CBSE Class 10 Maths Standard 2025 Question Paper 30/3/2 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/3/2 (2025),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
If tangents PA and PB drawn from an external point P to the circle with centre O are inclined to each other at an angle of 80∘ as shown in the given figure, then the measure of ∠POA is :
Assertion (A) : If two tangents are drawn to a circle from an external point, then they subtend equal angles at the centre of the circle. Reason (R): A parallelogram circumscribing a circle is a rhombus.
(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: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
If PA, PB are tangents from P to a circle with centre O, △OAP≅△OBP (RHS), so ∠AOP=∠BOP. A is true.
For a parallelogram ABCD circumscribing a circle, equal tangent lengths give AB+CD=AD+BC, so 2AB=2AD and it is a rhombus. R is true.
Assertion (A) : A ladder leaning against a wall, stands at a horizontal distance of 6 m from the wall. If the height of the wall up to which the ladder reaches is 8 m, then the length of the ladder is 10 m. Reason (R): The ladder makes an angle of 60∘ with the ground.
(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.
Length of ladder =62+82=100=10 m, so A is true.
If θ is the angle with the ground, cosθ=106=53=21, so θ=60∘; R is false.
The probability of guessing the correct answer of a certain test question is 12x. If the probability of not guessing the correct answer is 65, then find the value of x.
If 65% of the population has black eyes, 25% have brown eyes and the remaining have blue eyes, what is the probability that a person selected at random has : (a) blue eyes ? (b) brown or black eyes ?
In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find ∠CBA and ∠BCO, if ∠PCA=110∘.
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Answer:∠CBA=70∘, ∠BCO=70∘
∠OCP=90∘ (radius ⊥ tangent), so ∠OCA=110∘−90∘=20∘.
An AP consists of 'n' terms whose nth term is 4 and the common difference is 2. If the sum of 'n' terms of AP is −14, then find 'n'. Also, find the sum of the first 20 terms.
The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is 1:3. Calculate the first and the thirteenth terms of the AP.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
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Answer: 100
Volume of cone =31πr2h=31π(5)2(8)=3200π cm3.
Water that flows out =41×3200π=350π cm3 = total volume of lead shots.
A man lent a part of his money at 10% p.a. and the rest at 15% p.a. His income at the end of the year is ₹ 1,900. If he had interchanged the rate of interest on the two sums, he would have earned ₹ 200 more. Find the amount lent in both cases.
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Answer: ₹ 10,000 at 10% p.a. and ₹ 6,000 at 15% p.a.
Let ₹ x be lent at 10% and ₹ y at 15%.
10010x+10015y=1900, i.e. 2x+3y=38000 ... (1)
After interchanging: 10015x+10010y=2100, i.e. 3x+2y=42000 ... (2)
Adding: 5x+5y=80000, so x+y=16000. Subtracting (1) from (2): x−y=4000.
The India Meteorological Department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results. The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023. Rainfall (mm): 200 – 400, 400 – 600, 600 – 800, 800 – 1000, 1000 – 1200, 1200 – 1400 No. of Sub-divisions: 3, 4, 7, 4, 3, 3 Based on the information given above, answer the following questions : (i) Write the modal class. (1) (ii) (a) Find the median of the given data. (2) OR (b) Find the mean rainfall in the season. (2) (iii) If a sub-division having at least 800 mm rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall ? (1)
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Answer: (i) 600 – 800 (ii) (a) 75200≈742.86 mm OR (b) 775 mm (iii) 10
(i) Highest frequency is 7, for class 600 – 800; modal class is 600 – 800.
(ii)(a) N=24, 2N=12. Cumulative frequencies: 3, 7, 14, 18, 21, 24; median class 600 – 800.
(ii)(a) l=600, cf=7, f=7, h=200: Median =600+712−7×200=600+71000≈742.86 mm.
(ii)(b) Class marks: 300, 500, 700, 900, 1100, 1300; ∑fixi=900+2000+4900+3600+3300+3900=18600.
(ii)(b) Mean =2418600=775 mm.
(iii) Sub-divisions with at least 800 mm =4+3+3=10.
A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway. The total area of the lawn and the walkway is 360 square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are 12 metres by 10 metres. Based on the information given above, answer the following questions : (i) Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway = x m. (1) (ii) (a) Solve the quadratic equation to find the width of the walkway 'x'. (2) OR (b) If the cost of paving the walkway at the rate of ₹ 50 per square metre is ₹ 12,000, calculate the area of the walkway. (2) (iii) Find the perimeter of the lawn. (1)
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Answer: (i) (12+2x)(10+2x)=360, i.e. x2+11x−60=0 (ii) (a) x=4 m OR (b) 240 m2 (iii) 44 m
(i) Outer dimensions are (12+2x) m and (10+2x) m, so (12+2x)(10+2x)=360.
A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be 30∘ and 45∘ as it moves from point P to point Q. The height of the lighthouse is 50 metres. Based on the information given above, answer the following questions : (i) Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is 45∘. (1) (ii) Find the measures of ∠PBA and ∠QBA. (1) (iii) (a) Find the distance travelled by the ship between points P and Q. (2) OR (b) If the ship continues moving towards the shore and takes 10 minutes to travel from Q to A, calculate the speed of the ship in km/h, from Q to A. (2)
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Answer: (i) 50 m (ii) ∠PBA=60∘, ∠QBA=45∘ (iii) (a) 50(3−1) m ≈36.6 m OR (b) 0.3 km/h
(i) In right △QAB, ∠BQA=45∘, so tan45∘=AQAB, AQ=AB=50 m.
(ii) ∠PBA=90∘−30∘=60∘ and ∠QBA=90∘−45∘=45∘.
(iii)(a) In right △PAB, ∠BPA=30∘, so tan30∘=AP50, AP=503 m.
(iii)(a) PQ=AP−AQ=503−50=50(3−1)≈36.6 m.
(iii)(b) QA = 50 m = 0.05 km; time = 10 min = 61 h.