430/4/1 2022 Class 10 Maths Basic Question Paper Solution
Section – A
(1) Find the value of ‘k’ so that the quadratic equation 3x2 – 5x – 2k = 0 has real and equal roots.
Ans: D = b2 – 4ac = 25 + 24k
For real and equal roots
D = 0 = 25 + 24k = 0
K = – 25/24
(2)(a) Determine the 36th term of the A.P. whose first two terms are -3 and 4 respectively.
Ans: a1 = 3, a2 = 4, d = a2 – a1 = 4 +3 = 7
a36 = a + 35d = -3 + 35 (7)
= 242
Or
(b) Write the next two terms of the A.P. :
√27, √48, √75……
Ans: √27 = √3 × 3 × 3 = 3√3
√48 = √4 × 4 × 3 = 4√3
√75 = √5 × 5 × 3 = 5√3
d = 4√3 – 3√3 = √3
∴ Next two terms are 6√3, 7√3 or √108, √147.
(3) Using the empirical relationship between the three measures of central tendency, find the median of a distribution, whose mean is 169 and mode is 175.
Ans: Mode = 3 Median – 2 Mean
Median = 1/3 [Mode + 2 Mean] = 1/3 [175 + 338] = 171
(4) The frequency distribution table of agriculture holding in a village is given below:
Area of Land (in hectares) |
1 – 3 | 3 – 5 | 5 – 7 | 7 – 9 | 9 -11 | 11 – 13 |
Number of families | 20 | 45 | 80 | 55 | 40 |
12 |
Find the modal agriculture holding per family.
Ans: Modal class is 5 – 7
∴ Mode = l + (f1 – f0/ 2f1 – f0 – f2) × h
= 5 + ( 80 – 45/ 160 – 45 – 55) × 2
= 5 + 1.17 = 6.17
(5) The distance between two tangents parallel to each other of a circle is 13 cm. Find the radius of the circle.
Ans:
AOB is a diameter
∴ Distance between parallel tangents = Diameter = 13 cm
∴ Radius = 13/2 = 6.5 cm
(6) (a) Find the curved surface area of a right circular cone whose height is 15 cm and base ra. dm.s 1s 8 cm. [ Use π =7/22]
Ans: Getting l = 17 cm
Curved surface area of once = π r l
= 22/7 × 8 × 17 = 2992/7 cm2 or 427.4 cm2
Or
(b) The surface area of a sphere is 616 sq cm. Find its radius.[Use π = 22/7]
Ans: 4 π r2 = 616
4 × 22/7 r2 = 616
∴ r = 7 cm
Section – B
(7) (a) If x = 3 is one root of the quadratic equation 2x2 + px + 30 = 0, find the value of p and the other root of the quadratic equation.
Ans: x = 3 is one root of the quadratic equation
∴ 2 (3)2 + p (3) + 30 = 0 = 18 + 3p + 30 = 0
P = – 16
∴ 2x2 – 16x + 30 = 0
X2 – 8x + 15 = 0 = (x – 5) (x -3) = 0
X = 5,3
∴ Other Root = 5
Or
(b)The length of a rectangular park is 5 metres more than twice its breadth. If the area of the park is 250 sq m, find the length and breadth of the park.
Ans: Let Breath of the park = x m
∴ Length = (2x + 5) m
Area = 250 = (2x + 5) x = 250
2x2 + 5x = 250 = 2×2 + 5x – 250 = 0
(x – 10) (2x + 25) =0
X – 10 =0, 2x + 25 = 0
X = 10
X = – 25/2 (Not possible)
∴ Length = 2x + 5 = 25 m and Breadth = x = 10 m
(8) Draw a line segment of length 7 cm and divide it in the ratio 5:3.
Ans: Drawing a line segment of length 7 cm and making acute angle (s)
Dividing the line segment in the ratio 5:3 correctly
(9) Find the sum of first 16 terms of the A.P. whose nth term is given by an= 5n – 3.
Ans: an = 5n -3
n = l, a1 = 5(1) -3 = 2 (1st term)
n = 2, a2 = 5 (2) – 3 = 7 (2nd term)
∴ d = a2 – a1 = 7 – 2 = 5
S16 = 16/2 [2 (2) + (16 – 1) 5]
= 8 [4 + 75] = 8 × 79 = 632
(10) Two poles of heights 25 m and 35 m stand vertically on the ground. The tops of two poles are connected by a wire, which is inclined to the horizontal at an angle of 30°. Find the length of the wire and the distance between the poles.
Ans:
In right △BDE, sin 30° = DE/BD
1/2 = 10/BD = BD = 20
∴ Length of wire = 20 m
10/BE = tan 30° = 1/√3 = BE = 10√3 m
Distance between the poles = 10√3m
Section – C
(11) As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are found to be 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. [Use √3 = 1.732]
Ans:
In right △ABC, tan 45° = 100/AC
AC = 100 m
In right △ABD, tan 30° = 100/AD = AD = 100√3 m
∴ CD = Distance between ships = AD – AC
CD = 100√3 – 100 = 73.2m
Note: If √3 = 1.732 is not used in calculation of the distance CD, 1/2 mark to be deducated.
(12) (a)Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Ans:
△OBP ≅ △OBQ (SSS)
∠1 = ∠2 cpct
Similarly ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8
As ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
2∠8 + 2∠1 + 2∠4 + 2∠5 = 306° = ∠8 + ∠1 + ∠4 + ∠5 = 180°
∠AOB + ∠COD = 180° similarly ∠BOC + ∠AOD = 180°
Or
(b) If a circle is touching the side BC of MBC at P and is touching AB and AC produced at Q and R respectively (see the figure). Prove that AQ = 1/2 (perimeter of △ABC).
Ans:
We know that tangents drawn from the external point to the circle are equal
∴ BP = BQ
CP = CR
AQ = AR
2AQ = AQ + AR
= (AB + BQ) + (AC + CR)
= AB +BP + AC +CP
= AB + AC + BP +CP
= AB + AC + BC
AQ = 1/2 (AB + AC + BC)
Case Study – 1
(13) Electric buses are becoming popular nowadays. These buses have the electricity stored in a battery. Electric buses have a range of approximately 280 km with just one charge. These buses are superior to diesel buses as they reduce brake wear and also reduce pollution. Transport department of a city wants to buy some electric buses for the city. So, the department wants to know the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 50 existing public transport buses in a day.
Daily distance travelled (in km) | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |
Number of buses | 12 | 12 | 8 | 6 | 10 |
Ans:
Daily Distance (in km) |
x | Number of Buses (f) | cf | fx |
100 – 120 | 110 | 12 | 12 |
1320 |
120 – 140 |
130 | 14 | 26 | 1820 |
140 – 160 | 150 | 8 | 34 |
1200 |
160 – 180 |
170 | 6 | 40 | 1020 |
180 – 200 | 190 | 10 | 50 |
1900 |
|
50 |
7260 |
(a) Find the ‘median’ distance travelled by a bus.
Ans: N/2 = 50/2 = 25
∴ Median Class → 120 – 140
Median = l + (N/2 – C/f) × h = 120 + (25 – 12/14) × 20
Median = 120 + 130/7 = 120 + 18.57 = 138.57 km
(b) Find the ‘mean (average)’ distance travelled by a bus.
Ans: Mean = Σ fx/Σ f = 7260/50 = 145.2 km
Case Study – 2
(14) A company deals in casting and moulding of metal on orders received from its clients.
In one such order, company is supposed to make 50 toys in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of hemisphere. If the radius of the base of the cone is 21 cm and height is 28 cm,
(a) find the volume of 50 toys;
Ans:
r = 21 cm
h = 28 cm
Volume of 1 toy = Volume of hemisphere + volume of cone
= 2/3 πr3 + 1/3 πr2h
= 1/3 πr2 [2r + h]
= 1/3 × 22/7 × 21 × 2[2 × 21 + 28]
= 22 × 21 × 70 = 32340 cm3
Volume of 50 toys = 50 × 32340
= 1617000 cm3
(b) find the ratio of the volume of hemisphere to the volume of cone.
Ans: Volume of hemisphere/Volume of cone = 2/3 πr3/1/3 πr2h
= 3/2
Required ratio is 3:2
CBSE Class 10 Previous Question Paper 2020 Solution
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