CBSE Class 10 Maths Basic Previous Question Paper 2022 Solution
Section – A
(1) (a) In an AP, if a = 50, d = 4 and Sn = 0, then find the value of n.
Ans: 0 = n/2 [100+ (n – 1) (-4)]
=> n = 26
Or
(b) Find the sum of the first twelve 2-digit multiples of 7, using an AP.
Ans: Here a =14, n = 12, d = 7 or A.P. is 14, 21, 28, …..
S12 = 12/2 [28 + 11 × 7] = 6 [28 + 77] = 6 × 105
= 630
(2) A solid metallic sphere of radius 3 cm is melted and recast into the shape of a solid cylinder of radius 2 cm. Find the height of the cylinder.
Ans: Volume of sphere = Volume of cylinder
4/3 × π × 27 = π × 4 × h
h = 9 cm
(3) (a) Find the nature of the roots of the quadratic equation
X2 – 5x + 9 = 0.
Ans: D = 25 – 36 = -11<0
Therefore, equation has no real roots
Or
(b) Write a quadratic equation with roots 3 and 5.
Ans: Required equation is (x + 3) (x -5) = 0
=> X2 -2x – 15 = 0
(4) Find the mode of the following frequency distribution:
Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 |
Frequency | 8 | 7 | 12 | 5 | 3 |
Ans: Model class is 40 -60
Mode = 40 + 20 × 12 – 7/ 24 – 7 – 5
= 48.3
(5) Solve the quadratic equation 2x2 – 5x – 1 = 0 for x.
Ans: D = 33
X = -(-5) ± √33/2 × 2
X = 5 + √33/4, 5 – √33/4
(6) In Figure 1, if tangents PA and PB drawn from a point P to a circle with centre O, are inclined to each other at an angle of 70°, then find the measure of ∠POA.
Ans: ∠OAP = ∠OBP = 90°
Using angle sum property of the quadrilateral, ∠AOB = 110°
△APO ≅ △BPO = ∠POA = 55°
Section – B
(7) The frequency distribution given below shows the weight of 40 students of a class. Find the median weight of the students.
Weight (in kg) |
Number of Students |
40 – 45 |
9 |
45 – 50 |
5 |
50 – 55 |
8 |
55 – 60 |
9 |
60 – 65 |
6 |
65 – 70 |
3 |
Ans:
Class |
f | cf |
40 – 45 | 9 |
9 |
45 – 50 |
5 | 14 |
50 – 55 | 8 |
22 |
55 – 60 |
9 | 31 |
60 – 65 | 6 |
37 |
65 – 70 |
3 |
40 = N |
Median class is 50 – 55
Median = 50 + 5/8 (20 -14)
= 53.75
(8) (a) Draw a circle of radius 4 cm. Construct a pair of tangents to the circle from a point 6 cm away from its centre.
Ans: Constructing a circle of radius 4cm and marking a point at a distance of 6cm from the centre of the circle.
Constructing a pair of tangents correctly.
Or
(b) Draw a line segment PQ = 7·5 cm. Divide it in the ratio 3:1.
Ans: Drawing a line segment PQ = 7.5 cm and making acute angle(s)
Dividing PQ in the ratio 3:1 correctly
(9) In Figure 2, the angles of elevation of the top of a tower AB of height ‘h’ m, from two points P and Q at a distance of x m and y m from the base of the tower respectively and in the same straight line with it, are 60° and 30°, respectively. Prove that h2 = xy.
Ans: tan 60° = h/x = √3 = h = x√3 … (i)
Tan 30° = 1/√3 = h/y = h = y/√3 …. (ii)
Multiplying (i) & (ii) h2 = x√3 × y√3 = xy
(10) The following table shows the age of patients admitted in a hospital during a particular week:
Age (in years) |
5 – 15 | 15 – 25 | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 |
Number of Patients | 5 | 12 | 20 | 24 | 15 |
4 |
Find the mean age of the patients.
Ans:
Class |
x | f | d | fd |
5 – 15 | 10 | 5 | -30 |
-150 |
15 – 25 |
20 | 12 | -20 | -240 |
25 – 35 | 30 | 20 | -10 |
-200 |
35 – 45 |
40 | 24 | 0 | 0 |
45 – 55 | 50 | 15 | 10 |
150 |
55 – 65 |
60 | 4 | 20 | 80 |
80 |
-360 |
Mean = 40 – 360/80 = 35.5
Section – C
(11) (a) A spherical glass vessel has a cylindrical neck 8 cm long and 1 cm in radius. The radius of the spherical part is 9 cm. Find the amount of water (in litres) it can hold, when filled completely.
Ans:
Volume of cylindrical part
= π × 1 × 8 = 8 π cm3
Volume of spherical part
= 4/3 π × 9 × 9 × 9 = 972 π cm3
Total volume of glass is
8 π + 972 π = 980 × 22/7 = 3080 cm3
Amount of water = 3.08 l
Or
(b) From a solid cylinder, whose height is 2·4 cm and diameter 1·4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.
Ans: Getting l = 2.5 cm
Surface area of the remaining solid = 2 π (0.7) (2.4) + π (0.7)2 + π (0.7) (2.5).
= π (3.36 + 0.49 + .75)
= 17.6 cm2 or 5.6 π cm2
(12) In Figure 3, the tangent l is parallel to the tangent m drawn at points A and B respectively to a circle centred at O. PQ is a tangent to the circle at R. Prove that ∠POQ = 90°.
Ans:
△OBP ≅ △ORP (SSS)
∠1 = ∠2 (cpct)
△OAQ ≅ ORQ (SSS)
∠3 = ∠4 (cpct)
As ∠AQP and ∠BPQ are consecutive interior angles and l || m
∠1 + ∠2 + ∠3 + ∠4 = 180°
2 ∠2 + ∠3 = 180°
∠2 + ∠3 = 90°
∴ ∠POQ = 90° (Using angle sum property in △ POQ)
Case Study 1
(13) Do you know old clothes which are thrown as waste not only fill the landfill site but also produce very harmful greenhouse gas. So, it is very important that we reuse old clothes in whatever way we can. The picture given below on the right, shows a footmat (rug) made out of old t-shirts yarn. Observing the picture, you will notice that a number of stitches in circular rows are making a pattern : 6, 12, 18, 24,…
Based on the above information, answer the following questions:
(a) Check whether the given pattern forms an AP. If yes, find the common difference and the next term of the AP.
Ans: 12 – 6 = 18 – 12 = 24 – 18 = 6
Since difference of consecutive terms is same every time, hence it is an A.P.
∴ d = 6 and a5 = 24 + 6 = 30
(b) Write the nth term of the AP. Hence, find the number of stitches in the 10th circular row.
Ans: an = 6 + (n – 1) 6 = 6n
∴ a10 = 60
Case Study – 2
(14) The following TV Tower was built in 1988 and is located in Pitampura, Delhi. It has an observation deck. Observe the picture given below:
The TV Tower stands vertically on the ground. From a point ‘A’ on the ground. The angle of elevation of top of the tower (point ‘B’) is 60°. There is a point ‘C’ on the tower which is 78 m (approx.) above the ground. The angle of elevation of the point C from point A is found to be 30°.
(a) Draw a well-labelled figure, based on the information given above.
Ans:
Well labelled correct figure:
(b) Find the height of the tower and the distance of the tower from point A.
Ans: tan 30° = 78/x = 1/√3
=> x = 78 √3 m … (i) (distance of the tower from point A)
tan 60° = √3 = BD/x
BD = √3x … (ii)
Using (i) & (ii), height of the tower = BD = 3 × 78 = 234 m
CBSE Class 10 Previous Question Paper 2022 Solution
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