CBSE Class 10 Maths Standard 2026 Question Paper 30/1/3 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/1/3 (2026),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
The string of a flying kite is tied to a point on the ground. The length of the string between the kite and the point on the ground is 80 m. The string makes an angle of 30∘ with the ground. The height of the kite above the ground is :
In the given figure, O is the centre of circle. XYZ is an arc of the circle subtending an angle of 45∘ at the centre. If the radius of the circle is 32 cm, then the length of the arc XYZ is :
Assertion (A) : The polynomial p(y)=y2+4y+3 has two zeroes. Reason (R) : A quadratic polynomial can have at most two zeroes.
(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).
y2+4y+3=(y+1)(y+3), so the zeroes are −1 and −3. A is true.
A quadratic polynomial has at most two zeroes, so R is true.
R only gives an upper limit; it does not show that this polynomial actually has two zeroes (that comes from the factorisation). So R does not explain A.
Do the points P (1,0), Q (−5,0) and R (−2,5) form a triangle ? If so, name the type of triangle formed.
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Answer: Yes; an isosceles triangle (PR=QR=34, PQ=6).
PQ=(1+5)2+02=6.
QR=(−2+5)2+(5−0)2=9+25=34.
PR=(−2−1)2+(5−0)2=9+25=34.
The sum of any two sides exceeds the third (e.g. 234>6; also P, Q lie on the x-axis but R does not), so the points are not collinear and form a triangle.
In the given figure, if a circle touches the side QR of △PQR at S and extended sides PQ and PR at M and N respectively, then prove that : PM=21(PQ+QR+PR)
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Answer: Proved.
Tangents from an external point are equal: PM=PN, QM=QS, RN=RS.
A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surface areas are in the ratio 8 : 5, then find the ratio between the radius of their bases to their height.
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Answer: 3 : 4
Let the common radius be r and height h; slant height of the cone l=r2+h2.
Two different coins are tossed simultaneously. What is the probability of getting : (i) at least one head ? (ii) at most one tail ? (iii) a head and a tail ?
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In △ABC, DE∥BC, with D on AB and E on AC. To prove: DBAD=ECAE.
Construction: 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.
Similarly ar(ADE)=21AE⋅DM and ar(DEC)=21EC⋅DM, so ar(DEC)ar(ADE)=ECAE.
△BDE and △DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE)=ar(DEC).
A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains.
A brooch is crafted from silver wire in the shape of a circle with a diameter of 35 cm. The wire is also used to create 5 diameters, dividing the circle into 10 equal sectors as shown in figure. Based on the above information, answer the following questions : (i) What is the radius of circle ? (1) (ii) What is the circumference of the brooch ? (1) (iii) (a) What is the total length of silver wire required ? (2) OR (iii) (b) What is the area of each sector of the brooch ? (2)
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Answer: (i) 17.5 cm (ii) 110 cm (iii) (a) 285 cm OR (b) 96.25 cm2
(i) Radius =235=17.5 cm.
(ii) Circumference =πd=722×35=110 cm.
(iii) (a) Wire = circumference + 5 diameters =110+5×35=285 cm.
(iii) (b) Each sector has central angle 36∘, i.e. 101 of the circle.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato. The other potatoes are arranged 3 m apart in a straight line, with a total of 10 potatoes, as shown in the figure : A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket. Based on the above information, answer the following questions : (i) What is the distance covered to pick up the first potato and drop it in bucket ? (1) (ii) What is the distance covered to pick up the second potato and drop it in bucket ? (1) (iii) (a) What is the total distance the competitor has to run ? (2) OR (iii) (b) If average speed of competitor is 5 m/s, then find the average time taken by competitor to put all the potatoes in the bucket. (2)
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Answer: (i) 10 m (ii) 16 m (iii) (a) 370 m OR (b) 74 s
The potatoes are 5 m, 8 m, 11 m, ..., from the bucket; the 10th is 5+9×3=32 m away.
(i) Distance = 2×5=10 m.
(ii) Distance = 2×8=16 m.
The round-trip distances 10, 16, 22, ... form an AP with a=10, d=6, n=10.
Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O' (as shown in figure). Distance between the base of the tower and point 'O' is 6 m. From point 'O', the angle of elevation of the top of the section 'B' is 30∘ and the angle of elevation of the top of section 'A' is 60∘. Based on the above information, answer the following questions : (i) Find the length of the wire from the point 'O' to the top of section 'B'. (1) (ii) Find the length of the wire from the point 'O' to the top of section 'A'. (1) (iii) (a) Find the distance AB. (2) OR (iii) (b) Find the area of △OPB. (2)
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Answer: (i) 43 m (ii) 12 m (iii) (a) 43 m OR (b) 63 m2
P is the base of the tower, OP=6 m, ∠BOP=30∘, ∠AOP=60∘, ∠OPA=90∘.
(i) cos30∘=OBOP, so OB=3/26=312=43 m.
(ii) cos60∘=OAOP, so OA=1/26=12 m.
(iii) (a) AP=6tan60∘=63 m and BP=6tan30∘=36=23 m.
AB=63−23=43 m.
(iii) (b) Area of △OPB=21×OP×BP=21×6×23=63 m2.