CBSE Class 10

CBSE Sample Papers for Class 10 Maths Paper 3

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 3

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections — A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
Vertices of a triangle are (- 4, 0), (4, 0), (0, 3). What type of triangle is it ? [1]

Question 2.
If two positive integers p and q can be expressed as p = ab² and q = a²b; a, b being prime
numbers, then find LCM (p, q).  [1]

Question 3.
D and E are respectively the points on the sides AB and AC of a triangle ABC such that
AD = 2 cm, BD = 4 cm, BC = 9 cm and DE || BC. Then, find the length of DE.  [1]

Question 4.
Find a quadratic polynomial, the sum and product of whose zeroes are 3 and – 2 respectively.  [1]

Question 5.
For what value of p are 2p + 1, 12, 5p – 3 are three consecutive terms of an A.P. ?  [1]

Question 6.

SECTION-B

Question 7.
The decimal expansion of the rational number \(\frac { 33 }{ { 2 }^{ 2 }.{ 5 }^{ n } } \) terminates after 3 places of decimal. Then find the value of n.  [2]

Question 8.
Find the value of a, if the distance between the points A(- 3, – 14) and B(a, – 5) is 9 units  [2]

Question 9.
17 cards numbered 1, 2, 3, …, 17 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is :  [2]

• Odd
• A prime
• Divisible by 3
• Divisible by 3 and 2 bot

Question 10.
What is the probability that an ordinary year has 53 Sundays.  [2]

Question 11.
Determine the sum of first 100 odd natural number.  [2]

Question 12.  
Find the zeroes of the polynomial 2x² + x – 6.  [2]

SECTION-C

Question 13.
Find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.  [3]

Question 14.
The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC.  [3]
OR
A(6,1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ∆BCD. If E is the mid-point of DC, find the area of ∆ADE.

Question 15.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [3]
OR
If BL and CM are median of a triangle ABC right angled at A then prove that:
4(BL² + CM²) = 5BC².

Question 16.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.  [3]

Question 17.
From a circular piece of cardboard of radius 3 cm, two sectors of 90° have been cut off. Find the perimeter of the remaining portion to nearest hundredth centimetres. (Take π = \(\frac { 22 }{ 7 } \))  [3]
OR
Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60°.
(Use π = 3.14)

Question 18.
Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44m wide. In what time will the level of water in pond
rise by 21 cm ?  [3]
OR
How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.

Question 19.
Find the zeroes of the polynomial 4x² – x – 3 and verify the relationship between the zeroes and the coefficients.  [3]

Question 20.
Solve graphically : 4x + 6y = 9; 7x + 5y = 2.  [3]
OR
Sum of a two-digit number and the number formed by reversing the order of digits is 88. If difference of digits is 2 and the unit digit is greater, determine the number.

Question 21.
If sec θ = x + \(\frac { 1 }{ 4x } \), prove that : sec θ + tan θ = 2x or \(\frac { 1 }{ 2x } \) [3]

Question 22.
Find the mean of the following data :  [3]

SECTION-D

Question 23.
If the angle of elevation of a cloud from a point h metre above a lake is α and the angle of depression of its reflection in the lake is β, prove that the distance of the cloud from the point of observation is   [4]

OR
A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.

Question 24.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents of each circle from the centre of the other circle.  [4]

Question 25.
Show that the tangent at any point of a circle is perpendicular to the radius through the point of contact. In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Find the length of AT. [4]

Question 26.
A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.  [4]
OR
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Question 27.
A boat can go 24 km downstream and return in 5 hours. If the speed of the stream is 2 km/hr, find the speed of the boat in still water.  [4]
OR
Product of digits of a two-digit number is 14. When 45 is added to the number then the digits interchanged their places. Find the number.

Question 28.
Shalini gets pocket money from her father every day. Out of the pocket money, she saves ₹30 on the first day and on each succeeding day, she increases her savings by 500 paise.  [4]

1. Find the amount saved by Shalini on 10th day.
2. Find the total amount saved by Shalini in 30 days.

Question 29.

Question 30.
Draw “more than ogive” for the following data :  [4]

SOLUTIONS
SECTION-A

Answer 1.
Given vertices of triangle are A(- 4, 0), B(4, 0) and C(0, 3).
We know that distance formula is,

Answer 2.
We have, p = ab² and q = a²b
Now, p = a × b × b
and q = a × a × b
∴ H.C.F. (p,q) = ab
We know,
Product of two numbers = H.C.F × L.C.M

Answer 3.
Given : AD = 2 cm, BD = 4 cm, BC = 9 cm
and DE || BC
In ΔABC and ΔADE,
DE || BC
∴ ∠B = ∠D [Corresponding angles]
and ∠C = ∠E [Corresponding angles]
∴ By AA similarly axiom

Answer 4.
Given : Sum of zeroes = 3 and Product of zeroes = – 2
We know that, equation of quadratic polyomial is given as
= k[x² – (Sum of zeroes)x + Product of zeroes]
= k[x² – 3x – 2] where k ≠ 0.

Answer 5.
Given : 2p + 1, 12, 5p – 3 are three consecutive terms of A.P.
12 – (2p + 1) = (5p – 3) – 12
12 – 2p – 1 = 5p – 3 – 12
11 – 2p = 5 p – 15
-2p – 5p = -15 – 11
– 7p = – 26
p = \(\frac { 26 }{ 7 } \).

Answer 6.

SECTION-B

Answer 7.
Given : Decimal expansion of rational number \frac { 33 }{ { 2 }^{ 2 }.{ 5 }^{ n } } terminate after 3 decimals.
∴ Denominator of decimal number will be in the form = (2 x 5)n and n = 3 to terminate after 3 decimals.

Answer 8.
Given : A(- 3, – 14), B(a, – 5) and AB = 9 units
By distance formula,

On squaring both sides, we get
81 = (a + 3)² + (9)²
81 – 81 = (a + 3)²
⇒ (a + 3)² = 0
On taking square root both side, we get
a + 3 = 0
a = – 3.

Answer 9.
Total number of cards = 1, 2, 3, …, 17.
n(S) = 17
(i) Probability that the number is odd.
Number of odd number cards = 1, 3, 5, 7, 9, 11, 13, 15, 17.
∴ n(E) = 9
Probability that number on the card is odd,
P(E) = \(\frac { n(E) }{ n(S) } \)
P(E) = \(\frac { 9 }{ 17 } \).

(ii) Probability that the number is a prime.
Number of prime number cards = 2, 3, 5, 7, 11, 13, 17
∴ n(E) =7
Probability that a card is prime number,
n(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 7 }{ 17 } \)

(iii) Probability that the number is divisible by 3.
Number of cards divisible by 3 = 3, 6, 9, 12, 15
∴ n(E) = 5
Probability that the number is divisible by 3,
P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 5 }{ 17 } \)

(iv) Probability the number is divisible by 3 and 2 both.
Number of cards are divisible by 3 and 2 both = Number of cards are divisible by 6 = 6,12
∴ n(E) = 2
Probability that the number is divisible by 3 and 2 both,
P(E) = \(\frac { n(E) }{ n(S) } \) = \(\frac { 2 }{ 17 } \)

Answer 10.
1 year has 365 days = (52 x 7 + 1) days
Ordinary year has 52 Sunday + 1 day
1 day may be {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(S) = 7
That 1 day may be Sun = {Sun}
∴ n(E) = 1
Probability of 53 Sundays in ordinary year
P(E) = \(\frac { n(E) }{ n(S) } =\frac { 1 }{ 7 } \)

Answer 11.
Sum of first 100 odd natural numbers i.e.,
1 + 3 + 5 + 7 + 9+ …
These are in arithmetic progression, with
a = 1, d = 3 – 1 = 2, n = 100
We know that

Answer 12.
Given polynomial is,
2x² + x – 6 = 0
2x² + 3x – 3x – 6 = 0
2x(x + 2) – 3(x + 2) = 0
(x + 2) (2x – 3) = 0
Either x + 2 = 0 ⇒ x = -2 or 2x – 3 = 0 ⇒ x = \(\frac { 3 }{ 2 }\)
∴ Zeroes of the polynomial are – 2 and \(\frac { 3 }{ 2 }\)

SECTION-C

Answer 13.
Given : 1251, 9377 and 15628 leave remainders 1, 2 and 3 respectively.
Required number = H.C.F. of (1251 – 1), (9377 – 2) and (15628 – 3)
= H.C.F. of 1250, 9375 and 15625
By Euclid’s lemma,
15625 = 1 x 9375 + 6250
9375 = 1 x 6250 + 3125
6250 = 2 x 3125 + 0
∴ H.C.F. of 15625 and 9375 = 3125.
and H.C.F. of 3125 and 1250 = 625.
Again by Euclid’s lemma
3125 = 2 x 1250 + 625
1250 = 2 x 625 + 0
∴ H.C.F. of 1250, 9375 and 15625 = 625
Hence, 625 is largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

Answer 14.
Given : Vertices of triangle are A(2, 9), B(a, 5) and C(5, 5).
By distance formula,

Answer 15.

Answer 16.
Given : PA and PB are tangents to a circle having centre O.
To prove : ∠AOB + ∠APB = 180°.

Proof : OA is the radius of circle and PA is the tangent to the circle.
∴ ∠OAP = 90° …(i)
Similarly, OB is the radius of circle and PB is the tangent to the circle.
∴ ∠OBP = 90°
In quadrilateral APBO,
∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
90° + ∠APB + 90° + ∠BOA = 360°
∠APB + ∠BOA = 360° – 180°
∠APB + ∠BOA = 180°.

Answer 17.
We have, radius of cardboard, r = 3 cm.
Two sectors AOB and COD have been cut off in the circular cardboard.
∴ Perimeter of remaining cardboard = OA + length of arc APD + OD + OB + Length of arc BQC + OC
= r + Length of arc APD + r + r + Length of arc BQC + r
= 4 x r + length of arc APD + length of arc BQC
|
OR
Given : Radius of circle, r =12 cm
Central angle of segment = 60°
Area of segment = Area of sector OACD – Area of ∆AOB

Answer 18.
Given : Diameter of cylindrical pipe = 14 cm


OR
Given : Diameter of lead ball = 4.2 cm
∴ Radius, r = \(\frac { 4.2 }{ 2 }\) = 2.1 cm
Dimensions of lead piece = 66 cm × 42 cm × 21 cm
Volume of rectangular lead piece = l × b × h = 66 × 42 × 21 cm³

Answer 19.
Given polynomial is
4x² – x – 3 = 0
⇒ 4x² – 4x + 3x – 3 =0
⇒ 4x(x – 1) + 3(x – 1) =0
⇒ (x – 1) (4x + 3) = 0
Either x – 1 = 0 ⇒ x = 1 or x + 3 = 0 ⇒ x = – \(\frac { 1 }{ 4 }\)
∴ Zeroes of the polynomial are 1 and – \(\frac { 3 }{ 4 }\)
Now,

Answer 20.



The point of intersection of these lines is (- 1.5, 2.5).
∴ x = -1.5 and y = 2.5
OR
Let the ten’s place of digit be x.
Therefore, unit place of digit will be (x + 2)
Two-digit number = 10 × x + (x + 2) = 11x + 2
Reverse of the number = 10(x + 2) + x = 11x + 20
According to the question,
11x + 2 + 11x + 20 = 88
22x = 88 – 22
x = \(\frac { 66 }{ 22 }\) = 3
∴ Two-digit number = 11x + 2 = 11 × 3 + 2 = 35.

Answer 21.

Answer 22.

SECTION-D

Answer 23.




Putting the value of x in equation (i), we get
h = 8 + √3 . 8√3
h = 8 + 24 = 32 m
and x = 8 × 1.732 = 13.856 m
Thus, height of the hill = 32 m
and distance between hill and ship = 13.856 m.

Answer 24.
Steps of construction :

• Draw a line segment AB = 8 cm.
• With A as centre and radius 4 cm, draw a circle.
• With B as centre and radius 3 cm, draw another circle.
• Draw perpendicular bisector of AB which meets AB at M.
• With M as centre and MA as radius, draw a circle intersecting the circle with centre A at P and Q and the circle with centre B at R and S.
• Join AR, AS, BP, BQ which are the required tangents.

Answer 25.
Given : A circle C(0, r) and a tangent l at point A.
To prove : OA ⊥ l.
Construction : Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle at C.

Proof : We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.
Clearly, OA = OC
Now, OB = OC + BC
∴ OB > OC
⇒ OB > OA
⇒ OA < OB
Thus, OA is shortest than any other line segment joining O to any point on l.
Thus, OA ⊥ l.
Given : OT = 4 cm
∠OTA = 30°
Construction : Join OA.
Now, OA is the radius of circle and AT is the tangent to the circle.

∴ OA ⊥ AT
In ΔOAT, ∠A = 90°
∴ cos T = \(\frac { AT }{ OT } \)
cos 30° = \(\frac { AT }{ 4 } \)
\(\frac { \sqrt { 3 } }{ 2 } \) = \(\frac { AT }{ 4 } \)
∴ AT = 2√3 = 2 × 1.732 = 3.464 cm

Answer 26.
Given :
Radius of cylinder = Radius of cone = r = 60 cm
Height of cylinder (h) =180 cm
and Height of cone (H) =120 cm

Now,

OR
Given :
Diameter of well = 3 m
∴ Radius, r = \(\frac { 3 }{ 2 } \)
Depth of well, h = 14 m
and Width of embankment = 4 m
Volume of sand dug out = πr²h

Answer 27.
Given : Speed of stream = 2 Km/h
Let the speed of the boat in still water = x km/hr
∴ Speed of boat in downstream = (x + 2) km/hr
and speed of boat in upstream = (x – 2) Km/hr
We know
Time = \(\frac { Distance }{ Speed } \)
∴ Time taken to go 24 Km Up stream = \(\frac { 24 }{ x-2 } \)
and time taken to go 24 Km downstream = \(\frac { 24 }{ x+2 } \)
According to the question,



OR
Let the unit digit be x and tens digit be y
Number = 10y + x
and, reverse of the number = 10x + y
According to the question,
xy = 14
and, 10y + x + 45 = 10x + y
⇒ 9x – 9y = 45
⇒ x-y =5
⇒ x = 5 + y
Substituting the value of x in equation (i), we get
(5 + y)y = 14
y² + 5y – 14 = 0
y² + 7y – 2y – 14 = 0
y(y + 7)-2(y + 7) =0
(y + 7) (y-2) =0
y + 7 = 0 ⇒ y = – 7 (digit is not negative)
y – 2 = 0 ⇒ y = 2
x =5 + y = 7
When y = 2, then
∴ Required two-digit number = 10y + x = 10 × 2 + 7 = 27

Answer 28.
Money saved on 1st day = ₹ 30
Money saved on IInd day = ₹ 30 + 500 p = ₹ 35
Money saved on IIIrd day = ₹ 30 + 500 p + 500 p = ₹ 35
and so on.
Amount of money saved on successive days is an AP with
First term, a = 30 and common difference, d = 5.
(i) Money saved on 10th day,
We know that
∴ a_n = a + (n – 1 )d
a₁₀ = a + 9d = 30 + 9 x 5 = ₹ 75
(ii) Money saved in 30 days

Answer 29.

Answer 30.

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CBSE Sample Papers for Class 10 Maths Paper 5

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 5

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
AOBC is a rectangle whose three vertices are A(0, 3), 0(0, 0) and B(5, 0). Find the length of its diagonal CO.

Question 2.
In the given figure, if ∠AOB = 125°, then find the measure of ∠DOC.

Question 3.
A vessel has 136 lit. of oil. Another vessel has 92 lit. of oil. What is the capacity of the largest possible ladle which can measure out all the oil in each vessel in exact number of times ?

Question 4.
For what value of k, the system : kx + 3y = 11; 2x + 5y = 3, has unique solution.

Question 5.
If one root of 2x² + kx – 6 = 0 is 2, find the value of k.

Question 6.
Evaluate : sin 30° cos 60° + sin 60° cos 30°.

SECTION-B

Question 7.
If the HCF of 85 and 153 is expressible in the form 85m – 153, find the value of m.

Question 8.
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle or not.

Question 9.
A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is :
(i) Black ball
(ii) Red ball
(iii) Not a green ball
(iv) Green or white ball

Question 10.
A two digit number is selected at random. What is the chance that will be :
(i) A composite number which is odd
(ii) Successor of a prime number

Question 11.
Can x – 1 be the remainder on division of a polynomial p(x) by 2x + 3 ? Justify your answer.

Question 12.
Find k if the sum of roots of equation x² – x + k(2x – 1) is Zero.

SECTION-C

Question 13.
Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Question 14.
If P(9a – 2, – b) divides line segment joining A(3a + 1,- 3) and B(8a, 5) in the ratio 3 : 1, find the value of a and b.

Question 15.
In the given figure, D and E trisect BC. Prove that 8AE² = 3AC² + 5AD²

Question 16.
In the given figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then find the measure of ∠BAT.

A triangle ABC is drawn to circumscribe 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 8 cm and 6 cm respectively. Find the sides AB and AC.

Question 17.
A path of 4 m width runs round a semi-circular grassy plot whose circumference is \(163 \frac { 3 }{ 7 } \) m.
Find :
(i) The area of the path.
(ii) The cost of gravelling the path at the rate of Rs 1.50 per m².
(iii) The cost of turfing the plot at the rate of 45 paise per m².

OR

All the vertices of a square lie on a circle. Find the area of the square, if area of the circle is 1256 cm². (Use π = 3.14)

Question 18.
500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m3 ?

OR

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm ?

Question 19.
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

OR

If sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Question 20.
Solve for x :\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -9\left( x+\frac { 1 }{ x } \right) +14=0\)

Question 21.
If tan (A + B) = √3 and tan (A – B) = \(\frac { 1 }{ \sqrt { 3 } } \);0° < A + B ≤ 90°; A > B, find sin 2A and cos 6B.

Question 22.
Find the mean of the following data :

SECTION-D

Question 23.
The angle of elevation of a stationary cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level ?

OR

The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45°. Calculate the height of the tower.

Question 24.
Construct a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Question 25.
In a class test, the sum of Kamal’s marks in Maths and English is 40. Had he got 3 marks more in Maths and 4 marks less in English, the product of their marks would have been 360. Find his marks in two subjects separately.

Question 26.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel of its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

OR

A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cuboidal depression to hold the pens and pins, respectively. The dimensions of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

Question 27.
Solve graphically x – y = 0, 2x + 3y – 30 = 0. Also find coordinates of points where 2x + 3y – 30 = 0 meets axis of X and Y.

OR

Students of a class are made to stand in rows. If 4 students are extra in each row, there would be two rows less. If 4 students are less in each row, there would be 4 more rows. Find the number of students in the class.

Question 28.
Two pipes running together can fill a tank in \(5 \frac { 5 }{ 21 } \) minutes. If one pipe takes 1 minute more than the other, then find the time in which each pipe would individually fill the tank.

Question 29.
If sin α = a sin β and tan α = b tan β, then prove that cos² α = \(\frac { { a }^{ 2 }-1 }{ { b }^{ 2 }-1 } \)

Question 30.
Draw more than ogive’ and ‘less than ogive’ for the following distribution on the same graph. Also find the median from the graph.

SOLUTIONS

SECTION-A

Solution 1:
We know, diagonals of rectangle are equal.
OC = AB
=> \(OC=\left| \sqrt { { (5-0) }^{ 2 }+{ (0-3) }^{ 2 } } \right| \)
=> \(OC=\left| \sqrt { 25+9 } \right| =\left| \sqrt { 34 } \right| \)
=> \(OC=\sqrt { 34 } \) units

Solution 2:
Given
∠AOB = 125°
We know, opposite sides of a quadrilateral circumscribing a circle subtend supplements angles at the centre of a circle.
∴∠AOB +∠DOC = 180°
=> 125° + ∠DOC = 180°
=> ∠DOC = 180° – 125°
= 55° Ans.

Solution 3:
Largest capacity of the ladle which can measure out all the oil
= H.C.F. of 136 and 92
Now, 136 = 2 x 2 x 2 x 17
and 92 = 2 x 2 x 23
H.C.F. (136, 92) = 4
∴ 4 litre capacity ladle can measure out all the oil in each vessel in exact number of times. Ans.

Solution 4:
Given equations are,
kx + 3y – 11 = 0
and 2x + 5y – 3 =0

Solution 5:
Given equation is,
2x² + kx – 6 = 0
One root of the equation is 2.
x = 2
2(2)² + k(2) – 6 = 0
8 + 2k – 6 = 0
2k = – 2
k = – 1.

Solution 6:
We have,
sin 30° cos 60° + sin 60° cos 30°
=> \(\frac { 1 }{ 2 } .\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } .\frac { \sqrt { 3 } }{ 2 } \)
=> \(\frac { 1 }{ 4 } +\frac { 3 }{ 4 } \)
=> 1

SECTION-B

Solution 7:
85 = 5 x 17
153 = 3 x 3 x 17
∴H.C.F. (85, 153) = 17
But
∴H.C.F. = 85m – 153
17 = 85m – 153
17 + 153 = 85m
\(m= \frac { 170 }{ 85 } \)
= 2

Solution 8:
Let the given points be A(5, – 2), B(6, 4), C(7, – 2).
Now,

Here, we observe that
AB = BC
and AB+BC≠AC
∴Points A,B,C forms an isosceles triangle with AB = BC

Solution 9:
We have,
Number of red balls = 5
Number of white balls = 8
Number of green balls = 4
Number of black balls = 7
Total balls = 24

Solution 10:
Total two-digit numbers 10 to 99 = 90
∴n(S) = 90
(i) Composite numbers which is odd = {15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}
∴ n(E) = 24
Probability of odd composite number
= \(\frac { n(E) }{ n(S) } =\frac { 24 }{ 90 } =\frac { 4 }{ 15 } \)
(ii) Successor of a prime number = {12, 14, 18, 20, 24, 30, 32, 38, 42, 44, 48, 54, 60, 62, 68, 72, 74, 80, 84, 90, 98}
∴ n(E) = 21
Probability of a successor of a prime number
= \(\frac { n(E) }{ n(S) } =\frac { 21 }{ 90 } =\frac { 7 }{ 30 } \)

Solution 11:
No, degree of the remainder is always less than degree of the divisor.

Solution 12:
Given equation is,
x² – x + k(2x – 1) = 0
=> x² – x + 2kx – k = 0
=> x² + (2k – 1)x – k = 0

SECTION-C

Solution 13:
Let q be the quotient and r be the remainder when n is divisible by 3.
Therefore, n = 3q + r, where r = 0, 1, 2
=> n = 3q or n = 3q + 1 or n = 3q + 2
Case I: If n = 3q, then n is divisible by 3 but n + 2 = 3q + 2 and n + 4 = 3a + 4 are not divisible by 3.
Case II: If n = 3q + 1, then n + 2 = 3q + 3 = 3(q + 1), which is divisible by 3 and n + 4 = 3q + 5, which is not divisible by 3.
So, only (n + 2) is divisible by 3.
Case III: If n = 3q + 2, then n + 2 = 3q + 4, which is not divisible by 3 and n + 4 = 3q + 6 = 3(q + 2) is divisible by 3.
So, only (n + 4) is divisible by 3.
Hence, one and only one out of n, (n + 2), (n + 4) is divisible by 3.

Solution 14:
By section formula,

Solution 15:
Here, D and E trisect BC.
=>BD = DE = EC = \(\\ \frac { 1 }{ 3 } \) BC
or BE = 2 BD,
BC = 3BD
In right ∆ABD,
we have
AD² = AB² + BD²….(i)

Solution 16:
In the given circle,
∠ABC = 90°
∠ACB + ∠CAB = 90°
50° + ∠CAB = 90°
∠CAB = 90° – 50°
∠CAB = 40°
Now, OA is radius of the circle and AT is the tangent of circle
∠OAT = 90°
∠OAB + ∠BAT = 90°
40° + ∠BAT = 90°
∠BAT = 90° – 40°
∠BAT = 50°

Solution 17:
We have

Solution 18:
Given
length of pond, l = 80m
breadth of pond, b = 50m

Solution 19:
Odd integers between 1 and 1000 divisible by 3 are 3, 9,15, 21, … 999
The above series form an A.P.
where a = 3, d = 9 – 3 = 6 and an = 999
We know

Solution 20:
We have
\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -9\left( x+\frac { 1 }{ x } \right) +14=0\)
\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2 \right) -9\left( x+\frac { 1 }{ x } \right) +14-4=0\)

Solution 21:
We have
tan(A+B) = √3
=> tan(A+B) = tan 60°

Solution 22:

SECTION-D

Solution 23:
Let B be the point of oservation, P be the cloud and P’ be its reflection in the take.
AB = 2500 m
∠PBC = 15°
and ∠CBP = 45°
Let the height of cloud from lake be h m.
OP = OP’= h m

Solution 24:
Steps, of construction :
(1) Draw a line segment AB = 4 cm.
(2) Construct ∠XAB = 90°.
(3) With A as centre, draw an arc of radius 3 cm intersecting AX at C.
(4) Join BC. Thus, ∆ABC is obtained.
(5) Draw an acute angle ∠BAY below of AB.
(6) Mark points A1, A2, A3, A4, A5 on AY such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5.
(7) Join A3B.
(8) Draw a line through A5, parallel to A3B intersecting extended line segment AB at B’.
(9) Through B’, draw a line parallel to BC intersecting AX at C’.
(10) Thus, ∆AB’C’ is obtained whose sides are \(\\ \frac { 5 }{ 3 } \) times the corresponding sides of ∆ABC.

Solution 25:
Let the marks obtained in Maths be x, then marks obtained in English = 40 – x
According to the question,
(x + 3) (40 – x – 4) = 360
=> (x + 3) (36 – x) = 360
=> – x² + 36x + 108 – 3x = 360
=> – x² + 33x – 252 = 0
=> x² – 21x – 12x + 252 = 0
=> x(x – 21) – 12(x – 21) = 0
(x – 21) (x – 12) = 0
Either x – 21 = 0 or x – 12 = 0
=> x = 21 or x = 12
If x = 21, then marks obtained in Maths = 21
and marks obtained in English = 40 – 21 = 19
If x = 12, then marks obtained in Maths = 12
and marks obtained in English = 40 – 12 = 28

Solution 26:
Given
Height of the cone = 20 cm
PQ (h) = 10 cm

Solution 27:
Given equations are
x – y = 0
y = x
=> x – y = 0

Solution 28:
Let one pipe fill the tank in x minute
then other will fill the tank in (x + 1) minutes.
According to the question,

Solution 29:
Given : sin α = a sin β
sin² α = > a² sin² β

Solution 30:

We hope the CBSE Sample Papers for Class 10 Maths Paper 5 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 5, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 8

These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 8

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
If two integers a and b are written as a = x³y² and b = xy⁴; x, y are prime numbers, then find H.C.F. (a, b).

Question 2.
Find the perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0).

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then find the measure of ∠POA.

Question 4.
Determine the pair of linear equations from 2x + 3y = 5, y + \(\\ \frac { 2 }{ 3 } \) x = 5, 4x = 6y + 10, which has infinite solution.

Question 5.
The first term of an A.P. is p and its common difference is q. Find its 10th term.

Question 6.
If 9 sec A = 41, find cos A and cot A.

SECTION-B

Question 7.
Find a point which is equidistant from the points A(- 5, 4) and B(- 1, 6). How many such points are there ?

Question 8.
Which term of A.P. 129, 125, 121, … is its first negative term ?

Question 9.
Write the denominator of the rational number \(\\ \frac { 257 }{ 5000 } \) in the form 2^m x 5ⁿ, where m, n are non – negatives. Hence, write its decimal expansion without actual division.

Question 10.
It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb ?

Question 11.
In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.

Question 12.
For what value of k the quadratic equation (2k + 3) x² + 2x – 5 = 0 has equal roots ?

SECTION-C

Question 13.
Find the smallest number which when divided by 161, 207 and 184 leaves remainder 21 in each case.

Question 14.
Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

OR

If the points A( 1, – 2), B(2, 3), C(a, 2) and D(- 4, – 3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Question 15.
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

Question 16.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Question 17.
The diameter of coin is 1 cm (fig.). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take π = 3.1416).

OR
Area of a sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc of the sector.

Question 18.
A hemispherical tank of diameter 3 m, is full of water. It is being emptied by a pipe at the rate of \(3 \frac { 4 }{ 7 } \) litres per second. How much time will it take to make the tank half empty ?

OR

A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar occupies 1/10 of the volume of the wall, then find the number of bricks used in constructing the wall.

Question 19.
The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

OR

Find the sum of first 17 terms of an A.P. whose 4th and 9th terms are – 15 and – 30 respectively.

Question 20.
Solve for x :
\({ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 4 }-13{ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }+36=0\)

Question 21.
If cosec A + cot A = p, then prove that sec A = \(\frac { { p }^{ 2 }+1 }{ { p }^{ 2 }-1 } \)

Question 22.
Find mode of the following data :

SECTION-D

Question 23.
From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be a and p. If the height of the lighthouse is h metres and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac { h\left( tan\alpha +tan\beta \right) }{ tan\alpha \quad tan\beta } \) meters

OR

A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.

Question 24.
Construct a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the first triangle.

Question 25.
State and prove the converse of Pythagoras theorem.

Question 26.
A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl ?

OR

A gulab jamun, contains sugar syrup about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Question 27.
A boat goes 12 km downstream and 26 km upstream in 8 hours. It can go 16 km upstream and 32 km downstream in same time. Find the speed of boat in still water and the speed of stream.

OR

Solve graphically x – 2y = 1; 2x + y = 7. Also find the coordinates of points where the lines meet X-axis.

Question 28.
A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

Question 29.
Show that :
\(\frac { sinA-cosA+1 }{ sinA+cosA-1 } =\frac { 1 }{ secA-tanA } \)

Question 30.
The median of the following data is 20.75. Find the missing frequencies ‘x’ and ‘y’, if the total frequency is 100.

SOLUTIONS

SECTION-A

Solution 1.
Given : a = x³y²
b = xy⁴
H.C.F.(a, b) = xy².

Solution 2.
Given vertices are A(3, 0), B(0, 4) and 0(0, 0).
OA = 3 units
OB = 4 units
In ∆AOB, ∠O = 90°
∴From Pythagoras theorem,
AB² = OA² + OB² = 3² + 4²
AB = √9+16 = 5 unit
Perimeter of ∆OAB = OA + OB + AB
= 3 + 4 + 5 = 12 unit.

Solution 3.
Given :
∠APB = 80°
In ∆OAP and ∆OBP,
∠B = ∠A [each 90°]
OA = OB [radii]
OP = OP [common]
By RHS congruency rule

Solution 4.
Given equations are,
2x + 3y – 5 = 0
y + \(\\ \frac { 2 }{ 3 } \)x – 5=0
and 4x + 6y = 10
=> 2x + 3y – 5 = 0
2x + 3y – 15 = 0
and 4x + 6y – 10 =0

Solution 5.
Given : The first term of A.P. = a
and common difference = q
we know, a_n = a + (n – 1)d
∴ a₁₀ = p + (10 – 1)d
=> a₁₀ = P + 9d.

Solution 6.
Given : 9 sec A = 41
=> sec A = \(\\ \frac { 41 }{ 9 } \)
cos A = \(\\ \frac { 1 }{ sec A } \)
cos A = \(\\ \frac { 9 }{ 41 } \)

SECTION-B

Solution 7:
Let the point P(x, y) be equidistant from the points A (- 5, 4) and B(- 1, 6).

\(\left| \sqrt { { (x+5) }^{ 2 }+{ (y-4) }^{ 2 } } \right| =\left| \sqrt { { (x+1) }^{ 2 }+{ (y-6) }^{ 2 } } \right| \)
Squaring both sides
(x + 5)² + (y – 4)² = (x + 1)² + (y – 6)²
x² + 10x + 25 + y² – 8y + 16 = x² + 2x + 1 + y² – 12y + 36
10x – 2x – 8y + 12y + 41 – 37 = 0
8x + 4y + 4 = 0
2x + y + 1 =0
The points lying on 2x + y + 1 = 0 are equidistant from A and B.

Solution 8:
Given A.P. is 129, 125, 121,…
Here, a = 129, d = 125 – 129 = 121 – 125 = – 4
Let nth term be the first negative term.
Then T_n < 0.
We know, nth term, T_n = a + (n – 1)d
= 129 + (n -1) (- 4)
= 129 – 4n + 4
= 133 – 4n
∴T_n< 0
∵133 – 4n < 0
=> 133 < 4n
=> 4n > 133
=> n =\(\\ \frac { 133 }{ 4 } \)
= \(33 \frac { 1 }{ 4 } \)
Hence, 34th term will be the first negative term.

Solution 9:
We have,

Solution 10:
Total bulbs = 600
Number of defective bulbs = 12
Number of non-defective bulbs = 600 – 12 = 588
probability of a non defective bulb = \(\frac { Number\quad of\quad non\quad defective\quad bulbs }{ total\quad bulbs } \)
= \(\\ \frac { 588 }{ 600 } \)
= \(\\ \frac { 147 }{ 150 } \)

Solution 11:
Total tickets 1, 2, 3, 4, …, 50
=> n(S) = 50
Prime number tickets = 2, 3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
n(E) = 15
Probability that the ticket bears a prime number
P(E) = \(\\ \frac { n(E) }{ n(S) } \)
= \(\\ \frac { 15 }{ 50 } \)
= \(\\ \frac { 3 }{ 10 } \)
= 0.3

Solution 12:
Given quadratic equation is,
(2k + 3)x² + 2x – 5 = 0
On comparing the equation with ax² + bx + c = 0, we get
a = 2k + 3, b = 2, c = – 5
For equal roots,
b² – 4ac = 0
(2)² – 4(2k + 3) ( – 5) = 0
4 + 20(2k + 3) = 0
4 + 40k + 60 = 0
40k = – 64
k = \(\\ \frac { -64 }{ 40 } \)
k = \(\\ \frac { -8 }{ 5 } \)

SECTION-D

Solution 13:
Given numbers are 161, 207 and 184.
161 = 7 x 23
207 = 3 x 3 x 23
∴ 184 = 2 x 2 x 2 x 23
L.C.M. (161, 207, 1841) = 2 x 2 x 2 x 3 x 3 x 23 x 7
= 11592
Smallest number which when divided by 161, 207 and 184, leaves remainder 21 in each case
= 11592 + 21
= 11613. Ans.

Solution 14:
Given points are A(k + 1, 2k), B(3k, 2k + 3), C(5k – 1, 5k)
If the points A, B and C are collinear, then
ar (∆ABC) = 0

Solution 15:
Given, ABCD is trapezium in which
AB || CD || PQ

Solution 16:
Let O be the common centre of two concentric circles and let AB be a chord of large circle touching the smaller circle at P.
Construction : Join OP
Since OP is the radius of the smaller circle and AB is tangent to this circle at P,
∴ OP ⊥ AB
We know that the perpendicualr drawn from the centre of a circle to any chord of the circle bisects the chord.

Solution 17:
Given : Diameter of coin = 1 cm
Radius of coin, r = 0.5 cm
PQ = QR = RS = PS
= 2 x 0.5 = 1 cm

Solution 18:
Given : Diameter of hemispherical tank = 3 cm
Radius, r = \(\\ \frac { 3 }{ 2 } \)
Now, Volume of hemispherical tank = \(\frac { 2 }{ 3 } { \pi r }^{ 3 } \)

Solution 19:
Given, a = 2, a_n = 50, S_n = 442
We know that

Solution 20:
We have
\({ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 4 }-13{ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }+36=0\)
let
\({ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }=a\)

Solution 21:
Given : cosec A + cot A = p ….(i)
We know,
cosec² A – cot² A = 1
=> (cosec A + cot A) (cosec A – cot A) = 1 [∵ a² – b² = (a+b)(a-b)]
=> (p) (cosec A – cot A) = 1

Solution 22:

SECTION-D

Solution 23:
Let PO be the lighthouse and A, B be the position of the two ships.
PO = h m
Angle of depression of ship at the point A = a°.
Angle of depression of ship at the point B = P°.
In ∆OPA, ∠P = 90°

Solution 24:
Steps of construction :
(1) Draw a line segment BC = 8 cm and make a right angle (∠DBC) at the point B.
(2) Taking B as centre and radius 6 cm, draw an arc which intersect BD at the point A.
(3) Join AC. Thus, ∆ABC is obtained.
(4) Draw a ray BX such that ∠CBX is an acute angle.
(5) Make four points B₁ B₂, B₃ and B₄ such that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄
(6) Join B₄C.
(7) Through the point B₃, draw a line B₃C’ parallel to B₄C which intersect the line segment BC at the point C’.
(8) Through the point C’, draw a parallel line C’A’ parallel to CA which intersect the line segment BA at the point A’.
(9) Thus, ∆A B’C’ is the required triangle whose sides are \(\\ \frac { 3 }{ 4 } \) of corresponding sides of ∆ABC.

Solution 25:
Converse of Pythagoras Theorem : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the side is a right angle.
Given : A triangle ABC such that AC² = AB² + BC²
Construction : Construct a triangle DEF such that
DE = AB, EF = BC and ∠E = 90°.
Proof : Since, ∆DEF is a right angled triangle with right angle at E, therefore, by Pythagoras theorem, we have

Solution 26:
Given :
Radius of bowl, r = 9 cm
For cylindrical bottle :
Radius, R = 1.5 cm
Height, H = 4 cm

Solution 27:
Let the speed of boat in still water be x km/h and speed of the stream be y km/hr.
.’. Speed of the boat in upstream = x – y
Speed of the boat in downstream = x + y

Solution 28:
Given :
Total amount = Rs 700
No. of cash prizes = 7
Each prize is Rs 20 less than preceding prize.
Therefore, the list of value of prizes form an A.P.
Let the first cash prize = Rs a
Difference between two consecutive prize = – Rs 20

Solution 29:
Given
L.H.S = \(\frac { sinA-cosA+1 }{ sinA+cosA-1 } \)

Solution 30:

We hope the CBSE Sample Papers for Class 10 Maths Paper 8 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 8, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 2

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 2.

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections – A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
What are the coordinates of the centroid of the triangle ABC if A(2,3), B(5, 8) and C(2, 1) ? [1]

Question 2.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle. [1]

Question 3.
Express 4 as a sum of two irrational numbers.  [1]

Question 4.
Check if the system 3x – y – 8 = 0, 4x – 12y + 16 = 0 has infinite solution.  [1]

Question 5.
Check if x = 1 is a solution of 3x² – 2x – 1 = 0.  [1]

Question 6.
∆PQR is right angled at Q. If tan R = \(\frac { 24 }{ 7 } \), find sec R and sin R.  [1]

SECTION-B

Question 7.
If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then find the relationship between
‘a’ and ‘b’.  [2]

Question 8.
There is a circular path around a sports field. Sania takes 24 minutes to drive one round of the field, while Ravi takes 16 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point ?  [2]

Question 9.
What is the probability that a number selected from the numbers 1, 2, 3,…., 25 is a prime number, when each of the given numbers is equally likely to be selected ?  [2]

Question 10.
If a number x is chosen at random from the numbers -2,-1, 0, 1, 2, what is the probability that x² < 2 ? [2]

Question 11.
Find the values of for which x² – 2x + k = 0 has real roots.  [2]

Question 12.
Write the value of a₃₀ – a₁₀ for the A.P. 4, 9, 14, 19, 24,….  [2]

SECTION-C

Question 13.
If the point A(2, – 4) is equidistant from P(3, 8) and Q(- 10, y), find the value of y. Also find the distance PQ. [3]

Question 14.
If ∆ABC ~ ∆DEF and AL and DM are their corresponding angle bisector segments then show that
\(\frac { AL }{ DM } =\frac { AB }{ DE } =\frac { BC }{ EF } =\frac { AC }{ DF } \)  [3]

Question 15.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.  [3]

Question 16.
Find the LCM of 396, 429 and 561 by prime factorization method.  [3]

Question 17.
The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make to cover the distance which the front wheel covers in 140 revolutions.  [3]
OR
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.

Question 18.
A rectangular water tank of base 11 m x 6 m contains water up to a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.  [3]
OR
A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.  [3]

Question 19. 
Find the zeroes of the quadratic polynomial x² + 13x + 30, and verify the relationship between the zeroes and the coefficients.  [3]

Question 20.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?  [3]
OR
How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ?

Question 21.

Question 22.
Find median for the following data :  [3]

SECTION-D

Question 23.
Construct a triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°, then constuct a triangle similar to ∆ABC with its sides equal to (\(\frac { 5 }{ 4 } \))th of the corresponding sides of ∆ABC.  [4]

Question 24.
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metre towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the
point A.  [4]
OR
As observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship during the period of observation. [4]

Question 25.
In figure, P is the mid-point of BC and Q is the mid-point of AP. If BQ when produced meets AC at R, prove that RA = \(\frac { 1 }{ 3 } \)CA  [4]

Question 26.
A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ? 22 per litre which the container can hold.  [4]
OR
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 to the nearest cm².

Question 27.
In a rectangle if length is increased and breadth is reduced by 2 metre, the area is reduced by 28 sq. m. If length is reduced by 1 metre and breadth is increased by 2 metre, the area increased by 33 sq. m. Find the length and breadth of the rectangle.  [4]
OR
Out of group of swans, \(\frac { 7 }{ 2 } \) times the square root of the total number are playing on the shore of a pond. The remaining two are swimming in water. Find the total number of swans.  [4]

Question 28.
The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ?  [4]

Question 29.  
Show \(\frac { { cot }^{ 2 }A }{ 1+cosecA } +1=cosec A\)  [4]

Question 30.
Find the value of p, if the mean of the following distribution is 7.5.  [4]

SOLUTIONS
Section – A

Answer 1.
Here A(2,3) B(5,8) and C(2,1).

Answer 2.
Let, radius of circle OP = r
∵ Tangent to a circle is perpendicular to the radius
∴ In ∆POQ, using Pythagoras theorem

OQ² = PQ² + OP²
25² = 24² + r²
⇒ r² = 25² – 24²
⇒ r² = (25 + 24)(25 – 24)
⇒ r² = 49
⇒ r = 7 cm

Answer 3.
We know that (2 + √3) and (2 – √3) are irrational numbers
(2 + √3) + (2 – √3)
2 + √3 + 2 – √3 = 4

Answer 4.
Given system of equations is,
3x – y – 8 =0
and 4x – 12y + 16 = 0
For infinite solution

Answer 5.
We have
3x² – 2x – 1 = 0
If x = 1,
Then,
3 × (1)² – 2 × 1 – 1 = 0
3 – 2 – 1 = 0
3 – 3 = 0
0 = 0
L.H.S = R.H.S
Hence, x = 1 is a solution of the given equation.

Answer 6.
We have,
tan R =\(\frac { 24 }{ 7 } \)
From ∆PQR, tan R = \(\frac { PQ }{ RQ } \)
In ∆PQR, PR² = PQ² + RQ²
PR² = 24² + 7²
PR² = 576+ 49
PR² = 625
PR = √625
PR 25

sec R = \(\frac { PR }{ RQ } =\frac { 25 }{ 7 }\)
sin R = \(\frac { PQ }{ PR } =\frac { 24 }{ 25 } \)

SECTION-B

Answer 7.
We have, A( 1, 2), O(0, 0) and C(a, b) are collinear.
Since, the given points are collinear, therefore, area of triangle formed by them is zero.
i.e., ar (∆AOB) = 0
⇒ \(\frac { 1 }{ 2 } \) x1(y2 – y3) + x2(y3 – yx) + x3{yx – y2) 1=0
⇒ \(\frac { 1 }{ 2 } \) 1(0 – b) + 0(b – 2) + a(2 – 0) i =0
⇒ – b + 2a = 0
⇒ b = 2a.

Answer 8.
Given, Sania takes 24 minutes for one round and Ravi takes 16 minutes for one round. It means that Ravi takes lesser time than Sonia to drive one round of field.
Now,
Required time = L.C.M. of the time taken by Sonia and Ravi for completing one round
= L.C.M. of 24 and 16
∵ 24 = 2³ x 3 and 16 = 2⁴
∴ L.C.M. = 2⁴ x 3 = 16 x 3 = 48
Hence, Ravi and Sonia will meet each other at the starting point after, 48 minutes.

Answer 9.
Numbers are 1, 2, 3,…., 25.
∴ Total numbers n(S) = 25
Prime numbers between 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19, 23
⇒ Total prime numbers = n(E) = 9
∴ P(Prime Number) = \(\frac { n(E) }{ n(S) } \quad \frac { 9 }{ 25 }\)

Answer 10.
Given numbers are -2,-1, 0, 1, 2
⇒ Total numbers = 5
Here, (-2)² = 4, (-1)² = 1, 0² = 0, 1² = 1, 2² = 4
For x = – 1, 0, 1  x² < 2
⇒ Favourable outcomes = 3
p(x² < 2) = \(\frac { Favourable\quad outcomes }{ Total\quad outcomes } = \frac { 3 }{ 5 } \)

Answer 11.
Given equation is,
x² – 2x + k = 0
Here a = 1, b = -2, c = k
We know, D = b² – 4ac
⇒ D = (-2)² – 4 × 1 × k
⇒ D = 4 – 4k

For real roots,
D =0
⇒ 4 – 4 k =0
⇒ – 4k = – 4
⇒ k = \(\frac { -4 }{ -4 } \)
⇒ k = 1

Answer 12.
Given A.P. is, 4, 9, 14, 19, 24
Here, a = 4
d = 9 – 4=5
We know
a_n = a + (n – 1) d
a₃₀ = 4 + (30 – 1) 5
⇒ a₃₀ = 4 + 29 x 5
⇒ a₃₀ = 4 + 145
⇒ a₃₀ = 149
And a_n = a + (10 – 1) d = 4 + (30 -1 )5
= 4 + (9 x 5) = 4 + 45
⇒ a₃₀ = 49
∴ a₃₀ – a₁₀ = 149 – 49
= 100

SECTION-C

Answer 13.

AP =AQ
⇒ AP² = AQ²
Let (x₁, y₁) = (2, – 4)
(x₂, y₂) = (3, 8)
(x₃, y₃) = (-10, y)
Then,
(x₂ – x₁)² + (y₂ – y₁)² = (x₃ – x₁ )² + (y₃ – x₁)²
(3 – 2)² + (8 + 4)² = (-10 – 2)² + (y + 4)²
(1)² + (12)² = (12)² + (y + 4)²
1 + 144 = 144 + (y + 4)²
1 = (y + 4)²
±1 = y+4
⇒ y = -1 – 4 or y = 1 – 4
⇒ y=-5 or y = -3

Answer 14.

Answer 15.
Given : A circle with centre at O. AB and CD are tangents drawn at the end of diameter.
To prove : AB || CD
Proof : We know that a tangent at any point of a circle is perpendicular to the radius throught the point of contact.
∴ OM ⊥ AB and ON ⊥ CD

⇒ ∠OMB = 90°
∠OMB = 90°
∠OND = 90°
∠ONC = 90°
∠OMA = ∠OND
∠OMB = ∠ONC
These are the pair of alternate interior angles.
Since, alternate angles are equal, the line AB and CD are parallel to each other.
i.e, AB || CD

Answer 16.

Prime factorisation of the given numbers are,
396 = 2 × 2 × 3 × 3 × 11 = 2² × 3² × 11
429 = 3 × 143
561 = 3 × 187
∴ LCM (396, 429, 561) = 2² × 3² × 143 × 187 × 11 = 10589436

Answer 17.
Given, Diameter of front wheel = 80 cm
∴ Radius, r = \(\frac { 80 }{ 2 }\) = 40 cm
and Diameter of rear wheel = 2 m
∴ Radius, R = \(\frac { 2 }{ 2 }\) = 1 m = 100 cm
Number of revolutions covered by front wheel = 140.
Let the number of revolutions covered by rear wheel be n.
According to the question,
Distance covered by real wheel in n revolutions = Distance covered by front wheel in 140 revolutions
⇒ n x 2πR = 140 x 2πr
n × R = 140 x r
n x 100 = 140 x 40
n = \(\frac { 140\times 40 }{ 100 }\) = 56
Thus, the number of revolutions covered by rear wheel is 56.
OR
Given : Radius, OA = 3.5 cm
and OD = 2 cm
Area of shaded region = Area of quadrant OACB – Area of ∆DOB

Hence, area of shaded region is 6-125 cm².

Answer 18.
Given : Length of cuboid, l = 6 m
Breadth of cuboid, b = 11 m
Height of cuboid, h = 5 m
Radius of cylinder, r = \(\frac { 3.5 }{ 10 } \) m = \(\frac { 7 }{ 2 } \) m
Let height of water level in the cylindrical tank be H m
Since, water in rectangular tank is transferred to cylindrical tank.
∴ Volume of cuboid = Volume of cylinder of height H
l ×b × h =πr²H
6 x 11 x 5 = \(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 }\) × H
\(\frac { 30\times 2 }{ 7 } \) = H
H = \(\frac { 60 }{ 7 }\) m = 8.57 m.
OR
Given : Diameter of the copper rod = 1 cm
So, Radius (r) of the copper rod = 1/2 cm
Length of copper rod, h = 8 cm
Length of wire, H = 18 m = 18 x 100 cm
Let R be the radius of the cross-section (circular) of the wire.
We know that volume of cylinder = πr²h Now, according to the question
Now, according to the question,

Answer 19.
Let p(x) = x² + 13x + 30
= x² + 10x + 3x + 30
= x(x + 10) + 3(x + 10)
=(x + 3) (x + 10)
⇒ x + 3 = 0; x + 10 = 0
⇒ x = – 3, x = – 10
Let α = – 3
and β = – 10
Now, α + β = \(-\frac { b }{ a }\)
– 3 – 10 = \(-\frac { (-13) }{ 1 } \)
– 13 = – 13
and αβ = \(\frac { c }{ a } \)
(-3) (-10) = \(\frac { 30 }{ 1 }\)
30 = 30
Thus, the releationship between the zeroes and coefficients is verified.

Answer 20.
Given : First term of A.P., a = 17
Last term, l = 350
Common difference, d = 9
We know,
a_n = a + (n – 1) d
∵ 350 is the nth term of given A.P.,
∵ 350 = 17 + (n – 1) 9
350 – 17 = (n – 1) 9
333 = (n – 1) 9
\(\frac { 333 }{ 9 } \) = n – 1
37 = n – 1
37 + 1 =n
38 = n
We know that sum of n terms of A.P is given as,
S_n = \(\frac { n }{ 2 } \)[a + l]
⇒ S₃₈ = \(\frac { 38 }{ 2 }\)[17 + 350]
= 19 [367] = 6973
n = 38 and S_n = 6973.
OR
The first number greater than 10 which when divided by 4 leaves a remainder 3 is 11.
So, the next number will be 11 + 4 = 15
The other numbers will be 15 + 4 = 19, 19 + 4 = 23….
Thus, A.P. = 11, 15, 19, 23 …..
The last term of this A.P. will be 299.
We have to find n. i.e., number of terms in the A.P.
We know,
a + (n – 1 )d = a_n
⇒ 11 + (n – 1)4 =299
⇒ 11 + 4n – 4 = 299
⇒ 7 + 4n =299
⇒ 4n = 299 – 7
⇒ 4n = 292
⇒ n = \(\frac { 292 }{ 4 } \)
⇒ n = 73
∴ numbers lie between 10 and 300 which divided by 4 leaves a remainder 3.

Answer 21.

Answer 22.

N = 40 ⇒ \(\frac { N }{ 2 } \) = 20.
Frequency just greater than 20 is 22.
So, 30 – 49 will be median class.
Here, l = 30
ƒ = 4
c ƒ = 18
h = 9

SECTION-D

Answer 23.
Steps of construction :

• Draw a line segment AB = 4-5 cm
• Construct ∠YAB = 90°
• Taking A as centre and radius 4-5 cm, draw an arc intersecting AY at C.
• Join BC. Thus, ∆ABC is obtained. Y
• Draw any ray AX making acute angle with AB on the side opposite to the vertex A.
• Mark 5 points namely A_1, A₂, A_3, A_4, A₅ on AX such that, AA₁ = A₁A₂ = A₂A₃ = A₃ A₄ = A₄ A₅
• Join A₄ to B and draw a line through A₅ parallel to A₄B intersecting extended AB at B’
• Draw a line B’C’ parallel to BC to intersect AY at C’.

Thus, ∆AB’C’ is the required triangle whose sides are \(\frac { 5 }{ 4 }\) times of the corresponding sides of ∆ABC.

Answer 24.
Let CD be the tower of height y m.
Let BD = x.
In ∆CAD, we have

tan 30° = \(\frac { CD }{ AD } \)
\(\frac { 1 }{ \sqrt { 3 } } \) = \(\frac { y }{ 20+x } \)
20 + x = y√3
x = y√3 – 20
In ∆CBD, we have
tan 60°= \(\frac { CD }{ BD } \)
√3 = \(\frac { y }{ x } \)
⇒ x√3 = y
Putting the value of y in equation (i), we get
x = (x√3) (√3) – 20
x = 3x – 20
20 = 3x – x
20 = 2x
\(\frac { 20 }{ 2 } \) = x
10 = x
Putting the value of x in equation (ii), we get
10√3 = y
⇒ y = 10 × 1.732 = 1.732
Hence, distamce of the foot of tower from point A
= 20 + x = 20 + 10 = 30 m
and Height of tower = 17.32 m

OR

Let A and B be the two positions of the ship. Let d be the distance travelled by ship during the period of observation i.e. AB = d metres.
Period of observation i.e. AB = d meters.
Let the observer be at O, the top of the light house PO.
It is given that PO = 100 m and the angles of depression from O of A and B are 30° and 45° respectively.
∴ ∠OAP = 30° and ∠OBP = 45°

In ∆OPA, we have
tan 45° = \(\frac { OP }{ BP } \)
⇒ 1 = \(\frac { 100 }{ BP }\)
⇒ BP = 100 m
In ΔOPA, we have
⇒ tan 30° = \(\frac { OP }{ AP } \)
⇒ \(\frac { 1 }{ \sqrt { 3 } } \) = \(\frac { 100 }{ d+BP }\)
⇒ d + BP = 100√3
⇒d + 100 = 100√3 [∵ BP = 100 m]
⇒ d = 100√3 – 100
⇒ d = 100(√3 – 1)
= 100(1.732 – 1) = 73.2 m
Hence, the distance travelled by the ship from A to B is 73.2 m

Answer 25.
Given : ∆ABC in which P is the mid-point of BC, Q is the mid-point of AP, such that BQ produced meets AC at R.
To prove : RA = \(\frac { 1 }{ 3 } \) CA.

Construction : Draw PS || BR, meeting AC at S.
Proof : In ∆BCR, P is the mid-point of BC and PS || BR.
∴ By basic proportionality theorem, S is the mid-point of CR.
⇒ CS = SR
Similarly, in ∆APS, Q is the mid-point of AP and QR || PS.
∴ R is the mid-point of AS.
⇒ AR=RS
From equations (i) and (ii), we get
AR =RS = SC .
⇒ AC =AR +RS +SC = 3 AR
AR = \(\frac { 1 }{ 3 } \) AC = \(\frac { 1 }{ 3 } \) CA

Answer 26.
Given :
Radii of upper end of frustum, R = 20 cm
Radii of lower end of frustum, r = 8 cm
Height of frustum, h = 16 cm

We know that
1 cm³ = 0.001 litre
⇒ Volume of container = 10.46 litre
Cost of 1 litre milk = ₹ 22
⇒ Total cost = 22 × 10.46 = ₹ 230.12
OR
Let r cm be the radius and h cm be height of the cone .
Then r = 0.7 cm and h = 2.4 cm
let r₁ cm be the radius, l cm be the slant height and h₁ cm be the height of the cone.
Then r₁ = 0.7 cm and h₁ = 2.4 cm


Now, total surface area of the remaining solid
= C.S.A. of cylinder + C.S.A. of cone + Area of base of the cylinder
= 2πrh + πr₁l + πr₂
= 2πrh + πr₁ + πr²   [∵ r = r1]
= [ 2 × \(\frac { 22 }{ 7 } \) × 0.7 × 2.4 + \(\frac { 22 }{ 7 } \) × 0.7 × 2.5 + \(\frac { 22 }{ 7 } \) × 0.7 × 0.7 ] cm²
= ( 10.56 + 5.5 + 1.54 )cm2 = 17.6 cm²

Answer 27.
Let Length of rectangle = x
and Breadth of rectangle = y
∴ Area of rectangle = xy
According to question,
(x + 2)(y – 2) = xy – 28
and (x – 1) (y + 2) = xy + 33
From equation (i), we have
xy – 2x + 2y – 4 = xy – 28
– 2x + 2y = – 28 + 4
– 2x + 2y = – 24
2x – 2y = 24
x – y =12
x =12 + y
From equation (ii), we have
(x -1) (y + 2) = xy + 33
xy + 2x – y – 2 = xy + 33
2x – y =33 + 2
2x – y =35
Putting the value of x from equation (iii) to equation (iv)
2(12 + y) – y =35
24 + 2y – y =35
24 + y = 35
y = 35 – 24 =11
Putting the value y in equation (i), we get
x = 12 + 11 = 23
Hence, Length of rectangle = 19 m
Breadth of rectangle = 11 m
OR
Let the total number of swans be x².
Then, number of swans playing at the pond shore
= \(\frac { 7 }{ 2 } \sqrt { { x }^{ 2 } } =\frac { 7 }{ 2 } x\)
Number of swans playing in water = 2
Total number of swans = No. of swans playing on the shore + No.of swans playing in water
⇒ x² = \(\frac { 7 }{ 2 } \)x + 2
⇒ 2x² = 7x + 4
⇒ 2x² – 7x – 4 = 0
⇒ 2x² – 8x + x – 4 = 0
⇒ 2x(x – 4) + (x – 4) = 0
⇒ (2x + 1) + (x – 4) = 0
⇒2x + 1 = 0 or x – 4 = 0
⇒ x = \(\frac { -1 }{ 2 } \) or x = 4
But x = \(\frac { -1 }{ 2 } \) is not feasible. Thus, x = 4
Hence, number of swans = x² = 4² = 16.

Answer 28.

Let A be the middle-most point and 13 flags are fixed at points A₁,A₂,A₃,…. A₁₃, to the left of A and remaining to the right of A.
Distance travelled for fixing and coming back to A for
(i) first flag = (2 + 2)m = 4m = a
(ii) second flag = (4 + 4) m = 8 m = a₂
(iii) third flag = (6 + 6) m = 12 m = a₃
∴ a_1, a₂, a₃…., form an AP in which
a = 4, a₂ = 8, a₃ = 12……n = 13
and common difference, d =8 – 4 = 4
S₁₃ = \(\frac { 13 }{ 2 } \)[2 x 4 + (13-1) 4]
= \(\frac { 13 }{ 2 } \)(8 + 48) = \(\frac { 13 }{ 2 } \) × 56 = 364
∴ Distance travelled to fix 13 flags to the left of A = 364 m.
Similarly, the distance travelled to fix remaining 13 flags to right of A = 364 m
∴ Total distance travelled = 2 × 364 m = 728 m.

Answer 29.

Answer 30.

We have, ∑f_i = 41 + p, ∑f_ix_i = 303 + 9p
But Mean = 7.5
∴ Mean = \(\frac { \Sigma { f }_{ i }{ x }_{ i } }{ \Sigma { f }_{ i } }\)
⇒ 7.5 = \(\frac { 303+9p }{ 41+p } \)
⇒ 7.5 × (41 + p) = 303 + 9p
⇒ 307.5 + 7.5p = 303 + 9p
⇒ 9p – 7.5p = 307.5 – 303
⇒ 1.5p = 4.5
⇒ p = 3

We hope the CBSE Sample Papers for Class 10 Maths Paper 2 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 2, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 English Communicative Set 1

These Sample papers are part of CBSE Sample Papers for Class 10 English Communicative. Here we have given CBSE Sample Papers for Class 10 English Communicative Set 1

Time allowed: 3 hours
Maximum marks: 80

General Instructions

❖ The question papers divided into three sections :
Section A : Reading 20 marks
Section B : Writing and Grammar 30 marks
Section C : Literature 30 marks
❖ All questions are compulsory.
❖ You may attempt any section at a time.
❖ All questions of that particular section must be attempted in the correct order.

SECTION-A : READING
(Attempt all question from this section)

Question 1.
Read the passage given below and answer the questions that follow :

Ship of the Desert

1. Known as “ships of the desert”, camels have been used for transporting goods across deserts for thousands of years. In fact, camels are the only desert animals that can carry heavy loads of goods and travel for a long period of time without food or water. Transportation, however, is not the only benefit that camels can offer us. Desert people also rely on camels for their milk, meat and fur. Even camels’ droppings are useful—desert people use camels’ manure as fuel.

2. Weighing more than 1,500 pounds, adult camels can reach the height of six feet at their shoulders and seven feet at their humps. Camels have two hoofed toes on each foot, under which a leathery pad links the two toes. When camels walk, they spread their toes as wide apart as possible to prevent their feet from sinking into the sand. The tough, leathery pads under their feet also allow camels to walk on a stony and rough ground. Camels are nicknamed as the “ships of the desert” because they walk like the motion of a rolling boat and move both their feet on one side of their bodies together, then both feet on the other.

3. There are two different species of camels—Dromedary and Bactrian. To differentiate one from the other is easy; just count how many humps a camel has ! With just one hump on their backs, Dromedary camels live in North Africa and the Middle East. Bactrian camels have two humps and they exist only in China and Central Asia. Interestingly, camels’ humps are like their secret safe. When there is plenty of food any vegetation in deserts, including thorny twigs and salty plants that other desert animals cannot have, camels overeat and store the extra as fat in their humps. The excessive fat in their humps is like a safety net allowing camels to have enough energy to walk extra miles until they find something to eat.

1.1. Attempt any eight of the following questions on the basis of the passage you have read:  [1 x 8 = 8]
(i) Who are known as the ‘ships of the desert ?’
(ii) What are the other benefits of the camel that desert people rely on ?
(iii) What is the height and the weight of a camel ?
(iv) How do camels prevent themselves from sinking in the sand ?
(v) What are the two different species of camels ? .
(vi) Where do Dromedary camels live and how many humps do they have ?
(vii) Where do Bactrian camels exist ?
(viii) Where do camels store their extra fat ?
(ix) How does the extra fat help the camels ?

Question 2.
Read the passage given below and answer the questions that follow:  [12]

Tools of Persuasion

1. Persuasion is the art of convincing someone to agree with your point of view. According to the ancient Greek philosopher Aristotle, there are three basic tools of persuasion: ethos, pathos and logos. Ethos is a speaker’s way of convincing the audience that he is a credible source. An audience will consider a speaker credible if he seems trustworthy, reliable and sincere. This can be done in many ways. For example, a speaker can develop ethos by explaining how much experience or education he has in the field. After all, you are more likely to listen to an advice about how to take care of your teeth from a dentist than a firefighter. A speaker can also create ethos by convincing the audience that he is a good person who has his best interests at heart. If an audience cannot trust you, you will not be able to persuade them.

2. Pathos is a speaker’s way of connecting with the emotions of the audience. For example, a speaker who is trying to convince an audience to vote for him might say that he alone can save the country from a terrible war. These words are intended to fill the audience with fear, thus making them want to vote for him. Similarly, a charity organisation that helps animals might show an audience the pictures of injured dogs and cats. If the audience feels bad for the animals, they will be more likely to donate money.

3. Logos is the use of facts, information, statistics or other evidences to make your argument more convincing. An audience will be more likely to believe you if you have data to back up your claims. For example, a commercial for soap might tell you that laboratory tests have shown that their soap kills all 100% of the bacteria living on your hands. Presenting evidence is much more convincing than simply saying, “our soap is the best!”

4. Use of logos can also increase a speaker’s ethos; the more facts a speaker includes in his argument, the more likely you are to think that he is educated and trustworthy. Although ethos, pathos and logos all have their strengths, they are often most effective when they are used together. Indeed, most speakers use a combination of ethos, pathos and logos to persuade their audiences. The next time you listen to a speech, watch a commercial, or listen to a friend trying to convince you to lend him some money, be on the lookout for these ancient Greek tools of persuasion.

2.1. On the basis of your reading of the passage, answer any four of the following questions in
about 30-40 words each:                                                                                                              [2 x 4 = 8]
(i) What are the tools one can use for persuasion ?
(ii) How can a speaker evoke fear in the audience ?
(iii) How do numbers help in convincing people to do something ? Give an example.
(iv) Why is it that people often give importance to education ?
(v) How can a speech be very effective ?

2.2. On the basis of your reading of the passage, fill any two of the following blanks with
appropriate words/ phrases.                  [1 x 2 = 2]
(i) An audience will consider a speaker credible if he seems ……….
(ii) Logos is the use of facts …………
(iii) Most speakers use a combination of……….

2.3. Attempt any two of the following. Find out the words that mean the same as below:
[1 x 2 = 2]
(i) Trustworthy (Paragraph 2)
(ii) Influence (Paragraph 3)
(iii) Indication (Paragraph 4)

SECTION-B : WRITING AND GRAMMAR
(Attempt all question from this section)

Question 3.
(a) You are Rajesh Kumar living at 51,7th Cross Street, OMBR Layout and Bengaluru 43. The
residents of your area are facing a lot of inconvenience due to poor maintenance of the public park of your locality. Write a letter to the editor of a local daily drawing the attention of the concerned higher authorities towards the problem and requesting them to solve it.         [8]

OR

(b) Write the letter to Jasmeet Traders & Co., Furniture Manufacturers, Haryana as the Principal, Vaidik Kanya Public School, Haryana placing an order for school furniture.

Question 4.
Write a short story, in about 200-250 words, with any one set of the cues given below. Give a suitable title to the story.  [10]
No one would believe what happened that night ……….

OR

Storms were certain that night but it wasn’t the weather that caused the turmoil…….

Question 5.
Fill in any four of the following blanks choosing the most appropriate option from the ones given below. Write the answers in your answer-sheet against the correct blank numbers.  [1 x 4 = 4]
(a) We went out…………….the fact that it was raining.
(i) despite
(ii) in spite
(iii) even as

(b) ………..she has got the right qualifications, she lacks experience.
(i) in spite of
(ii) although
(iii) even as

(c) He told me…………he loved me, but he was lying.
(i) that
(ii) what
(iii) so

(d) ………..you know, I work hard.
(i) as
(ii) that
(iii) since

(e) you focus on your studies; you will not pass.
(i) unless
(ii) if
(iii) whether

Question 6.
In the following passage one word has been omitted in each line. Write the missing word, in any four sentences of the given paragraph, along with the word that comes before and the word that comes after it in the space provided.    [1 x 4 = 4]

Question 7.
Rearrange any four of the following word clusters to make meaningful sentences. [1 x 4 = 4]
(a) ship/ violently/ the/ storm/ rocked/ the
(b) masterpiece/ artist/ painstakingly/ the/ his/ worked/ at
(c) gift/ free/ a/dad/ offered/ firm/ by/ the/ was.
(d) i/ read/ paper/ in/ the/ burglar/ been/ caught/ had/ the/ that.
(e) nightfall/ began/ cricketers/ the/ at/ their/ piercing/ calls.

SECTION-C : LITERATURE
(Attempt all question from this section)

Question 8.
Read the extract given below and answer the questions that follow. Write the answers in your
answer sheets in one or two lines only.  [1 x 4= 4]
(a) The ice was here, the ice was there,
The ice was all around :
It cracked and growled, and roared and howled,
i Like noises in a swound.
(i) What are the sailors surrounded by ?
(ii) List the adjectives used to describe the crunching noise of ice.
(iii) What are the sailors experiencing ? ‘
(iv) Find the synonym of’rumbled.’

OR

(b) Caesar: I am ashamed I did yield to them.
(i) What is Caesar ashamed about ?
(ii) What does Caesar decides to do ?
(iii) What made Caesar decide to go out to the Senate ?
(iv) Find the synonym of the word’embarrassed.’

Question 9.
Answer any four of the following questions in 30-40 words each:    [2 x 4 = 8]
(i) How did the author react when he drove Nicola and Jacopo back to the city ?
(ii) Why Lavinia did not get upset on seeing Helen ‘the ghost’ ?
(iii) How can you say that the frog was very determined in his singing ?
(iv) What role does Victoria play in ‘The Dear Departed ?’
(v) What were Michael’s thoughts on how Sebastian entered the computer games ?

Question 10.
Attempt any one out of the two following long answer type questions in 100-120 words.  [8]
(a) The frog has a good self-image. Discuss how he thrives even after the nightingale dies.

OR

(b) Discuss the fickleness of the mob in the play ‘Julius Caesar’.

Question 11.
(a) Answer the following question based on prescribed novel text for extended reading in
about 200-250 words.        [10]

The Diary of a Young Girl

(i) Discuss Anne Frank’s early entries that reveal a bubbly girl of thirteen and her awareness of reality.

OR

(ii) Discuss the theme of loneliness in adolescence in Anne Frank’s Diary.

(b) Answer the following question based on the prescribed novel text for extended reading in about 200-250 words.

The Story of My Life

(i) Discuss the “The Frost King Incident” as given in the ‘The Story of My Life’ by Helen Keller.

OR

(ii) How did Dr. Alexander Graham Bell influence Helen Keller’s life ?

ANSWERS
SECTION-A

Answer 1.
1.1
(i) Camels are known as the ships of the desert.
(ii) Desert people also rely on camels for their milk, meat and fur.
(iii) Adult camels weigh more than 1,500 pounds and can reach a height of six feet at their shoulders and seven feet at their humps.
(iv) When camels walk, they spread their toes as wide apart as possible to prevent their feet from sinking into the sand.
(v) The two different species of camels are Dromedary and Bactrian.
(vi) Dromedary camels live in North Africa and the Middle East and have only one hump.
(vii) Bactrian camels have two humps and they exist only in China and Central Asia.
(viii) Camels store their extra fat in their humps.
(ix) The extra fat in their humps gives them the energy to walk extra miles until they find food.

Answer 2.
2.1
(i) One can use any of the three tools of persuasion according to the ancient Greek Philosopher Aristotle. They are ethos, pathos and logos.
(ii) The speaker can use pathos while speaking, to evoke emotions of fear and to convince the audience of his cause. His words are intended to create fear in the audience.
(iii) Use of data in logos can convince people to do or buy something as there are proofs in numbers. We might buy a soap that kills 100% bacteria based on an advertisement which claims that they have done the laboratory research.
(iv) Education gives credibility to any person who is educated and the people believe him based on the fact that he is knowledgeable and one can trust him.
(v) Ethos, pathos and logos all have their strengths, however, a speech can be effective when they are used together. Indeed, most speakers use a combination of ethos, pathos and logos to persuade their audiences.

2.2
(i) An audience will consider a speaker credible if he seems trustworthy, reliable and sincere.
Logos is the use of facts, information, statistics, or other evidences to make your argument more convincing.

2.3
(i) credible
(ii) convince
(iii) evidence

SECTION-B

Answer 3.
(a)
51,7th Cross Street,
OMBR Layout,
Bengaluru- 43
23rd June, 20xx
The Editor,
The Hindustan Times,
Brigade Gardens,
Church Street,
Bengaluru – 560001

Sub: About poor maintenance of the Public Park of our locality.

Respected Sir,

Through the columns of your esteemed newspaper, I would like to express my serious concern on the poor maintenance of the public park of our locality.

The public park of our area is in a pitiable condition. The gardeners are not regular so, the plants are not pruned regularly, as a result of which, the park gives an unpleasant appearance. The park has become a thoroughfare for stray animals as there is no gate at the entrance of the park. It has become a haven for the anti-social elements to gather in the park in the evening making it difficult for the women folk to have a walk and also for children to play in the evening. Our regular complaints to the local authorities have failed to bring any change in the situation.

Therefore, I would like to request the concerned higher authorities to improve the condition of the park for the convenience of the local residents. I shall be highly obliged to you for publishing my letter in your newspaper.

Thanking you,

Yours faithfully,
Rajesh Kumar

OR

(b)
Principal,
Vaidik Kanya Public School.
Haryana.
21st March, 20 xx
M/s. Jasmeet Traders & Co.
Furniture Manufacturers,
Haryana.

Sub: Order for school furniture

Sir,

On behalf of the Vaidik Kanya Public School, I am placing an order for the following items of school furniture:

I would be grateful if you supply the above articles at the earliest before the session starts. Please send the bill along with the furniture and we will dispatch a demand draft for the amount of the bill in your favour within a week after receiving the articles.

Thanking You,

Yours Faithfully,
Sidharth Sinha (Principal)

Answer 4.

Believe It or Not

No one would believe what happened that night when I went to the village festival. I was just ten years old. I had been to my grandparents’ house which is in a very small village. We used to go there every summer when we were young, until our grandparents came to live with us.

It was the night of the temple festival, when the whole village came to life in many colours and sounds. Music blared in every street, the streets were decorated with streamers and the front of every house was adorned with rangoli and it was a resplendent sight. In the evening, there was a cultural programme organized by the village heads. It was held in the open grounds outside the village.

I participated in a dance competition and was having such a good time that I did not see time going by quickly. I had made friends in the village and they planned to go out into the fields, after everything was over, for a walk. I went along with them and as we walked, I heard the tinkling of bells and saw a ghost-like figure. I was terrified and my heart palpitated and I soon realized that my friends were off the path and I was alone.

That was the last time I went with friends or anybody to the fields. Believe it or not but I am sure, I really saw a ghost that night.

OR
Stormy Night

Storms were certain that night but it was not the weather that caused the turmoil. It was getting stormy with the winds howling through the window. The weather was getting terrible by every minute. I heard the branches snap. Electricity was cut off and I could see through the windows that the power lines were cut and hanging dangerously on the poles and branches of the trees.

The storm was nothing at all when we realised that our father had got a wheezing attack which got severe because of the weather. My mother gave him the usual medication but that did not seem to help. We took turns to sit with him and rub his back to ease the pain but it did not seem to get better with time.

Usually, in such situations, we used to take him to the hospital and he would be given oxygen that would ease his condition. We were in turmoil because we could not take him out in such bad weather. Every moment was like being on pins and needles.

The long and disturbed wait for relief got over when the rains let up and a neighbour helped us in taking our father to the hospital. Though it was just for a night, it was unforgettable and turbulent. The sun shone through in morning and the storm had passed.

Answer 5.
(a) despite
(b) Although
(c) that
(d) As
(e)Unless

Answer 6.

Answer 7.
(a) The storm rocked the ship violently.
(b) The artist painstakingly worked at his masterpiece.
(c) Dad was offered a free gift by the firm.
(d) I read in the paper that the burglar had been caught.
(e) The cricketers began their piercing calls at nightfall.

SECTION-C

Answer 8.
(a) (i) The sailors are surrounded by the ice as far as their eyes can see.
(ii) Cracked, growled, roared and howled.
(iii) The sailors are experiencing severe storm that has driven them to icy seas which is dismal as they can see neither human nor a beast.
(iv) The synonym of the word ‘rumbled’ is roared.

OR

(b) (i) Caesar is ashamed that he listened to Calpumia’s dreams and wished to stay at home.
(ii) Caesar decides to go out to the Capitol against the wishes of Calpumia.
(iii) Caesar had been coaxed and convinced by Decius Brutus who said that the dream meant positive things for Caesar and not as Calpumia feared. Caesar decided to go based on this counsel.
(iv) The synonym of the word ’embarrassed’ is ashamed.

Answer 9.
(i) The author waited outside until the boys returned and did not say a word until they reached the city. He wanted the boys to feel that their secret was safe and he admired them for their nobility in their hearts.

(ii) Lavinia, who had already decided to leave her husband as he was supposedly having an affair with a lady, comes to see him before she leaves and finally meets Helen, who is just as a ghost should be, not so pretty, Lavinia just smiles knowing that there is no need for doubting John, leaves and reconciles with him.

(iii) Though none liked the frog singing, he continued to croak from morning till evening and no amount of stones, bricks or complaints could stop him from doing what he did. This shows that the frog was very determined to keep on singing, in spite of criticism.

(iv) Victoria portrays the innocence against the manipulative and scheming adults in the play, who wish that their grandfather was dead. They use to get their work done and want her to behave properly when they themselves are not doing the right thing. She is a loving girl and still has the innocence which is not seen in the grown-ups.

(v) Michael feels that Sebastian’s memory must have plunged into the computer when he banged his head in the accident. The computer must have immediately stored his memory upon immediate contact. A little later, he also realises that the memory must have got transferred when Michael and his father picked the game at the fair.

Answer 10.
(a) The frog merily croaks which is disliked by all in the bog but that does not deter the frog from belting out in rain or shine, day or night. His determination is so extreme that neither sticks nor bricks, complaints or insults stopped his elated heart; until a nightingale came to the bog and every bird and animal was enraptured with her singing. The frog with his confidence takes the opportunity to deride the nightingale’s singing, makes money out of her talents and pushes the nightingale to extreme limits that it burst a vein and killed it. The frog was not concerned but blamed the nightingale for being stupid and influenced. The frog had the audacity to say that he can sing with panache and continue to out blare out loud in the bog.

OR

(b) The citizens of Rome clearly play an important role in the play ‘Julius Caesar’. The characters in the play are timed to express themselves and are very passionate. Throughout the play these emotions are communicated through various events.

An example of expression by the citizens occur during the funeral oration by Mark Antony. Brutus logially gives his reasons that required Caesar’s death. He informs them that he acted out of love of Rome and his desire to prevent tyrants from controlling it. The citizens embrace his words with cheers and understanding. However, their mood alters when Antony offers his interpretation of the situation. He passionately described the deeds Caesar performed on behalf of the citizens of Rome, which clearly contradicted the opinions of the conspirators that Caesar was too ambitious. This alteration in the minds of the crowd fells us about their ficklemindedness in Julius Caesar.

Answer 11.
(a) (i) The first entries in the diary tell us about the lively existence of a young Jewish girl who has just turned thirteen. She seems to bubble with laughter and joy, playing Ping-Pong, participating in pranks and flirting with young men. Although she seems popular with her school friends and is doted upon by her parents, Anne feels loneliness. Not having a best friend with whom she can share her emotions, she decides to write her thoughts and feeling into her diary, which she names “Kitty.”

Anne is particularly close to her father, Otto Frank. His father’s birthday gift to her, the red- checkered diary, is her favourite. He also trusts her enough to tell her about his plans to take the family into hiding. Anne is also close to her sister, Margot, even though she seems a bit jealous of her. She states that Margot is the smart one in the family while she struggles to do well in school to keep up with Margot.

In spite of her youth and happiness, Anne cannot disregard what is going on around her. She had to quit her Dutch school and attend a Jewish one. Additionally, she gives a long list of other restrictions that were put on the Jews. They cannot ride trains, own bicycles, work in most businesses, go for movies or visit Church. Anne finds the restrictions difficult, but she says that life is bearable. It is obvious that this optimistic and positive young lady tries to make the best of any situation.

OR

(ii) Anne Frank’s perpetual feeling of being lonely and misunderstood provides the impetus for her dedicated diary writing and colours many of the experiences she recounts. Even in her early diary entries, she writes about her friends and her lively social life. Anne expresses gratitude that the diary can act as a confidant with whom she can share her innermost thoughts. This might seem an odd sentiment from such a playful, amusing and social young girl but Anne explains that she is never comfortable discussing her inner emotions, even around close friends. Notwithstanding her excitement over developing into a woman and despite the specter of war surrounding her, Anne nonetheless finds that she and her friends talk only about trivial topics.

We learn later in the diary that neither Mrs. Frank nor Margot offers much emotional support to Anne. Though Anne feels very connected to her father and derives strength and encouragement from him, he is not a fitting confidant for a thirteen-year-old girl. Near the end of her diary, Anne shares a quotation she once read with which she strongly agrees : “Deep down, the young are lonelier than the old.” Because young people are less able than adults to define or express their needs clearly, they are more likely to feel lonely, isolated and misunderstood.

Feelings of loneliness and isolation also play out in the larger scheme of the annex. All the inhabitants feel anxious, fearful, and stressed because of their circumstances, yet no one wants to burden the others with such depressing feelings. As an outcome, the residents become impatient with one another over trivial matters and never address their deeper fears or worries. This continual masking and repression of serious emotions creates isolation and misunderstanding between all the residents of the annex.

(b) (i) The winter of 1892, was a troubling time for Helen. During an autumn at Fern Quarry she wrote a story called “The Frost King,” seemingly inspired by the fall foliage around her and sent it to Mr. Anagnos at the Perkins Institute as a gift. Mr. Anagnos loved it and was very impressed, so he published the story in one of the Institute’s reports. However, it was soon discovered that a very similar story called “The Frost Fairies” existed in a book that was published long before Helen wrote her story. Helen could not remember hearing such a story but Mr. Anagnos came to think he had been deceived and that Helen and Miss Sullivan had deliberately taken the words from “The Frost Fairies.”

Helen was investigated by a court comprised of the teachers and officers of the Perkins Institute and it took a toll on her morale, as she felt horrible that such a thing had happened. Miss Sullivan had never heard of “The Frost Fairies,” either. Eventually, it comes out that Mrs. Sophia C. Hopkins had read books to Helen during the summer she spent at Brewster while Miss Sullivan was away on vacation and “The Frost Fairies” must have been one of those. The strange words had remained in Helen’s mind, though she did not realise it. Helen was forgiven but the experience left her petrified to write for a long time. Even when writing a letter she felt that the words were not her own.

OR

(ii) As Helen grew older, so did her frustration of not being able to properly express herself. She had many terrible fits of anger that deeply troubled her parents. When she was six, her t parents took her to an occultist, Dr. Chisolm, in Baltimore to see if he could help her but it went in vain. Instead, the doctor advised to find Helen a teacher so that she could be educated. He suggested them to consult Dr. Alexander Graham Bell in Washington for recommendations. Finally, after writing to Dr. Anagnos at the Perkins Institution in Boston, they received confirmation that a teacher for Helen had been found.

Alexander Graham Bell helped Helen by suggesting her father the school that might be able f to provide Helen with a teacher. Helen dedicated ‘The Story of My Life’ to Bell because of his work with the deaf and also because he was the first to start her journey towards communicating and learning.
He had wonderful stories of his inventions and experiments to tell her and every subject he
touched, kept her fascinated. He made her feel confidant about herself and also dream that she could be an inventor one day. He was humorous and poetic and his love for children , was above all else. Helen Keller’s steps toward greater things, learning and experiences
were guided by Dr. Bell.

We hope the CBSE Sample Papers for Class 10 English Communicative Set 1 help you. If you have any query regarding CBSE Sample Papers for Class 10 English Communicative Set 1, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Hindi A Set 3

These Sample Papers are part of CBSE Sample Papers for Class 10 Hindi A. Here we have given CBSE Sample Papers for Class 10 Hindi A Set 3

निर्धारित समय : 3 घण्टे
अधिकतम अंक : 80

सामान्य निर्देश

* इस प्रश्न-पत्र में चार खण्ड हैं
खण्ड (क) : अपठित अंश -15 अंक
खण्ड (ख) : व्यावहारिक व्याकरण -15 अंक
खण्ड (ग) : पाठ्य पुस्तक एवं पूरक पाठ्य पुस्तक -30 अंक
खण्ड (घ) : लेखन -20 अंक
* चारों खण्डों के प्रश्नों के उत्तर देना अनिवार्य है।
* यथासंभव प्रत्येक खण्ड के प्रश्नों के उत्तर क्रमश: दीजिए।

खण्ड (क) : अपठित अंश

प्र.1. निम्नलिखित गद्यांश को ध्यानपूर्वक पढ़िए और नीचे लिखे प्रश्नों के उत्तर लिखिए-
सामाजिक समानता का अभिप्राय है कि सामाजिक क्षेत्र में जाति, धर्म, व्यवसाय, रंग आदि के आधार पर किसी प्रकार का भेदभाव न हो। सबको एक समान समझा जाए और सबको समान सुविधाएँ दी जाएँ। हमारे देश में सामाजिक समानता का अभाव है। जाति-प्रथा के कारण करोड़ों लोग समाज में अछूत के रूप में रहते हैं। उन्हें समाज से बहिष्कृत समझा जाता है और सामाजिक अधिकारों से वंचित कर दिया जाता है। हमारे समाज में लड़कियों के साथ भी भेदभाव किया जाता है। माता-पिता भी उन्हें वे सुविधाएँ नहीं देते जो वे अपने लड़कों को देते हैं। इस प्रकार की असमानता के कारण बहुत-सी लड़कियों का शारीरिक और मानसिक विकास नहीं हो पाता है। इससे समाज की उन्नति में बाधा पड़ती है। इस प्रकार की असमानता का दूर होना अत्यधिक आवश्यक है। नागरिक समता का अर्थ है कि राज्य में नागरिकों को समान अधिकार प्राप्त हों । कानून और न्यायालयों में गरीब-अमीर और ऊँच-नीच का कोई भेद न किया जाए। दंड से कोई अपराधी बेच न सके। उसी प्रकार राज्य के प्रत्येक नागरिक को राज्य-कार्य में समान रूप से भाग लेने का, मत देने का, सरकारी नौकरी प्राप्त करने को तथा राज्य के ऊँचे-से-ऊँचे पद को अपनी योग्यता के बल पर प्राप्त करने का अधिकार राजनीतिक समानता का द्योतक है।
(i) सामाजिक समानता से क्या अभिप्राय है?
(ii) हमारे देश में सामाजिक असमानता किन रूपों में दिखाई देती है?
(ii) नागरिक समानता से क्या अभिप्राय है?
(iv) लड़कियों का शारीरिक और मानसिक विकास सुचारु रूप से क्यों नहीं हो पाता है?
(v) इस गद्यांश का उपयुक्त शीर्षक दीजिए।

प्र. 2. निम्नलिखित काव्यांश को ध्यानपूर्वक पढ़कर पूछे गये प्रश्नों के उत्तर लिखिए
लहरों से डरकर नौका पार नहीं होती,
कोशिश करने वालों की कभी हार नहीं होती।
नन्हीं चींटी जब दाना लेकर चलती है,
चढ़ती दीवारों पर सौ बार फिसलती है।
मन का विश्वास रगों में साहस भरता है,
चढ़कर गिरना, गिरकर चढ़ना, ना अखरता है।
आखिर उसकी मेहनत बेकार नहीं होती,
कोशिश करने वालों की कभी हार नहीं होती।
डुबकियाँ सिंधु में गोताखोर लगाता है,
जा-जा कर खाली हाथ लौटकर आता है।
मिलते नहीं सहज ही मोती गहरे पानी में,
बढ़ता दुगुना उत्साह इसी हैरानी में,
मुट्ठी उसकी खाली हर बार नहीं होती,
कोशिश करने वालों की कभी हार नहीं होती।
असफलता एक चुनौती है, इसे स्वीकार करो,
क्या कमी रह गई, देखो और सुधार करो।
जब तक न सफल हो, नींद-चैन को त्यागो तुम,
संघर्ष का मैदान छोड़ कर मत भागो तुम।
कुछ किये बिना ही जय-जयकार नहीं होती,
कोशिश करने वालों की कभी हार नहीं होती।
(i) असफलता एक चुनौती है स्वीकार करो।’-पंक्ति द्वारा कवि क्या कहना चाहता है?
(ii) ‘मिलते नहीं सहज ही मोती गहरे पानी में।’ इस पंक्ति के आधार पर बताइए कि मोती कौन और कब प्राप्त कर पाता है?
(iii) चींटी से प्रेरणा लेकर हमें क्या लाभ प्राप्त होगा?
(iv) कविता में कवि क्या प्रेरणा दे रहा है?
(v) नन्हीं चींटी के उदाहरण द्वारा कवि ने कैसे लोगों को प्रेरणा दी है?

खण्ड (ख) : व्यावहारिक व्याकरण

प्र. 3. निम्नलिखित प्रश्नों के निर्देशानुसार उत्तर दीजिए
(क) तुम घर गए और वह रोने लगी। (सरल वाक्य में परिवर्तित कीजिए)
(ख) वे बोलते जा रहे थे और पिताजी का चेहरा गर्व में बदलता जा रहा था। (रचना के आधार पर वाक्य भेद बताइए)
(ग) घंटी बजी, छात्र पुस्तकें लेकर कक्षा से बाहर निकले। छात्र घर चले गए। (संयुक्त वाक्य में परिवर्तित कीजिए)

प्र. 4. निम्नलिखित प्रश्नों के निर्देशानुसार उत्तर दीजिए
(क) उसके द्वारा उछलकर पतंग पकड़ ली गई। (सरल वाक्य में परिवर्तित कीजिए)
(ख) स्त्रियों को पढ़ाने से अनर्थ होते हैं। (वाच्य का प्रकार बताइए)
(ग) दिलीप दौड़ा। (भाववाच्य में परिवर्तित कीजिए)
(घ) कवयित्री कविता पढ़ती है। (कर्मवाच्य में परिवर्तित कीजिए)

प्र. 5. निम्नलिखित वाक्यों रेखांकित पदों का पद परिचय लिखिए
(क) मैं अपनी मातृभूमि पर मर मिटॅगी।
(ख) जल्दी भागो, शेर आने वाला है।
(ग) अब हम क्या करें, मरते दम तक न यह शहनाई छूटेगी न काशी।
(घ) माँ ने नौकर से सब्जी मँगाई।

प्र. 6. निम्नलिखित प्रश्नों के उत्तर निर्देशानुसार लिखिए
(क) “नेक विलोकि धौं रघुबरनि ।
बाल-भूषन बसन, तन सुंदर रुचिर रज भरनि।
परस्पर खेलनि अजिर उठि चलनि गिरि परनि ।’
-उपर्युक्त काव्य-पंक्तियों में कौन-सा रस निहित है?
(ख) ‘अनुराग/ईश्वर विषयक रति’ स्थायी भाव किस रस का है?
(ग) निम्न काव्य-पंक्तियों में कौन-सा स्थायी भाव है?
“कहुँ सुलगत कोउ चिता कहुँ कोउ जाति बुझाई ।
एक लगाई जाति, एक की राख बहाई।।”
(घ) रस निष्पत्ति में सहायक अवयवों के नाम बताइए।

खण्ड (ग) : पाठ्य पुस्तक एवं पूरक पाठ्य पुस्तक

प्र. 7. निम्नलिखित गद्यांश को ध्यानपूर्वक पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए
फादर को याद करना एक उदास शांत संगीत को सुनने जैसा है। उनको देखना करुणा के निर्मल जल में स्नान करने जैसा था और उनसे बात करना कर्म के संकल्प से भरना था। मुझे ‘परिमल’ के वे दिन याद आते हैं जब हम सब एक पारिवारिक रिश्ते में बँधे जैसे थे जिसके बड़े फादर बुल्के थे। हमारे हँसी-मजाक में वह निर्लिप्त शामिल रहते, हमारी गोष्ठियों में वह गंभीर बहस करते, हमारी रचनाओं पर बेबाक राय और सुझाव देते और हमारे घरों के किसी भी उत्सव और संस्कार में वह बड़े भाई और पुरोहित जैसे खड़े हो हमें अपने आशीषों से भर देते। मुझे अपना बच्चा और फादर का उसके मुख में पहली बार अन्न डालना याद आता है और नीली आँखों की चमक में तैरता वात्सल्य भी-जैसे किसी ऊँचाई पर देवदारु की छाया में खड़े हों।
(क) करुणा के निर्मल जल में स्नान करना’ का आशय क्या है?
(ख) फादर बुल्के संकल्प से संन्यासी थे, मन से संन्यासी नहीं थे। कैसे?
(ग) फ़ादर बुल्के के व्यक्तित्व की विशेषताएँ बताइए।

प्र. 8. निम्नलिखित प्रश्नों के उत्तर संक्षेप में लिखिए
(क) बालगोबिन भगत की किन विशेषताओं के कारण उन्हें साधु कहा जाता था? अपने विचार लिखिए।
(ख) हालदार साहब बार-बार क्या सोचते और क्यों ?
(ग) गिरती आर्थिक दशा ने मन्नू भंडारी के पिता के व्यक्तित्व पर क्या प्रभाव डाला?’एक कहानी यह भी’ पाठ के आधार पर बताइए।
(घ) आपके विचार से बुल्के ने भारत आने का मन क्यों बनाया होगा? ‘मानवीय करुणा की दिव्य चमक’ पाठ के आधार पर उत्तर दीजिए।

प्र.9. निम्नलिखित पद्यांश से संबंधित प्रश्नों के उत्तर दीजिए
बादल, गरजो!
घेर घेर घोर गगन, धाराधर ओ!
ललित ललित, काले घुघराले ,
बाल कल्पना के-से पाले ,
विद्युत-छबि उर में, कवि, नवजीवन वाले!
वज्र छिपा, नूतन कविता
फिर भर दो-
बादल, गरजो!
(क) कवि बादल को क्या घेरने के लिए कह रहा है और क्यों ?
(ख) ललित काले धुंघराले बालों की कल्पना किसके लिए की गई है?
(ग) कवि बादल को गरजने के लिए क्यों कह रहा है?

प्र. 10. निम्नलिखित में से किन्हीं चार प्रश्नों के उत्तर दीजिए
(क) कवि ने ‘श्रीब्रजदूलह’ किसके लिए प्रयुक्त किया है तथा उन्हें ‘संसार रूपी मंदिर का दीपक’ क्यों कहा है?
(ख) गायकों को गायन के दौरान कौन-कौनसी कठिनाइयाँ आती हैं?
(ग) क्या आप इस बात से पूरी तरह सहमत हैं कि बेटी अपनी माँ के सबसे निकट और उसके सुख-दुःख की साथी होती है? ‘कन्यादान’ कविता के आधार पर उत्तर दीजिए।
(घ) कवि बादल से फुहार, रिमझिम तथा बरसने के स्थान पर गरजने के लिए क्यों कह रहे हैं?

प्र. 11. पहाड़ों पर पुरुषों की अपेक्षा स्त्रियों का जीवन अधिक कठिनाइयों से भरा है। उन कठिनाइयों का निवारण वे कर्तव्यप्रियता से ही करती हैं-सोदाहरण स्पष्ट कीजिए।

अथवा

“लड़के और बंदर पराई पीर नहीं समझते।” क्या आप इस बात से सहमत हैं? क्या यह उचित है? स्पष्ट कीजिए।

खण्ड (घ) : लेखन

प्र. 12. निम्नलिखित में से किसी एक विषय पर दिए गए संकेत-बिन्दुओं के आधार पर लगभग 200 से 250 शब्दों में निबंध लिखिए। [10]

(क) नोटबंदी : राष्ट्रहित की ओर एक बड़ा कदम

• नोटबंदी की घोषणा
• कालेधन पर वार
• राष्ट्रहित के लिए देश की जनता का ज़बरदस्त समर्थन

(ख) समाचार-पत्रों की उपयोगिता

• भूमिका
• आरम्भ एवं प्रसार
• महत्व एवं उपयोगिता
• सामाजिक परिवर्तन में समाचार-पत्रों की भूमिका
• उपसंहार

(ग) कन्या भ्रूण हत्या : कारण और निवारण

• भारतीय समाज का कलंक
• कन्याओं के प्रति समाज की दूषित सोच
• इसके कारण
• इस बुराई को रोकने के उपाय
• सरकारी प्रयास
• उपसंहार

प्र. 13. विद्यालय में दसवीं और बारहवीं कक्षा के अच्छे परिणामों पर प्रधानाचार्य को पत्र लिखकर सुझाइए कि उन्हें और अच्छा कैसे बनाया जा सकता है?

अथवा

अपने क्षेत्र में एक नया पुस्तकालय स्थापित करने हेतु सांसद महोदय को एक पत्र लिखिए।

प्र. 14. आपकी कम्पनी ने मोटापे की रोकथाम के लिये एक आयुर्वेदिक चूर्ण/सीरप/कैप्सूल बनाया है। 25-50 शब्दों में उसका एक विज्ञापन तैयार करें।

अथवा

विद्यालय के वार्षिकोत्सव के अवसर पर विद्यार्थियों द्वारा निर्मित हस्तकला की वस्तुओं की प्रदर्शनी के प्रचार हेतु लगभग 50 शब्दों में एक विज्ञापन लिखिए।

उत्तरमाला

खण्ड (क)

उत्तर 1. (i) सामाजिक क्षेत्र में जाति, धर्म, व्यवसाय, रंग आदि के आधार पर भेदभाव न करनी तथा सभी को एक समान सुविधाएँ एवं अधिकार प्रदान कराना सामाजिक समानता कहलाता है।
(ii) हमारे देश में सामाजिक असमानता कई रूपों में दिखाई देती है। जैसे-जाति-प्रथा यह सामाजिक असमानता का एक कारण है। इसके कारण करोड़ों व्यक्तियों को अछूत समझा जाता है और उन्हें सामाजिक अधिकारों से भी वंचित कर दिया जाता है। लड़कियों के साथ भेदभाव किया जाता है। उन्हें प्रायः वे सुविधाएँ नहीं दी जातीं जो लड़कों को दी जाती हैं।
(iii) राज्य में सभी नागरिकों को समान अधिकार प्राप्त हों। उनके साथ जाति, धर्म आदि के आधार पर कोई भेदभाव न किया जाए। गरीब-अमीर, ऊँच-नीच सभी को समान रूप से न्यायालयों से न्याय मिले। कोई भी अपराधी दंड से बच न पाए। इस प्रकार की समानता को ही नागरिक समानता कहते हैं।
(iv) माता-पिता प्राय: लड़कियों को वे सुविधाएँ नहीं देते हैं जो वे अपने लड़कों को देते हैं, जिसके कारण लड़कियों का शारीरिक और मानसिक विकास सुचारु रूप से नहीं हो पाता है।
(v) सामाजिक व नागरिक समानता।

उत्तर 2. (i) ‘असफलता एक चुनौती है स्वीकार करो।’ इस पंक्ति द्वारा कवि कहना चाहता है कि कोई फर्क नहीं पड़ता कि हम कितनी बार हार जाते हैं। वह कहता है कि हमें चुनौती के रूप में विफलता को स्वीकार करना चाहिए। हमें निराश नहीं होना चाहिए और आशा खोनी नहीं चाहिए, परन्तु हमें देखना चाहिए कि वास्तव में क्या कमी रही और क्या गलती की ? वह हमें सलाह देता है कि हमें प्रयास करते रहना चाहिए, जब तक कि सफलता न मिल जाये।
(ii) ‘मिलते नहीं सहज ही मोती पानी में।’
इस पंक्ति में सफलता को ‘मोती’ कहा गया है। इस सफलता रूपी मोती को वही प्राप्त कर सकता है जो मेहनत करता है और असफलता से नहीं डरता है।
(iii) चींटी से प्रेरणा लेकर हमें यह लाभ प्राप्त होगा कि हम उससे सीख सकेंगे कि जीवन में कितनी बार भी असफलताएँ आ जाएँ, प्रयास करना नहीं छोड़ना चाहिए। जब तक सफलता न मिल जाए तब तक मेहनत करते रहना चाहिए और दृढ़ता से आगे बढ़ते रहना चाहिए।
(iv) कविता में कवि प्रेरणा दे रहा है कि कोशिश करते रहना चाहिए, एक या उससे अधिक विफलता से निराश नहीं होना चाहिए। अंत में सफलता जरूर प्राप्त होगी।
(v) नन्हीं चींटी के उदाहरण द्वारा कवि ने उन लोगों को प्रेरणा दी है जो असफलता से डरते हैं, निराशा में रहते हैं और जो प्रयास नहीं करते हैं।

खण्ड (ख)

उत्तर 3.
(क) तुम्हारे घर जाते ही वह रोने लगी।
(ख) संयुक्त वाक्य।।
(ग) घंटी बजी और छात्र पुस्तकें लेकर कक्षा से बाहर निकलकर घर चले गए।

उत्तर 4. (क) उसने उछलकर पतंग पकड़ ली।
(ख) कर्तृवाच्य।
(ग) दिलीप से दौड़ा गया।
(घ) कवयित्री के द्वारा कविता पढ़ी जाती है।

उत्तर 5. (क) मर मिटूगी-अकर्मक क्रिया, स्त्रीलिंग, एकवचन, भविष्यत काल, उत्तम पुरुष, कर्तृवाच्य, ‘मैं’ क्रिया की कर्ता।
(ख) जल्दी-अव्यय, कालवाचक, क्रिया-विशेषण, क्रिया का विशेषण’ भागो’ ।
(ग) हम-पुरुषवाचक सर्वनाम, उत्तम पुरुष, पुल्लिग, कर्ता कारक।
(घ) नौकर-जातिवाचक संज्ञा, पुल्लिग, एकवचन, अपादान कारक में।

उत्तर 6. (क) वात्सल्य रस।
(ख) भक्ति ।
(ग) घृणा (जुगुप्सा) वीभत्स रस।
(घ) रस निष्पत्ति में सहायक अवयव हैं-स्थायी भाव, विभाव, अनुभाव तथा संचारी भाव।

खण्ड (ग)

उत्तर 7. (क) लेखक ने फादर बुल्के के स्वभाव की कल्पना ऐसी की है कि उनको देखकर ही मन में करुणा का भाव जाग्रत हो जाता था मानो निर्मल जल में स्नान कर रहे हों। उनको सुनकर मन कर्म करने के लिए दृढ़ हो जाता था।
(ख) फादर बुल्के संकल्प से संन्यासी थे। उन्होंने ईसाई पादरी बनकर घर-गृहस्थी से संन्यास तो ले लिया था, पर उनके मन में अपनों-आत्मीयों के प्रति गहरा लगाव था। सांसारिक लगाव संन्यासियों का लक्षण नहीं है। पर फादर बुल्के लेखक के प्रति, उनके परिवारीजनों के प्रति या अन्य परिचितों के प्रति गहरा स्नेह रखते थे। वे लेखक को गले लगाकर मिलते थे। वे जन्म, मृत्यु जैसे समारोहों में पूरे स्नेहभाव से सम्मिलित होते थे।
(ग) फादर बुल्के का व्यक्तित्व वात्सल्यमय और करुणापूर्ण था। उनकी बाँहें हर किसी को गले लगाने के लिए तैयार रहती थीं। वे अपरिचितों को भी परिचित की तरह प्रेम करते थे।

उत्तर 8. (क) बालगोबिन भगत गृहस्थ जीवन जीते हुए मोह-माया में नहीं बँधे थे। वे कबीर को अपना इष्ट मानते थे तथा अपने खेत की पैदावार को भी कबीरपंथी मठ को अर्पित कर देते थे। वहाँ से प्रसाद रूप में जो भी प्राप्त होता था, उसी से वे गुजारा करते थे। वे कभी झूठ नहीं बोलते थे और सबसे खरा व्यवहार करते थे। वे कभी भी किसी की चीज नहीं छूते थे। उनका आचरण साधु की। परिभाषा पर पूरी तरह खरा उतरता था। इसी कारण उन्हें साधु कहा जाता था।
(ख) हालदार साहब बार-बार यही सोचते थे कि कस्बे में सुभाषचंद्र बोस की मूर्ति पर चश्मा क्यों नहीं था? किसकी कमी के कारण चश्मा बनने से रह गया। फिर उन्हें याद आता था कि कस्बे वाले जैसे-तैसे कोई-न-कोई चश्मा नेताजी की मूर्ति पर लगा देते हैं। उन्हें कस्बे वालों का यह प्रयास प्रशंसनीय लगता था। हालदार साहब भी देशभक्त थे। नेताजी के प्रति उनके मन में गहरा सम्मान था। इसलिए वे आते-जाते उनकी मूर्ति के बारे में सोचा करते थे।
(ग) गिरती आर्थिक दशा ने मन्नू भंडारी के पिता के व्यक्तित्व पर बहुत गहरा प्रभाव डाला। उन्हें सदा शीर्ष पर रहने की आदत थी, परन्तु गिरती आर्थिक दशा ने उनकी स्थिति में बहुत परिवर्तन ला दिया था। अंग्रेजी-हिन्दी शब्दकोश पूरा करने पर उन्हें यश और प्रतिष्ठा तो खूब मिली, किन्तु धन नहीं। इससे उनकी आर्थिक स्थिति पहले से और बुरी हो गई थी। इससे उनके व्यक्तित्व के सारे सकारात्मक पहलू समाप्त होते चले गए। इसी कारण उन्हें अपना शहर इंदौर भी छोड़ना पड़ा था।
(घ) बुल्के जब इंजीनियरिंग के अंतिम वर्ष में थे, तब अचानक ही उन्होंने संन्यासी बनने एवं भारत आने का निर्णय लिया। भारत आने का विचार अचानक ही बुल्के के मन में नहीं उठा होगा। शायद उनके अवचेतन मन में भारतभूमि और भारतीय संस्कृति के प्रति पहले से ही प्रेम की भावना रही होगी, जो संन्यास लेने के समय उभरकर सामने आ गई।

उत्तर 9. (क) कवि बादल को पूरा आकाश घेरने के लिए कह रहा है। ऐसा प्रतीत होता है जैसे आकाश धरती का संरक्षक है।
(ख) बादल का रूप काला, घना और फैलाव भरा होता है। उनका रूप ऐसा होता है मानो किसी बच्चे के काले धुंघराले सुंदर केश हों। रूप, रंग और फैलाव की समानता के कारण बादलों की कल्पना ललित काले हुँघराले बालों के लिए की गई है।
(ग) कवि बादल को गरजने के लिए इसलिए कह रहा है क्योंकि वह वातावरण में जोश, पौरुष और क्रान्ति चाहता है। निराला को बादल की गड़गड़ाहट बहुत प्रिय है।

उत्तर 10. (क) कवि ने ‘श्रीब्रजदूलह’ का प्रयोग श्रीकृष्ण जी के लिए किया है। जिस प्रकार दीपक मंदिर को अपनी पवित्र लौ से प्रकाशित कर देता है, वैसे ही उसकी पवित्र लौ भक्तों के अंदर श्रद्धा की भावनाओं को और गहरा करती है और उन्हें प्रेरणा देती है, उसी प्रकार श्रीकृष्ण अपने तेज से संसार रूपी मंदिर को प्रकाशित करते हैं तथा उनके ज्ञान तथा तेजस्वी व्यक्तित्व से प्रेरणा एवं मार्गदर्शन पाकर लोग कर्मपथ पर आगे बढ़ते हैं।
(ख) गायकों को गायन के दौरान अनेक कठिनाइयाँ आती हैं। कभी-कभी ऊँचा स्वर उठाते समय उनका गला बैठ जाता है। कभी गाने की शक्ति समाप्त हो जाती है। कभी उत्साह मंद पड़ जाता है। कभी स्वर टूटने लगता है। कभी आलाप करते-करते गायक मस्ती में इतना खो जाता है कि वह गीत का मूल स्वर भूल जाता है। इस प्रकार वह भटक जाता है।
(ग) हम इस बात से पूर्णतः सहमत हैं कि बेटी अपनी माँ के सबसे निकट होती है। बेटी माँ के साथ सबसे ज्यादा समय बिताती है। बेटी का माँ के साथ अधिक लगाव होता है और उसके सुख-दुःख को भली-भाँति समझ सकती है। वह अपनी माँ की भावनाओं का सम्मान करना जानती है।
(घ) “फुहार’, ‘रिमझिम’ तथा ‘बरसना’ वास्तव में कोमलता व मृदुल सोच के प्रतीक हैं, किन्तु जब नवीन परिवर्तन लाना हो तो ‘गर्जन’ यानी विद्रोह और क्रान्ति की आवश्यकता होती है। इसलिए कवि बादल से फुहार, रिमझिम या बरसने के स्थान पर गरजने के लिए कह रहा है क्योंकि कवि का मानना है कि नवीनता लाने के लिए विध्वंस, विप्लव और क्रान्ति आवश्यक हैं। कवि
ने बादल से गरजने का आह्वान करके एक प्रकार से कविता द्वारा नूतन विद्रोह का आह्वान किया है।

उत्तर 11. पहाड़ों पर पुरुषों की अपेक्षा स्त्रियों का जीवन अधिक कठिनाइयों से भरा होता है, क्योंकि घरेलू जिम्मेदारियों का भार स्त्रियों को ही वहन करना पड़ता है। घर के सभी सदस्यों के लिए पीने के पानी का प्रबंध करना, खाना बनाने के लिए ईंधन इकट्ठा करना, मवेशियों को चराना आदि काम स्त्रियों को ही करने पड़ते हैं। इसके लिए उन्हें काफी परिश्रम करना पड़ता है। अपने परिवार की आर्थिक मदद के लिए वे सड़कें बनाने जैसा दुस्साध्य कार्य भी करती हैं।

इन सभी कामों के साथ-साथ उनकी मातृत्व साधना भी चलती रहती है, हमारे समाज में बच्चों के पालन-पोषण की प्राथमिक जिम्मेदारी स्त्रियों को ही निभानी पड़ती है। उन्हें पत्थर तोड़ने और सड़क बनाने जैसे खतरनाक कामों को करते समय अपने बच्चों को भी पीठ पर बाँधकर सँभालना पड़ता है। वे इन सभी कठिनाइयों का निवारण अत्यंत कर्तव्यप्रियता से करती हैं। भूख, मौत, दैन्य और जिंदा रहने की जंग में भी वे मुस्कुराती रहती हैं और अपने कर्तव्यों का सहज भाव से पालन करती रहती हैं। इस प्रकार की जीवन-शैली को उन्होंने स्वाभाविक तथा सहज रूप से अपना लिया है।

अथवा

यह बात पूर्णत: सत्य है कि जब लड़के खेल खेलते हैं, तो कई बार वे ऐसे खेल भी खेलने लगते हैं, जिससे बेवजह मासूम पक्षियों और पशुओं को कष्ट होता है। कई बार तो पक्षियों, तितलियों, चटियों आदि को अपनी जान से हाथ धोना पड़ता है। इसी प्रकार बच्चे गली में खेलते हुए कुत्तों, गधों आदि को बहुत तंग करते हैं, जिससे कई बार इन पशुओं को चोट भी लग जाती है।

इसी प्रकार बंदर किसी का भी कीमती सामान उठाकर ले जाते हैं तथा तोड़-फोड़ देते हैं, वे नहीं जानते कि ये किसी के लिए कितना आवश्यक है। मेरे विचार से ऐसा करना किसी भी दृष्टि से उचित नहीं है, क्योंकि पशु-पक्षियों को सताना, तंग करना और दु:ख देना किसी भी दृष्टि से सही नहीं कहा जा सकता है। ऐसे पशु-पक्षियों को तंग करना, जो बच्चों को तंग भी नहीं करते, पूरी तरह अनुचित है।

खण्ड (घ)

उत्तर 12. (क) नोटबंदी : राष्ट्रहित की ओर एक बड़ा कदम

नोटबंदी की घोषणा-नोटबंदी कालेधन पर करारा वार सिद्ध हुआ और धीरे-धीरे कालाधन बाहर आने लगा। नोटबंदी काले धन से देश को मुक्ति दिलाने के लिए देशहित में सरकार द्वारा उठाया गया एक बड़ा कदम था। इस घोषणा के तहत सरकार ने पाँच सौ व एक हजार के नोटों को बंद करके पाँच सौ तथा दो हजार के नए नोट छापने का निर्णय लिया। नोटबंदी की घोषणा से काला धन लिए बैठे लोगों की हवाइयाँ उड़ गईं और साथ ही आम आदमी को भी थोड़ी दिक्कतों का सामना करना पड़ा, किन्तु तमाम परेशानियों के बावजूद इस ऐतिहासिक निर्णय को देश की जनता का जबरदस्त समर्थन मिला क्योंकि जनता जानती थी कि देश को कालेधन से मुक्ति दिलाने के लिए ये परेशानियाँ झेली जा सकती हैं।

काले धन पर वार-विमुद्रीकरण का सही तात्पर्य यह है कि जब किसी देश कि सरकार अपनी पुरानी मुद्रा को कानूनी रूप से बंद कर देती है तो इस प्रक्रिया को विमुद्रीकरण (Demonetization) कहते हैं। 8 नवम्बर, 2016 को प्रधानमंत्री श्री नरेन्द्र मोदी जी ने भारत में 500 व 1000 के नोटों को अचानक बंद करने की घोषणा की। कालेधन का उपयोग आतंकवाद, अपराध और तस्करी जैसे आपराधिक कार्यों में भी बड़े पैमाने पर नगद लेन-देन के रूप में होता है।

अलगाववादियों, आतंकवादियों में नक्सलियों को काले धन से मिलने वाली फंडिंग बंद होने से उनकी कमर टूटने लगी है और उनकी क्रूर योजनाओं पर रोक लगाना संभव हो सका है।

राष्ट्रहित के लिये देश की जनता का जबरदस्त समर्थन-कुछ विपक्षी नेताओं ने ऐसे साहस भरे कदम की आलोचना भी की, किन्तु देश की जनता ने उनके कथनों पर ध्यान न देकर इस कदम का व्यापक स्तर पर समर्थन किया। इस कदम के नतीजे देश में दिखाई देने लगे हैं। अत: निष्कर्ष रूप में कहा जा सकता है कि नोटबंदी की घोषणा सरकार द्वारा देशहित में लिया गया एक उचित निर्णय था।

(ख) समाचार-पत्रों की उपयोगिता

भूमिका-समाचार-पत्र जनसंचार का एक सशक्त तथा प्रभावशाली माध्यम है। यद्यपि आज का युग प्रौद्योगिकी और संचार क्रांति का युग है, जिसमें टेलीविजन, इंटरनेट जैसे नवीन जनसंचार माध्यम भी उपलब्ध हैं, लेकिन समाचार-पत्रों की उपयोगिता तथा सार्थकता ज्यों-की-त्यों ही बनी हुई है। इनकी निष्पक्षता, निर्भीकता, प्रामाणिकता तथा विश्वसनीयता में लगातार वृद्धि हो रही है।

आरम्भ एवं प्रसार-समाचार-पत्रों का उद्भव सोलहवीं शताब्दी में प्रिंटिंग प्रेस के आविष्कार के साथ ही हो गया था, परन्तु इनका विकास अठारहवीं शताब्दी में ही रफ्तार पकड़ पाया। भारत में समाचार-पत्रों की शुरुआत ‘बंगाल गजट’ के प्रकाशन के साथ हुई थी। 1780 ई. में जेम्स ऑगस्टस हिकी द्वारा अंग्रेजी भाषा में समाचार-पत्र प्रकाशित किया गया था। हिन्दी का पहला समाचार-पत्र ‘उदंत मार्तंड’ था। इस समय भारत में कई भाषाओं में लगभग तीस हजार से भी अधिक त्रैमासिक, मासिक, पाक्षिक तथा दैनिक समाचार-पत्र प्रकाशित हो रहे हैं, इनमें ‘द टाइम्स ऑफ इंडिया’, ‘द हिंदू’, ‘नवभारत टाइम्स’, ‘दैनिक जागरण’, ‘जनसत्ता’, ‘हिन्दुस्तान टाइम्स’ आदि प्रमुख हैं।

महत्व एवं उपयोगिता- भारतवर्ष में स्वतंत्रता संग्राम के समय से ही समाचार-पत्रों की महत्ता तथा उपयोगिता बनी हुई है। देश के प्रसिद्ध नेताओं ने भारतीय जनता में देशप्रेम की भावना जगाने के लिए समाचार-पत्रों को ही माध्यम बनाया था। वर्तमान समय में इनकी महत्वपूर्ण भूमिका को देखते हुए इसे ‘लोकतंत्र का चौथा स्तंभ’ नाम से अलंकृत किया गया है। समाचार-पत्रों को लोकमत निर्माण, सूचनाओं का प्रसार, भ्रष्टाचार एवं घोटालों का पर्दाफाश तथा समाज की सच्ची तस्वीर प्रस्तुत करने के लिए जाना जाता है। देश के प्रथम नागरिक से लेकर एक आम आदमी तक इनकी पहुँच है, क्योंकि समाचार-पत्र जनसंचार का सबसे सस्ता, परन्तु विश्वसनीय माध्यम है। समाचार-पत्र सरकार एवं जनता के बीच एक सेतु का कार्य करते हैं।

सामाजिक परिवर्तन में समाचार-पत्रों की भूमिका-समय के साथ-साथ समाचार-पत्रों का कार्यक्षेत्र भी बढ़ गया है। अब इनका मुख्य उद्देश्य केवल सूचनाएँ उपलब्ध करवाना ही नहीं है, बल्कि सामाजिक परिवर्तन में इनकी भूमिका उल्लेखनीय हो गई है। यहाँ तक कि कभी-कभी सरकार को गिराने में भी ये सफल रहते हैं। इसीलिए एक कवि ने कहा है
“झुक जाती है तलवार भी अखबार के आगे,
झुक जाती है सरकार भी अखबार के आगे।”

उपसंहार-किसी भी देश में जनता का मार्गदर्शन करने के लिए निष्पक्ष तथा निर्भीक समाचार-पत्रों का होना आवश्यक है। समाचार-देश की राजनीतिक, सामाजिक, आर्थिक एवं सांस्कृतिक गतिविधियों की सच्ची तस्वीर प्रस्तुत करते हैं। समाचार-पत्रों को सामाजिक एवं नैतिक मूल्यों से जन साधारण को अवगत कराने की जिम्मेदारी भी वहन करनी पड़ती है। अत: समाचार-पत्रों को पत्रकारिता के मूल्यों पर चलते हुए स्वस्थ लोकतंत्र के निर्माण में अपना योगदान देते रहना होगा।

(ग) कन्या भ्रूणहत्या : कारण और निवारण

भारतीय समाज का कलंक-कन्या भ्रूणहत्या जैसी बुराई भारतीय समाज पर कलंक है। ‘यत्र नार्यस्तु पूज्यन्ते, रमन्ते तत्र देवता’ जैसे आदर्शों का शंखनाद करने वाले भारतीय परिवेश में कन्याएँ जिस सम्मान व स्नेह की अधिकारी हैं, वह उन्हें आज तक प्राप्त नहीं हुआ है। आज जब देश चाँद पर पहुँचने के साथ-साथ मंगल ग्रह पर भी अपनी दस्तक दे चुका है, तब कन्या भ्रूणहत्या जैसे कृत्य को अंजाम देना यही दर्शाता है कि हम आज भी लकीर के फकीर बने हुए हैं और हम आज भी पुत्र को ही वंश चलाने के लिए आवश्यक मानते हैं।

कन्याओं के प्रति समाज की दूषित सोच-प्रतिवर्ष न जाने कितनी बेटियाँ कोख में ही कत्ल कर दी जाती हैं। इस जघन्य कृत्य के पीछे बहुत से कारण हैं, जिनमें प्रमुख कारण हैं-दकियानूसी मानसिकता, अशिक्षा, मूंछ नीची हो जाने का भय, निर्धनता, एक ही जगह दो या अधिक पुत्रों की अभिलाषा, कानून का भय न होना आदि। कई राज्यों में स्थिति इतनी भयावह है कि कन्या के जन्म लेने के बाद उसकी इतनी उपेक्षा की जाती है या इतनी प्रताड़ना दी जाती है कि वह कुछ ही घंटों या दिनों में दम तोड़ देती है।

इसके कारण-कन्या भ्रूण हत्या का एक बड़ा कारण दहेज प्रथा भी है। लोग लड़कियों को पराया धन समझते हैं, उनकी शादी के लिए दहेज की व्यवस्था करनी पड़ती है। दहेज जमा करने के लिए कई परिवारों को कर्ज भी लेना पड़ता है, इसलिए भविष्य में इस प्रकार की समस्याओं से बचने के लिए लोग गर्भावस्था में ही लिंग परीक्षण करवाकर कन्या भ्रूण होने की स्थिति में उसकी हत्या करवा देते हैं। हमारे समाज में महिलाओं से अधिक पुरुषों को महत्व दिया जाता है।

इस बुराई को रोकने के उपाय-इसके निवारण के लिए भारतीयों को कन्याओं के प्रति अपनी दकियानूसी सोच को पूरी तरह बदलना होगा और बेटा-बेटी में अंतर करने की भावना को छोड़ना होगा।

सरकारी प्रयास- भारत में वर्ष 2004 में पीसीपीएनडीटी एक्ट लागू कर भ्रूण हत्या को अपराध घोषित कर दिया गया। इसके बाद भी कन्या भ्रूण हत्या पर पूर्ण नियंत्रण नहीं हो सका। लोग चोरी छिपे पैसे के बल पर इस कुकृत्य को अंजाम देते हैं। कन्या भ्रूण हत्या एक सामाजिक अभिशाप है और इसे रोकने के लिए लोगों को जागरूक करना होगा। महिलाओं को आत्मनिर्भर बनाकर ही इस कृत्य को रोका जा सकता है।

उपसंहार-किसी भी देश की प्रगति तब तक संभव नहीं है, जब तक वहाँ की महिलाओं को प्रगति के पर्याप्त अवसर न मिलें । जिस देश में महिलाओं का अभाव हो, उसके विकास की कल्पना भला कैसे की जा सकती है। कन्या भ्रूण हत्या पर नियंत्रण कर इसे समाप्त करने में महिलाओं की भूमिका सर्वाधिक महत्वपूर्ण हो सकती है, किन्तु साक्षर महिला ही अपने अधिकारों की रक्षा कर पाने में सक्षम होती है, इसलिए हमें महिला शिक्षा पर विशेष ध्यान देना होगा।

उत्तर 13. सेवा में,
प्रधानाचार्य जी,
राजकीय माध्यमिक विद्यालय,
नेहरू मार्ग,
नई दिल्ली।
दिनांक 8 मई 20xx
विषय-अच्छे परिणाम हेतु सुझाव-पत्र।
महोदय,
नम्र निवेदन है कि हमारा विद्यालय अपने क्षेत्र के विशेष विद्यालयों में गिना जाता है, शिक्षा का और अधिक विस्तार करने के लिए। हमें हमारे विद्यालय में परिश्रमी एवं योग्य शिक्षकों को नियुक्त करना चाहिए जिससे शिक्षा व्यवस्था और अच्छी, सुव्यवस्थित एवं नियमित रूप से हो सके। जिससे कक्षा दसवीं एवं बारहवीं के परिणाम अच्छे आ सकेंगे तथा शिक्षकों द्वारा उन्हें नया मार्गदर्शन प्राप्त होगा जिसके कारण हमारे विद्यालय को ख्याति प्राप्त होगी। हमारा उद्देश्य विद्यार्थियों को उचित शिक्षा मुहैया करवाना है।
धन्यवाद!
आपका आज्ञाकारी शिष्य
विनोद शर्मा

अथवा

परीक्षा भवन,
फरीदाबाद।
दिनांक : 19 मई, 20xx
सेवा में,
माननीय सांसद महोदय,
फरीदाबाद ।
विषय : क्षेत्र में नया पुस्तकालय खोलने हेतु।
महोदय,
निवेदन यह है कि मैं सेक्टर-20 से सेक्टर-65 तक बहुत आबादी बस चुकी है। यहाँ की जनसंख्या पिछले पाँच सालों की तुलना में तीन गुना अधिक हो चुकी है, किन्तु इस क्षेत्र में पुस्तकालय की व्यवस्था अभी तक नहीं है। पुस्तकालय से संबंधित किसी भी काम के लिए लोगों को यहाँ से बहुत दूर सेक्टर-14 में जाना पड़ता है, जिससे बहुत परेशानी होती है। विशेषकर बूढ़ों और महिलाओं को बहुत कष्ट उठाना पड़ता है। सिर्फ विद्यार्थी ही इसका लाभ ले पाते हैं।

मेरा आपसे विनम्र अनुरोध है कि इस क्षेत्र की आवश्यकता और परेशानी को देखते हुए इस क्षेत्र में शीघ्र ही एक पुस्तकालय खुलवाने की व्यवस्था करें, जिससे यहाँ के नागरिकों को होने वाली पठन-पाठन की परेशानियों से छुटकारा मिल सके।
धन्यवाद!
भवदीय
क. ख. ग.

उत्तर 14.

अथवा

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