CBSE Class 12 Maths 2025 Question Paper 65/7/2 with Solutions
All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/7/2 (2025),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
A coin is tossed and a card is selected at random from a well shuffled pack of 52 playing cards. The probability of getting head on the coin and a face card from the pack is :
(A)132
(B)263
(C)2619
(D)133
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Answer: (B) 263
P(head) =21; there are 12 face cards, so P(face card) =5212=133.
The events are independent: required probability =21×133=263.
A student tries to tie ropes, parallel to each other from one end of the wall to the other. If one rope is along the vector 3i^+15j^+6k^ and the other is along the vector 2i^+10j^+λk^, then the value of λ is :
Q212 marksVery Short AnswerProbabilityNot in current syllabus
10 identical blocks are marked with ‘0’ on two of them, ‘1’ on three of them, ‘2’ on four of them and ‘3’ on one of them and put in a box. If X denotes the number written on the block, then write the probability distribution of X and calculate its mean.
In a village of 8000 people, 3000 go out of the village to work and 4000 are women. It is noted that 30% of women go out of the village to work. What is the probability that a randomly chosen individual is either a woman or a person working outside the village ?
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Answer:4029
Let W: woman, O: works outside the village.
P(W)=80004000=21, P(O)=80003000=83.
Women working outside =30% of 4000 =1200, so P(W∩O)=80001200=203.
In a Linear Programming Program (LPP) for objective function Z=14x−10y subject to constraints x+y≤8 3x−2y≥−6 x,y≥0 shade the feasible region and mark the corner points in a neatly drawn graph.
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Answer: Feasible region is the quadrilateral with corner points (0, 0), (8, 0), (2, 6) and (0, 3).
Draw x+y=8 through (8, 0) and (0, 8); draw 3x−2y=−6 through (−2, 0) and (0, 3).
(0, 0) satisfies both inequalities, so the feasible region is on the origin side of both lines, in the first quadrant.
The lines meet where 3x−2(8−x)=−6, i.e. x=2, y=6.
Corner points: (0, 0), (8, 0), (2, 6), (0, 3); shade the quadrilateral formed by them.
Let 2x+5y−1=0 and 3x+2y−7=0 represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.
A shopkeeper sells 50 Chemistry, 60 Physics and 35 Maths books on day I and sells 40 Chemistry, 45 Physics and 50 Maths books on day II. If the selling price for each such subject book is ₹ 150 (Chemistry), ₹ 175 (Physics) and ₹ 180 (Maths), then find his total sale in two days, using matrix method. If cost price of all the books together is ₹ 35,000, what profit did he earn after the sale of two days ?
Let R be a relation defined on a set N of natural numbers such that R={(x,y):xy is a square of a natural number, x,y∈N}. Determine if the relation R is an equivalence relation.
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Answer: Yes, R is an equivalence relation.
Reflexive: x⋅x=x2 is a square, so (x,x)∈R.
Symmetric: if xy is a square then yx=xy is a square, so (y,x)∈R.
Transitive: let xy=a2 and yz=b2. Then xz⋅y2=a2b2, so xz=(yab)2.
xz is a natural number whose square root yab is rational, so the square root is a natural number; (x,z)∈R.
Find dimensions of a rectangle of perimeter 12 cm which will generate maximum volume when swept along a circular rotation keeping the shorter side fixed as the axis.
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Answer: 2 cm (axis) and 4 cm
Let the shorter side (axis) be x cm; the other side is 6−x cm.
Rotating gives a cylinder of height x and radius 6−x: V=πx(6−x)2.
dxdV=π(6−x)(6−3x)=0 gives x=2 (x = 6 gives zero volume).
dx2d2V=π(6x−24)=−12π<0 at x=2, so V is maximum.
Dimensions: 2 cm and 4 cm (maximum volume 32π cm³).
The scalar product of the vector a=i^−j^+2k^ with a unit vector along sum of vectors b=2i^−4j^+5k^ and c=λi^−2j^−3k^ is equal to 1. Find the value of λ.
Find the image of the point (−1, 5, 2) in the line 22x−4=2y=32−z. Find the length of the line segment joining the points (given point and the image point).
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Answer: Image (6, −3, −1); length 122 units
The line is 1x−2=2y=−3z−2; general point M(λ+2,2λ,2−3λ).
A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle 4π anticlockwise along the positive direction of x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.
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Answer:8π sq cm
Table top: x2+y2=64; scratch: y=x.
They meet in the first quadrant at (42,42).
Area =∫042xdx+∫42864−x2dx.
First part =[2x2]042=16.
Second part =[2x64−x2+32sin−18x]428=16π−(16+8π)=8π−16.
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record. Let A1 : People with good health, A2 : People with average health, and A3 : People with poor health. During a pandemic, the data expressed that the chances of people contracting the disease from category A1, A2 and A3 are 25%, 35% and 50%, respectively. Based upon the above information, answer the following questions : (i) A person was tested randomly. What is the probability that he/she has contracted the disease ? (2) (ii) Given that the person has not contracted the disease, what is the probability that the person is from category A2 ? (2)
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Answer: (i) 0.295 (ii) 14126
P(A1)=0.7, P(A2)=0.2, P(A3)=0.1; let E: contracted the disease.
Three friends A, B and C move out from the same location O at the same time in three different directions to reach their destinations. They move out on straight paths and decide that A and B after reaching their destinations will meet up with C at his predecided destination, following straight paths from A to C and B to C in such a way that OA=a, OB=b and OC=5a−2b respectively. Based upon the above information, answer the following questions : (i) Complete the given figure to explain their entire movement plan along the respective vectors. (1) (ii) Find vectors AC and BC. (1) (iii) (a) If a⋅b=1, distance of O to A is 1 km and that from O to B is 2 km, then find the angle between OA and OB. Also, find ∣a×b∣. (2) OR (iii) (b) If a=2i^−j^+4k^ and b=j^−k^, then find a unit vector perpendicular to (a+b) and (a−b). (2)
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Answer: (i) Join A to C and B to C, with arrows from A to C and from B to C (ii) AC=4a−2b, BC=5a−3b (iii) (a) 3π; ∣a×b∣=3 OR (iii) (b) ±171(3i^−2j^−2k^)
(i) A walks along AC and B along BC, so draw the segments AC and BC directed towards C.
Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature. (Cylindrical-shaped Camphor tablets) A cylindrical camphor tablet whose height is equal to its radius (r) evaporates when exposed to air such that the rate of reduction of its volume is proportional to its total surface area. Thus, dtdV=kS is the differential equation, where V is the volume, S is the surface area and t is the time in hours. Based upon the above information, answer the following questions : (i) Write the order and degree of the given differential equation. (1) (ii) Substituting V=πr3 and S=2πr2, we get the differential equation dtdr=32k. Solve it, given that r(0) = 5 mm. (1) (iii) (a) If it is given that r = 3 mm when t = 1 hour, find the value of k. Hence, find t for r = 0 mm. (2) OR (iii) (b) If it is given that r = 1 mm when t = 1 hour, find the value of k. Hence, find t for r = 0 mm. (2)
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Answer: (i) Order 1, degree 1 (ii) r=32kt+5 (iii) (a) k=−3; t=25 hours OR (iii) (b) k=−6; t=45 hours
(i) The highest derivative is dtdV (first order) and it appears to power 1: order 1, degree 1.
(ii) dr=32kdt gives r=32kt+C; r(0)=5 gives C=5, so r=32kt+5.
(iii) (a) 3=32k+5 gives k=−3; then r=5−2t, and r=0 when t=25 hours.
(iii) (b) 1=32k+5 gives k=−6; then r=5−4t, and r=0 when t=45 hours.