CBSE Class 10 Maths Standard 2023 Question Paper 30/4/3 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Standard board paper, Set 30/4/3 (2023),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
Assertion (A) : The probability that a leap year has 53 Sundays is 72. Reason (R) : The probability that a non-leap year has 53 Sundays is 75.
(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 and 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: (C) Assertion (A) is true but Reason (R) is false.
A leap year has 366 days = 52 weeks + 2 days. The 2 extra days can be (Sun, Mon), (Mon, Tue), ..., (Sat, Sun): 7 cases, 2 of which contain a Sunday. P =72, so A is true.
A non-leap year has 365 days = 52 weeks + 1 day; the extra day is a Sunday in 1 of 7 cases. P =71, so R is false.
From an external point, two tangents are drawn to a circle. Prove that the line joining the external point to the centre of the circle bisects the angle between the two tangents.
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Answer: Proved.
Let PA and PB be tangents from external point P to a circle with centre O, touching it at A and B. Join OA, OB and OP.
In △OAP and △OBP:
OA = OB (radii); ∠OAP=∠OBP=90∘ (radius ⊥ tangent); OP is common.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
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Answer: 374 cm²
r=3.5 cm, h=10 cm
TSA = CSA of cylinder + CSA of two hemispheres =2πrh+2(2πr2)
The monthly expenditure on milk in 200 families of a Housing Society is given below : Monthly Expenditure (in ₹): 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000 Number of families: 24, 40, 33, x, 30, 22, 16, 7 Find the value of x and also, find the median and mean expenditure on milk.
A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high tower observes two cars at angles of depression of 30∘ and 60∘, which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use 3=1.73)
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Answer:503 m = 86.5 m
Let AB = 75 m be the tower, and C, D the cars with angles of depression 60∘ and 30∘ (C nearer).
Angle of elevation from each car equals the angle of depression.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60∘ and the angle of depression of its foot is 30∘. Determine the height of the tower.
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Answer: 28 m
Let AB = 7 m be the building and CD the tower; draw AE ⊥ CD, so ED = AB = 7 m and AE = BD.
Angle of depression of D is 30∘: tan30∘=BDAB⇒BD=73 m
Angle of elevation of C is 60∘: tan60∘=AECE⇒CE=73×3=21 m
Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking. After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats. Based on the above information, answer the following questions : (i) What is the total perimeter of the parking area ? (1) (ii) (a) What is the total area of parking and the two quadrants ? (2) OR (b) What is the ratio of area of playground to the area of parking area ? (2) (iii) Find the cost of fencing the playground and parking area at the rate of ₹ 2 per unit. (1)
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Answer: (i) 18 units (ii) (a) 28715≈25.54 sq units OR (b) 56 : 11 (iii) ₹92
Parking area is a semicircle of diameter 7 units, radius 27 units.
(i) Perimeter =πr+2r=722×27+7=11+7=18 units
(ii) (a) Parking area =21×722×449=477=19.25 sq units
Two quadrants =2×41×722×22=744≈6.29 sq units
Total =477+744=28715≈25.54 sq units
(ii) (b) Playground area =14×7=98; ratio =98:477=392:77=56:11
(iii) Outer boundary =14+7+14+11=46 units (the side shared with the parking area is not fenced)
Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey ₹ x per student and Cricket ₹ y per student. School ‘P’ decided to award a total of ₹ 9,500 for the two games to 5 and 4 students respectively; while school ‘Q’ decided to award ₹ 7,370 for the two games to 4 and 3 students respectively. Based on the given information, answer the following questions : (i) Represent the following information algebraically (in terms of x and y). (1) (ii) (a) What is the prize amount for hockey ? (2) OR (b) Prize amount on which game is more and by how much ? (2) (iii) What will be the total prize amount if there are 2 students each from two games ? (1)
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Answer: (i) 5x+4y=9500 and 4x+3y=7370 (ii) (a) ₹980 OR (b) Cricket, by ₹170 (iii) ₹4,260
(i) School P: 5x+4y=9500; School Q: 4x+3y=7370
(ii) (a) Multiply the first equation by 3 and the second by 4: 15x+12y=28500 and 16x+12y=29480
Jagdish has a field which is in the shape of a right angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O. Based on the above information, answer the following questions : (i) Taking O as origin, coordinates of P are (−200,0) and of Q are (200,0). PQRS being a square, what are the coordinates of R and S ? (1) (ii) (a) What is the area of square PQRS ? (2) OR (b) What is the length of diagonal PR in square PQRS ? (2) (iii) If S divides CA in the ratio K:1, what is the value of K, where point A is (200,800) ? (1)
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Answer: (i) R(200,400), S(−200,400) (ii) (a) 160000 sq units OR (b) 4002 units (iii) K = 1
(i) Side PQ =200−(−200)=400, so R =(200,400) and S =(−200,400).
(ii) (a) Area =4002=160000 sq units
(ii) (b) PR=(200+200)2+(400−0)2=2×4002=4002 units
(iii) C lies on the x-axis, so C =(c,0). S divides CA in K : 1, so its y-coordinate is K+1K×800+1×0=400
800K=400K+400⇒K=1
(Check with the figure: C =(−600,0), and the x-coordinate K+1200K−600=−200 also gives K = 1.)