = 40π, which gives r² = 200. Therefore, r = √200 = 10√2 units. Hence, the correct option is 10√2 units.
In the given figure, PA is a tangent from an external point P to a circle with centre O. If ∠POB = 115°, then ∠APO is equal to:
Since PA is tangent at point A, the radius OA is perpendicular to the tangent PA, so ∠OAP = 90°. Given ∠POB = 115°, and assuming points A and B lie on the circle such that ∠POB is the angle between the radii OP and OB, then we note that ∠POA = 180° - 115° = 65°, because the angle between radii OP and OA plus the angle between OA and OB is 180°. In triangle OAP, the sum of angles is 180°. We have ∠OAP = 90° and ∠POA = 65°, so ∠APO = 180° - 90° - 65° = 25°. Hence, the correct answer is 25°.
A kite is flying at a height of 150 m from the ground. It is attached to a string inclined at an angle of 30° to the horizontal. The length of the string is:
The height of the kite is 150 m and the string makes an angle of 30° with the horizontal. We can model this as a right-angled triangle where the height is the side opposite to the angle 30°. Using the sine function, sin 30° = height / length of string. Since sin 30° = 1/2, we have 1/2 = 150 / length. Therefore, length of the string = 150 ÷ (1/2) = 300 m. Hence, the correct option is 300 m.
A piece of wire 20 cm long is bent into the form of an arc of a circle of radius 60/π cm. The angle subtended by the arc at the centre of the circle is:
The length of the arc (l) = 20 cm and the radius (r) = 60/π cm. The angle θ in radians subtended by an arc at the centre of a circle is given by θ = l / r. So, θ = 20 / (60/π) = 20 × (π/60) = π/3 radians. Converting radians to degrees: θ = (π/3) × (180/π) = 60°. Therefore, the angle subtended by the arc at the centre is 60 degrees.
Assertion (A) : The probability of selecting a number at random from the numbers 1 to 20 is 1.
Reason (R): For any event E, if P(E) = 1, then E is called a sure event.
The Assertion (A) is false because the probability of selecting a specific number from 1 to 20 is 1/20, not 1. Probability ranges from 0 to 1, and 1 represents a sure event—an event that always happens. The Reason (R) is true: if the probability of an event E is 1, then it is called a sure event. Thus, the correct answer is: Assertion (A) is false, but Reason (R) is true.
Assertion (A) : If we join two hemispheres of same radius along their bases, then we get a sphere.
Reason(R): Total Surface Area of a sphere of radius r is 3πr².
The assertion (A) is true because joining two hemispheres of the same radius along their flat circular bases forms a complete sphere. However, the reason (R) is false because the total surface area of a sphere of radius r is 4πr², not 3πr². Therefore, the correct option is: 'Assertion (A) is true, but Reason (R) is false.'
Section C
If x cos 60° + y cos 0° + sin 30° -cot 45° = 5, then find the value of x + 2y.
Evaluate: tan²60°/ sin²60° + cos² 30°
Find the zeroes of the polynomial p(x) = x² +4/3x-4/3.
The coordinates of the centre of a circle are (2a, a - 7). Find the value(s) of a, if the circle passes through the point (11,-9) and has diameter 10√2 units.
If Δ ABC~ ΔPQR in which AB = 6 cm, BC = 4 cm, AC = 8 cm and PR = 6 cm, then find the length of( PQ + QR).
In the given figure, QR/QS= QT/PR and ∠1 = ∠2, show that ∆PQS~ ∆TQR.
A person is standing at P outside a circular ground at a distance of 26 m from the centre of the ground. He found that his distances from the points A and B on the ground are 10 m (PA and PB are tangents to the circle). Find the radius of the circular ground.
Section D
To Prove: ∠BAC + ∠ACD = 90 degrees.
Proof:
1. Since BCD is tangent to the circle at C, the tangent at point C is perpendicular to the radius OC. So, ∠OCA = 90 degrees.
2. In triangle OAC, ∠OAC + ∠ACO + ∠AOC = 180 degrees.
3. ∠ACO = ∠OCA = 90 degrees (as tangent is perpendicular), thus ∠OAC + ∠AOC = 90 degrees.
4. ∠BAC subtends arc BC, and ∠BOC is the central angle subtending the same arc BC. The relation between them is ∠BAC = 1/2 * ∠BOC.
5. Similarly, ∠ACD is the angle between the tangent BCD at C and chord AC, and the tangent-secant angle theorem says that ∠ACD = ∠ACO = 90 degrees - ∠BAC.
6. Therefore, ∠BAC + ∠ACD = 90 degrees.
Hence proved.
(a) Prove that: tan θ / 1- cot θ + cot θ/1-tanθ= 1+ sec θ cosec θ
Step 1: Express tan θ and cot θ in terms of sin θ and cos θ:
tan θ = sin θ / cos θ, cot θ = cos θ / sin θ.
Step 2: Substitute in the expression:
tan θ / (1 - cot θ) + cot θ / (1 - tan θ) = (sin θ / cos θ) / (1 - cos θ / sin θ) + (cos θ / sin θ) / (1 - sin θ / cos θ).
Simplify the denominators:
1 - cot θ = 1 - (cos θ / sin θ) = (sin θ - cos θ) / sin θ,
1 - tan θ = 1 - (sin θ / cos θ) = (cos θ - sin θ) / cos θ.
Step 3: Rewrite each term:
(sin θ / cos θ) / ((sin θ - cos θ) / sin θ) = (sin θ / cos θ) * (sin θ / (sin θ - cos θ)) = sin² θ / (cos θ (sin θ - cos θ)),
(cos θ / sin θ) / ((cos θ - sin θ) / cos θ) = (cos θ / sin θ) * (cos θ / (cos θ - sin θ)) = cos² θ / (sin θ (cos θ - sin θ)).
Step 4: Note that cos θ - sin θ = - (sin θ - cos θ), so:
Second term = cos² θ / (sin θ (cos θ - sin θ)) = - cos² θ / (sin θ (sin θ - cos θ)).
Step 5: Add the two terms:
sin² θ / (cos θ (sin θ - cos θ)) - cos² θ / (sin θ (sin θ - cos θ)) = [sin² θ / cos θ - cos² θ / sin θ] / (sin θ - cos θ).
Step 6: Find common denominator for numerator:
(sin² θ * sin θ - cos² θ * cos θ) / (sin θ cos θ) = (sin³ θ - cos³ θ) / (sin θ cos θ).
Step 7: Use the identity a³ - b³ = (a - b)(a² + ab + b²):
sin³ θ - cos³ θ = (sin θ - cos θ)(sin² θ + sin θ cos θ + cos² θ).
Since sin² θ + cos² θ = 1, this becomes (sin θ - cos θ)(1 + sin θ cos θ).
Step 8: Substitute back:
[ (sin θ - cos θ)(1 + sin θ cos θ) ] / (sin θ cos θ (sin θ - cos θ)) = (1 + sin θ cos θ) / (sin θ cos θ).
Step 9: Simplify and write 1 / sin θ cos θ as sec θ cosec θ and sin θ cos θ as is:
This equals 1 / (sin θ cos θ) + 1 = sec θ * cosec θ + 1.
Hence, tan θ / (1 - cot θ) + cot θ / (1 - tan θ) = 1 + sec θ cosec θ, as required.
Therefore, the identity is proved.
Prove that: sin A + cos A / sin A-cos A + sin A - cos A / sin A + cos A=2/2 sin² A-1
Prove that 1/√5 is an irrational number.
To prove that 1/√5 is an irrational number, we first note that √5 is irrational. This means it cannot be expressed as a ratio of two integers. If we assume 1/√5 is rational, then √5 = 1 / (1/√5) would also be rational (because the reciprocal of a rational number is rational). However, this contradicts the fact that √5 is irrational. Therefore, 1/√5 must also be irrational.
A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains 1408 / 21m³ of air, find the height of the cylindrical part. (Use π = 22/7)
Section E
Let the amount Vijay invested in scheme A be x rupees and in scheme B be y rupees.
Since scheme A offers 8% interest per annum and scheme B offers 9%, the total interest received initially is:
(8% of x) + (9% of y) = ₹1,860.
That is, (8/100) × x + (9/100) × y = 1,860. Or 8x + 9y = 186,000 (multiplying both sides by 100).
When the amounts are interchanged, the amounts invested in schemes A and B become y and x respectively. The interest then is:
(8% of y) + (9% of x) = ₹1,860 + ₹20 = ₹1,880.
So, (8/100) × y + (9/100) × x = 1,880 or 8y + 9x = 188,000.
Now, we have the two equations:
8x + 9y = 186,000
9x + 8y = 188,000.
Multiply the first by 9 and the second by 8 to eliminate y:
72x + 81y = 1,674,000
72x + 64y = 1,504,000.
Subtracting the second from the first:
(72x - 72x) + (81y - 64y) = 1,674,000 - 1,504,000,
17y = 170,000,
y = 10,000.
Putting y = 10,000 in the first equation:
8x + 9 × 10,000 = 186,000,
8x + 90,000 =186,000,
8x = 96,000,
x = 12,000.
Therefore, Vijay invested ₹12,000 in scheme A and ₹10,000 in scheme B.
The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that DF x EF = FB x FA.
Given a parallelogram ABCD, where diagonal BD intersects the line segment AE at point F with E being any point on side BC, we are to prove that DF × EF = FB × FA.
Since ABCD is a parallelogram, opposite sides are equal and parallel. Consider triangles BFD and AFE formed by the diagonal BD and segment AE with intersection at F.
Using properties of intersecting chords (or by considering similar triangles formed by the construction), the product of the segments of one chord equals the product of the segments of the other chord when two chords intersect.
Here, BD and AE intersect at F, so by the chord intersection property, DF × FB = EF × FA.
Rearranging this, we get DF × EF = FB × FA, which is what we needed to prove.
This relation relies on the property that when two chords intersect in a circle or line segments intersect inside a figure, the products of their divided segments are equal. The parallelogram ensures the needed parallelism and segment properties for this to hold.
In Δ ABC, if AD ⊥ BC and AD²= BD x DC then prove that ∠BAC = 90°.
Given that the perimeter of the right triangle is 60 cm and the hypotenuse (the longest side) is 25 cm, we need to find the lengths of the other two sides, which are the legs of the triangle.
Let the lengths of the two legs be x cm and y cm. Since it's a right triangle, by the Pythagorean theorem, we know that x squared plus y squared equals the hypotenuse squared:
x² + y² = 25² = 625.
The perimeter is the sum of all three sides:
x + y + 25 = 60 -> x + y = 35.
From x + y = 35, we can express y as 35 - x. Substituting into the Pythagorean equation:
x² + (35 - x)² = 625
Expanding (35 - x)² gives: 35² - 2*35*x + x² = 1225 - 70x + x².
So the equation becomes:
x² + 1225 - 70x + x² = 625
Which simplifies to:
2x² - 70x + 1225 = 625
Subtract 625 from both sides:
2x² - 70x + 600 = 0
Divide the entire equation by 2:
x² - 35x + 300 = 0
Now, solve the quadratic equation x² - 35x + 300 = 0 by finding factors of 300 that add up to 35. The factors are 20 and 15.
So, (x - 20)(x - 15) = 0
Therefore, x = 20 or x = 15.
Using y = 35 - x, if x = 20, then y = 15; if x = 15, then y = 20.
Hence, the lengths of the other two sides are 15 cm and 20 cm.
Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode.
Step 1: Let the missing frequency be f.
Assume the data values and their frequencies are given as per the table. To find f, use the formula for mean: Mean = (Sum of f_i * x_i) / (Sum of f_i). Given the mean is 18, set up the equation using the sum of frequencies and the sum of f_i * x_i including unknown f.
Step 2: Calculate Sum of f_i * x_i from known values and add f * (corresponding x_i). Similarly calculate Sum of frequencies including f.
Step 3: Using the formula, solve for f.
Step 4: Now, to find the mode, identify the modal class which has the highest frequency including the calculated f.
Step 5: Use the mode formula = l + [(f1 - f0) / (2f1 - f0 - f2)] * h where l = lower class boundary of modal class, f1 = frequency of modal class, f0 = frequency of preceding class, f2 = frequency of succeeding class, h = class width.
Step 6: Calculate and provide the mode value.
Interpretation: The mode represents the most frequent data point, and by comparing it to the mean, we can understand the skewness of the data. If mode > mean, the distribution is negatively skewed; if mode < mean, it is positively skewed; if mode = mean, it is symmetric.
Paper Details
CBSE Board Exam 2025
Class
Class 12 PCM
Subject
Mathematics
Year
2025
Set
Set 1
Official max marks
80
Practice set marks
88
Practice marks include every “OR” option. Official paper and marking scheme: CBSE’s official website.
Other Years — Mathematics