CBSE Class 10 Maths Basic 2025 Question Paper 430/4/3 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/4/3 (2025),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
Assertion (A) : Median marks of students in a class test is 16. It means half of the class got marks less than 16. Reason (R) : Median divides the distribution in two equal parts.
(A)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of 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 Assertion (A).
The median is the middle value of the arranged data, so it divides the distribution into two equal parts: R is true.
Hence median 16 means half the students scored below 16 (and half above): A is true.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
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Answer: Proved.
Let PA and PB be tangents from an external point P to a circle with centre O, touching it at A and B.
The radius is perpendicular to the tangent at the point of contact, so ∠OAP=∠OBP=90∘.
Show that 45n can not end with the digit 0, n being a natural number. Write the prime number ‘a’ which on multiplying with 45n makes the product end with the digit 0.
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Answer:45n=32n×5n has no factor 2, so it cannot end with 0; a=2.
A number ends with 0 only if its prime factorisation contains both 2 and 5.
45n=(32×5)n=32n×5n, which has no factor 2.
By the uniqueness of prime factorisation, 45n can never end with the digit 0.
Multiplying by the prime 2 gives 2×32n×5n, which has both 2 and 5 and ends with 0. So a=2.
A coin is dropped at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with radius 0.7 m ?
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Answer:30077 (about 0.257)
Area of rectangle =3×2=6 m2.
Area of circle =722×0.7×0.7=1.54 m2.
P(coin lands inside the circle) =61.54=600154=30077.
If points A(−5,y), B(2,−2), C(8,4) and D(x,5) taken in order, form a parallelogram ABCD, then find the values of x and y. Hence, find lengths of sides of the parallelogram.
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Answer:x=1, y=−1; AB = CD = 52 units, BC = AD = 62 units
Diagonals of a parallelogram bisect each other, so the mid-points of AC and BD coincide.
Mid-point of AC =(2−5+8,2y+4); mid-point of BD =(22+x,2−2+5).
The angle of elevation of the top of a tower, 300 m high, from a point on the ground is observed as 30∘. At an instant a hot air balloon passes vertically above the tower and at that instant its angle of elevation from same point on the ground is 60∘. Find height of the balloon from the ground and distance of tower from point of observation. (Use 3=1.73)
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Answer: Height of balloon = 900 m; distance of tower = 3003=519 m
Let the tower be AB = 300 m, the point of observation P, and PB = x m.
tan30∘=x300, so x=3003=300×1.73=519 m.
The balloon C is vertically above B, so tan60∘=xBC.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then prove that the other two sides are divided in the same ratio.
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Answer: Proved.
Given: in △ABC, DE ∥ BC with D on AB and E on AC. To prove: DBAD=ECAE.
Join BE and CD. Draw EN ⊥ AB and DM ⊥ AC.
ar(ADE) =21AD×EN and ar(BDE) =21DB×EN, so ar(BDE)ar(ADE)=DBAD.
ar(ADE) =21AE×DM and ar(DEC) =21EC×DM, so ar(DEC)ar(ADE)=ECAE.
Triangles BDE and DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
It is given that p2x2+(p2−q2)x−q2=0 ; (p=0) (i) Show that the discriminant (D) of above equation is a perfect square. (ii) Find the roots of the equation.
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Answer: (i) D=(p2+q2)2 (ii) x=p2q2, x=−1
(i) a=p2, b=p2−q2, c=−q2.
D=(p2−q2)2−4p2(−q2)=p4−2p2q2+q4+4p2q2=(p2+q2)2, a perfect square.
Three consecutive positive integers are such that the sum of the square of smallest and product of other two is 67. Find the numbers, using quadratic equation.
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Answer: 5, 6, 7
Let the integers be x, x+1, x+2.
x2+(x+1)(x+2)=67
2x2+3x+2=67, i.e. 2x2+3x−65=0.
(2x+13)(x−5)=0, so x=5 or x=−213.
x is a positive integer, so x=5; the numbers are 5, 6, 7.
Playing in a ball pool is good entertainment for kids. Suhana bought 600 new balls of diameter 7 cm to fill in the pool for her kids. The cuboidal box containing 600 balls has dimensions 42 cm × 91 cm × 50 cm (l×b×h). Based on above information, answer the following questions : (i) Find the volume of one ball. (1) (ii) 10 balls are painted with neon colours. Determine the area of painted surface. (1) (iii) (a) Find the volume of empty space in the box. (2) OR (b) The lowermost layer of the balls covers the base of the box edge to edge when balls are placed evenly adjacent to each other. (A) How much area is covered by one ball? (B) How many balls are there in lowermost layer? (2)
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Answer: (i) 3539≈179.67 cm3 (ii) 1540 cm2 (iii) (a) 83300 cm3 OR (b) (A) 49 cm2 (B) 78
Radius of a ball r=3.5 cm.
(i) Volume =34×722×3.53=3539≈179.67 cm3.
(ii) Surface area of one ball =4×722×3.52=154 cm2; for 10 balls =1540 cm2.
(iii) (a) Volume of box =42×91×50=191100 cm3. Volume of 600 balls =600×3539=107800 cm3. Empty space =191100−107800=83300 cm3.
(iii) (b) (A) Each ball occupies a square of side equal to its diameter: 7×7=49 cm2. (B) Number of balls =742×791=6×13=78.
Rahim and Nadeem are two friends whose plots are adjacent to each other. Rahim’s son made a drawing of the plots with necessary details. It is decided that Rahim will fence the triangular plot ABC and Nadeem will fence along the sides AF, FE and BE. Observe the diagram carefully and answer the following questions : (Use 2=1.41 and 3=1.73) (i) Find length BC. (1) (ii) Find length AG. (1) (iii) (a) Calculate perimeter of △ABC. (2) OR (b) Calculate length of (AF + FE + EB). (2)
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Answer: (i) 502=70.5 m (ii) 38 m (iii) (a) 298.5 m OR (b) 152.82 m (approx.)
BD = BH + HD = 40 + 10 = 50 m; AB = AG + GD + DH + HB.
(i) In right △CDB, ∠CBD=45∘, so BC=cos45∘BD=502=50×1.41=70.5 m.
(ii) In right △AGF, ∠GAF=45∘, so AG = GF = 38 m (check: AF =382=53.58 m).
(iii) (a) CD = BD tan45∘ = 50 m. In right △ADC, ∠ACD=60∘, so AC=cos60∘CD=100 m. AB = 38 + 40 + 10 + 40 = 128 m.
Perimeter =128+70.5+100=298.5 m.
(iii) (b) In right △BHE, ∠HBE=30∘, so EB=cos30∘BH=380=1.7380≈46.24 m.
A telecommunication company came up with two plans– plan A and plan B for its customers. The plans are represented by linear equations where ‘t’ represents the time (in minutes) bought and ‘C’ represents the cost. The equations are : Plan A : 3C=20t Plan B : 3C=10t+300 Based on above information, answer the following questions : (i) If you purchase plan B, how much initial amount you have to pay ? (1) (ii) Charu purchased plan A. How many minutes she bought for ₹ 250 ? (1) (iii) (a) At how many minutes, do both the plans charge the same amount? What is that amount? (2) OR (b) Which plan is better if you want to buy 60 minutes? Give reason for your answer. (2)
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Answer: (i) ₹ 100 (ii) 37.5 minutes (iii) (a) 30 minutes, ₹ 200 OR (b) Plan B, as it costs ₹ 300 against ₹ 400 for plan A
(i) Initial amount is the cost at t=0: 3C=300, so C=₹100.
(ii) 3×250=20t, so t=20750=37.5 minutes.
(iii) (a) 20t=10t+300, so t=30 minutes; C=320×30=₹200.
(iii) (b) For t=60: plan A, C=31200=₹400; plan B, C=3600+300=₹300. Plan B is cheaper, so it is better.