The document Sample Question Paper (2020-21) (Standard) - 1 Notes | EduRev is a part of the Class 10 Course CBSE Sample Papers For Class 10.

All you need of Class 10 at this link: Class 10

**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**

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

**Q.1. If xy = 180 and HCF (x, y) = 3, then find the LCM (x, y). (1 Mark)****ORThe decimal representation of 14587 / 2 ^{1 }x 5^{4} will terminate after how many decimal places? (1 Mark)**

► "LCM × HCF = Product of two numbers"

(LCM) (3) = 180

LCM = 60

Four decimal places

= 11.6696

Thus, the given rational number terminates after four decimal places.

► So we have,

α + β = k / 3

3 = k / 3

k = 9

a

3 / 6 = 1 / k ≠ 3 / 8

3 / 6 = 1 / k

k = 2

k not equal to 8 / 3.

and the cost of 1 table = ₹ y

According to question,

3x + y = 1500

and 6x + y = 2400

OR

In an Arithmetic Progression, if d = – 4, n = 7, a

► For l = 0

0 = 27 + (n – 1)(–3)

30 = 3n

n = 10

10

a

4 = a + 6 × (– 4)

a = 28

Ans.

► For equal roots b

► Discriminant D = 0 so,

(6k)

36k

k

k = ±2

OR

For what value(s) of ‘a’ quadratic equation 3ax

x

(x + 5)(x + 2) = 0

Either x = – 5 or x = – 2

Given that,

3ax

► For no real roots b

► Discriminant D < 0 so,

(– 6)

12a > 36

a > 3

PL + LQ = PM + MT

PL + LN = PM + MN

(LQ = LN, MT = MN)

(Tangents to a circle from a common point)

Perimeter(ΔPLM)

= PL + LM + PM

= PL + LN + MN + PM

= 2(PL + LN)

= 2(PL + LQ)

= 2 × 28 = 56 cm

∵ LQ = LN

MT = MN

Tangents to a circle from a common point.

OR

PQ is a tangent to a circle with centre O at point P. If ΔOPQ is an isosceles triangle, then find ∠OQP. (1 Mark)

In ΔPAO,

tan 30° = AO / PA

(Using trigonometry)

1 / √3 = 3 / PA

PA = 3√3 cm**OR**

In ΔOPQ,

∠P + ∠Q + ∠O = 180°

(∠P = ∠Q isosceles triangle)

2∠Q + ∠P = 180°

2∠Q+90° =180°

2∠Q =90°

∠Q = 45°**Q.10. In the DABC, D and E are points on side AB and AC respectively such that DE || BC. If AE = 2 cm, AD = 3 cm and BD = 4.5 cm, then find CE. (1 Mark)****Ans. **__Using basic proportionality theorem__:

AD / BD = AE / CE [By using BPT]

3 / 4.5 = 2 / CE

CE = 3 cm**Q.11. In the figure, if B _{1}, B_{2}, B_{3},…... and A_{1}, A_{2}, A_{3},….. points have been marked at equal distances. On lines X and Y in what ratio does C divides AB? (1 Mark)**

1 / 2 + cos B = 1

cos B = 1 / 2 = cos 60°

Ans.

x + y = 2 sin

= 2(sin

(As sin2x +cos2x = 1)

= 3

60º / 360º = (2 x 22 / 7 x 21)

= 22 cm

H = height of cylinder

r = radius of sphere

πR

1 × 1 × 16 =

r

r = 1

d = 2 cm.

OR

Find the probability of getting a black queen when a card is drawn at random from a well-shuffled pack of 52 cards. (1 Mark)

Doublet means same number on both dice.

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

► Total number of outcomes = n(S) = 36

► Favourable outcomes are : (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) i.e., n(E) = 6

► Required Probability = n(E) / n(S) = 6 / 36 = 1 / 6

► Probability of getting a black queen = 2 / 52 = 1 / 26

► Total no. of cards = 52

► No. of black queens = 2

► So, Probability of black queen = 2 / 52 = 1 / 26

**Section - II**

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

**Q.17. Case Study Based-1 ****Sun Room**

The diagrams show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sun room are square clear glass panels. The roof is made using

- Four clear glass panels, trapezium in shape, all of the same size
- One tinted glass panel, half a regular octagon in shape

**(i) Find the mid-point of the segment joining the points J (6, 17) and I (9, 16). (1 Mark)****(a) 33 / 2, 15 / 2****(b) 3 / 2, 1 / 2****(c) 15 / 2, 33 / 2****(d) 1 / 2, 3 / 2****Ans. **(c)**Solution. **Mid- point of J(6, 17) and I(9, 16) is

x = 15 / 2 and y = 33 / 2

**(ii) The distance of the point P from the y-axis is (1 Mark)**

(a) 4

(b) 15

(c) 19

(d) 25**Ans. **(a)

**(iii) The distance between the points A and S is (1 Mark)**

(a) 4

(b) 8

(c) 16

(d) 20**Ans. **(c)

**(iv) Find the co-ordinates of the point which divides the line segment joining the points A and B in the ratio 1 : 3 internally. (1 Mark)**

(a) (8.5, 2.0)

(b) (2.0, 9.5)

(c) (3.0, 7.5)

(d) (2.0, 8.5)**Ans. **(d)

**(v) If a point (x, y) is equidistant from the Q(9, 8) and S(17, 8), then (1 Mark)****(a) x + y = 13 ****(b) x – 13 = 0 ****(c) y – 13 = 0 ****(d) x – y = 13****Ans. **(b)**Solution. **Let point be P(x, y)

PQ^{2} = PS^{2}

or, (x – 9)^{2} +(y – 8)^{2} = (x – 17)^{2} + (y – 8)^{2}

or, x – 13 = 0**Q.18. Case Study Based- 2 Scale Factor And Similarity**

__Scale Factor:__

A scale drawing of an object is of the same shape as the object but of a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The value of scale is written as a ratio.__Similar Figures:__

The ratio of two corresponding sides in similar figures is called the scale factor.

If one shape can become another using Resizing then the shapes are **Similar**

Hence, two shapes are Similar when one can become the other after a resize, flip, slide or turn.

**(i) A model of a boat is made on the scale of 1:4. The model is 120 cm long. The full size of the boat has a width of 60 cm. What is the width of the scale model? (1 Mark)****(a) 20 cm ****(b) 25 cm ****(c) 15 cm ****(d) 240 cm****Ans. **(c)**Solution. **width of the scale model = 60 / 4 = 15 cm

**(ii) What will effect the similarity of any two polygons? (1 Mark)****(a) They are flipped horizontally ****(b) They are dilated by a scale factor ****(c) They are translated down ****(d) They are not the mirror image of one another****Ans. **(d)**Solution. **They are not the mirror image of one another

**(iii) If two similar triangles have a scale factor of a:b. Which statement regarding the two triangles is true? (1 Mark)****(a) The ratio of their perimeters is 3a:b ****(b) Their altitudes have a ratio a:b ****(c) Their medians have a ratio a/2:b****(d) Their angle bisectors have a ratio a ^{2}: b^{2}**

**(iv) The length of shadow of a stick 5 m long is 2 m. At the same time the shadow of a tree 12.5 m high is (1 Mark)****(a) 3 m (b) 3.5 m (c) 4.5 m (d) 5 m**

By the property of similar triangle

we have 5 / 2 =12.5 / x

or, x = (12.5 × 2) / 5 = 5 cm

**(v) Below you see a student's mathematical model of a farmhouse roof with measurements. The attic floor, ABCD in the model, is a square. The beams that support the roof are the edges of a rectangular prism, EFGHKLMN. E is the middle of AT, F is the middle of BT, G is the middle of CT, and H is the middle of DT. All the edges of the pyramid in the model have length of 12 m.****What is the length of EF, where EF is one of the horizontal edges of the block? (1 Mark)****(a) 24 m ****(b) 3 m ****(c) 6 m ****(d) 10 m****Ans. **(c)**Solution.** Length of the horizontal edge EF

= half of the edge of pyramid

= 12 / 2

= 6 cm (as E is the mid point of AT)**Q.19. Case Study Based- 3 **

Applications of Parabolas-Highway Overpasses/Underpasses

A highway underpass is parabolic in shape.

__Parabola__

A parabola is the graph that results from p(x) = ax^{2} + bx + c.

Parabolas are symmetric about a vertical line known as the **Axis of Symmetry.**

The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.

**(i) If the highway overpass is represented by x ^{2} – 2x – 8. Then its zeroes are (1 Mark)**

or, x

or, x(x – 4) + 2(x – 4) = 0

or, (x – 4)(x + 2) = 0

or, x = 4, x = – 2

**(ii) The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial (1 Mark)**

(a) Intersects x-axis

(b) Intersects y-axis

(c) Intersects y-axis or x-axis

(d) None of the above**Ans.** (a)

**(iii) Graph of a quadratic polynomial is a (1 Mark)**

(a) straight line

(b) circle

(c) parabola

(d) ellipse**Ans. **(c)

**(iv) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is (1 Mark)****(a) x ^{2} – 6x + 2 **

or, x = 6, – 6

**(v) The number of zeroes that polynomial f(x) = (x – 2) ^{2} + 4 can have is : (1 Mark)**

(a) 1

(b) 2

(c) 0

(d) 3

A stopwatch was used to find the time that it took a group of students to run 100 m.

**(i) Estimate the mean time taken by a student to finish the race. (1 Mark)****(a) 54****(b) 63 ****(c) 43 ****(d) 50****Ans. **(c)**Solution. **

Mean = 1720 / 40 = 43

**(ii) What will be the upper limit of the modal class? (1 Mark)****(a) 20 ****(b) 40 ****(c) 60 ****(d) 80****Ans. **(c)**Solution. **Modal class = 40 - 60

Upper limit = 60

**(iii) The construction of cumulative frequency table is useful in determining the (1 Mark)**

(a) Mean

(b) Median

(c) Mode

(d) All of the above**Ans. **(b)

**(iv) The sum of lower limits of median class and modal class is (1 Mark)****(a) 60 ****(b) 100 ****(c) 80 ****(d) 140****Ans. **(c)**Solution. **Median class = 40 - 60

Modal class = 40 - 60

Therefore, the sum of the lower limits of median and modal class = 40 + 40 = 80

**(v) How many students finished the race within 1 minute? (1 Mark)****(a) 18 ****(b) 37 ****(c) 31 ****(d) 8****Ans. **(c)**Solution. **Number of students are: 8 + 10 + 13 = 31

**Part - B**

**Section - III**

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

**Q.21. 3 bells ring at an interval of 4, 7 and 14 minutes. All three bells rang at 6 am, when the three bells will the ring together next? (2 Mark)****Ans. **We know that,

The three bells again ring together on that time which is the LCM of individual time of each bell

4 = 2 × 2

7 = 7 × 1

14 = 2 × 7

LCM = 2 × 2 × 7 = 28

The three bells will ring together again at 6 : 28 am**Q.22. Find the point on x-axis which is equidistant from the points (2, – 2) and (– 4, 2). (2 Mark)ORP(–2, 5) and Q (3, 2) are two points. Find the co-ordinates of the point R on PQ such that PR = 2QR. (2 Mark)**

PA = PB

PA

(x – 2)

x

= 8x + 16

x = –1

Hence co-ordinate of point P(–1, 0)

PR : QR = 2 : 1

R(4 / 3, 3)

► Product of zeroes = (5 − 3√2)(5+ 3√2) = 7

► Polynomial is given by

x

P(x) = x

**Steps of Construction****1.** Draw a line segment AB of 9 cm.**2.** Taking A and B as centres draw two circles radii of 5 cm and 3 cm respectively.**3.** Bisect the line AB. Let mid-point of AB be C.**4.** Taking C as centre draw a circle of radius AC which intersects the two circles at point P, Q, R and S.**5.**** **Join BP, BQ, AS and AR.

BP, BQ and AR, AS are the required tangents.**Q.25. If tan A = 3 / 4, find the value of 1 / sin A + 1 / cos A (2 Mark)****OR****If √3 sin θ – cos θ = 0 and 0˚< θ < 90˚, then find the value of θ. (2 Mark)****Ans. **Given that, tan A = 3 / 4 = 3k / 4k

sin A = 3k / 5k = 3 / 5

cos A = 4k / 5k = 4 / 5

1 / sin A + 1 / cos A = 5 / 3 + 5 / 4

= 20 + 15 / 12

= 35 / 12**OR**

√3 sin θ = cos θ

sin θ / cos θ = 1 / √3

tan θ = 1 / √3

θ = 30°**Q.26. In the figure, quadrilateral ABCD is circumscribing a circle with centre O and AD ⊥ AB. If radius of incircle is 10 cm, then find the value of x. (2 Mark)****Ans. **∠A = ∠OPA = ∠OSA = 90°

∠SOP = 90°

Also, AP = AS

Hence, OSAP is a square.

AP = AS = 10 cm

CR = CQ = 27 cm

BQ = BC – CQ

= 38 – 27 = 11 cm

BP = BQ = 11 cm

x = AB = AP + BP

= 10 + 11 = 21 cm

**Section - IV**

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

**Q.27. Prove that 2 − √3 is irrational, given that √3 is irrational. (3 Mark)****Ans. **Let 2 − √3 be a rational number

We can find co-primes numbers a and b (b ≠ 0) such that

2 − √3 = a / b

2 - a / b = √3

So, we get 2b - a / b = √3

Since a and b are integers, we get 2b - a / b = √3 is irrational and so √3 is rational. But √3 is an irrational number.

Which contradicts our statement

Therefore 2 − √3 is irrational.**Q.28. If one root of the quadratic equation 3x ^{2} + px + 4 = 0 is 2 / 3, then find the value of p and the other root of the equation. (3 Mark)ORThe roots α and β of the quadratic equation x^{2} – 5x + 3(k – 1) = 0 are such that α – β = 1. Find the value k. (3 Mark)**

∵ 2 / 3 is a root so it must satisfy the given equation

On solving we get

p = –8

3x

3x

3x(x – 2) – 2(x – 2) = 0

x = 2 / 3 or x = 2

Hence, x = 2

So the other root is 2.

α – β = 1 ...(ii)

Solving (i) and (ii), we get

α = 3 and β = 2

also αβ = 6

or 3(k – 1) = 6

k – 1 = 2

k = 3

= 14 cm

► Area of 8 segments = 8 × 14 = 112 cm

► Area of the shaded region = 14 × 14 – 112

= 196 – 112 = 84 cm

(each petal is divided into 2 segments)

Ans.

In ΔABM, AB

= AM

= AM

= AD

Hence, 7AB

55 + a + b = 70

a + b = 15

1 = (11 – a) / 3

a = 8

55 + a + b = 70

55 + 8 + b = 70

b = 7

► Let AB = candle C and D are Points

tan 60° = AB / BD = h / b

√3 = h / b

h = b√3 ...(i)

tan 30° = AB / BD

1 / √3 = h / b

h = a / √3

► Multiplying (i) and (ii), we get

h

h = √ab m

7 × (18 – x) = 10(15 – x)

126 – 7x = 150 – 10x

3x = 150 – 126

3x = 24

x = 8

**Section - V**

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

**Q.34. ****The two palm trees are of equal heights and are standing opposite to each other on either side of the river, which is 80 m wide. From a point O between them on the river the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O from the trees. (5 Mark)ORThe angles of depression of the top and bottom of a building 50 meters high as observed from the top of a tower are 30˚ and 60˚ respectively. Find the height of the tower, and also the horizontal distance between the building and the tower. (5 Mark)**

► Let BD = river

AB = CD = palm trees = h

BO = x

OD = 80 – x

In ΔABO,

tan 60° = h / x

√3 = h / x

h = √3x

In ΔCDO,

...(ii)

► Solving (i) and (ii), we get

x = 20

h = 3x [From eqn. (i)]

= 34.6

► The height of the trees = h = 34.6 m

BO = x = 20 m

DO = 80 – x

= 80 – 20

= 60 m**OR**

► Let AB = Building of height 50 m

RT = tower of height = h m

BT = AS = x m

AB = ST = 50 m

RS = TR – TS = (h – 50)m

In ΔARS, tan 30° = RS / AS

..(i)

► In ΔRBT, tan 60° = RT / BT

√3 = h / x

► Solving (i) and (ii), we get

h = 75

► From (ii), x = h / √3 = 75 / √3

= 25√3

► Hence, height of the tower = h = 75 m

► Distance between the building and the tower

= 25√3

= 43.25 m**Q.35. Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour? (5 Mark)****Ans. **For pipe, r = 1 cm

Length of water flowing in 1 sec,

h = 0.7 m = 7 cm

Cylindrical Tank, R = 40 cm

rise in water level = H

Volume of water flowing in 1 sec

= πr^{2}h = π × 1 × 1 × 70

= 70π

Volume of water flowing in 60 sec =70π × 60

Volume of water flowing in 30 minutes

= 70π × 60 × 30

Volume of water in tank = πr^{2}H

= π × 40 × 40 × H

Volume of water in tank

= Volume of water flowing in 30 minutes

π × 40 × 40 × H = 70π × 60 × 30

H = 78.75 cm**Q.36. A motorboat covers a distance of 16 km upstream and 24 km downstream in 6 hours. In the same time it covers a distance of 12 km upstream and 36 km downstream. Find the speed of the boat in still water and that of the stream. (5 Mark)****Ans. **Let speed of the boat in still water = x km/hr, and speed of the current = y km/hr

Downstream speed = (x + y) km/hr

Upstream speed = (x − y) km/hr

...(i)

...(ii)

Let

and

Put in the above equation, we get

24u + 16v = 6

or, 12u + 8v = 3 ...(iii)

36u + 12v = 6

or, 6u + 2v = 1 ...(iv)

Multiplying (iv) by 4, we get

24u + 8v = 4 ...(v)

Subtracting (iii) by (v), we get

12u = 1

⇒ u = 1 / 12

Putting the value of u in (iv), we get, v = 1 / 4

⇒

and

⇒ x + y = 12 and x − y = 4

Thus, speed of the boat in still water = 8 km/hr,

Speed of the current = 4 km/hr

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

116 docs