430/3/1 2022 Class 10 Maths Basic Question Paper Solution
Section – A
1.) (a) For what value of p, does the quadratic equation px2 + 2x + p = 0 have real and equal roots?
Ans: For equal roots 4 – 4p2 = 0
=> P = ± 1
Or
(b) Solve the quadratic equation for x: 6 – x – x2 = 0
Ans: Equation is x2 + x – 6 = 0 => (x + 3) (x – 2) = 0
=> X = -3, 2
2.) For an AP with common difference 6, the sum of first ten terms is same as four times the sum of first five terms. Determine the first term of the AP.
Ans: Here d = 6, S10 = 4 × S5
=> 10/2 [2a + 9 × 6] = 4 × 5/2 [2a + 4 × 6]
2a + 54 = 4a + 48 => a = 3
3.) Find mode of the following frequency distribution:
Class | 100 – 110 | 110 – 120 | 120 – 130 | 130 – 140 | 140 – 150 |
Frequency | 5 | 9 | 8 | 11 | 7 |
Ans: Modal class is 130 – 140
Mode = 130 + 10 × 11 – 8/ 22 – 8 – 7
= 134.3
4.) (a) A cubical block of side 7 cm is surmounted by a hemisphere of largest possible diameter as shown in Figure 1. Find the total surface area of the solid.
Ans: r = 7/2 cm
Total surface area = 6l2 – π r2 + 2 π r2
= 6 × 72 + 22/7 × 7/2 × 7/2
= 332.5 cm2
Or
(b) How many solid cones of height 3 cm and radius 2 cm can be formed by melting a solid sphere of radius 3 cm?
Ans: n × Volume of one cone = Volume of solid sphere
=> n × π/3 × 22 × 3 = 4/3 π × 33
=> n = 3 × 3 = 9
5.) Which term of AP: 11/7, 7/2, 17/4,…. is 53/4?
Ans: Here, a =11/4, d =3/4, an = 53/4
∴ 53/4 = 11/4 + (n – 1) 3/4
=> n = 15
6.) In Figure 2, PA and PB are tangents to the circle with centre at O. If ∠APB = 70°, then find m ∠AQB.
Ans: ∠AOB = 180° – 70° = 110°
= ∠AQB = 110°/2 = 55°
Section – B
7.) (a) Draw a circle of radius 2·5 cm. From a point P lying outside the circle at a distance of 6 cm from the centre of the circle, construct tangents PA and PB to the circle.
Ans: Constructing a circle of radius 2.5 cm and marking a point P at a distance of 6 cm from the centre of the circle.
Constructing a pair of tangents correctly.
Or
(b) Draw a line segment of length 9·5 cm and divide it in the ratio 2:3.
Ans: Drawing a line segment 9.5 cm and making acute angle (s)
Dividing PQ in the ratio 2:3 correctly.
8.) Two poles of equal heights are standing opposite each other on either side of the road of width 60 m. From a point P between them on the road, the angles of elevation of the top of the poles are 60 and 30 respectively, as shown in Figure 3. Find the height of the poles and distances of the point from the poles.
Ans: tan 60° = √3 => h/x √3
tan 30° = 1/√3 = h/y => y = h √3
y = 3x (from (i) )
Now x + y = 60 = x = 15m,
∴ h = 15 √3 m, y = 45 m
9.) The mileage (km / l) of 50 cars was recorded by a dealer and tabulated as given below:
Mileage (in km / l) | Numbers of Cars |
10 – 12 | 13 |
12 – 14 | 18 |
14 – 16 | 10 |
16 – 18 | 7 |
18 – 20 | 2 |
Find mean of the above distribution.
Ans:
Class | x | f | d = x – 15 | fd |
10 – 12 | 11 | 13 | -4 | -52 |
12 – 14 | 13 | 18 | -2 | -36 |
14 – 16 | 15 | 10 | 0 | 0 |
16 – 18 | 17 | 7 | 2 | 14 |
18 – 20 | 19 | 2 | 4 | 8 |
50 | -66 |
x̄ = 15 – 66/50 = 13.68
10.) Determine median of the following frequency distribution:
Class | Frequency |
15 – 20 | 8 |
20 – 25 | 13 |
25 – 30 | 21 |
30 – 35 | 12 |
35 – 40 | 5 |
40 – 45 | 4 |
Ans:
Class | F | cf |
15 – 20 | 8 | 8 |
20 – 25 | 13 | 21 |
25 – 30 | 21 | 42 |
30 -35 | 12 | 54 |
35 – 40 | 5 | 59 |
40 – 45 | 4 | 63 = N |
Median class is 25 – 30
Median = 25 + 5/21 (63/2 – 21) = 25 +5/21 × 21/2
= 27.5
Section – C
11.) (a) In Figure 4, quadrilateral ABCD circumscribes a circle centred at O. Prove that AD + BC = AB + CD.
Ans:
Tangents to a circle from external point are equal in length
∴ AR = AS, BS = BP, CP = CQ, DQ = DR
∴ AB + CD = x + u + y + z = BC + AD
Or
(b) In Figure 5, two concentric circles are drawn with centre O. PQ and RS are two chords of the larger circle which are tangents to the smaller circle. Prove that PQ = RS.
Ans:
Let r and r’ be the radii of the smaller and larger circles respectively
Since PM is tangent to smaller circle ∴ OM ⊥ PQ
=> PM = MQ (Similarly RN = NS)
Also PM2 = OP2 – OM2 = (r)2 – r2 …… (1)
Similarly RN2 = (r)2 – r2 ….. (2)
From (1) and (2), PM = RN
= 2 PM = 2 RN = PQ = RS
Alternative method
OM ⊥ PQ and ON ⊥ PQ
(Tangent to a circle is perpendicular to the radius through the point of contact)
Since OM = ON (Radii of the smaller circle)
∴ PQ = RS (chords equidistant from the centre of a circle are equal in length)
12.) The angles of depression of the top and bottom of a 6 m tall building from the top of a multi – storeyed building are 30° and 45° respectively. Find the height of the multi – storeyed building and the distance between the two buildings. (Use √3 = 1·73)
Ans:
Here, XY/BY = tan 45° = 1 = XY = BY = 6 + XZ = BY … (1)
Also, XZ/AZ = tan 30° = 1/√3 = BY = √3 XZ …. (2)
From (1) and (2), 6 = (√3 -1) XZ
XZ = 6 (√3 + 1)/ 2 = 3 (√3 + 1) = 8.19 m
Hence, XY =BY = 6 + 8.19 = 14.19 m
(Note: If √3 = 1.73 is not used in calculation of the either of the distances XZ or XY, 1/2 mark to be deducted.)
Case Study – 1
13.) The tradition of pottery making in India is very old. In fact, it is older than Indus Valley Civilization. The shaping and baking of clay articles has continued through the ages. The picture of a potter is shown below:
A potter makes a certain number of pottery articles in a day. It was observed on a particular day the cost of production of each article (in ₹ ) was one more than twice the number of articles produced on that day. The total cost of production on that day was ₹ 210.
(a) Taking number of articles produced on that day as x, form a quadratic equation in x.
Ans: Let number of articles produced = x
and cost of each article be ₹ y
(a) ∴ y = 1 + 2x and xy = 210
=> X (1 +2x) = 210
=> 2x2 + x – 210 = 0
(b) Find the number of articles produced and the cost of each article.
Ans: 2x2 + 21x – 20x -210 = 0 => (2x + 21) (x – 10) = 0
=> X = 10 and y = 21
Case Study – 2
14.) The technique of Rainwater harvesting through Recharge pit is very useful. Rainwater is collected on the roof and then flowing through the Recharge pit it goes to the ground. Observe the picture given below:
The surface area of the roof floor is 100 m2. The cuboidal pit measures 3 m × 3 m × 2 m.
(a) Water standing on the roof is released into the cuboidal pit. If the cuboidal pit is filled completely by the roof water, then find the height of standing water on the roof.
Ans: Area of roof × height of standing water = volume of the cuboidal pit
=> 100 × h = 3 × 3 × 2
=> h = 0.18 m
(b) Instead of a cuboidal pit, if a cylindrical pit with diameter 3 m and height 2 m had been built, then which tank would hold more water?
Ans: r = 3/2, h = 2. Volume of cylindrical pit = 22/7 × 3/2 × 3/2 × 2 = 14.14 m3
As 18> 14.14 therefore cuboidal pit holds more water.
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