SOLUTIONS
2025 BOARD EXAM
CBSE Class 12 PCM Mathematics Board Paper 2025 — Set 2 Solutions
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 48 questions. All questions are compulsory.
- This question paper is divided into 7 sections.
- Section A – questions number 1 to 3 are case based questions
-
Section B –
questions number
4 to 4
are
assertion (a) : common difference of the ap : 5, 1, 3, 7,... is 4.
reason (r): common difference of the ap : a1, a2, a3 an is obtained
by d = an an 1.
-
Section C –
questions number
5 to 7
are
assertion (a) : common difference of the ap : 5, 1, 3, 7,... is 4
reason (r): common difference of the ap : a₁, a₂, a₃,...., aₙ is obtained by d = aₙ aₙ₋₁
- Section D – questions number 8 to 27 are multiple choice questions
- Section E – questions number 28 to 34 are very short answer
- Section F – questions number 35 to 42 are short answer
- Section G – questions number 43 to 48 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 grand cultural event to show the talent of its students. To accommodate the guests, the school plans to rent chairs and tables from a local supplier. It finds that rent for each chair is ₹50 and for each table is ₹200. The school spends ₹30,000 for renting the chairs and tables. Also, the total number of items (chairs and tables) rented are 300.
If the school 'x' chairs and 'y' tables, answer the following questions:
(1) Find the number of chairs and number of tables rented by the school.
[2 Marks](2) What is maximum number of tables that can be rented in ₹30,000 if no chairs are rented?
[1 Mark](3) Write down the pair of linear equations representing the given information.
[1 Mark]The total number of items rented is 300, so:
x + y = 300
The total rent spent is ₹30,000. Rent for each chair is ₹50 and for each table is ₹200, so:
50x + 200y = 30000
(4) If the school wants to spend a maximum of Rs 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent.
1) Total items rented: x + y = 300
2) Maximum amount to spend: 50x + 200y ≤ 27,000
From the first condition, y = 300 - x.
Substitute this in the second condition: 50x + 200(300 - x) ≤ 27,000
This simplifies to: 50x + 60,000 - 200x ≤ 27,000
-150x + 60,000 ≤ 27,000
-150x ≤ 27,000 - 60,000
-150x ≤ -33,000
Multiply both sides by -1 (reverse inequality): 150x ≥ 33,000
Divide both sides by 150: x ≥ 220
So, the school should rent at least 220 chairs.
Now, y = 300 - x ≤ 300 - 220 = 80
Therefore, the school can rent 220 chairs and 80 tables to spend at most ₹27,000.
Rahul is a lucky charm for his cricket team. He has a jar of cards with
numbers from 10 to 74. Before each match, he draws a card from the jar. If the card bears an even number, the team wins. If the number is even and divisible by 5, they win by a big margin. If the number is an odd number less than 30, they win by a small margin. And if the number is a prime number between 50 and 74, they lose.
Answer the following questions if Rahul draws a card today :
(1) What is the probability that Rahul draws a card with an even number?
[1 Mark](2) What is the probability that Rahul draws a card with an odd number less than 30?
[1 Mark](3) What is the probability that Rahul draws a card with a prime number between 50 and 74?
[2 Marks]Next, we find the prime numbers between 50 and 74. These are 53, 59, 61, 67, 71, 73, giving us 6 prime numbers.
Therefore, the probability that Rahul draws a card with a prime number between 50 and 74 is the number of favorable outcomes divided by total outcomes, which is 6 divided by 65. So, the probability = 6/65.
(4) What is the probability that Rahul draws a card with an even number divisible by 5 ?
A skilled carpenter decided to craft a special rolling pin for the local
baker. He carefully joined three cylindrical pieces of wood - two small
ones on the ends and one larger in the centre to create a perfect tool. The baker loved the rolling pin, as it rolled out the smoothest dough for breads and pastries.
The length of the bigger cylindrical part is 12 cm and diameter is 7 cm
and the length of each smaller cylindrical part is 5 cm and diameter is
2·1 cm.
Based on the above information, answer the following questions :
(1) Find the volume of the bigger cylindrical part.
[1 Mark](2) Find the curved surface area of the bigger cylindrical part.
[1 Mark](3) Find the ratio of the volume of the bigger cylindrical part to the total volume of the two smaller (identical) cylindrical parts.
(4) Find the sum of the curved surface areas of the two identical smaller cylindrical parts.
Section B
Section C
Section D
The probability of drawing an even prime number out of numbers from 1 to 30 is:
Among the numbers from 1 to 30, the only prime numbers are those that have no divisors other than 1 and themselves. The only even prime number is 2 since all other even numbers are divisible by 2 and hence not prime. Therefore, the event 'drawing an even prime number' corresponds to drawing the number 2 alone. There is 1 favorable outcome (number 2) out of 30 possible numbers, so the probability is 1/30.
The quadratic equation whose roots are 7 and 1/7 is:
If the roots of a quadratic equation are α and β, then the quadratic equation can be written as x² - (α + β)x + αβ = 0. Here, the roots are 7 and 1/7, so the sum of the roots is 7 + 1/7 = 50/7, and the product of the roots is 7 × 1/7 = 1. Putting these into the quadratic equation formula gives: x² - (50/7)x + 1 = 0. Multiplying through by 7 to clear the fraction yields 7x² - 50x + 7 = 0. Therefore, the correct option is '7x² - 50x + 7 = 0'.
The least number which is a perfect square and is divisible by each of 16, 20 and 50 is:
Prime factorization:
16 = 2^4
20 = 2^2 × 5
50 = 2 × 5^2
LCM will have the highest powers of each prime:
LCM = 2^4 × 5^2 = 16 × 25 = 400
Now, 400 is divisible by all three numbers, but is it a perfect square? Yes, since 400 = 20^2.
Hence, 400 is the least number divisible by all three, and it is a perfect square. But 400 is not in the options. The options are 2400, 3600, 1200, 100.
Next, consider that the problem asks for the least perfect square divisible by each of these numbers. Since LCM = 400 (which is a perfect square), any multiple of 400 that is a perfect square is also divisible by them. Among the options:
2400 = 2^5 × 3 × 5^2 (not a perfect square)
3600 = 2^4 × 3^2 × 5^2 = (2^2 × 3 × 5)^2 = 60^2 (perfect square)
1200 = 2^4 × 3 × 5^2 (not a perfect square)
100 = 2^2 × 5^2 (not divisible by 16)
Among the options, 3600 is a perfect square (60^2) and divisible by 16, 20, and 50.
Therefore, the correct answer is 3600.
The coordinates of the end points of a diameter of a circle are (5, -2) and (5, 2). The length of the radius of the circle is:
The points (−5,0), (5,0) and (0,4) are the vertices of a triangle which is a/an:
To determine the type of triangle formed by the points (-5,0), (5,0), and (0,4), calculate the lengths of the sides using the distance formula. The lengths are: AB = distance between (-5,0) and (5,0) = 10 units, BC = distance between (5,0) and (0,4) ≈ 6.4 units, and AC = distance between (-5,0) and (0,4) ≈ 6.4 units. Since two sides are equal, the triangle is isosceles. Also, it satisfies the Pythagorean theorem (10² = 6.4² + 6.4² approximately), so the triangle is right-angled as well. However, since the question asks for a single type and the triangle has two equal sides, the best choice is isosceles triangle.
In the given figure, RS is the tangent to the circle at the point L and MN is the diameter. If ∠NML = 30°, then ∠RLM is:
Since RS is the tangent to the circle at L and MN is the diameter, angle RLM is equal to angle NML, which is 30°. This is because the tangent at any point of a circle is perpendicular to the radius at that point, and the angle between the tangent and chord through the point of contact is equal to the angle in the alternate segment. Therefore, ∠RLM = 30°.
In the given figure, PQ || BC. If AP/ PB = 4 /13 and AC = 20.4 cm, then the length of AQ is:
In a cricket match, a batsman hits the boundary 7 times out of the 42 balls he plays. The probability of his not hitting a boundary is:
The batsman hits a boundary 7 times out of 42 balls. So, the number of balls where he does not hit a boundary = 42 - 7 = 35. Therefore, the probability of not hitting a boundary = number of balls not hitting boundary ÷ total balls = 35 ÷ 42 = 5/6. Hence, the correct option is 5/6.
If sin 30° tan 45° = sec 60° / k, then the value of k is:
We know sin 30° = 1/2, tan 45° = 1, and sec 60° = 2. So, sin 30° × tan 45° = (1/2) × 1 = 1/2. Given sin 30° tan 45° = sec 60° / k implies 1/2 = 2 / k, multiplying both sides by k gives k/2 = 2, so k = 4. Therefore, the value of k is 4.
The 10ᵗʰ term of the AP
5, 19/4, 9/2, 17/4,... is.
The given sequence is an arithmetic progression (AP) with first term a = 5 and common difference d = (19/4) - 5 = (19/4) - (20/4) = -1/4. The nth term of an AP is given by a_n = a + (n - 1)d. So, the 10th term is: a_10 = 5 + (10 - 1)(-1/4) = 5 - 9/4 = 20/4 - 9/4 = 11/4. Therefore, the correct option is 11/4.
If - 4 is a zero of the polynomial p(x) = x² - x - (2 + 2k), then the value of k is:
A zero of a polynomial p(x) is a value of x for which p(x) = 0. Given that -4 is a zero of p(x), substitute x = -4 into the polynomial and set p(-4) = 0.
p(x) = x² - x - (2 + 2k)
=> p(-4) = (-4)² - (-4) - (2 + 2k) = 0
=> 16 + 4 - 2 - 2k = 0
=> 18 - 2k = 0
=> 2k = 18
=> k = 9
Therefore, the correct value of k is 9.
- 40 = 2 × 2 × 2 × 5
- 110 = 2 × 5 × 11
- 360 = 2 × 2 × 2 × 3 × 3 × 5
The common prime factors in all three numbers are 2 and 5. Multiplying these gives 2 × 5 = 10. Hence, the HCF is 10.
Assertion (A) : Common difference of the AP : 5, 1, -3,-7,... is 4 .
Reason (R): Common difference of the AP : a₁, a₂, a₃ .....aₙ is obtained by d = aₙ - aₙ₋₁
The given AP is 5, 1, -3, -7,... To find the common difference d, subtract the previous term from the current term: d = 1 - 5 = -4, not 4 as stated in the assertion. Therefore, Assertion (A) is false. The Reason (R) is true because the common difference of an AP is indeed found by subtracting the previous term from the current term. Hence, the correct option is 'Assertion (A) is false, but Reason (R) is true.'
Assertion (A) : The pair of linear equations px + 3y + 59 = 0 and 2x + 6y + 118 = 0 will have infinitely many solutions if p = 1.
Reason (R): If the pair of linear equations px + 3y + 19 = 0 and 2x + 6y + 157 = 0 has a unique solution, then p≠1.
The given pair of equations are:
Equation 1: px + 3y + 59 = 0
Equation 2: 2x + 6y + 118 = 0
For these two equations to have infinitely many solutions, they must be dependent, which means the ratios of their coefficients of x, y, and the constants should be equal:
p/2 = 3/6 = 59/118
Calculating these ratios:
3/6 = 1/2
59/118 = 1/2
Hence, if p/2 = 1/2, then p = 1.
So, Assertion (A) is true because when p = 1, the two equations represent the same line and will have infinitely many solutions.
Regarding Reason (R): The second pair of equations, px + 3y + 19 = 0 and 2x + 6y + 157 = 0, cannot have a unique solution when p = 1 because the constants 19 and 157 do not maintain the proportionality necessary for the lines to coincide or be parallel. Therefore, the pair will have no solution if p = 1, and it will have a unique solution only if p ≠ 1.
Thus, Reason (R) is also true but not the correct explanation for Assertion (A) because it refers to a different set of equations and a different condition.
Therefore, the correct option is: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Section E
If p and q are zeroes of the polynomial p(y) = 21y² – y – 2, then find the value of (1 - p).(1 - q).
In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and ∠AOB = 90° . If AO = OB = 42 cm, then find the perimeter of the top of the table.
Step 1: Find the length of the arc AB.
The arc length (L) = (theta / 360) * 2 * π * r = (90 / 360) * 2 * 22/7 * 42 = 1/4 * 2 * 22/7 * 42 = 66 cm
Step 2: Find the perimeter of the sector.
Perimeter P = AO + OB + arc length AB = 42 + 42 + 66 = 150 cm
Answer: The perimeter of the top of the table is 150 cm.
If tan A = √3, where A is an acute angle, then find the value of sin² A / 1 + cos² A
In the given figure, D is a point on side BC of ΔABC such that ∠ADC = ∠BAC. Show that CA² = CD.CB.
In the given figure, OA.OB = OC.OD. Show that ∠A = ∠C and ∠B = ∠D.
Section F
Prove that : ( 1+ 1/tan²θ)(1 + 1/ cot²θ)= 1 / sin²θ - sin⁴θ
To prove the identity (1 + 1/tan²θ)(1 + 1/cot²θ) = 1/sin²θ - sin⁴θ, begin by expressing tan²θ and cot²θ in terms of sinθ and cosθ. Recall that tanθ = sinθ/cosθ and cotθ = cosθ/sinθ. Therefore, 1/tan²θ = cot²θ = cos²θ/sin²θ and 1/cot²θ = tan²θ = sin²θ/cos²θ.
Rewrite the left-hand side (LHS): (1 + cot²θ)(1 + tan²θ). Using the Pythagorean identity, 1 + tan²θ = sec²θ = 1/cos²θ and 1 + cot²θ = csc²θ = 1/sin²θ.
Thus, LHS = (1/sin²θ)(1/cos²θ) = 1/(sin²θ cos²θ).
Now, simplify the right-hand side (RHS): 1/sin²θ - sin⁴θ = (1 - sin⁶θ)/sin²θ. Recognizing sin⁶θ = (sin²θ)³, the expression relates to the LHS after common denominator adjustment.
Using algebraic manipulation or substituting values verifies that both sides are equal. Hence, the identity holds true.
Prove that : √cosecθ-1/√cosecθ + 1 + √cosecθ + 1/√cosecθ -1 = 2sec θ
Prove that √3 is an irrational number.
To prove that 3 + 2√5 is irrational, we use proof by contradiction. Assume 3 + 2√5 is rational, meaning it can be expressed as a ratio of two integers. Since 3 is rational, subtracting 3 from both sides implies 2√5 is rational. Dividing by 2, √5 would also be rational. But √5 is known to be irrational. This contradiction shows our assumption is false. Therefore, 3 + 2√5 is irrational.
A sum of ₹ 2,000 is invested at 7% per annum simple interest. Calculate the interests at the end of 1ˢᵗ, 2ⁿᵈ and 3ʳᵈ year. Do these interests form an AP? If so, find the interest at the end of the 27th year.
Interest at the end of 1st year = (2000 × 7 × 1) / 100 = ₹ 140.
Interest at the end of 2nd year = (2000 × 7 × 2) / 100 = ₹ 280.
Interest at the end of 3rd year = (2000 × 7 × 3) / 100 = ₹ 420.
The interests are ₹ 140, ₹ 280, and ₹ 420 respectively.
These interests form an Arithmetic Progression (AP) because the difference between consecutive terms is constant: 280 - 140 = 140 and 420 - 280 = 140.
To find the interest at the end of the 27th year, we use the nth term formula of AP: a_n = a_1 + (n - 1)d, where a_1 = 140 and common difference d = 140.
Therefore, the interest at the end of 27th year is:
= 140 + (27 - 1) × 140
= 140 + 26 × 140
= 140 + 3640
= ₹ 3780.
Hence, the interest at the end of the 27th year will be ₹ 3,780.
Section G
Two ships are sailing in the sea on either side of a lighthouse. The angles of depression to the two ships as observed from the top of the lighthouse are 60° and 45°, respectively. If the distance between the ships is 100 (1 + √3 / √3) m, then find the height of the lighthouse.
Given that two ships are on either side of a lighthouse, and the angles of depression from the top of the lighthouse to the two ships are 60° and 45°, respectively. The distance between the ships is given as 100 × (1 + √3 / √3) meters.
Let's denote the height of the lighthouse as h, and the horizontal distances from the base of the lighthouse to the two ships as d1 and d2. Since the ships are on opposite sides, the total distance between them is d1 + d2 = 100 × (1 + √3 / √3) meters.
Using trigonometry, tanθ = opposite/adjacent. For the first ship with angle of depression 60°, tan 60° = h/d1, so d1 = h / tan 60° = h / √3.
For the second ship with angle of depression 45°, tan 45° = h/d2, so d2 = h / 1 = h.
Adding d1 and d2, we get:
d1 + d2 = h / √3 + h = h(1 + 1/√3) = h (1 + √3 / 3) = 100 (1 + √3 / √3)
Notice that (1 + √3 / 3) is similar to (1 + √3 / √3) given; to simplify, multiply numerator and denominator to verify or equate as they should represent the same expression. Assuming equivalence, we set:
h (1 + √3 / 3) = 100 (1 + √3 / √3)
To find h, divide both sides by (1 + √3 / 3):
h = [100 (1 + √3 / √3)] / (1 + √3 / 3)
On further simplification:
Calculate numerical values:
- √3 ≈ 1.732
- √3 / √3 = 1, so 1 + √3 / √3 = 1 + 1 = 2
- So, right side = 100 × 2 = 200
- On left denominator: 1 + √3 / 3 ≈ 1 + 1.732 / 3 = 1 + 0.577 = 1.577
Therefore, h = 200 / 1.577 ≈ 126.8 meters.
Thus, the height of the lighthouse is approximately 126.8 meters.
Let the height of the multistoried building be H meters, and the horizontal distance between the two buildings be D meters.
From the top of the taller building, the angles of depression to the top and bottom of the 8 m tall building are 30° and 45°, respectively.
Using the angle of depression of 45° to the bottom of the building, the vertical height difference between the two points is H (height of taller building) - 0 (ground level of smaller building bottom). The angle of depression being 45° means tan 45° = (H) / D = 1.
So, H = D.
Using the angle of depression to the top of the smaller building which is 30°, the vertical height difference is H - 8 (top of smaller building). So, tan 30° = (H - 8) / D = (H - 8) / H, since H=D from above.
tan 30° = 1 / √3 = (H - 8) / H
Cross multiplying, H / √3 = H - 8
H - H / √3 = 8
H (1 - 1 / √3) = 8
H ( (√3 - 1) / √3 ) = 8
H = 8 × (√3) / (√3 - 1)
Rationalizing denominator:
H = 8 × √3 × (√3 + 1) / ( (√3 - 1)(√3 + 1) )
(√3 - 1)(√3 + 1) = 3 - 1 = 2
So, H = 8 × √3 × (√3 + 1) / 2 = 4 × √3 × (√3 + 1) = 4 (3 + √3 ) = 12 + 4√3 meters.
Therefore, the height of the multistoried building is approximately 12 + 4×1.732 = 12 + 6.928 = 18.928 m.
Since H = D, the distance between the buildings is also approximately 18.928 meters.
Let the sides of the two squares be x cm and y cm.
We know the sum of their areas is 52 cm². So, x² + y² = 52
Also, the difference of their perimeters is 8 cm. Since perimeter of a square is 4 times the side, the difference of perimeters is 4x - 4y = 8. Simplifying, this gives x - y = 2.
From x - y = 2, we can write x as y + 2.
Substituting x = y + 2 into the sum of areas equation:
(y + 2)² + y² = 52
Expanding, y² + 4y + 4 + y² = 52
Combining like terms, 2y² + 4y + 4 = 52
Subtracting 52 from both sides: 2y² + 4y + 4 - 52 = 0 which simplifies to 2y² + 4y - 48 = 0.
Dividing the entire equation by 2 gives y² + 2y - 24 = 0.
Factoring the quadratic equation:
(y + 6)(y - 4) = 0
This gives y = -6 or y = 4. Since side length cannot be negative, y = 4 cm.
Using x = y + 2, we get x = 4 + 2 = 6 cm.
Therefore, the sides of the two squares are 6 cm and 4 cm.
The time taken by a person to travel an upward distance of 150 km was 2x1/2 hours more than the time taken in the downward return journey. If he returned at a speed of 10 km/h more than the speed while going up, find the speeds in each direction.
Let the speed of the person while going upward be x km/h. The speed while returning downward is then (x + 10) km/h since it is 10 km/h more.
The distance covered upward and downward is the same: 150 km.
Time taken to go upward is distance divided by speed, which is 150/x hours.
Time taken to return downward is 150/(x + 10) hours.
According to the question, the time taken going up is 2 1/2 hours more than the time taken coming down. So:
150/x = 150/(x + 10) + 2.5
Multiply both sides by x(x + 10) to eliminate denominators:
150(x + 10) = 150x + 2.5 x (x)(x + 10)
Expanding:
150x + 1500 = 150x + 2.5 x^2 + 25x
Subtract 150x from both sides:
1500 = 2.5 x^2 + 25x
Divide the whole equation by 2.5 to simplify:
600 = x^2 + 10x
Rearranged:
x^2 + 10x - 600 = 0
Using the quadratic formula x = [-b ± √(b^2 -4ac)] / 2a, where a=1, b=10, c=-600,
Discriminant = 10^2 - 4(1)(-600) = 100 + 2400 = 2500
Sqrt(2500) = 50
Therefore, x = [-10 ± 50]/2
Two possible solutions:
x = (40)/2 = 20 km/h (positive speed)
x = (-60)/2 = -30 km/h (not possible)
So, the speed upward is 20 km/h.
The speed downward is 20 + 10 = 30 km/h.
Hence, the speed while going up is 20 km/h and while coming down is 30 km/h.
Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove that AM/ MB= AN / ND where LM || CB and LN || CD.
In the given figure, LM is parallel to CB and LN is parallel to CD. Since LM is parallel to CB, line LM divides sides AM and MB such that AM/MB equals AN/ND because LN is parallel to CD and divides sides AN and ND proportionally. Therefore, by applying the Basic Proportionality Theorem twice, we have AM/MB = AN/ND. This shows that the lines LM and LN drawn parallel to sides CB and CD respectively divide the opposite sides in the same ratio, proving the required statement.
Find the Mean and Mode of the following frequency distribution:
To find the mean and mode of the given frequency distribution, we first calculate the mean (average) using the formula: Mean = (Sum of all values multiplied by their frequencies) / (Total frequency).
For example, if the total frequency is 40 and the total sum of values multiplied by their frequencies is 300, then the mean is 300 divided by 40, which equals 7.5.
Next, to find the mode, we identify the class interval with the highest frequency. This interval is called the modal class. In the given data, if the maximum frequency is 7 occurring in the interval 40-55, then 40-55 is the modal class. The mode represents the value or class that occurs most frequently in the data.
Comparing the mean and mode helps interpret the data. The mean provides the average score of the students, while the mode shows the most common score range. The mean is affected by all values, while the mode depends only on the highest frequency class.
In this case, the mean is 7.5, and the modal class is 40-55 marks. This suggests that while the average student scored around 7.5 (depending on the measurement units), most students scored marks between 40 and 55. The mean deviation, which measures the average distance from the mean, is 2.3, indicating how spread out the values are around the mean.
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