CBSE Class 10 Maths Basic 2024 Question Paper 430/2/3 with Solutions
All 44 questions from the CBSE Class 10 Mathematics Basic board paper, Set 430/2/3 (2024),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
The annual rainfall record of a city for 66 days is given in the following table : Rainfall (in cm) : 0–10, 10–20, 20–30, 30–40, 40–50, 50–60 Number of days : 22, 10, 8, 15, 5, 6 The difference of upper limits of modal and median classes is :
(A)10
(B)15
(C)20
(D)30
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Answer: (C) 20
Highest frequency is 22, so modal class is 0–10 (upper limit 10).
Assertion (A) : A line drawn parallel to any one side of a triangle intersects the other two sides in the same ratio. Reason (R) : Parallel lines cannot be drawn to any side of a triangle.
(A)Both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are correct 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: (C) Assertion (A) is true, but Reason (R) is false.
By the Basic Proportionality Theorem, a line parallel to one side of a triangle meeting the other two sides in distinct points divides them in the same ratio. So A is true.
A line parallel to any side of a triangle can always be drawn, so R is false.
15 defective pens are accidentally mixed with 145 good ones. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of 120∘. Find the area cleaned at each sweep of the blades.
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Answer: 924 cm2
Each blade sweeps a sector of radius 21 cm and angle 120∘.
Area of one sector =360120×722×21×21=462 cm2
The wipers do not overlap, so total area =2×462=924 cm2
All the kings and queens are removed from a deck of 52 playing cards. Remaining cards are well shuffled and then a card is drawn at random. Find the probability that the drawn card is (a) an ace of hearts (b) a black card (c) a jack of spades
In two concentric circles, a chord of length 24 cm of larger circle touches the smaller circle, whose radius is 5 cm. Find the radius of the larger circle.
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Answer: 13 cm
Let the chord AB touch the smaller circle at P. Then OP⊥AB and OP = 5 cm.
The perpendicular from the centre bisects the chord, so AP = 12 cm.
In right △OPA, OA2=OP2+AP2=25+144=169
OA = 13 cm, so the radius of the larger circle is 13 cm.
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 external point P to a circle with centre O, touching it at A and B.
Radius is perpendicular to tangent, so ∠OAP=90∘ and ∠OBP=90∘.
In quadrilateral OAPB, ∠OAP+∠APB+∠PBO+∠BOA=360∘.
90∘+∠APB+90∘+∠AOB=360∘
∠APB+∠AOB=180∘, so the two angles are supplementary.
As observed from the top of a 75 m light house from the sea-level, the angles of depression of two ships are 30∘ and 45∘. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use 3=1.732]
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Answer:75(3−1)≈54.9 m
Let the light house be AB = 75 m. The nearer ship (depression 45∘) is at distance x, the farther one (depression 30∘) at distance y.
tan45∘=x75⇒x=75 m
tan30∘=y75⇒y=753 m
Distance between ships =753−75=75(1.732−1)=75×0.732=54.9 m
A textile industry runs in a shed. This shed is in the shape of a cuboid surmounted by a half cylinder. If the base of the industry is of dimensions 14 m × 20 m and the height of the cuboidal portion is 7 m, find the volume of air that the industry can hold. Further, suppose the machinery in the industry occupies a total space of 400 m3. Then, how much space is left in the industry ?
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Answer: Volume of air = 3500 m3; space left = 3100 m3
Volume of cuboid =14×20×7=1960 m3
Half cylinder: radius =7 m, length =20 m. Volume =21×722×7×7×20=1540 m3
From a solid cylinder of height 8 cm and radius 6 cm, a conical cavity of the same height and same radius is carved out. Find the total surface area of the remaining solid. (Take π=3.14)
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Answer: 602.88 cm2
Slant height of cone l=82+62=10 cm
TSA = curved surface of cylinder + top circular face + curved surface of cone
The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice the breadth. Find the length and breadth of the plot. Also, find the cost of levelling the plot at the rate of ₹ 80 per square metre.
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Answer: Length = 33 m, breadth = 16 m; cost = ₹ 42240
The top of a table is hexagonal in shape. On the basis of the information given above, answer the following questions : (i) Write the coordinates of A and B. (1) (ii) Write the coordinates of the mid-point of line segment joining C and D. (1) (iii) Find the distance between M and Q. (2) OR (iii) Find the coordinates of the point which divides the line segment joining M and N in the ratio 1:3 internally. (2)
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Answer: (i) A(1, 9), B(5, 13) (ii) (11, 11) (iii) 45 units; OR (iii) (6, 11)
From the graph: A(1, 9), B(5, 13), C(9, 13), D(13, 9), M(5, 11), N(9, 11), Q(9, 3).
Saving money is a good habit and it should be inculcated in children right from the beginning. Rehan’s mother brought a piggy bank for Rehan and puts one ₹ 5 coin of her savings in the piggy bank on the first day. She increases his savings by one ₹ 5 coin daily. Based on the above information, answer the following questions : (i) How many coins were added to the piggy bank on 8th day ? (1) (ii) How much money will be there in the piggy bank after 8 days ? (1) (iii) If the piggy bank can hold one hundred twenty ₹ 5 coins in all, find the number of days she can contribute to put ₹ 5 coins into it. (2) OR (iii) Find the total money saved, when the piggy bank is full. (2)
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Answer: (i) 8 coins (ii) ₹ 180 (iii) 15 days; OR (iii) ₹ 600
Coins added each day: 1, 2, 3, ... form an AP with a=1, d=1.
(i) a8=1+7×1=8 coins
(ii) Coins after 8 days =28(2+7)=36; money =36×5=₹180
(iii) 2n(2+(n−1))=120⇒n(n+1)=240⇒n=15 days
OR (iii) Full piggy bank holds 120 coins, so money =120×5=₹600
Heart Rate : The heart rate is one of the ‘vital signs’ of health in the human body. It measures the number of times per minute that the heart contracts or beats. While a normal heart rate does not guarantee that a person is free of health problems, it is a useful benchmark for identifying a range of health issues. Thirty women were examined by doctors of AIIMS and the number of heart beats per minute were recorded and summarized as follows : Number of heart beats per minute : 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86 Number of Women : 2, 4, 3, 8, 7, 4, 2 Based on the above information, answer the following questions : (i) How many women are having heart beat in the range 68 – 77 ? (1) (ii) What is the median class of heart beats per minute for these women ? (1) (iii) Find the modal value of heart beats per minute for these women. (2) OR (iii) Find the median value of heart beats per minute for these women. (2)
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Answer: (i) 15 (ii) 74–77 (iii) Mode = 76.5; OR (iii) Median = 76.25
(i) Classes 68–71, 71–74, 74–77: 4+3+8=15 women
(ii) Cumulative frequencies: 2, 6, 9, 17, 24, 28, 30; 2N=15, which first lies in cf 17, so median class is 74–77
(iii) Modal class 74–77: l=74, f1=8, f0=3, f2=7, h=3
Mode =74+16−3−78−3×3=74+65×3=76.5
OR (iii) Median =l+f2N−cf×h=74+815−9×3=74+2.25=76.25