CBSE Class 10 Maths Standard 2024 Question Paper 30/4/1 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/4/1 (2024),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
In the given figure, QR is a common tangent to the two given circles touching externally at A. The tangent at A meets QR at P. If AP = 4.2 cm, then the length of QR is :
(A)4.2 cm
(B)2.1 cm
(C)8.4 cm
(D)6.3 cm
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Answer: (C) 8.4 cm
Tangents from P to the first circle: PQ = PA = 4.2 cm.
Tangents from P to the second circle: PR = PA = 4.2 cm.
Assertion (A) : Mid-point of a line segment divides the line segment in the ratio 1 : 1. Reason (R) : The ratio in which the point (−3, k) divides the line segment joining the points (−5, 4) and (−2, 3) is 1 : 2.
(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.
Assertion: the mid-point is equidistant from both ends, so the ratio is 1 : 1. True.
Reason: let the ratio be m:n. Using x-coordinates, m+nm(−2)+n(−5)=−3.
−2m−5n=−3m−3n gives m=2n, so the ratio is 2 : 1, not 1 : 2. False.
A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a : (i) 2-digit number less than 40. (ii) number divisible by 5 and greater than 50. (iii) a perfect square number.
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Answer: (i) 31 (ii) 454 (iii) 101
Total outcomes = 90.
(i) 2-digit numbers less than 40 are 10 to 39: 30 numbers. P = 9030=31.
(ii) Multiples of 5 greater than 50: 55, 60, 65, 70, 75, 80, 85, 90 (8 numbers). P = 908=454.
Rehana went to a bank to withdraw ₹ 2,000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Rehana got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 did she receive.
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Answer: 10 notes of ₹ 50 and 15 notes of ₹ 100
Let the number of ₹ 50 notes be x and of ₹ 100 notes be y.
x+y=25 and 50x+100y=2000, i.e. x+2y=40.
Subtracting: y=15, so x=10.
She received 10 notes of ₹ 50 and 15 notes of ₹ 100.
Two pillars of equal lengths stand on either side of a road which is 100 m wide, exactly opposite to each other. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are 60∘ and 30∘. Find the length of each pillar and distance of the point on the road from the pillars. (Use 3 = 1.732)
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Answer: Each pillar is 253 m = 43.3 m long; the point is 25 m from the pillar seen at 60∘ and 75 m from the other pillar.
Let each pillar have height h and let the point be x m from the pillar seen at 60∘, so 100−x m from the other.
A train travels a distance of 90 km at a constant speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
The following table shows the ages of the patients admitted in a hospital during a year : Age (in years): 5–15, 15–25, 25–35, 35–45, 45–55, 55–65 Number of patients: 6, 11, 21, 23, 14, 5 Find the mode and mean of the data given above.
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Answer: Mode = 36119≈36.82 years; Mean = 35.375 ≈ 35.38 years
Mode: the modal class is 35–45 (highest frequency 23). l=35, f1=23, f0=21, f2=14, h=10.
Mode = l+2f1−f0−f2f1−f0×h=35+46−21−142×10=35+1120≈36.82 years.
Mean: class marks 10, 20, 30, 40, 50, 60; fixi = 60, 220, 630, 920, 700, 300.
Ryan, from a very young age, was fascinated by the twinkling of stars and the vastness of space. He always dreamt of becoming an astronaut one day. So he started to sketch his own rocket designs on the graph sheet. One such design is given below : Based on the above, answer the following questions : (i) Find the mid-point of the segment joining F and G. (1) (ii) What is the distance between the points A and C ? (2) OR Find the coordinates of the point which divides the line segment joining the points A and B in the ratio 1 : 3 internally. (2) (iii) What are the coordinates of the point D ? (1)
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Answer: (i) (−1, 2) (ii) 213 units; OR (3,27) (iii) (−2, −5)
From the graph: A(3, 4), B(3, 2), C(−1, −2), D(−2, −5), F(−3, 0), G(1, 4).
(i) Mid-point of FG = (2−3+1,20+4)=(−1,2).
(ii) AC = (3+1)2+(4+2)2=16+36=52=213 units.
OR: Point dividing AB in 1 : 3 = (41×3+3×3,41×2+3×4)=(3,414)=(3,27).
Treasure Hunt is an exciting and adventurous game where participants follow a series of clues/numbers/maps to discover hidden treasures. Players engage in a thrilling quest, solving puzzles and riddles to unveil the location of the coveted prize. While playing a treasure hunt game, some clues (numbers) are hidden in various spots collectively forming an A.P. If the number on the nth spot is 20+4n, then answer the following questions to help the players in spotting the clues : (i) Which number is on first spot ? (1) (ii) Which spot is numbered as 112 ? (2) OR What is the sum of all the numbers on the first 10 spots ? (2) (iii) Which number is on the (n−2)th spot ? (1)
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Answer: (i) 24 (ii) 23rd spot; OR 420 (iii) 4n+12
an=20+4n, so the A.P. is 24, 28, 32, ... with a=24, d=4.
(i) a1=20+4=24.
(ii) 20+4n=112 gives 4n=92, so n=23: the 23rd spot.
Tamper-proof tetra-packed milk guarantees both freshness and security. This milk ensures uncompromised quality, preserving the nutritional values within and making it a reliable choice for health-conscious individuals. 500 mL milk is packed in a cuboidal container of dimensions 15 cm × 8 cm × 5 cm. These milk packets are then packed in cuboidal cartons of dimensions 30 cm × 32 cm × 15 cm. Based on the above given information, answer the following questions : (i) Find the volume of the cuboidal carton. (1) (ii) Find the total surface area of a milk packet. (2) OR How many milk packets can be filled in a carton ? (2) (iii) How much milk can the cup (as shown in the figure) hold ? (1)
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Answer: (i) 14400 cm3 (ii) 470 cm2; OR 24 packets (iii) 550 cm3, i.e. 550 mL
(i) Volume of carton = 30×32×15=14400 cm3.
(ii) TSA of a packet = 2(lb+bh+hl)=2(15×8+8×5+5×15)=2(120+40+75)=470 cm2.
OR: Volume of a packet = 15×8×5=600 cm3; number of packets = 60014400=24 (they fit exactly: 1530×832×515=2×4×3=24).
(iii) The cup is a cylinder of radius 5 cm (marked from the centre of the base) and height 7 cm.