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**Class X****Mathematics – Standard****Sample Question Paper 2020-21****Max. Marks : 80****Duration : 3 hrs.**

**General Instructions : **

1. This question paper contains two parts A and B.

2. Both Part A and Part B have internal choices.

**Part – A : **

1. It consists two sections - I and II.

2. Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.

3. Section II has 4 questions on case study. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 sub-parts.

**Part – B : **

1. Section III, Question No. 21 to 26 are Very short answer Type questions of 2 marks each.

2. Section IV, Question No. 27 to 33 are Short Answer Type questions of 3 marks each.

3. Section V, Question No. 34 to 36 are Long Answer Type questions of 5 marks each.

4. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

**Part – A**

**Sections - I**

**Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.**

**Q.1. ****Find the LCM of smallest two digit composite number and smallest single digit composite number. (1 Mark)ORThe decimal representation of 11 / 2**

The smallest 2 digit composite number = 10

∴ LCM of 4 and 10.

4 = 2 × 2

10 = 5 × 2

L.C.M. = 2 × 5 × 2

= 20

∴ LCM of 4 and 10 = 20**OR**

Since 11 / 2^{3} x 5 = 11 x 5^{2} / 2^{3} x 5^{3}

= 275 / 10^{3} = 0.275

Thus, 11 / 2^{3} x 5 will terminate after 3 decimal places.**Q.2. If the zeroes of the quadratic polynomial ax ^{2} + bx + c, a ≠ 0 are equal, then what are the signs of a and c? (1 Mark)ORShow that the sum of roots of quadratic equation –x^{2} + 3x – 3 = 0 is 3. (1 Mark)**

–x

On comparing with

ax

∴ Sum of the roots = -b / a = -3 / -1 = 3

m = 1, n = 2

Given, (x

(x

(x, y) = [18 / 3, -15 / 3]

(x, y) = (6, – 5)

Hence, the point (6, – 5) lies in IV quadrant.

Given, HCF of 144 and 180 is in form of 13m – 16,

Now, HCF of 144 and 180.

144 = 2 × 2 × 2 × 2 × 3 × 3

180 = 2 × 2 × 3 × 3 × 5

∴ HCF (144, 180) = 2 × 2 × 3 × 3

= 36

∵ HCF of 144 and 180 is given in the form of 13m - 16,

∴ 13m - 16 = 36

13m = 36 + 16

13m = 52

m = 52 / 13

m = 4**Q.4. If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then DE / PQ = FE / RP is true? Justify (1 Mark)****Ans. **In D DEF and DPQR, ∠D = ∠Q and ∠R = ∠E. By AA similarity, we get DDEF ~ DQRP.

Hence, DE / QR = EF / RP = DF / QP.**Q.5. At one end A of diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. Find the length of the chord CD which is at a distance 8 cm from A and parallel to XY. (1 Mark)Ans. **XAY is tangent and AO is radius at contact point A of circle.

AO = 5 cm

∴ ∠OAY = 90°

CD is another chord at distance (perpendicular) of 8 cm from A and CMD || XAY meets AB at M.

Join OD.

OD = 5 cm

OM = 8 – 5 = 3 cm

∠OMD = ∠OAY = 90°

Now, in right angled ΔOMD

MD^{2} = OD^{2} – MO^{2}

= 52 – 32

= 25 – 9

= 16

⇒ MD = 4 cm

We know that, Perpendiculars from centre O of circle bisect the chord.

∴ CD = 2

= 2 × 4

= 8 cm.

Hence, length of chord, CD = 8 cm.**Q.6. Find the area of a sector of angle p (in degrees) of a circle with radius R. (1 Mark)****Ans. **We know that area of sector of angle θ

Area of sector of angle, p =**Q.7. Find HCF of the number k, 2k, 3k, 4k and 5k, where k is a positive integer. (1 Mark)Ans. **HCF of

Ans.

a

or 4 / a

Hence, one of the possible, second parallel line is 12x + 9y = 5.

Ans.

7t

7 (a + 6d) = 11 (a + 10d)

4a + 68d = 0

4 (a + 17d) = 0

a + 17d = 0

t

OR

Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, –3) and B is the point (1, 4). (1 Mark)

∴ The coordinates of P are (k + 6 / k + 1, -7k + 4 / k + 1)

Since, y coordinates of every point on the x-axis is zero. -7k + 4 / k + 1 = 0 ⇒ – 7k + 4 = 0

⇒ k = 4 / 7

So, required ratio is 4 : 7.**OR**

Since, AB is the diameter, centre C must be the mid point of the diameter of AB.

Let the co-ordinates of point A be (x, y). x-coordinate of C = x + 1 / 2

⇒ 2 = x + 1 / 2

⇒ 4 = x + 1

⇒ x = 3

and y-coordinate of C = y + 4 / 2

⇒ - 3 = y + 4 / 2

⇒ –6 = y + 4

⇒ y = –10

Hence, coordinates of point A is (3, –10).**Q.12. If the angle between two tangent drawn from an external point 'p' to a circle of radius 'r' and centre O is 60°, then find the length of OP. (1 Mark)ORIn the figure, QR is a common tangent to a given circle which meet at T. Tangent at T meets QR at P. If QP = 3.8 cm, then find length of QR. (1 Mark)**

In ΔOBP, OB / OP = sin 30°

∴ OP = 2r

OR

QP = 3.8

QP = PT

(Length of tangents from external points are equal)

or, PT = 3.8 cm

PR = PT = 3.8 cm

or, QR = QP + PR

= 3·8 + 3·8 = 7.6 cm.

or, k + 1 = sec

or, k + 1 = sec

or, k + 1 =

or, k + 1 = 1

or, k = 1 – 1

∴ k = 0.

or, 3 : 1 : 2

Side of square = √121 = 11 cm

Perimeter of square = 4 × 11 = 44 cm.

Circumference of the circle = Perimeter of the square

= 44 cm

Change the above data to a continuous grouped frequency distribution.

**Section - II**

**Case study based questions are compulsory. Attempt any four sub parts of each question. Each subpart carries 1 mark**

**Q.17. Case Study based - 1 : **

Small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of 1 / 4 m and a tread of 1 / 2 m (see figure below).

**(a) What is the volume of the required concrete for step 1? (1 Mark)(i) 25 / 4(ii) 50 / 4(iii) 75 / 4(iv) 150 / 4Ans. **(i)

Solution.

1 / 4 x 1 / 2 x 50 = 25 / 4

**(b) What is the volume of the required concrete for step 2? (1 Mark)****(i) 25 / 4****(ii) 50 / 4****(iii) 75 / 4****(iv) 150 / 4****Ans. **(ii)**Solution. **For the 2^{nd} step, the volume of required concrete is

2 / 4 x 1 / 2 x 50 = 50 / 4

**(c) What is the volume of the required concrete for step 3? (1 Mark)****(i) 25 / 4(ii) 50 / 4(iii) 75 / 4(iv) 150 / 4**

3 / 4 x 1 / 2 x 50 = 75 / 4

**(d) The given problem is based on which mathematical concept? (1 Mark)(i) AP (ii) Surface areas and volumes (iii) Height and Distances (iv) None of these**

a = 25 / 4, d = 25 / 4, n = 15

(i) 750 m

(ii) 500 m

(iii) 570 m

(iv) 1000 m

we have

= 15 / 2 x 400 / 4

= 750

Thus, the total volume of concrete required to build the terrace is 750 m

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in Figure. Niharika runs 1/4th the distance AD on the second line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag.

**(a) At what distance Niharika posted the green flag from the starting point of second line? (1 Mark)(i) 20 m (ii) 25 m (iii) 100 m (iv) 50 mAns. **(ii)

Solution.

**(b) At what distance Preet posted the green flag from the starting point of eighth line? (1 Mark)(i) 20 m (ii) 25 m (iii) 100 m (iv) 50 mAns. **(i)

Solution.

**(c) What is the distance between both the flags? (1 Mark)(i) √61 m (ii) √101 m (iii) √51 m (iv) √11 m**

= √61 m

**(d) If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag? (1 Mark)(i) (5, 22.5) (ii) (22.5, 5) (iii) (5, 5) (iv) None of these**

x = 10 / 2 = 5 y = 45 / 2 = 22.5

Hence, A(x, y) = (5, 22.5)

Therefore, Rashmi should post her blue flag at 22.5 m on fifth line.

**(e) The given problem is based on which mathematical concept? (1 Mark)**(i) Lines

(ii) surface areas and volumes

(iii) Triangles

(iv) None of these

A carpenter made a wooden pen stand. It is in the shape of cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. (See Figure).

**(a) What is the volume of cuboid? (1 Mark)(i) 525 cm ^{3} (ii) 225 cm^{3} (iii) 552 cm^{3} (iv) 255 cm^{3}Ans. **(i)

Solution.

l = 15 cm, b = 10 cm and h = 3.5 cm

Volume of the cuboid = l × b × h

= 15 × 10 × 3.5

= 525 cm

**(b) What is the volume of cone? (1 Mark)(i) 11 / 3cm ^{3}(ii) 11 / 30cm^{3}(iii) 3 / 11cm^{3}(iv) 30 / 11cm^{3}Ans.** (ii)

r = 0.5 cm, h = 1.4 cm

Volume of four conical depressions

**(c) What is the total volume of conical depressions? (1 Mark)(i) 1.74 cm ^{3}(ii) 1.44 cm^{3}(iii) 1.47 cm^{3}(iv) 1.77 cm^{3}Ans. **(iii)

Solution.

**(d) What is the volume of wood in the entire stand? (1 Mark)(i) 522.35 cm3 (ii) 532.53 cm3 (iii) 523.35 cm3 (iv) 523.53 cm3Ans. **(iv)

Solution.

= Volume of cuboid – Volume of 4 conical depression

= 525 – 1.47

= 523.53 cm

**(e) The given problem is based on which mathematical concept? (1 Mark)**(i) Triangle

(ii) Lines

(iii) Height and Distances

(iv) None of these

The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows.

**(a) What is the modal class of the given data? (1 Mark)(i) 85 - 100 (ii) 100 - 115 (iii) 115 - 130 (iv) 130 - 145Ans. **(i)

Solution.

**(b) What is the value of class interval for the given data set? (1 Mark)(i) 10 (ii) 15 (iii) 5 (iv) 20Ans. **(ii)

Solution.

= 115 – 100

= 130 – 115 = 145 – 130

**(c) What is the median class of the given data? (1 Mark)(i) 85 - 100 (ii) 100 - 115 (iii) 115 - 130 (iv) 130 - 145**

Median of 33 observations = 17th observation, which lies in class 100 – 115

**(d) What is the median blowing speed? (1 Mark)(i) 109.17 km/hr(ii) 109.71 km/hr(iii) 107.17 km/hr(iv) 109.19 km/hrAns.** (i)

Median =

l = 100, f = 9, c.f. = 11, h = 100 – 85 = 15

Median =

= 100 + 9.166

= 109.17 km/h

Hence, the median bowling speed is 109.17 km/h

**(e) What is the sum of lower limit of modal class and upper limit of median class? (1 Mark)(i) 100 (ii) 200 (iii) 300 (iv) 400**

and upper limit of median class = 115

sum = 85 + 115 = 200

**Part - B**

**Section - III**

**All questions are compulsory. In case of internal choices, attempt any one.**

**Q.21. Given that √5 is irrational, prove that 2√5 - 3 is an irrational number. (2 Mark)****Ans. **Let us assume, to the contrary, that 2√5 − 3 is a rational number

∴ 2√5 − 3 = p / q , where p and q are integers and q ≠ 0

⇒ .....(1)

Since p and q are integers

∴is a rational number

∴ √5 is a rational number which is contradiction as √5 is an irrational number

Hence our assumption is wrong and hence 2√5 − 3 is an irrational number.**Q.22. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. (2 Mark)****Ans. **

AD ⊥ BC ∴ In ΔABD, AB^{2} = AD^{2} + BD^{2}

⇒ AB^{2} = AD^{2} + BC^{2} / 4 or 4AB^{2} = 4AD^{2} + BC^{2}

⇒ 3AB^{2} = 4AD^{2}**Q.23. Find the value of : sin 30°. cos 60° + cos 30°. sin 60° Is it equal to sin 90° or cos 90°? (2 Mark)ORProve that := cosec A – cot A (2 Mark)**

= 1 / 2 x 1 / 2 + √3 / 2 x √3 / 2

= 1 / 4 + 3 / 4

= 4 / 4 = 1

It is equal to sin 90° = 1 but not equal to cos 90° as cos 90° = 0.

LHS =

=

=

= cosec A – cot A = RHS

Hence proved.

Let the tree be AC and is broken at B. The broken part touches at the point D on the ground.

In right angled ΔBCD,

cos 30° = CD / BD

⇒ √3 / 2 = 8 / BD

⇒ BD = 16 / √3

and tan 30° = BC / CD

⇒ 1 / √3 = BC / 8

⇒ BC = 8 / √3

∴ Height of tree= BC + BD

= 8 / √3 + 16 / √3

=

=

=

Hence, the height of the tree = 8√3 m.**Q.25. A paper is in the form of a rectangle ABCD in which AB = 20 cm and BC = 14 cm. A semi-circular portion with BC as diameter is cut off. Find the area of the remaining part. (Use π = 22 / 7) (2 Mark)ORIf the perimeter of a protractor is 72 cm, calculate its area. (Use π = 22 / 7) (2 Mark)**

=

= 280 – 77

Hence, area of remaining part = 203 cm

Perimeter of semi-circle,

(π + 2)r = 72 cm

or,

or,

or, 36 / 7 r = 72

r = 14 cm

∴ Area of protractor =

= 308 cm

Mean =

= 125 / 5 = 25

**Section - IV**

**All questions are compulsory. In case of internal choices, attempt any one.**

**Q.27. Find the zeroes of the following polynomial : 5√5x ^{2} + 30x+ 8√5 (3 Mark)**

= 5√5x

= 5x(√5x + 4) + 2√5 (√5x + 4)

= (√5x + 4)( 5x + 2√5)

Thus, zeroes are -4 / √5 = -4√5 / 5 and -2√5 / 5

p(y) =

=

∴ Zeroes are 2 / 3, -1 / 7

Sum of zeroes = 2 / 3 - 1 / 7 = 11 / 21

-b / a = 11 / 21 ∴ sum of zeroes = -b / a

Product of zeroes = (2 / 3)(-1 / 7) = -2 / 21

c / a = -2 / 3(1 / 7) = -2 / 21 ∴ Product = c / a

Given, polynomial is...(i)

⇒ 21y

⇒ 21y

⇒ 7y (3y – 2) +1 (3y – 2) = 0

⇒ (3y – 2)(7y + 1) = 0

⇒ y = 2 / 3 or y = -1 / 7

Hence, zeros of given polynomial are

y = 2 / 3 and y = -1 / 7

On comparing eq(i) with ax

a = 21, b = –11 and c = –2

Now, Sum of roots = 2 / 3 + (-1 / 7)

= 2 / 3 - 1 / 7

= 14 - 3 / 7

= 11 / 21

= -b / a

and Product of roots = 2 / 3 x (-1 / 7)

= -2 / 21

= c / a

In the figure, ABCDE is a pentagon with BE || CD and BC || DE. BC is perpendicular to CD. AB = 5 cm, BE = 7 cm, BC = x – y and CD = x + y. If the perimeter of ABCDE is 27 cm. find the value of x and y, given x, y ≠ 0.

and 2x – 10y = 12 ⇒ y = x - 6 / 5

Since, the lines are coincident, so the system of linear equations is consistent with infinitely many solutions.

x + y = 7 and 2 (x – y) + x + y + 5 + 5 = 27

∴ x + y = 7 and 3x – y = 17

Solving, we get, x = 6 and y = 1

For no real roots, D < 0

i.e., D ⇒ b

b

= 4[a

= 4[a

= – 4[a

= – 4(ad – bc)

Since, ad ≠ bc

Therefore, D < 0

Hence, the equation has no real roots.

Let the common difference of the A.P. be d.

Since,

∴

= 1050

20 + 13d = 1050 / 7 = 150

13d = 130

d = 130 / 13 = 10

a

a

Hence, a

⇒ ∠1 = ∠2

Similarly ∠4 = ∠3

⇒ ∠1 + ∠4 = ∠2 + ∠3 = 1 / 2 (180°)

⇒ ∠2 + ∠3 = 90° or ∠AOB = 90°

In ΔDOA and ΔCOA

DA = AC

[Tangents drawn from common point]

∠ODA = ∠OCA = 90°

[angle between tangent and radius]

OD = OC ( radius of circle)

∴ ΔDOA ≅ ΔCOA (By SAS)

Hence, ∠1 = ∠2 i.e., ∠DOA = ∠COA (By cpct) ...(i)

Similarly, ΔBOC ≅ ΔBOE (By SAS)

∴ ∠3 = ∠4 i.e., ∠COB = ∠BOE (By cpct.) ...(ii)

Now, ∠1+ ∠2 + ∠3 + ∠4 = 180° (angles on a straight line)

2∠2 + 2 ∠3 = 180°

[from eq(i) & eq(ii)] ∠2 + ∠3 = 90°

i.e., ∠AOC + ∠BOC = 90° or ∠AOB = 90°

Hence Proved.**Q.32. sinθ + cosθ = √3 , then prove that tanθ + cotθ = 1 (3 Mark)**

⇒ 1 + 2sinθ cosθ = 3 ⇒ sinθ cosθ = 1

∴ tanθ + cotθ = sinθ / cosθ + cosθ / sinθ = 1

Given sinθ + cosθ = √3

Squaring on both sides,

(sinθ + cosθ)

sin

1 + 2sinθ cosθ = 3

2sinθ cosθ = 2

sinθ cosθ = 1 ... (i)

∴ tanθ + cotθ = sinθ / cosθ + cosθ / sinθ

Now using equation (i) =

= 1 / 1 = 1

What is the probability that :

(i) Ramesh will buy the selected shirt?

(ii) ‘Kewal ‘ will buy the selected shirt?

P (Ramesh buys the shirt) = 88 / 100 or 22 / 25

P (Kewal buys a shirt) = 96 / 100 or 24 / 25

**Section - V**

**All questions are compulsory. In case of internal choices, attempt any one.**

**Q.34. ****Amit, standing on a horizontal plane, find a bird flying at a distance of 200 m from him at an elevation of 30°. Deepak standing on the roof of a 50 m high building, find the angle of elevation of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak. (5 Mark)ORThe angle of depression of two ships from an aeroplane flying at the height of 7500 m are 30° and 45°. If both the ships are in the same line that one ship is exactly behind the other, find the side such distance between the ships. (5 Mark)**

In ΔAPQ

PQ / AP = sin 30° = 1 / 2

PQ = (200) (1 / 2) = 100 m

PQ = 100 – 50 = 50 m

In ΔPRD, PR / PD = sin 45° = 1 / √2

PD = (PR) (√2)

= 50√2 m**Detailed Solution :**

Let O be the position of the bird, A be the position for Amit, D be the position for Deepak and FD be the building at which Deepak is standing at height 50m.

Given, AO = 200m and FD = 50m

In ΔOLA, ∠L = 90°

sin 30° = OL / OA

⇒ 1 / 2 = OL / 200

⇒ OL = 200 / 2

= 100m

∴ OM = OL – LM

= OL – FD

= (100 – 50)m = 50m

In ΔOMD, ∠M = 90°

sin 45° = OM / OD

⇒ 1 / √2 = 50 / OD

⇒ OD = 50√2

= 50 × 1.414

= 70.7m

Thus, the distance of the bird from Deepak is 70.7 m.**OR**

In ΔABC, AB / BC = tan 45°

⇒ 7500 / y = 1

⇒ y = 7500

In ΔABD, AB / BD = tan 30°

7500 / x + y = 1 / √3

⇒ x + y = 7500√3

x + 7500 = 7500√3

x = 7500√3 − 7500

= 7500 (√3 −1)

= 7500(1·73 – 1)

= 7500 × 0·73

= 5475 m

Hence, the distance between two ships

= 5475 m**Q.35. A hemispherical portion is cut from one face of a cubical block, such that diameter ‘l’ of hemisphere is equal to the edge of cube. Find the surface area of the remaining solid. (5 Mark)**

∴ Therefore, r = l / 2

Now, the required surface area = Surface area of cubical block – Area of base of hemisphere + Curved surface area of hemisphere.

= 6(side)

=

=

=

Required surface area =

= 184.18 l

Ans.

Let 2x / (x - 5) = y

∴ y

⇒ (y + 8) (y – 3) = 0

y = 3 or – 8

Putting y = 3 we get

2x / x - 5 = 3

2x = 3x – 15

⇒ x = 15

Again, for y = – 8, 2x / x - 5 = -8

2x = – 8x + 40

10x = 40

x = 4

Hence, x = 15 or 4

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