CBSE Class 10 Maths Standard 2023 Question Paper 30/6/1 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/6/1 (2023),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball ?
3 chairs and 1 table cost ₹ 900; whereas 5 chairs and 3 tables cost ₹ 2,100. If the cost of 1 chair is ₹ x and the cost of 1 table is ₹ y, then the situation can be represented algebraically as
In the given figure, PA and PB are tangents from external point P to a circle with centre C and Q is any point on the circle. Then the measure of ∠AQB is
(A)6221∘
(B)125∘
(C)55∘
(D)90∘
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Answer: (A) 6221∘
CA ⊥ PA and CB ⊥ PB, so in quadrilateral PACB, ∠ACB=180∘−55∘=125∘.
The angle subtended by arc AB at a point Q on the remaining part of the circle is half the angle at the centre.
Statement A (Assertion) : If 5+7 is a root of a quadratic equation with rational co-efficients, then its other root is 5−7. Statement R (Reason) : Surd roots of a quadratic equation with rational co-efficients occur in conjugate pairs.
(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 A and R are true and R is the correct explanation of A.
By the quadratic formula, roots with rational coefficients are 2a−b±D, so irrational (surd) roots occur in conjugate pairs. R is true.
Hence if 5+7 is a root, 5−7 is the other root. A is true and follows from R.
A line intersects y-axis and x-axis at point P and Q, respectively. If R(2, 5) is the mid-point of line segment PQ, then find the coordinates of P and Q.
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Answer: P(0, 10) and Q(4, 0)
Let P = (0, b) on the y-axis and Q = (a, 0) on the x-axis.
In the given figure, PA is a tangent to the circle drawn from the external point P and PBC is the secant to the circle with BC as diameter. If ∠AOC=130∘, then find the measure of ∠APB, where O is the centre of the circle.
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Answer:∠APB=40∘
∠AOB=180∘−∠AOC=180∘−130∘=50∘ (linear pair).
OA ⊥ PA (radius is perpendicular to tangent), so ∠OAP=90∘.
In the given figure, E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, then prove that △ABD∼△ECF.
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Answer: Proved.
AB = AC, so ∠ABC=∠ACB (angles opposite equal sides).
In a circle of radius 21 cm, an arc subtends an angle of 60∘ at the centre. Find the area of the sector formed by the arc. Also, find the length of the arc.
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Answer: Area of sector = 231 cm2; length of arc = 22 cm
A circle touches the side BC of a △ABC at a point P and touches AB and AC when produced at Q and R respectively. Show that AQ = 21 (Perimeter of △ABC).
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Answer: Proved.
Tangents from an external point are equal:
AQ = AR, BQ = BP, CP = CR.
Perimeter of △ABC = AB + BC + CA = AB + BP + PC + CA
A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid.
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Answer:143731 cm3 (about 1437.33 cm3)
r = 7 cm, height of cone h = 14 cm.
Volume of cone =31πr2h=31×722×49×14=32156 cm3
Volume of hemisphere =32πr3=32×722×343=32156 cm3
The ratio of the 11th term to the 18th term of an A.P. is 2 : 3. Find the ratio of the 5th term to the 21st term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms.
250 apples of a box were weighed and the distribution of masses of the apples is given in the following table : Mass (in grams): 80 – 100, 100 – 120, 120 – 140, 140 – 160, 160 – 180 Number of apples: 20, 60, 70, x, 60 (i) Find the value of x and the mean mass of the apples. (3) (ii) Find the modal mass of the apples. (2)
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Answer: (i) x = 40, mean mass = 134.8 g (ii) modal mass = 125 g
(i) 20 + 60 + 70 + x + 60 = 250, so x = 40.
Class marks: 90, 110, 130, 150, 170.
∑fixi=1800+6600+9100+6000+10200=33700
Mean =25033700=134.8 g
(ii) Modal class is 120 – 140 (frequency 70); l=120, f1=70, f0=60, f2=40, h=20.
A coaching institute of Mathematics conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, there are 20 poor and 5 rich children, whereas in batch II, there are 5 poor and 25 rich children. The total monthly collection of fees from batch I is ₹ 9000 and from batch II is ₹ 26,000. Assume that each poor child pays ₹ x per month and each rich child pays ₹ y per month. Based on the above information, answer the following questions : (i) Represent the information given above in terms of x and y. (1) (ii) Find the monthly fee paid by a poor child. (2) OR Find the difference in the monthly fee paid by a poor child and a rich child. (iii) If there are 10 poor and 20 rich children in batch II, what is the total monthly collection of fees from batch II ? (1)
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. Distance between the base of the tower and point O is 36 cm. 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 45∘. 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 distance AB. (2) OR Find the area of △OPB. (iii) Find the height of the Section A from the base of the tower. (1)
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Answer: (i) 243 cm (ii) AB = 12(3−3) cm; OR: 2163 cm2 (iii) 36 cm
(i) In right △OPB, cos30∘=OBOP, so OB=2336=372=243 cm.
(ii) PB=36tan30∘=336=123 cm and PA=36tan45∘=36 cm.
“Eight Ball” is a game played on a pool table with 15 balls numbered 1 to 15 and a “cue ball” that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8 and seven are striped balls numbered 9 to 15. The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random. Based on the above information, answer the following questions : (i) What is the probability that the drawn ball bears number 8 ? (ii) What is the probability that the drawn ball bears an even number ? OR What is the probability that the drawn ball bears a number, which is a multiple of 3 ? (iii) What is the probability that the drawn ball is a solid coloured and bears an even number ?
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Answer: (i) 151 (ii) 157; OR: 31 (iii) 154
Total outcomes = 15.
(i) Only one ball bears 8: P = 151.
(ii) Even numbers: 2, 4, 6, 8, 10, 12, 14, i.e. 7 balls: P = 157.
OR: Multiples of 3: 3, 6, 9, 12, 15, i.e. 5 balls: P = 155=31.
(iii) Solid coloured balls are 1 to 8; even ones: 2, 4, 6, 8, i.e. 4 balls: P = 154.