CBSE Class 10 Maths Standard 2025 Question Paper 30/1/3 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/1/3 (2025),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
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 :
A piece of wire 20 cm long is bent into the form of an arc of a circle of radius π60 cm. The angle subtended by the arc at the centre of the circle is :
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.
(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
When a number is selected from 1 to 20, it is certain that the number selected is one of the numbers 1 to 20, so this is a sure event and its probability is 1. Assertion is true.
Reason states the definition of a sure event: if P(E)=1, E is a sure event. Reason is true.
The assertion holds because the event is a sure event, so R correctly explains A.
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.
(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.
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Answer: (C) Assertion (A) is true, but Reason (R) is false.
Two hemispheres of the same radius joined along their circular bases form a complete sphere. Assertion is true.
The surface area of a sphere of radius r is 4πr2, not 3πr2 (3πr2 is the total surface area of a solid hemisphere). Reason is false.
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.
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Answer:24 m
OA⊥PA (radius is perpendicular to tangent at point of contact), so △OAP is right-angled at A.
Two dice are rolled together. Find the probability of getting : (i) a multiple of 2 on one and a multiple of 3 on the other die. (ii) the product of two numbers on the top of the two dice is a perfect square number.
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Answer: (i) 3611 (ii) 92
Total outcomes =36.
(i) Multiples of 2: 2, 4, 6. Multiples of 3: 3, 6.
First die a multiple of 2 and second a multiple of 3: 3×2=6 outcomes: (2,3), (2,6), (4,3), (4,6), (6,3), (6,6).
First die a multiple of 3 and second a multiple of 2: 6 outcomes: (3,2), (3,4), (3,6), (6,2), (6,4), (6,6).
(6,6) is counted twice, so favourable outcomes =6+6−1=11. P=3611.
(ii) Product is a perfect square: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (1,4), (4,1), i.e. 8 outcomes.
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 211408 m3 of air, find the height of the cylindrical part. (Use π=722).
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Answer:4 m
Let the radius be r; then the height of the cylinder is h=2r.
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
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Answer:40 km/h
Let the speed be x km/h.
x−8480−x480=3
x(x−8)480×8=3, so x(x−8)=1280
x2−8x−1280=0
(x−40)(x+32)=0, so x=40 or x=−32.
Speed cannot be negative, so the speed of the train is 40 km/h.
A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of 24. If three times the number of red balls exceeds the number of blue balls by 20, find the number of red and blue balls.
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Answer: Red balls =40, blue balls =100
Let the number of red balls be x and blue balls be y.
10010x+10020y=24, i.e. x+2y=240 ... (1)
3x−y=20, i.e. y=3x−20 ... (2)
Substitute (2) in (1): x+6x−40=240, so 7x=280 and x=40.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table : Length (in mm): 118 – 126, 127 – 135, 136 – 144, 145 – 153, 154 – 162, 163 – 171, 172 – 180 Number of Leaves: 3, 5, 9, 12, 5, 4, 2 Find the median length of the leaves.
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Answer:146.75 mm
The classes are not continuous; subtract 0.5 from lower limits and add 0.5 to upper limits:
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 : (i) If CD is h metres, find the distance BD in terms of 'h'. (1) (ii) Find distance BC in terms of 'h'. (1) (iii) (a) Find the height CE of the lighthouse [Use 3=1.73] (2) OR (iii) (b) Find distance AE, if AC = 100 m. (2)
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Answer: (i) BD=h m (ii) BC=h2 m (iii) (a) CE=20(3+3)=94.6 m OR (iii) (b) AE=50 m
A is the original position, B the deck (AB=40 m), CE the lighthouse, BD∥AE, DE=40 m, CD=h.
(i) In right △BDC: tan45∘=BDCD, so BD=h m.
(ii) BC=BD2+CD2=h2+h2=h2 m.
(iii) (a) AE=BD=h and CE=h+40. In right △AEC: tan60∘=hh+40=3.
h(3−1)=40, so h=3−140=20(3+1)=20×2.73=54.6 m.
CE=h+40=54.6+40=94.6 m.
(iii) (b) In right △AEC: cos60∘=ACAE, so AE=100×21=50 m.
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 : (i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. (1) (ii) Determine the distance of the 8th round. (1) (iii) (a) Find the total distance run after completing all 10 rounds. (2) OR (iii) (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner ? (2)
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Answer: (i) 450 m, 500 m, 550 m (ii) 650 m (iii) (a) 5250 m OR (iii) (b) 2550 m
AP: a=300, d=50.
(i) a4=300+3(50)=450, a5=500, a6=550 (metres).
(ii) a8=300+7(50)=650 m.
(iii) (a) S10=210[2(300)+9(50)]=5(600+450)=5250 m.
(iii) (b) S6=26[2(300)+5(50)]=3(600+250)=2550 m.
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 : (i) Find the central angle of each sector. (1) (ii) Find the length of the arc ACB. (1) (iii) (a) Find the area of each sector of the brooch. (2) OR (iii) (b) Find the total length of the silver wire used. (2)
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Answer: (i) 36∘ (ii) 11 mm (iii) (a) 96.25 mm2 OR (iii) (b) 285 mm
Radius r=235=17.5 mm; circumference =2×722×17.5=110 mm.
(i) Central angle of each sector =10360∘=36∘.
(ii) Arc ACB is the arc of one sector: length =36036×110=11 mm.
(iii) (a) Area of each sector =36036×722×(17.5)2=101×962.5=96.25 mm2.
(iii) (b) Wire = circumference +5 diameters =110+5×35=110+175=285 mm.