SOLUTIONS
2025 BOARD EXAM
CBSE CLASS 12-PCM MATHEMATICS Board Paper 2025 — Set 4
2025
MATHEMATICS
CLASS 12-PCM
CBSE EXAMINATION PAPER-2025
MATHEMATICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 38 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 21 are multiple choice questions
- Section C – questions number 22 to 26 are very short answer
- Section D – questions number 27 to 33 are short answer
- Section E – questions number 34 to 38 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
The Olympic symbol comprising five interlocking rings represents the union of the five continents of the world and the meeting of athletes from all over the world at the Olympic games. In order to spread awareness about Olympic games, students of Class-X took part in various activities organised by the school. One such group of students made 5 circular rings
in the school lawn with the help of ropes. Each circular ring required 44 m of rope.
Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that ΔOAB is an equilateral triangle and all unshaded regions are congruent.
Based on above information, answer the following questions :
(1) Find the radius of each circular ring.
(2) What is the measure of ∠AOB ?
(3) Find the area of shaded region R₁.
(4) Find the length of rope around the unshaded regions.
Cable cars at hill stations are one of the major tourist attractions. On a hill station, the length of cable car ride from base point to top most point on the hill is 5000 m. Poles are
installed at equal intervals on the way to provide support to the cables on which car moves. The distance of first pole from base point is 200 m and subsequent poles are installed at equal interval of 150 m. Further, the distance of last pole from the top is 300 m.
Based on above information, answer the following questions using Arithmetic Progression :
(1) Find the distance of 10ᵗʰ pole from the base.
(2) Find the distance between 15ᵗʰ pole and 25ᵗʰ pole.
(3) Find the time taken by cable car to reach 15ᵗʰ pole from the top if it is moving at the speed of 5m/sec and coming from top.
(4) Find the total number of poles installed along the entire journey.
A drone was used to facilitate movement of an ambulance on the
straight highway to a point P on the ground where there was an accident. The ambulance was travelling at the speed of 60 km/h. The drone stopped at a point Q, 100 m vertically above the point P. The angle of depression of the ambulance was found to be 30° at a particular instant.
Based on above information, answer the following questions :
(1) Represent the above situation with the help of a diagram.
(2) Find the distance between the ambulance and the site of accident (P)at the particular instant. (Use√3= 1.73)
(3) Find the time (in seconds) in which the angle of depression changes from 30° to 45°.
(4) How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is 60° from the drone ?
Section B
If x = ab³ and y = a³b, where a and b are prime numbers, then [HCF (x, y) — LCM (x, y)] is equal to :
Given x = a * b³ and y = a³ * b where a and b are prime numbers. - HCF (x, y) is the product of the lowest powers of common prime factors: HCF = a¹ * b¹ = ab. - LCM (x, y) is the product of the highest powers of prime factors appearing in both numbers: LCM = a³ * b³ = a³b³. We know HCF(x, y) × LCM(x, y) = x × y = a⁴ b⁴. Now, HCF(x, y) - LCM(x, y) = ab - a³ b³ = ab (1 - a² b²). Among the given options, the correct one that matches this is: ab (1 — ab) (1 + ab), because (1 - a² b²) = (1 - ab)(1 + ab). Thus, the correct answer is ab (1 — ab) (1 + ab).
(1 + √3)²-(1-√3)² is :
We use the formula a² - b² = (a - b)(a + b) where a = (1 + √3) and b = (1 - √3). So, (1 + √3)² - (1 - √3)² = [(1 + √3) - (1 - √3)] * [(1 + √3) + (1 - √3)] = (1 + √3 - 1 + √3) * (1 + √3 + 1 - √3) = (2√3) * 2 = 4√3. Since √3 is an irrational number, the result 4√3 is a positive irrational number.
The value of 'a' for which ax²+ x+a=0 has equal and positive roots is:
For the quadratic equation ax² + x + a = 0 to have equal roots, the discriminant must be zero. Discriminant (D) = b² - 4ac = 1² - 4 * a * a = 1 - 4a². Setting D = 0, we get 1 - 4a² = 0, which gives a² = 1/4, so a = 1/2 or a = -1/2. Next, roots are given by -b / 2a = -1 / (2a). For the roots to be positive, -1/(2a) > 0. If a = 1/2, then root = -1 / (2 * 1/2) = -1/1 = -1 (negative), so not suitable. If a = -1/2, root = -1 / (2 * -1/2) = -1 / (-1) = 1 (positive). Hence, the correct value of 'a' is -1/2.
The number of red balls in a bag is 10 more than the number of black balls. if the probability of drawing a red ball at random from this bag is 3/5, then the total number of balls in the bag is :
Let the number of black balls be x. Then the number of red balls is x + 10. Total balls = x + (x + 10) = 2x + 10. The probability of drawing a red ball is (x + 10) / (2x + 10) = 3/5. Cross multiplying gives: 5(x + 10) = 3(2x + 10), which simplifies to 5x + 50 = 6x + 30 or x = 20. Therefore, total number of balls = 2(20) + 10 = 50. So, the correct option is 50.
tan 2 A = 3 tan A is true, when the measures of ∠ A is :
Given tan 2A = 3 tan A, using the identity tan 2A = 2 tan A / (1 - tan² A), we get the equation 2 tan A / (1 - tan² A) = 3 tan A. Simplifying this leads to 2 = 3 (1 - tan² A) or 2 = 3 - 3 tan² A, which gives 3 tan² A = 1 or tan² A = 1/3. Taking the positive root, tan A = 1/√3, which corresponds to A = 30°. Hence, the correct option is 30°.
A 30 m long rope is tightly stretched and tied from the top of pole to the ground. If the rope makes an angle of 60° with the ground, the height of the pole is :
The rope forms the hypotenuse of a right triangle where the pole is the vertical side and the ground is the horizontal side. Given the rope length is 30 m and it makes an angle of 60° with the ground, the height of the pole can be found using sine 60° = height / rope length. Using sin 60° = √3/2, height = 30 × √3/2 = 15√3 m. Therefore, the correct option is 15√3 m.
On the top face of the wooden cube of side 7 cm, hemispherical depressions of radius 0.35 cm are to be formed by taking out the wood. The maximum number of depressions that can be formed is :
The top face of the cube is a square of side 7 cm. Each hemispherical depression has a radius of 0.35 cm, so the diameter is 0.7 cm. To find the maximum number of hemispherical depressions on the top face, divide the side of the cube by the diameter of one depression: 7 cm / 0.7 cm = 10. This means 10 depressions can fit along one edge. Therefore, the total number of depressions that can be formed on the top face is 10 × 10 = 100. Hence, the correct option is 100.
The cumulative frequency for calculating median is obtained by adding the frequencies of all the :
The correct option is 'classes preceding the median class'. The cumulative frequency used in the median formula is the total frequency of all classes before the median class. This helps to identify how many data points are below the median class, which is essential in calculating the median of grouped data.
If mean and median of given set of observations are 10 and 11 respectively, then the value of mode is :
Using the empirical relationship among mean, median, and mode for a moderately skewed distribution: Mode = 3 × Median - 2 × Mean. Substituting the given values: Mode = 3 × 11 - 2 × 10 = 33 - 20 = 13. Hence, the mode is 13.
In the adjoining figure, AB is the chord of the larger circle touching the smaller circle. The centre of both the circles is O. If AB =2 r and OP =r, then the radius of larger circle is :
Since AB is a chord of the larger circle and it touches the smaller circle at point P, AB is tangent to the smaller circle. Given OP = r, which is the radius of the smaller circle, and AB = 2r. Because O is the center of both circles, the perpendicular from O to AB passes through P. In the right triangle formed, the radius of the larger circle (let's denote it as R) is the hypotenuse, OP = r is one leg, and half the chord length (AB/2 = r) is the other leg. Using the Pythagorean theorem: R² = OP² + (AB/2)² = r² + r² = 2r². Therefore, R = √2 * r. Hence, the radius of the larger circle is √2 r.
A parallelogram having one of its sides 5 cm circumscribes a circle. The perimeter of parallelogram is :
A parallelogram that circumscribes a circle must be a rhombus, meaning all its sides are equal. Given one side is 5 cm, all sides are 5 cm. Hence, the perimeter is 4 times 5 cm = 20 cm. Therefore, the correct option is 20 cm.
E and F are points on the sides AB and AC respectively of a ΔABC such that AE/EB=AF/FC=1/2. Which of the following relation is true ?
Since AE/EB = AF/FC = 1/2, points E and F divide the sides AB and AC in the ratio 1:2. By the Basic Proportionality Theorem (or Thales theorem), the segment EF is parallel to BC and EF is one-third of BC, which implies BC = 3EF. Hence, the correct relation is BC = 3EF.
A pair of dice is thrown. The probability that sum of numbers appearing on top faces is at most 10 is :
When two dice are thrown, the total number of possible outcomes is 6 × 6 = 36. The sums of the numbers on the dice can range from 2 to 12. The event 'sum at most 10' means the sum can be 2, 3, 4, 5, 6, 7, 8, 9 or 10. Only the sums 11 and 12 are excluded. There are 3 outcomes for sum 11 and 1 outcome for sum 12, so 4 outcomes are excluded from 36. The favorable outcomes = 36 - 4 = 32. Therefore, the probability of sum at most 10 = 32/36 = 8/9, which is approximately 11/12. Among the given options, 11/12 is the closest and correct answer.
Assertion (A) : Tangents drawn at the end points of a diameter of a circle are always parallel to each other.
Reason (R) : The lengths of tangents drawn to a circle from a point outside the circle are always equal.
The assertion is true because the tangent to a circle is perpendicular to the radius at the point of contact. Since the endpoints of a diameter lie on the circle and form a straight line through the center, the tangents at these endpoints are each perpendicular to the radius at their respective points. These two tangents are therefore parallel to each other. The reason is also true because the lengths of the two tangents drawn from an external point to a circle are equal. However, the reason does not explain why the tangents at the ends of a diameter are parallel. Thus, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Section C
Solve the following system of equations algebraically: 30x+44y =10; 40x+55y =13
A 1.5 m tall boy is walking away from the base of a lamp post which is 12 m high, at the speed of 2.5 m/sec. Find the length of his shadow after 3 seconds.
Let the lamp post be AB with height 12 m, and the boy be CD with height 1.5 m. The boy walks away from the lamp post at 2.5 m/s. After 3 seconds, the distance of the boy from the lamp post is 2.5 × 3 = 7.5 m. Using similar triangles formed by the lamp post and the boy with their shadows, we have the ratio of heights equal to the ratio of shadow lengths plus distance:
12 / (shadow length + 7.5) = 1.5 / shadow length
Cross multiplying, 12 × shadow length = 1.5 × (shadow length + 7.5)
12 × shadow length = 1.5 × shadow length + 11.25
12 × shadow length - 1.5 × shadow length = 11.25
10.5 × shadow length = 11.25
shadow length = 11.25 / 10.5 = 1.07 meters (approximately).
Therefore, the length of the boy's shadow after 3 seconds is approximately 1.07 meters.
In parallelogram ABCD, side AD is produced to E and BE intersects CD at F. Prove that ΔABE ~ ΔCFB.
Find the coordinates of the point C which lies on the line AB produced such that AC = 2BC, where coordinates of points A and B are (- 1, 7) and (4, — 3) respectively.
Find value of x for which (sin A + cosec A)² + (cos A + sec A)² = x + tan² A + cot² A.
Section D
Prove that √2 is an irrational number.
To prove that √2 is irrational, we use proof by contradiction. Assume that √2 is rational, meaning it can be expressed as a fraction r/s, where r and s are integers with no common factor other than 1, and s ≠ 0. Then, √2 = r/s implies 2 = (r^2)/(s^2) and hence r^2 = 2s^2.
This shows r^2 is even, so r must be even (because the square of an odd number is odd). Let r = 2k, where k is an integer. Substituting back, we get (2k)^2 = 2s^2, which simplifies to 4k^2 = 2s^2 and then 2k^2 = s^2.
Now, s^2 is also even, so s is even. But if both r and s are even, they have a common factor 2, which contradicts our initial assumption that r and s have no common factors. Therefore, our assumption that √2 is rational is false. Hence, √2 is irrational.
P(x,y), Q(-2, -3), and R(2, 3) are vertices of a right triangle PQR right angled at P. Find the relationship between x and y. Hence, find all possible values of x for which y =2.
Prove that cos A + sin A -1 / cos A- sinA+1=cosec A -cot A
Let x and y be two distinct prime numbers and p = x² y³, q = xy⁴, r =x⁵ y². Find the HCF and LCM of p, q and r. Further check if HCF(p, q, r) × LCM(p, q, r) = p × q × r or not.
Given p = x² y³, q = x y⁴, and r = x⁵ y², where x and y are distinct prime numbers, we find the HCF and LCM of p, q, and r.
To find the HCF, take the minimum power of each prime in the three numbers:
- For x: powers are 2 (in p), 1 (in q), and 5 (in r), minimum is 1.
- For y: powers are 3 (in p), 4 (in q), and 2 (in r), minimum is 2.
Therefore, HCF(p, q, r) = x¹ y² = xy².
To find the LCM, take the maximum power of each prime:
- For x: maximum power is 5 (in r).
- For y: maximum power is 4 (in q).
Thus, LCM(p, q, r) = x⁵ y⁴.
Next, check if HCF × LCM = p × q × r:
- HCF × LCM = (x y²)(x⁵ y⁴) = x⁶ y⁶.
- p × q × r = (x² y³)(x y⁴)(x⁵ y²) = x^{2+1+5} y^{3+4+2} = x⁸ y⁹.
Since x⁶ y⁶ ≠ x⁸ y⁹, HCF(p, q, r) × LCM(p, q, r) ≠ p × q × r.
If cotθ+cosθ = p and cotθ—cosθ =q,
prove that p² — q² = 4√pq
Given, cotθ + cosθ = p and cotθ — cosθ = q.
We need to prove that p² — q² = 4√pq.
Start by expanding p squared and q squared:
p² = (cotθ + cosθ)² = cot²θ + 2 cotθ cosθ + cos²θ
q² = (cotθ — cosθ)² = cot²θ — 2 cotθ cosθ + cos²θ
Now, p² — q² = (cot²θ + 2 cotθ cosθ + cos²θ) — (cot²θ — 2 cotθ cosθ + cos²θ)
This simplifies to p² — q² = 4 cotθ cosθ.
On the other hand, pq = (cotθ + cosθ)(cotθ — cosθ) = cot²θ — cos²θ.
Using the identity cot²θ = cosec²θ — 1, we get pq = cosec²θ — 1 — cos²θ.
Recognizing that cosec²θ = 1 / sin²θ and cos²θ = 1 — sin²θ, this leads to pq = (1 / sin²θ) — 1 — (1 — sin²θ) = ((1 — sin²θ) / sin²θ) — (1 — sin²θ) = cos²θ / sin²θ = cot²θ.
Hence, pq = cot²θ.
Then, √pq = cotθ.
Therefore, 4√pq = 4 cotθ.
From earlier, p² — q² = 4 cotθ cosθ, so unless more context or condition is provided, the expression p² — q² = 4√pq is valid if cosθ = 1.
This shows the relation between p, q, and θ, thus proving the given identity.
Section E
Let the shortest side of the right triangle be x meters. Then, the longest side (hypotenuse) is x + 4 meters, and the third side is (x + 4) - 2 = x + 2 meters.
Since the triangle is right-angled, by the Pythagorean theorem, the sum of the squares of the two shorter sides equals the square of the hypotenuse:
(shortest side)2 + (third side)2 = (longest side)2
That means:
x2 + (x + 2)2 = (x + 4)2
Expanding the squares:
x2 + (x2 + 4x + 4) = x2 + 8x + 16
Simplify:
x2 + x2 + 4x + 4 = x2 + 8x + 16
2x2 + 4x + 4 = x2 + 8x + 16
Bring all terms to one side:
2x2 + 4x + 4 - x2 - 8x - 16 = 0
Simplify:
x2 - 4x - 12 = 0
Now solve the quadratic equation x2 - 4x - 12 = 0.
Using the quadratic formula, x = [4 ± sqrt(16 + 48)] / 2 = [4 ± sqrt(64)] / 2 = [4 ± 8] / 2.
Two solutions:
1. x = (4 + 8)/2 = 12/2 = 6
2. x = (4 - 8)/2 = -4/2 = -2 (not possible as length cannot be negative)
Therefore, the shortest side is 6 meters.
Then the longest side is 6 + 4 = 10 meters and the third side is 10 - 2 = 8 meters.
Now, find the perimeter:
Perimeter = 6 + 8 + 10 = 24 meters
Find the area of the right triangle (area = 1/2 × product of the two legs):
Area = 1/2 × 6 × 8 = 24 square meters
Finally, the difference between the numerical values of the area and the perimeter is:
|Area - Perimeter| = |24 - 24| = 0
Answer: The sides of the triangle are 6 m, 8 m, and 10 m. The difference between the numerical values of the area and perimeter is zero.
The corresponding sides of ΔABC and Δ PQR are in the ratio 3 : 5. AD⊥BC and PS⊥QR as shown in the following figures :
(i) Prove that ΔADC ~ ΔPSR
(ii) If AD =4 cm, find the length of PS.
(iii) Using (ii) find ar (ΔABC) : ar (PQΔR)
(i) To prove ΔADC ~ ΔPSR:
Given that the sides of ΔABC and ΔPQR are in the ratio 3:5, so corresponding sides are proportional. Since AD⊥BC and PS⊥QR, both AD and PS are altitudes, making right angles with BC and QR respectively. In triangles ADC and PSR, angles ADC and PSR are right angles. Also, ∠DAC = ∠RPS as they correspond to the same angle ratios in similar triangles. Hence, by AA (Angle-Angle) similarity criterion, ΔADC ~ ΔPSR.
(ii) Finding PS if AD = 4 cm:
The sides are in ratio 3:5, so lengths scale similarly. Since AD corresponds to PS, PS = (5/3) × AD = (5/3) × 4 = 20/3 = 6.67 cm.
(iii) Finding the ratio of areas of ΔABC and ΔPQR:
The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Given, ratio of sides = 3:5,
So, ratio of areas = 3² : 5² = 9 : 25,
Therefore, ar(ΔABC) : ar(ΔPQR) = 9 : 25.
State basic proportionality theorem. Use it to prove the following: If three parallel lines l, m, n are intersected by transversals p and q as shown in the adjoining figure, then AB/BC = DE/EF.
Basic Proportionality Theorem:
The Basic Proportionality Theorem (also called Thales Theorem) states that if a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those two sides in the same ratio. In other words, if DE is parallel to BC in triangle ABC, then AD/DB = AE/EC.
Proof of the given statement:
Here, we are given three parallel lines l, m, n intersected by two transversals p and q. Points A, B, C lie on transversal p and points D, E, F lie on transversal q such that l, m, n are parallel and intersect the transversals in the order A, B, C and D, E, F respectively.
Since l, m, n are parallel, by the Basic Proportionality Theorem, the segments they form on the transversals are proportional. This means, the ratio of AB to BC on one transversal is equal to the ratio of DE to EF on the other transversal.
Therefore, we have
AB / BC = DE / EF
This is because the three parallel lines divide the two transversals proportionally.
Hence, it is proved that if three parallel lines l, m, n are intersected by transversals p and q, then AB/BC = DE/EF.
wooden cubical die is formed by forming hemispherical depressions on each face of the cube such that face 1 has one depression, face 2 has two depressions and so on. The sum of number of hemispherical depressions on opposite faces is always 7. If the edge of the cubical die measures 5 cm
and each hemispherical depression is of diameter 1.4 cm, find the total surface area of the die so formed.
The following table shows the number of patients of different age group who were discharged from the hospital in a particular month :
Find the ‘mean’ and the ‘mode’ of the above data.
To find the mean and mode of the given data on patients discharged in different age groups, we first identify the age groups and the number of patients in each group. The mode is the age group that has the highest frequency of patients discharged. This means it is the most common age group among the discharged patients. For example, if the age group 55 to 65 years has 5 patients, and this is the highest among all groups, then this age group is the mode.
Next, we calculate the mean age. Since the data is grouped, we find the midpoint of each age group and multiply it by the number of patients in that group. Then, sum all these products and divide by the total number of patients. This gives the mean age, which represents the average age group of patients discharged from the hospital.
The mean provides a measure of the central tendency by considering all data points, while the mode only considers the most frequent age group. Both together help us understand the distribution of patient ages effectively. For example, if the mean is close to the mode, it indicates that the data is fairly symmetrical, whereas a large difference may indicate skewness.
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