30/2/1 2022 Class 10 Maths Standard Question Paper Solution
Section – A
(1) Solve the quadratic equation: x2 + 2√2x – 6 = 0 for x.
Ans: D = 32
X = -2 √2 ± √8 + 24/2
X = √2, – 3√2
(2) (a) Which term of the A.P. – 11/2, – 3, – 1/2, …. is 49/2?
Ans: Here a = -11/2, d =5/2, an = 49/2
49/2 = -11/2 + (n – 1) 5/2
n = 13
Or
(b) Find a and b so that the numbers
a, 7, b, 23 are in A.P.
Ans: Numbers are in AP
Therefore, a + b = 14 and 2b = 30
b = 15, a = 1
(3) A solid piece of metal form of a cuboid of dimensions 11 cm × 7 cm × 7 cm is melted to form ‘n’ number of solid spheres of radii 7/2 cm each. Find the value of n.
Ans: n × 4/3 × 22/7 × (7/2)3 = 11 × (7)2
n = 3
(4) (a) In Fig. 1, AB is diameter of a circle centred at O. BC is tangent to the circle at B. If OP bisects the chord AD and ∠AOP = 60°, then find m ∠C.
Ans: ∵ AP = PD = OP ⊥ AD
∴ ∠OAP = 30°
Also ∠ABC = 90°
∠C = 60°
Or
(b) In Fig. 2, XAY is a tangent to the circle centred at O. If ∠ABO = 40°, then find m ∠BAY and m ∠AOB.
Ans: OA = OB = ∠OAB = 40°
OA ⊥ AY = ∠BAY = 50°
∠AOB = 180° – 80° = 100°
(5) If mode of the following frequency distribution is 55, then find the value of x.
Class: | 0 -15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
Frequency: | 10 | 7 | x | 15 | 10 | 12 |
Ans: Modal class is 45 – 60
Therefore, 55 = 45 + 15 × 15 – x/30 – x – 10
X = 5
(6) Find the sum of first 20 terms of an A.P. whose nth term is given as an = 5 – 2n.
Ans: a1 = 5 – 2 = 3
a20 = 5 – 40 = – 35
S20 = 20/2 (3 – 35) = -320
Section – B
(7) Draw two concentric circles of radii 2 cm and 5 cm. From a point on the outer circle, construct a pair of tangents to the inner circle.
Ans: Correct construction
(8) In Fig. 3, AB is tower of height 50 m. A man standing on its top, observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the two cars.
Ans:
tan 30° = 1/√3 = AB/BC
BC = AB √3m ….. (i)
tan 45° = AB/BD = 1
BD = AB …….. (ii)
Adding (i) and (ii) BC + BD = AB (√3 + 1)
CD = 50 (√3 + 1) m
(9) (a) The mean of the following frequency distribution is 25. Find the value of f.
Class: |
0 – 10 | 10 -20 | 20 – 30 | 30 – 40 | 40 -50 |
Frequency: | 5 | 18 | 15 | f |
6 |
Ans:
Class |
x | f | fx |
0 – 10 | 5 | 5 |
25 |
10 – 20 |
15 | 18 | 270 |
20 – 30 | 25 | 15 |
375 |
30 – 40 |
35 | f | 35f |
40 – 50 | 45 | 6 |
270 |
|
44 + f |
940 + 35f |
Correct table
x̄ = 25 = 940 + 35 f/ 44 + f
f = 16
Or
(b) Find the mean of the following data using assumed mean method:
Class: |
0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 |
Frequency: | 8 | 7 | 10 | 13 |
12 |
Ans:
Class |
x | f | d = x – 12.5 | fd |
0 – 5 | 2.5 | 8 | -10 |
-80 |
5 – 10 |
7.5 | 7 | -5 | -35 |
10 – 15 | 12.5 | 10 | 0 |
0 |
15 – 20 |
17.5 | 13 | 5 | 65 |
20 – 25 | 22.5 | 12 | 10 |
120 |
50 |
70 |
Correct table
x̄ = 12.5 + 70/50
= 13.9
(10) Heights of 50 students of class X of a school are recorded and following data is obtained:
Height
(in cm): |
130 – 135 | 135 – 140 | 140 -145 | 145 – 150 | 150 – 155 | 155 – 160 |
Number of Students: | 4 | 11 | 12 | 7 | 10 | 6 |
Find the median height of the students.
Ans:
Class |
f | cf |
130 – 135 | 4 |
4 |
135 – 140 |
11 | 15 |
140 – 145 | 12 |
27 |
145 – 150 |
7 | 34 |
150 – 155 | 10 |
44 |
155 – 160 |
6 |
50 = N |
Correct table
Median class is 140 – 145
Median = 140 + 5/12 (25 – 15)
= 144.1 (approx)
Hence, Median height is 144.1cm
(11) In Fig. 4. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q meet at a point T. Find the length of TP.
Ans:
OP = 5 cm, PR = 1/2 PQ = 4 cm
∴ △PRT ≅ △QRT
TR ⊥ PQ
∴ OR = √25 – 16 = 3 cm
Let TR = x = x2 + 16 = TP2 …… (i)
Also (3 + x)2 – 25 = TP2 ….. (ii)
Solving (i) and (ii) x = 16/3 cm
∴ TP = √256/9 + 16 = 20/3 cm
(12) (a) A 2 – digit number is such that the product of its digits is 24. If 18 is subtracted from the number, the digits interchange their places. Find the number.
Ans: Let the unit’s digit be y and ten’s digit be x
Therefore, the number is 10x + y
∴ XY = 24 ….. (i)
(10x + y) – 18 = 10y + x = x – y = 2 …… (ii)
Using (i) and (ii) we get x2 – 2x – 24 = 0
(x – 6) (x + 4) = 0
Therefore, x = 6 any y = 4
∴ number is 64.
Or
(b) The difference of the squares of two numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.
Ans: Let the two numbers be x and y, x>y
∴ x2 – y2 = 180
Y2 = 8x
Thus x2 – 8x – 180 = 0
(x – 18) (x + 10) = 0
Therefore, x = 18
Y = ± 12
13)
Case Study – 1
Kite Festival
Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites.
The picture given below, shows three kites flying together.
In Fig. 5, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be 30° and 60° respectively. Taking AD = 50 m and BE = 60 m, find
(1) the lengths of strings used (take them straight) for kites A and B as shown in the figure.
Ans: Sin 60° = √3/2 = 60/BC
BC = 40√3 m
Sin 30° = 1/2 = 50/AC
AC = 100 m
(2) the distance ‘d’ between these two kites
Ans: Since DE is a straight line therefore ∠ACB = 90°
∴ d2 = AC2 + BC2 = (100)2 + (40√3)2
d = √14800 or 20√37m
(14)
Case Study – 2
A ‘circus’ is a company of performers who put on shows of acrobats, clowns etc. to entertain people started around 250 years back, in open fields, now generally performed in tents.
One such ‘Circus Tent’ is shown below.
The tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part, than find
(1) the area of the canvas used in making the tent;
Ans:
l = √82 + 152 = 17 m
Area of canvas used = π r l + 2 π r h
= π r(l + 2h)
= 22/7 × 15 (17 + 18)
= 1650 m2
(2) the cost of the canvas bought for the tent at the rate ₹ 200 per sq m, if 30 sq m canvas was wasted during stiching.
Ans: Canvas used = 1650 + 30 = 1680 m2
∴ cost of canvas used = 200 × 1680
= ₹ 3,36,000
CBSE Class 10 Previous Paper 2022 Solution
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30/ 2/ 2 | Click here |
30/ 2/ 3 | Click here |