Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Class 10: Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

The document Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10 is a part of the Class 10 Course CBSE Sample Papers For Class 10.
All you need of Class 10 at this link: Class 10

Class-X

Time: 120 Minutes
M.M: 40

General Instructions:
Read the following instructions very carefully and strictly follow them:

  1. The question paper consists of 14 questions divided into 3 sections A, B, C.
  2. All questions are compulsory.
  3. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
  4. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
  5. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.
Section - A

Q.1. How many two digits numbers are divisible by 3?
OR
Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.

Lowest two digit number divisible by 3 is 12. and highest two digit number divisible by 3 is 99.

Hence, the sequence start with 12 and ends with 99 and common difference is 3.
So, the A.P. will be 12, 15, 18, ----, 96, 99.
Here, a = 12, d = 3, l = 99
∴ l = a + (n – 1)d
∴ 99 = 12 + (n – 1)3
⇒ 99 – 12 = 3(n – 1)
⇒ n – 1 = (87/3)
⇒ n – 1 = 29
⇒ n = 30
Therefore, there are 30, two digit numbers which are divisible by 3.
OR
The number which ends with 0 is divisible by 2 and 5 both.
∴ Such numbers between 102 and 998 are:
110, 120, 130, .........., 990.
Last term, an = 990
a + (n + 1)d = 990
110 + (n – 1) × 10 = 990 (∵ a = first term = 110)
110 + 10n – 10 = 990
10n + 100 = 990
10n = 990 – 100
10n = 890
n = 890/10 = 89.


Q.2. Solve the following quadratic equation for x:
9x2 – 9(a + b)x + 2a2 + 5ab + 2b2 = 0

Given,

9x2 – 9(a + b)x + 2a2 + 5ab + 2b2 = 0
First, we solve,
2a2 + 5ab + 2b2 = 2a2 + 4ab + ab + 2b2
Here, = 2a[a + 2b] + b[a + 2b]
= (a + 2b) (2a + b)
Hence, the equation becomes
9x2 – 9(a + b)x + (a + 2b)(2a + b) = 0
⇒ 9x2 – 3[3a + 3b]x + (a + 2b)(2a + b) = 0
⇒ 9x2 – 3[(a + 2b) + (2a + b)]x + (a + 2b)(2a +b) = 0
⇒ 9x2 – 3(a + 2b)x – 3(2a + b)x +(a + 2b) (2a + b)= 0
⇒ 3x[3x – (a + 2b)] – (2a + b) [3x – (a + 2b)] = 0
⇒ [3x – (a + 2b)][3x – (2a + b)] = 0
⇒ 3x – (a + 2b) = 0 or 3x – (2a + b) = 0
⇒ x = ((a + 2b)/3) or x = ((2a + b)/3)
Hence, the roots = Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10


Q.3. Prove that the lengths of two tangents drawn from an external point to a circle are equal.

Given, AP and BP are tangents of circle having centre O.

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10To Prove : AP = BP
Construction : Join OP, AO and BO
Proof : In DOAP and DOBP
OA = OB (Radius of circle)
OP = OP (Common side)
∠OAP = ∠OBP = 90° (Radius ⊥ tangent)
ΔOAP ≅ ΔOBP (RHS congruency rule)
AP = BP (By cpct) 1. 

Hence Proved.


Q.4. The volume of a right circular cylinder with its height equal to the radius, is 25(1/7) cm3

Find the height  of the  cylinder.  (Use π = 22/7)

Given,

Volume of a right circular  cylinder = 25(1/7) cm3
i.e., πr2h = 176/7
Since, r = h ...given
(22/7) x h2 x h = 176/7
⇒ h3 = 176/22 = 8 = 23.
Hence, height of the cylinder = 2 cm.


Q.5. Given below is the distribution of weekly pocket money received by students of a class. Calculate the pocket money that is received by most of the students.

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Here, Modal Class = 80 – 100
l = 80, f1 = 18, f2 = 5, f0 = 12 and h = 20
∴ Mode = l + Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
= 80 + (6/19) x 20
= 80 + 6.31 = 86.31 (approx.)
Hence, mode = 86.31.


Q.6. For what value(s) of 'a' quadratic equation 3ax2 – 6x + 1 = 0 has no real roots ?
OR
If one root of the equation (k – 1)x2 – 10x + 3 = 0 is the reciprocal of the other, then find the value of k.

Given that,

3ax2 – 6x + 1 = 0
For no real roots b2 – 4ac < 0
Discriminant D < 0 so,
(– 6)2 – 4(3a) (1) < 0
12a > 36
a > 3
Detailed Solution:
Given, 3ax2 – 6x + 1 = 0
On Comparing with AX2 + BX + C = 0,
we get A = 3a, B = – 6 and C = 1
Discriminant, D = B2 – 4AC
= (– 6)2 – 4 × 3a × 1
= 36 – 12a
For condition of 'no real roots',
B2 – 4AC < 0
⇒ 36 – 12a < 0
⇒ 12a > 36
⇒ a > 3.
OR
Let one root = α
and the other root = 1/α
Product of roots = α x (1/α) = 1
Given equation, (k – 1)x2 – 10x + 3 = 0
Comparing it with standard quadratic equation ax2 + bx + c = 0
we get, a = (k – 1), b = – 10 & c = 3
Product of roots = c/a = 3/(k - 1)
1 = 3/(k - 1)
or, 3 = k – 1
or, k = 4.

Section - B

Q.7. Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the median
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

n/2 = 250/2
= 125 ⇒ median class is 2 – 3,
l = 2, h = 1, cf = 120, f = 62
Median = Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
= 2 + (5/62)
= 129/62 = 2(5/62) m or 2.08 m
50% of students jumped below 2(5/62) m and 50% above it.
Detailed Solution:
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
Take N/2 = 250/2 = 125
∴ Median class = 2 – 3
Lower limit of median class (l) = 2
Size of median class (h) = 1
Frequency corresponding to median class (f) = 62
Total number of observations
Frequency (N) = 250
Cumulative frequency preceding
Median class (c.f.) = 120
Median = Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
= 2 + (5/62)
= 129/62
= 2(5/62) m or 2.08  m
∴ 50% of students jumped below 2.08 m and 50% of students jumped above 2.08 m.


Q.8. Construct a pair of tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°.

Draw circle of radius 4cm

Draw OA and  construct ∠AOB = 120°
Draw ∠OBP = ∠OBP = 90°
PA and PB are required tangents
Detailed Solution:
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Steps of construction:

  1. Draw a circle of radius 4 cm.
  2. Draw two radii having an angle of 120°.
  3. Let the radii intersect circle at A and B.
  4. Draw angle of 90° on both A and B.
  5. The point where both rays of 90° intersect is P.
  6. PA and PB are the required tangents.


Q.9. The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Mean Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10


Q.10. Two vertical poles of different heights are standing 20 m away from each other on the level ground. The angle of elevation of the top of the first pole from the foot of the second pole is 60° and angle of elevation of the top of the second pole from the foot of the first pole is 30°. Find the difference between the heights of two poles. (Take √3 = 1.73)
OR
A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is 60°. Calculate the height of the building. (Take √3 = 1.73)

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10In ∆PQS, tan60° = y/20
⇒ y = 20√3 m
In ∆RSQ, tan30° = x/20
⇒ x = (20/√3) m
y – x = Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
OR
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10Let PR be the building and AB be the boy
In ∆PQR, tan 60° = PQ/50 ⇒ PQ = 50√3m
Height of the building = ( 50√3 +1.7 )m = = 88.2 m

Detailed Solution:
Let the heights of two pole be y and x.
Distance between the poles is QS = 20 m.

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

In ∆PQS, ∠Q = 90°
tan 60° = PQ/QS
√3 =(y/20)
y = 20√3 m
In ∆RSQ, ∠S = 90°
tan 30° = RS/QS
1/√3 = x/20
20/√3 = x
x = (20/3)√3 m
Difference between their heights
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
= (20/3)√3 x 2
= (40/3)√3.
= 23.07m
OR
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Height of Boy = AB = 1.7 m
= QR = 1.7 m
Distance between Boy and building
BR = 50 m
⇒ AQ = 50 m
In ∆PQA, ∠Q = 90°
tan 60° = PQ/AQ
√3 = PQ/50
50√3 = PQ
Total height of the building = PQ + QR
= 86.5 + 1.7 = 88.2 m.

Section - C

Q.11. A solid is in the shape of a cone mounted on a hemisphere  of same base radius. If the curved surface areas of the hemispherical part and the conical part are equal, then find the ratio of the radius  and the height of the conical part. 

Let ABC be a cone, which is mounted on a hemisphere. 

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

Given, OC = OD = r cm
Curved surface area of the hemispherical part = (1/2)4πr2
= 2pr2
Slant height  of a cone, Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
and curved surface area of a cone = πrl
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
According to the problem,
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10 (Given)
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
on  squaring  both of the sides, we get
4r2 = h2 + r2
⇒ 4r2 – r2 = h2
⇒ 3r2 = h2
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
Hence, the ratio of the radius and the height of conical part = 1 : √3.


Q.12. PQ is a tangent to a circle with centre O at point P. If ΔOPQ is an isosceles triangle, then find ∠OQP.
OR
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P on producing. Find the length of AP.

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

In ΔOPQ,
∠P+∠Q+∠O =180°
(∠O = ∠Q, isosceles triangle)
2∠Q+∠P = 180°
2∠Q+90° = 180°
2∠Q = 90°
∠Q = 45°
Detailed Solution:
Since, ∠OPQ = 90° (Angle between tangent and radius)
Let ∠PQO be x°, then ∠QOP = x°
(Since OPQ is an isosceles triangle) (OP = OQ) (given)
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

In ΔOPQ,
∠OPQ + ∠PQO + ∠QOP = 180° (Sum of the angles of a triangle)
∴ 90° + x° + x° = 180°
⇒ 2x° = 180° – 90° = 90°
⇒ x = 90°/2 = 45°.
Hence, ∠OQP is 45°.
OR
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

∠APB = 90° (angle in semi-circle) and ∠ODB = 90° (radius is perpendicular to tangent)
ΔABP ~ ΔOBD
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
⇒ 26/13 = AP/8
Hence, AP = 16 cm.


Case Study-1

Q.13. ‘Skysails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The sky sails technology allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively.
Based on the following figure related to sky sailing answer the questions:

Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

(i) In the given figure, if tan θ = cot (30° + θ), where Q and 30° + θ are acute angles, then the value of θ.
(ii) Find the value of tan 30°. cot 60°.

(i) Given, tan θ = cot(30° + θ)
= tan[90° – (30° + θ)]
= tan(90° – 30° – θ)
⇒ tan θ = tan(60° –  θ)
⇒ θ = 60° – θ
⇒ 2θ = 60°
⇒ θ = 30°.
(ii) tan 30° = 1/√3
cot 60° = 1/√3
Now, tan 30° × cot 60° = Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
= 1/3.


Case Study-2
Q.14. A ladder has rungs 25 cm apart. (see the below).
Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10

The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. The top and the bottom rungs are 2(1/2)m apart.
(i) Find the number of the rungs:
(ii) What is the length of the wood required for the rungs?

(i) The distance between the two rungs is 25 cm.
Hence, the total number of rungs Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10
(ii) Here, length of first rung be a and last rung be l, then
S= (n/2)(a+l)
Here, a = 25, l = 45
and n = 11
Then, the required length of the wood,
S11 = (11/2)[25 + 45]
= (11/2) x 70
= 385 cm.

The document Class 10 Mathematics: CBSE Sample Question Paper- Term II (2021-22) - 2 Notes | Study CBSE Sample Papers For Class 10 - Class 10 is a part of the Class 10 Course CBSE Sample Papers For Class 10.
All you need of Class 10 at this link: Class 10

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