30/3/1 2022 Class 10 Maths Standard Question Paper Solution
Section – A
(1) (a) Solve the quadratic equation for x :
X2 – 2ax – (4b2 – a2) = 0
Ans: x2 – 2ax – (4b2 – a2) = 0 gives
X2 – 2ax + a2 – 4b2 = 0
(x – a)2 – (2b)2 = 0
∴ x = a – 2b and a + 2b
(b) If the quadratic equation
(1 +a2) x2 + 2abx + (b2 – c2) = 0
has equal and real roots, then prove that:
b2 = c2 (1 + a2)
Ans: The equation has equal roots
Therefore 4a2 b2 – 4a (1 + a2) (b2 – c2) = 0
b2 = c2 (1 + a2)
(2) Find the sum of first 20 terms of an AP in which d = 5 and a20 = 135.
Ans: a + 19 × 5 = 135 = a = 40
S20 = 20/2[80 + 19 × 5] = 1750
(3) Find the mode of the given frequency distribution:
Class |
Frequency |
15 – 25 |
6 |
25 – 35 |
11 |
35 – 45 |
22 |
45 – 55 |
23 |
55 – 65 |
14 |
65 – 75 |
5 |
Ans: Modal class is 45 – 55
Mode = 45 + 10 × 23 – 22/ 46 – 22 – 14
= 46
(4) (a) 150 spherical marbles, each of diameter 1·4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, and are completely immersed in water. Find the rise in the level of water in the cylindrical vessel.
Ans: Let h cm be the rise in the water level. Then
π(3.5)2 h = 4π/3 (0.7)3 (150)
h = 5.6 cm
Or
(b) Three cubes of side 6 cm each, are joined as shown in Figure 1. Find the total surface area of the resulting cuboid.
Ans: Length of cuboid = 18cm
Total surface area of solid = 2 (18×6 + 6×6 + 6×18)
= 504cm2
(5) For what value of ‘n’, are the nth terms of the APs: 9, 7, 5, … and 15, 12, 9, …. the same?
Ans: nth terms are 9 + (n – 1)(-2) and 15 + (n -1)(-3)
Thus, 9 -2(n – 1) = 15 – 3(n – 1) gives n = 7
(6) In Figure 2, PQ and PR are tangents to the circle centred at O. If ∠OPR = 45°, then prove that ORPQ is a square.
Ans: △OQP ≅ △ORP = ∠QPO = ∠RPO = 45°
∠QPR = 90°. Also ∠OQP = ∠QOR = 90°
Also OR = OQ. This implies ORPQ is a square.
Section – B
(7) (a) Draw a line segment AB of length 8 cm and locate a point P on AB such that AP : PB = 1: 5.
Ans: Correct construction
Or
(b) Draw a circle of radius 3 cm. From a point P lying outside the circle at a distance of 6 cm from its centre, construct two tangents PA and PB to the circle.
Ans: Correct construction
(8) The tops of two poles of heights 20 m and 28 m are connected with a wire. The wire is inclined to the horizontal at an angle of 30°. Find the length of the wire and the distance between the two poles.
Ans:
Correct Figure
AE = 8 m
Sin 30° = 1/2 = AE/AC
AC = 16 m (length of wire)
Cos 30° = √3/2 = x/16
X = 8√3 m
∴ The distance between two poles = 8√3 m
(9) The weights (in kg) of 50 wild animals of a National Park were recorded and the following data was obtained:
Weight (in kg) |
Number of animals |
100 – 110 |
4 |
110 – 120 |
12 |
120 – 130 |
23 |
130 – 140 |
8 |
140 -150 |
3 |
Find the mean weight (in kg) of animals, using assumed mean method.
Ans:
Class |
x | f | d = x – 125 | fd |
100 – 110 | 105 | 4 | -20 |
-80 |
110 – 120 |
115 | 12 | -10 | -120 |
120 – 130 | 125 | 23 | 0 |
0 |
130 – 140 |
135 | 8 | 10 | 80 |
140 – 150 | 145 | 3 | 20 |
60 |
50 |
-60 |
x̄ = 125 – 60/50 = 123.8
∴ Mean Weight of animals is 123.8kg
(10) For the following frequency distribution, find the median:
Class |
Frequency |
1400 – 1550 |
6 |
1550 – 1700 |
13 |
1700 – 1850 |
25 |
1850 – 2000 |
10 |
Ans:
Class |
f |
cf |
1400 – 1550 |
6 | 6 |
1550 – 1700 | 13 |
19 |
1700 – 1850 |
25 | 44 |
1850 – 2000 | 10 |
54 = N |
Correct Table
Median class is 1700 – 1850
Median = 1700 + 150/25 (27 – 19)
= 1748
Section – C
(11) (a) In Figure 3, two circles with centres at O and O of radii 2r and r respectively, touch each other internally at A. A chord AB of the bigger circle meets the smaller circle at C. Show that C bisects AB.
Ans: ∠OCA = 90°
In △s OCA and OCB, we have
OA = OB, ∠OCA = ∠OCB = 90°
and OC = OC
So, △OCA ≅ △OCB
AC = BC = C bisects AB
Or
(b) In Figure 4, O is centre of a circle of radius 5 cm. PA and BC are tangents to the circle at A and B respectively. If OP = 13 cm, then find the length of tangents PA and BC.
Ans:
PA2 = OP2 – OA2 = 169 – 25
PA = 12 cm
Let BC = x = AC = x
∴ PC = 12 – x
OP ⊥ BC = (12 – x)2 = x2 + 82
x = 10/3 cm
∴ BC = 10/3 cm
(12) A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Ten seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Ans:
Correct Figure
tan 30° = 1/√3 = h/x + y
h = x + y/√3 …. (i)
tan 60° = √3 = h/y
h = y√3 ….. (ii)
Using (i) and (ii), x + y/√3 = y√3
y = x/2
Distance x is covered in 10 sec.
∴ Distance y is covered in 5 sec.
∴ Time taken by car to reach the foot of the tower is 5sec
Case Study – 1
(13) In the picture given below, one can see a rectangular in-ground swimming pool installed by a family in their backyard. There is a concrete sidewalk around the pool of width x m. The outside edges of the sidewalk measure 7m and 12m. The area of the pool is 36 sq. m.
(a) Based on the information given above, form a quadratic equation in terms of x.
Ans:
Sides of pool are 7 – 2x and 12 – 2x
Area = 36 sq. m
(7 – 2x) (12 – 2x) = 36
4×2 – 38x + 48 = 0
(b) Find the width of the sidewalk around the pool.
Ans: 2×2 – 19x + 24 = 0
(x – 8) (2x – 3) = 0
x ≠ 8m
∴ x = 3/2m
Case Study – 2
(14) John planned a birthday party for his younger sister with his friends. They decided to make some birthday caps by themselves and to buy a cake from a bakery shop. For these two items, they decided the following dimensions:
Cake: Cylindrical shape with diameter 24 cm and height 14 cm.
Cap: Conical shape with base circumference 44 cm and height 24 cm.
Based on the above information, answer the following questions:
(a) How many square cm paper would be used to make 4 such caps?
Ans:
2 π r = 44
r = 7
Therefore, l = √242 +72 = 25 cm
Paper required for 4 caps = 4 × π r l
= 4 × 22/7 × 25
= 2200cm2 or 700 π cm2
(b) The bakery shop sells cakes by weight (0·5 kg, 1 kg, 1·5 kg, etc.).To have the required dimensions, how much cake should they order, if 650 cm3 equals 100 g of cake?
Ans: Volume of cake = 22/7 × 12 × 12 × 14
= 6336 cm3
6336 cm3 is nearest to 6500 cm3
Now, 650 cm3 = 100 g
6500 cm3 = 1 kg
∴ 1 kg cake should be orderd.
CBSE Class 10 Previous Paper 2022 Solution
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