CBSE Class 10 Maths Standard Previous Question Paper 2022 Solution
Section – A
(1) (a) Find the sum of first 30 terms of AP: – 30, – 24, – 18,….. .
Ans: Here
a = – 30, d = 6, n = 30
S30 = 30/2 [-60 + 29 × 6]
= 1710
Or
(b) In an AP if Sn = n (4n + 1), then find the AP.
Ans: a = S1 = 1 (4 × 1 + 1) = 5
a + (a + d) =S2 = 2 (4 × 2 + 1) = 18
∴ d = 8
Hence, AP is
5, 13, 21, …
(2) A solid metallic sphere of radius 10·5 cm is melted and recast into a number of smaller cones, each of radius 3·5 cm and height 3 cm. Find the number of cones so formed.
Ans: n × 1/3. π.(3.5)2 (3) = 4/3 π (10.5)3
n = 126
(3) (a) Find the value of m for which the quadratic equation
(m – 1) x2 + 2 (m – 1) x + 1 = 0
has two real and equal roots.
Ans: For real and equal roots
4 (m – 1)2 – 4 (m – 1) = 0
m = 1 or m = 2
m ≠ 1 = 2
Or
(b) Solve the following quadratic equation for x :
√3x2 + 10x + 7 √3 = 0
Ans: √3x2 + 10x + 7 √3 = 0
or √3x2 + 3x + 7x + 7 √3 = 0
or (√3x + 7) (x + √3) = 0
x = – 7/√3, – √3 or -7/3 √3, – √3
(4) Find the mode of the following frequency distribution:
Class |
10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
Frequency | 15 | 10 | 12 | 17 |
4 |
Ans: Modal class is 40 – 50
Mode – 40 + 10 × 17 – 12/34 -12 – 4
= 42.7 or 42 7/9
(5) The product of Rehan’s age (in years) 5 years ago and his age 7 years from now, is one more than twice his present age. Find his present age.
Ans: Let Rehan’s present age be x years
∴ (x – 5) (x +7) = 2x +1
X2 = 36
X = 6
(6) Two concentric circles are of radii 4 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Ans:
For correct figure
Here OB = 3 cm, OA = 4 cm
OB ⊥ AC
∴ AB = √42 – 32 = √7 cm
Hence AC = 2√7 cm
Section – B
(7) For what value of x, is the median of the following frequency distribution 34·5?
Class |
Frequency |
0 – 10 |
3 |
10 -20 |
5 |
20 -30 |
11 |
30 – 40 |
10 |
40 – 50 |
x |
50 – 60 |
3 |
60 – 70 |
2 |
Ans: Median class is 30 – 40
Class |
Frequency | c.f. |
0 – 10 | 3 |
3 |
10 – 20 |
5 | 8 |
20 – 30 | 11 |
19 |
30 – 40 |
10 | 29 |
40 – 50 | x |
29 + x |
50 -60 |
3 | 32 + x |
60 – 70 | 2 |
34 + x |
∴ 34.5 = 30 + 10/10 (34 + x/2 – 19)
X = 13
(8) Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Construct tangents to the circle from these two points P and Q.
Ans: Correct Construction
(9) (a) The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60° . If the tower is 50 m high, then find the height of the building.
Ans:
For correct figure
tan 30° = AB/AX and tan 60° = 50/AX
AB = 1/√3 AX and AX = 50/√3
∴ AB = 1/√3. 50/√3 = 50/3 m
Or
(b) From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, then find the width of the river.
Ans:
Here, PD/AD = tan 30° = 1/√3 = AD = 3√3 m
and PD/BD = tan 45° = 1 = BD = 3 m
So, AB = AD + BD = (3√3 + 3) m = 3 (√3 + 1) m
(10) Following is the daily expenditure on lunch by 30 employees of a company:
Daily Expenditure (in Rupees) |
Number of Employees |
100 – 120 |
8 |
120 – 140 |
3 |
140 – 160 |
8 |
160 – 180 |
6 |
180 – 200 |
5 |
Find the mean daily expenditure of the employees.
Ans:
Class |
x | f | d | f.d |
100 – 120 | 110 | 8 | -40 |
-320 |
120 – 140 |
130 | 3 | -20 | -60 |
140 – 160 | 150 | 8 | 0 |
0 |
160 – 180 |
170 | 6 | 20 | 120 |
180 – 200 | 190 | 5 | 40 |
200 |
30 |
-60 |
For correct table
Mean = 150 + -60/30 = 148
Therefore, mean expenditure = Rs. 148
Section – C
(11) (a) From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and same radius is hollowed out. Find the total surface area of the remaining solid.
Ans:
l = √576 + 49 = 25 cm
TSA = 2 π r h + πr2 + π r l
= 22/7 × [60 + 7 + 25]
= 2024 cm2
Or
(b) Water in a canal, 8 m wide and 6 m deep, is flowing with a speed of 12 km/hour. How much area will it irrigate in one hour, if 0·05 m of standing water is required?
Ans: Distance covered by water in 1 hr = 12000 m
∴ Volume of water flown in 1 hr
= 8 × 6 × 12000 m3
Hence area of field × 0.05 = 8 × 6 × 12000
Area of field = 1152 × 104 m2 or 11520000 m2
(12) In Figure 1, a triangle ABC with ∠B = 90 is shown. Taking AB as diameter, a circle has been drawn intersecting AC at point P. Prove that the tangent drawn at point P bisects BC.
Ans:
PR = RB (tangents from external point) …….. (i)
Proving ∠RPC = ∠RCP
PR = CR …… (ii)
Using equations (i) and (ii)
BR = RC
Hence the tangent drawn at point P bisects BC
Case Study – 1
(13) In Mathematics, relations can be expressed in various ways. The matchstick patterns are based on linear relations. Different strategies can be used to calculate the number of matchsticks used in different figures.
One such pattern is shown below. Observe the pattern and answer the following questions using Arithmetic Progression:
(a) Write the AP for the number of triangles used in the figures. Also, write the nth term of this AP.
Ans: Number of triangles in figures are 4, 6, 8, …
This is an A. P. with a = 4, d = 2
∴ an = 4 + (n -1) × 2 = 2n + 2
(b) Which figure has 61 matchsticks?
Ans: Number of matchsticks in figure are 12, 19, 26, ….
This is an A.P. with a = 12, d = 7
∴ 61 = 12 + (n – 1) × 7
n = 8
Case Study – 2
(14) Gadisar Lake is located in the Jaisalmer district of Rajasthan. It was built by the King of Jaisalmer and rebuilt by Gadsi Singh in 14th century. The lake has many Chhatris. One of them is shown below:
Observe the picture. From a point A h m above from water level, the angle of elevation of top of Chhatri (point B) is 45° and angle of depression of its reflection in water (point C) is 60°. If the height of Chhatri above water level is (approximately) 10 m, then
(a) draw a well-labelled figure based on the above information;
Ans:
(b) find the height (h) of the point A above water level. (Use √3 = 1·73)
Ans: tan 45° = 1 = 10 – h/x
x = 10 – h … (i)
tan 60° = √3 = 10 + h/x
x = 10 + h/√3 … (ii)
Solving (i) and (ii) 10 (√3 – 1) =h (√3 + 1)
h = 10 (√3 – 1)2/2
= 2.67 m or 2.7 m
CBSE Class 10 Previous Paper 2022 Solution
Others Set |
Solution |
30/ 1/ 2 | Click here |
30/ 1/ 3 | Click here |