30/4/1 2022 Class 10 Maths Standard Question Paper Solution
Section – A
1.) The mode of a grouped frequency distribution is 75 and the modal class is 65-80. The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8. Find the frequency of the modal class.
Ans: Mode = 75, Modal class = 65 – 80
l = 65 f0 = 6, f2 = 8
Mode = l + f1 – f0/2f1 – f0 – f2 × 6
75 = 65 + (f1 – 6/2f1 – 6 – 8) × 15
4f1 – 28 = 3f1 – 18 => f1 = 10
∴ Frequency of modal class is 10.
2.) How many natural numbers are there between 1 and 1000 which are divisible by 5 but not by 2?
Ans: Numbers divisible by 5 but not by 2:
5, 15, 25, 35, …., 995
a = 5, d = 10, an = 995 => a + (n -1)d = 995
5 + (n – 1) 10 = 995 => (n – 1) 10 = 990
=> n – 1 = 90 => n = 100
3.) (a) If the sum of the roots of the quadratic equation ky2 – 11y + (k – 23) = 0 is 13/21 more than the product of the roots, then find the value of k.
Ans: ky2 – 11y + (k – 23) = 0. Here a = k, b = -11, c = k – 23
Sum of roots = 11/k
Product of roots = k – 23/k
ATQ, 11/k = k – 23/k + 13/21
Solving, we get k = 21
Or
(b) If x = -2 is the common solution of quadratic equations ax2 + x – 3a = 0 and x2 + bx + b = 0, then find the value of a2b.
Ans: x = -2 is the common solution of ax2 + x – 3a = 0 and x2 + bx + b = 0.
∴ a(-2)2 + (-2) – 3a = 0 => 4a – 2 – 3a = 0
a = 2
And (-2)2 + b(-2) + b = 0 = 4 – 2b + b = 0 => b = 4
a2b = 4 × 4 = 16
4.) Find the mean of the following frequency distribution:
Class |
1 – 5 | 5 – 9 | 9 – 13 | 13 – 17 |
Frequency | 4 | 8 | 7 |
6 |
Ans:
Class |
xi | Frequency, fi |
xifi |
1 – 5 |
3 | 4 | 12 |
5 – 9 | 7 | 8 |
56 |
9 – 13 |
11 | 7 | 77 |
13 – 17 | 15 | 6 |
90 |
Σfi = 25 |
Σxifi = 235 |
∴ Mean x̄ = Σxifi/Σfi = 235/25 = 9.4
5.) In Fig. 1, there are two concentric circles with centre O. If ARC and AQB are tangents to the smaller circle from the point A lying on the larger circle, find the length of AC, if AQ = 5 cm.
Ans:
AQ = AR (tangents drawn from external point to the circle)
∴ AR = 5 cm
Join OR
∴ OR ⊥ AC (radius tangent)
Now AC is the chord of larger circle and we know that perpendicular from the centre bisects the chord
∴ AR = RC = 5 cm
AC => 5 + 5 = 10 cm
6.) (a) The curved surface area of a right circular cylinder is 176 sq cm and its volume is 1232 cu cm. What is the height of the cylinder?
Ans: Let h be the height of cylinder
CSA of cylinder = 176 = 2πrh = 176 … (i)
Volume of cylinder = 1232 = πr2h = 1232
On dividing, πr2h/2πrh = 1232/176
We get, r = 14 cm
∴ (i) => 2 × 22/7 × 14 × h = 176
=> h = 2 cm
Or
(b) The largest sphere is curved out of a solid cube of side 21 cm. Find the volume of the sphere.
Ans: Diameter of sphere = side of cube = 21 cm
∴ radius r = 21/2 cm
Volume of sphere = 4/3πr3 = 4/3 × 22/7 × 21/2 × 21/2 × 21/2
= 4851 cm3
Section – B
7.) Construct a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle of 60°.
Ans: Neat and accurate construction
8.) (a) Find the value of ‘p’ for which the quadratic equation p(x – 4) (x – 2) + (x – 1)2 = 0 has real and equal roots.
Ans: p(x – 4) (x -2) + (x -1)2 = 0
P(x2 – 6x + 8) + x2 – 2x + 1 = 0
(p +1)x2 – (6p + 2)x + (8p + 1) = 0
a = p + 1, b = 6p + 2, c = 8p + 1
For real and equal roots,
∴ D = 0 => b2 – 4ac = 0
=> (6p +2)2 -4(p + 1) (8p + 1) = 0
36p2 + 24p + 4 – 4(8p2 + 9p +1) = 0
4p2 -12p = 0 = 4p (p – 3) = 0
=> P = 0, 3
Or
(b) Had Aarush scored 8 more marks in a Mathematics test, out of 35 marks, 7 times these marks would have been 4 less than square of his actual marks. How many marks did he get in the test?
Ans: Let actual marks be x
ATQ 7(x + 8) = x2 -4
X2 -7x – 60 = 0
X2 – 12x + 5x – 60 = 0
(x -12) (x +5) = 0
X = 12, x = -5 (rejecting)
∴ Actual marks obtained by Aarush = 12
9.) An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.
Ans:
Correct figure
Let the aeroplanes be at positions C and D
Let
CD = x
AB = y
∠BAC = 30°, ∠BAC = 60°
In right angled △ABC, tan 30° = 3125/y
1/√3 = 3125/y => y = 3125√3 m
In right angled △ABD, tan 60° = x + 3125/y
√3y = x + 3125
√3(3125√3) = (x + 3125) => x = 2 (3125)
X = 6250 m
∴ Distance between two planes = 6250 m
10.) If the last term of an A.P. of 30 terms is 119 and the 8th term from the end (towards the first time) is 91, then find the common difference of the A.P. Hence, find the sum of all the terms of the A.P.
Ans: Last term an = 119 => a + 29d = 119 … (i)
8th term from end = 23rd term from the beginning
=> a + 22d = 91 …. (ii)
Solving (i) and (ii), we get
a = 3 and d = 4
∴ S30 = n/2 (a + 1)
= 30/2 (3 +119)
= 1830
Section – C
11.) (a) In Fig. 2, if a circle touches the side QR of △PQR at S extended sides PQ and PR at M and N, respectively, then
Prove that PM = 1/2 (PQ + QR + PR)
Ans:
We know that tangents drawn from the external point to the circle are equal
∴ QS = QM
RS = RN
PM = PN
Now 2 PM = PM +PN
= (PQ + QM) + (PR + PN)
= PQ + QS +PR + RS
= PQ + (QS + RS) + PR
= PQ + QR + PR
∴ PM = 1/2 (PQ + QR + PR)
Or
(b) In Fig. 3, a triangle ABC is drawn to circum scribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6 cm and 8 cm respectively. If the area of △ABC is 84 cm2, find the lengths of sides AB and AC.
Ans:
BF = BD = 6 cm
CE = DC = 8 cm
Let AF = AE = x cm
=> AB = (6 + x) cm, AC = (8 + x) cm & BC = 14 cm
Or △ABC = 1/2[p].r = 1/2 × (28 + 2x) × 4 = 84
14 +x = 21 => x = 7 cm
=> AB = 13 cm, AC = 15 cm
12.) From the top of an 8 m high building, the angle of elevation of the top a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. (Take √3 = 1.732)
Ans:
Correct figure
Let AB = height of building = 8 cm
Let CD = height of tower = h m
∠DBE = 60°
∠ACB = ∠EBC = 45°
AC = BE = y (let)
In right △ABC,
tan 45° = 8/AC
1 = 8/AC
=> AC = 8 m => y = 8 m
In right △BDE, tan 60° = h – 8/BE
√3 = h – 8/y => √3y = h – 8
√3(8) = h – 8
h = 8√3 + 8 = 8 (√3 + 1)
h = 8 (1.732 +1) = 8(2.732) = 21. 856 m
∴ Height of tower = 21.856 m
Case Study – 1
13.) Yoga is an ancient practice which is a form of meditation and exercise. By practising yoga, we not even make our body healthy but also achieve inner peace and calmness. The International Yoga Day is celebrated on 21st of June every year since 2015.
To promote Yoga, Green park society in Pune organized a 7-day Yoga camp in their society. The number of people of different age groups who enrolled for this camp is given as follows:
Age Group |
15 – 25 | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 | 75 -85 |
Number of People | 8 | 10 | 15 | 25 | 40 | 24 |
18 |
Based on the above, find the following:
(a) Find the meaning age of people enrolled for the camp.
Ans:
Age Group |
No. of people (f) | Cf |
15 – 25 | 8 |
8 |
25 – 35 |
10 | 18 |
35 – 45 | 15 |
33 |
45 – 55 |
25 | 58 |
55 – 65 | 40 |
98 |
65 – 75 |
24 | 122 |
75 – 85 | 18 |
140 |
N = 140, ∴ N/2 = 70, which corresponds to 55 – 65
∴ Median class = 55 – 65
∴ l = 55, f = 40, cf = 58, h = 10
= 55 + 70 -58/40 ×10 = 55 + 3 = 58
∴ Median = 58
Case Study – 2
14.) Khurja is a city in the Indian state of Uttar Pradesh famous for the pottery. Khurja pottery is traditional Indian pottery work which has attracted Indians as well as foreigners with a variety of tea – sets, crockery and ceramic tile works. A huge portion of the ceramics used in the country is supplied by Khurja and is also referred as “The Ceramic Town’.
One of the private schools of Bulandshahr organized an Educational Tour for class 10 students to Khurja. Students were very excited about the trip. Following are the few pottery objects of Khurja.
Students found the shapes of the objects very interesting and they could easily relate them with mathematical shapes viz sphere, hemisphere, cylinder etc. Maths teacher who was accompanying the students asked following questions:
(a) The internal radius of hemispherical bowl (filled completely with water) in I is 9 cm and radius and height of cylindrical jar in II is 1.5 cm and 4 cm respectively. If the hemispherical bowl is to be emptied in cylindrical jars, then how many cylindrical jars are required?
Ans: Cylinder – h = 4 cm, r = 1.5 cm = 3/2 cm
Volume of cylinder = πr2h
= π × (1.5)2 × 4 cm3
Radius of hemisphere R = 9 cm
Volume of hemisphere = 2/3 πR3
= 2/3 × π × (9)3 cm3
Let the number of cylindrical jars be n
∴ n × π × (1.5)2 × 4 = 2/3 × π × (9)3
=> n = 9 × 9 × 9 × 2/4 ×1.5 × 1.5 × 3 = 54
∴ Number of cylindrical jars required = 54
(b) If in the cylindrical jar full of water, a conical funnel of same height and same diameter is immersed, then how much water will flow out of the jar?
Ans: For conical funnel, r = 3/2 cm, h = 4 cm
∴ Volume of conical funnel = 1/3 πr2h = 1/3 × 22/7 × 3/2 × 3/2 × 4
= 66/7 cm3 of water will flow out.
CBSE Class 10 Previous Paper 2022 Solution
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