Paper: CBSE Sample paper 3 (Silver series) - X - Math -SA II

Total marks of the paper: 90

Total time of the paper: 3.5 hrs

General Instructions:

1. All questions are compulsory.

2. The question paper consists of 34 questions divided into four sections A, B, C, and D. Section – A comprises of 8 questions of 1 mark each, Section – B comprises of 6 questions of 2 marks each, Section – C comprises of 10 questions of 3 marks each and Section – D comprises of 10 questions of 4 marks each.

3. Question numbers 1 to 8 in Section – A are multiple choice questions where you are to select one correct option out of the given four.

4. There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions.

5. Use of calculator is not permitted.

6. An additional 15 minutes has been allotted to read this question paper only.

Questions:

1]

The 9th term of an AP is 449 and 449th term is 9. The term

which is equal to zero is

[Marks:1]

A. 459th

B. 502thC. 501thD. 458th

2]

Two circles touch each other externally at C and AB is a

common tangent to the circles. Then, ACB=

[Marks:1]

A. 300

B. 450

C. 600

D. 90°

3]

The length of an arc that subtends an angle of 24o at the

centre of a circle with 5 cm radius is

[Marks:1]

A.cm

B.cm

C.cm

D.cm

4]

An observer 1.5 m tall is 28.5 m away from a tower. Theangle of elevation of the top of the tower from his eyes is 450.

The height of the tower is

[Marks:1]

A. 40 mB. 20 m

C. 10 mD. 30 m

5]

The probability that a randomly chosen number from one to

twelve is a divisor of twelve is

[Marks:1]

A.

B.

C.

D.

6]

The value of x, for which the points (x,-1), (2,1) and (4,5) lie

on a line is

[Marks:1]

A. 3B. 2C. 0D. 1

7]

The ratio in which the point R divides the join of

P(-2,-2) and Q(2,-4) is

[Marks:1]

A. 1:2B. 2:1C. 4:3D. 3:4

8]

A pendulum swings through an angle of 300 and describes an

arc 8.8cm in length. The length of the pendulum is

[Marks:1]

A. 17 cmB. 15.8 cmC. 8.8 cmD. 16.8 cm

9]

If the centroid of the triangle formed by points P(a, b),Q(b, c)and R(c, a) is at the origin, then find the value of a + b + c.

ORFind the area of the quadrilateral ABCD whose vertices are

A(1,1), B(7,-3), C(12,2) and D(7,21) respectively.

[Marks:2]

10] An A.P. consists of 60 terms. If the first and the last term be 7 and 125 respectively, find the 32nd term.

[Marks:2]

11]

Find two consecutive positive integers, sum of whose squares is 25.

[Marks:2]

12] Prove that the tangents at the extremities of any chord make equal angles with the chord. [Marks:2]

13]

In given figure, find the area of the circle not included in therectangle whose sides are 8cm and 6cm respectively.

[Marks:2]

14]

The perimeter of a sector of a circle of radius 5.2 cm is 16.4

cm, find the area of the sector.

[Marks:2]

15]

If a b c, prove that the points (a,a2), (b,b2) and (c,c2)

can never be collinear.

[Marks:3]

16] A body falls 8 metres in the first second of its motion, 24 metres in the second, 40 metres in the third second and so on. How long will it take to fall 2048 metres?

[Marks:3]

17] Draw a pair of tangents to a circle of any convenient radius, which are inclined to the line joining the centre of the circle and the point at which they intersect at an angle of 45°. Also write the steps of construction.

[Marks:3]

18]

From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 450 and 600. If the cars are 100m apart, find the height of the balloon.

OR

The angles of elevation of the top of a tower from two points at a distance a and b metres from the base and in the same straight line with it are complementary. Prove that the height of the tower is metres.

[Marks:3]

19]

A bag contains 12 balls out of which x are white.(i) If one ball is drawn at random, what is the probability that it will be a white ball?(ii) If 6 more white balls are put in the bag, then the probability of drawing a white ball will be double than that in (i). Find x.OR

Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any day of the week. What is the probability that both will visit the shop on

(i) the same day?

(ii) two different days?(iii) consecutive days?

[Marks:3]

20]

In an equilateral triangle of side 24 cm, a circle is inscribedtouching its sides. Find the area of the remaining portion ofthe triangle. (take =1.732)

OR

[Marks:3]

A round table cover has six equal designs as shown in thefigure. If the radius of the cover is 28 cm, find the cost of making the design at the rate of Rs. 0.35 per cm2.

21] A canal is 300cm wide and 120cm deep. The water in the canal is flowing with a speed of 20km/h. How much area will it irrigate in 20 minutes if 8cm of standing water is desired?

[Marks:3]

22] From a solid cylinder whose height is 2.4 cm and diameter 1.4cm, 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.(use p=3.1416)

[Marks:3]

23]

Solve for x, using the quadratic formula:

abx2 _ (a2+b2)x + ab=0

[Marks:3]

24]

If D, E and F are the mid points of sides BC, CA and ABrespectively of a ∆ABC, whose vertices are A(-4,1), B(6,7) and

C(2,-9), then prove that,

[Marks:3]

Ar. ∆DEF = (Ar ∆ABC).

25]

Find the value of the middle most term (s) of the A.P. :-11, -7, -3, ...., 49

OR

The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, find the AP.

[Marks:4]

26] Two circles touch externally. The sum of their area is 130 p sq. cm and the distance between their centres is 14 cm. Find the radii of the circles.

[Marks:4]

27]

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is . On advancing p metres towards the foot of the tower, the angle of elevation becomes β. Show that the height h of the tower is given by

[Marks:4]

28]

A passenger train takes 2 hour less for a journey of 300 km, if its speed is increased by 5km/hr from its usual speed. Find its usual speed.

OR

In a class test, the sum of the marks obtained by a student P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have

[Marks:4]

been 180. Find the marks obtained in the two subjects separately.

29] Prove that the lengths of tangents drawn from an external point to a circle are equal.

[Marks:4]

30]

In given figure, a circle is inscribed in a quadrilateral ABCD in which B=900. If AD=23 cm, AB=29 cm, and DS=5 cm, find the radius r of the circle.

[Marks:4]

31] Kavita saved Rs 4 the first week of the year and then increased her weekly savings by Rs 1.75 each week. In what week will her weekly saving be Rs 19.75?

[Marks:4]

32] A solid consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having been given that the radius of the cylinder is 3 cm and its height is 6 cm, the radius of the hemisphere is 2 cm and the height of the cone is 4 cm.

[Marks:4]

33] A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) white or blue (ii) red or black (iii) not white

(iv) neither white nor black?

[Marks:4]

34] A beautiful park with flower plants is to be developed within the region joining the points A(0, -1), B(6, 7), C(-2,3) and D(8, 3) as vertices of a quadrilateral such that AB and CD are diagonals. Show that AB and CD bisects each other and AD2 + DB2 = AB2. Find the area of the park if all distances are in km.

As P.M. of your country, will you make a policy of creating green parks and gardens in every village and town of you country? Give reasons.

[Marks:4]

Paper: CBSE Sample paper 3 (Silver series) - X - Math -SA II

Total marks of the paper: 90

Total time of the paper: 3.5 hrs

General Instructions:

1. All questions are compulsory.

2. The question paper consists of 34 questions divided into four sections A, B, C, and D. Section – A comprises of 8 questions of 1 mark each, Section – B comprises of 6 questions of 2 marks each, Section – C comprises of 10 questions of 3 marks each and Section – D comprises of 10 questions of 4 marks each.

3. Question numbers 1 to 8 in Section – A are multiple choice questions where you are to select one correct option out of the given four.

4. There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You

have to attempt only one of the alternatives in all such questions.

5. Use of calculator is not permitted.

6. An additional 15 minutes has been allotted to read this question paper only.

Solutions:

1]

a+8d = 449 and,a+448d = 9on solving we get, 440d = -440 i.e. d = -1therefore, a-8 = 449 i.e. a = 457let its nth term be zero.an = a+(n-1)d 0 = 457+(n-1)(-1)457 = n-1

or n = 458

2]

Lengths of tangents drawn from an external point to a circle are equal.PA=PC (from P)Therefore, PAC= PCA = x (say)Also, PC=PB (from P)

PBC= PCB=yIn ∆ABC,

ABC+ ACB+ BAC=180y + (x + y) + x = 1802(x + y) = 180x+ y = 90

ACB = 900.

3]

The length of an arc that subtends an angle of 24o at thecentre of a circle with a 5 cm radius is

The length of the arc is cm.

4]

Let AC be the tower of height h metres and ED be the observer of height 1.5 m at a distance of DC=28.5 m from the tower AC.In right ∆AED,

Tan45 =

1= (as EB = DC = 28.5 m )

AB = 28.5 m

Height of the tower = h = AB + BC = AB+DE

= 28.5 + 1.5 = 30m

5]

Total number of outcomes = 12Numbers which are divisors of 12 are 1,2,3,4,6,12.Number of favourable outcomes = 6

P(divisor of 12) =

6]

Area of a triangle = 0[x x1 + 2x5 + 4x(-1)] – [2x-1 + 4x1 + xx5] = 0(x+10-4) - (-2+4+5x) = 0(x+6)-(5x+2)=0

-4x+4=0

x=1

7]

Let R divides PQ in the ratio K:1

and 14K-14 = -2K-216K = 12

K=

Thus R divides PQ in the ratio 3:4.

8]

Length of arc (l) = 2pr8.8 360 7 = 30 2 22 r

r= cm = 16.8 cm

9]

Let P(x1,y1)= P(a, b), Q(x2,y2) = Q(b, c)and R(x3,y3) = R(c, a) be the vertices of ∆PQR.We know that the coordinates of centroid of a triangle is given by

i.e.

Also, given that centroid is at origin i.e. its coordinates are

(0,0).

So, . OR

Area of quadrilateral ABCD = area of triangle ABC + area of triangle ACD

Area of triangle ABC = [1(-3-2) +7(2-1)+12(1+3)]

= (-5+7+48) = 25 sq. units

Area of triangle ACD = [1(2-21) + 12(21-1)+7(1-2)]

= (-19+240-7) = 107 sq. units

Area of quadrilateral ABCD = 25 + 107

= 132 sq. Units

10]

Here, n = 60, a = 7 and = 125 7+59d = 125d=2

Therefore, 32nd term (t32) = a + 31d = 7 + 31 x 2 = 69

11]

let the required numbers be x and x+1

Given x2 + (x+1)2 = 25

2x2 +2x-24 =0 Þ x2 + x – 12 = 0

(x+4) (x-3) = 0

X = -4 or x = 3

Reject x = -4

Therefore the given consecutive positive integers areX = 3 and x + 1 = 3 + 1 = 4

12]

Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively. Suppose the tangents meet at P. join OP. suppose OP meets A at C.We have to prove that PAC= PBC In triangles PCA and PCB, we havePA = PB (tangents from an external point are equal)

APC = BPC (PA and PB are equally inclined to OP)And, PC = PC (common)∆PAC ∆PBC ( by SAS )

PAC = PBC ( by cpct )

13]

Diameter of the circle = diagonal BD of the rectangle ABCD

In right triangle BCD,

BD2 = BC2+BD2 = 62+82 = 100

BD2 = 100 i.e.BD = 10 cm

Radius of the circle = = 5cm

Area of the circle = pr2 = 3.14(5)2 = 78.50 cm2

Area of rectangle ABCD = 6 8 = 48 cm2

Area of the circle not included in the rectangle

= area of the circle – area of rectangle ABCD

= 78.50 – 48 = 30.50 cm2

14]

Let OAB be the given sector. Then,

Perimeter of sector OAB = 16.4 cm

OA+OB+ arc AB = 16.4 cm

5.2+5.2+arc AB =16.4

Arc AB = 6 cm i.e. l =6cm

Area of sector = lr = 6 5.2 = 15.6 cm2

15]

If the area of the triangle formed by joining the given points is zero, then only the points are collinear.

Area of triangle = Here, x1 = a, y1 = a2; x2 = b, y2 = b2; x3 = c, y3 = c2

Substituting the values in the formula for area of a triangle, we get

Area of triangle =

It is given that a b c

Area of the triangle can never be 0

16]

The AP is 8, 24, 40, ….., and the sum is 2048.We have to determine n.Here a = 8, d = 16 Since, Sn = 2048, we have

2048 = [2x8+(n-1)16] 2048 = n[8+8n-8]2048 = 8n2

n2 = 256 i.e. n = 16(as n being number of terms, it can't be negative)

Hence, the body will take 16 seconds to fall 2048 metres.

17]

The steps of construction are as follows:(i) Draw a circle of any convenient radius with O as centre.(ii) Take a point A on the circumference of the circle and join OA.

Draw a perpendicular to OA at point A.(iii) Draw a radius OB, making an angle of 90° with OA.(iv) Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. (v) Join OP. PA and PB are the required tangents, which make an angle of 45° with OP.

18]

Let the height of the balloon at P be h metres. Let A and B be the two cars. Thus AB=100m.

From right ∆PAQ, AQ = PQ = h (as tan 45 =1)From right ∆PBQ,

= tan60 = or or h = (h-100)

Therefore, h = = 50(3+ )i.e. the height of the balloon is 50(3+ ) m.

OR

Let AB be the tower of height h metres and C and D be two points at a distance of a and b respectively from the base of the tower.AC = a mAD = b m (a>b)In right ∆CAB,

tanx = tanx = ………………….(i) In right ∆DAB,

Tan(90-x) = cotx = ……………(ii) From (i) and (ii),

tanx.cotx =

1 = Þ h2 = ab or h = metres.

Thus, the height of the tower is metres.

19]

There are 12 balls in the bag. Out of these 12 balls, one ball

can be chosen in 12 ways.Therefore, total number of outcomes = 12There are x white balls out of which one can be chosen in x ways.Therefore, favourable number of outcomes = x

Hence, p1=P(Getting a white ball) = If 6 more white balls are put in the bag, thenTotal number of balls in the bag = 12+6 = 18Number of white balls in the bags = x + 6

Therefore p2=P(Getting a white ball) = It is given that

p2 = 2p1

6(x+6) = 18x 6x+36 = 18x

12x = 36

x = = 3 OR

Total number of equally likely outcomes for visiting shop in

the same week = 6x6=36

(i) Same days are (M,M),(T,T),(W,W),(Th,Th),(F,F),(SAT,SAT)

Number of favourable outcomes = 6

P(same day) =

(ii) Number of different days = 36-6 = 30

Number of favourable outcomes = 30

P(two different days) =

(iii) Favourable outcomes for (I customer,II customer)

= (M,T),(T,W),(W,Th),(Th,F),(F,SAT) and for

(II customer,I customer)

= (M,T),(T,W),(W,Th),(Th,F),(F,SAT)

Number of favourable outcomes = 10

P(consecutive days) =

20]

Let ABC be an equilateral triangle of side 24cm, and let AD be an altitude from A on BC. Since the triangle is equilateral, so AD bisects BC.Therefore BD=CD=12 cm The centre O of the inscribed circle will coincide with the centroid of ∆ABC

Therefore OD= (as centroid divides each median in the ratio 2:1) In right ∆ABD, we haveAB2 = AD2+BD2 [Using Pythagoras Theorem]242 = AD2+122

AD = 12 cm

Therefore OD = Area of the incircle = p(OD)2

=

Area of the triangle ABC = (Side)2

= (24)2 = 249.4 cm2 Therefore, area of the remaining portion of the triangle = (249.4-150.85)cm2

= 98.55 cm2

OR

Area of one design = – area of ∆OAB

= - = 282(0.5238-0.433)

=282x0.0908= 71.1872=71.19 cm2 (approx.)

Total cost = 6 71.19x 0.35

= Rs149.50

21]

Volume of water flows in the canal in one hour = width of the canal x depth of the canal x speed of the canal water

= 3x1.2x20x1000m3 = 72000m3.

In 20 minutes the volume of water = m3

= 24000 m3.

Area irrigated in 20 minutes, if 8cm, i.e. 0.08 m standing

water is required = = 300000 m2 = 30 hectares.

22]

Radius of the cylinder = 0.7cmHeight of the cylinder = 2.4 cmSurface area of the cylinder = 2prh+pr2

= p[2 0.7 2.4+(0.7)2] = 3.85 p cm2

Slant height of the cone is given byl2 = (0.7)2+(2.4)2 = 6.25 žÃ l = 2.5 cm

surface area of interior conical portion = prl = px0.7 2.5 = 1.75p Total surface area = 3.85p +1.75p = 5.60p cm2

= 5.6 3.1416 = 17.59 cm2.

23]

Here, abx2-(a2+b2)x + ab = 0

X =

=

=

Therefore, x =

24]

Let A = (x1,y1) = (-4,1), B = (x2,y2) = (6,7) and C = (x3,y3)= (2,-9)

Area of ?ABC = [x1 (y2 - y3 )+x2 ( y3- y1) +x3 (y1 - y2)]

Area of ?ABC = [ -4(7+9) + 6(-9-1) + 2(1-7)]

Area of ?ABC = [ -64-60-12] ? = 68 sq. Units. Coordinates of D, E and F are

D and F i.e. D(4,-1), E(-1,-4) and F(1,4)

Area of ?DEF = [ 4(-4-4) -1(4+1) + 1(-1+4)]

Area of ?DEF = ? [ -32-5+3] ? = 17 sq. Units.

(Area of ?ABC) = (68) = 17 = Area of ?DEF

25]

Here, a =-11, d =-7-(-11) =4,an = 49 We have an = a+(n-1)d So, 49 = -11+(n-1) 4

i.e., 60 = (n-1) 4i.e., n = 16 As n is an even number, there will be two middle terms which

are

th and th i.e., 8th term and the 9th term.

a8 = a+7d = -11+7 4 =17 a9 = a+8d = -11+8 4 = 21 So, the values of the two middle most terms are 17 and 21,

respectively. OR

Let the three terms in AP be d, a, a + d

so, a-d+a+a+d=33or a = 11 Also, (a-d)(a+ d) = a+29i.e., a2-d2 = a+29

i.e., 121-d2 = 11+29i.e., d2 = 81i.e., d = ± 9 So there will be two APs and they are:2, 11, 20,...

And 20, 11, 2, ...

26]

Let r1, r2 be the radii of the given circles respectively. Since the circles touch externally/\\ distance between their centres = (r1 + r2)\\ r1 + r2 = 14 …(1)Sum of their areas = pr12 = pr22 = p(r12 + r12) = 130 m (Given)\\ r12 + r22 = 130 …(2)Now, (r1 + r2)2 + (r1 – r2)2 = 2 (r12 + r22) Þ (14)2 + (r1 – r2)2 = 2(130) = 260Þ (r1 – r2)2 = 260 – 196 = 64 Þ r1 – r2 = 8 …(3)(1) + (2) gives, 2r1 = 22 Þ r1 = 11 (1) – (2) gives, 2r2 = 6 Þ r2 = 3

Thus, radii of the given circles are 11 cm and 3 cm.

27]

Let CQ be the tower of height h metres and A and B be the points of observation. AB = p metres and let BC = x metres.In right ∆ACQ,

= tan Þ h = (p+ x)tan or x = - p ………………(i)In right ∆BCQ,

= tanβ Þ h = xtanβ or x = …………………………(ii)

From (i) and (ii),

- p =

P = h

P =h

or h =

28]

Let usual speed = x km / hr andincreased speed = (x+5) km/hr

According to question,

(Because Distance/Speed = time)

2 1500 = 2x(x + 5)x2+5x = 750

x2+5x-750 = 0 (x+30) (x-25) = 0X = -30 or x = 25 But x being speed can't be negative.Therefore original speed = x = 25 km/hr

OR

Let the marks obtained by P in Mathematics be x.Therefore marks obtained by P in Science = 28 – xNew marks in Mathematics = x + 3New marks in science = 28-x-4 = 24-x

ATQ,(x+3) (24-x) = 180 24x+72-x2-3x = 180-x2+21x = 180-72x2-21x+108 = 0x2-12x-9x+108 = 0x(x-12)-9(x-12) = 0(x-12)(x-9) = 0x = 12,9 Therefore marks in Mathematics = 12Marks in Science = 28-12 = 16

and marks in Mathematics = 9

Marks in Science = 28-9 = 19

29]

Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.To prove: PA = PBConstruction: Join OA, OB, and OP.

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.\\ OA ^ PA and OB ^ PB ... (1)In DOPA and DOPB:ÐOAP = ÐOBP (Using (1))OA = OB (Radii of the same circle)OP = PO (Common side)Therefore, DOPA @ DOPB (RHS congruency criterion)\\PA = PB (Corresponding parts of congruent triangles are equal)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.

30]

Since tangents drawn from an exterior point to the circle are equal.

Therefore, DR = DS = 5 cm Now AR = AD-DR = 23-5 = 18 cm But AR = AQ Therefore AQ = 18 cm Also BQ = AB-AQ = 29-18 = 11 cm But BP = BQTherefore BP = 11 cm Also OQB = OPB = 900

[Because OQ and OP are perpendiculars to AB and BC respectively.]

In quadrilateral BPOQ,QOP+ OPB+ OQB+ B = 3600

QOP = 3600-(900+900+900)= 900

Hence, OQBP is a square. BQ = OQ = OP = BP = 11 cm

Hence, radius of circle is 11 cm.

31] The sequence of her saving (in Rs) is 4, 5.75, 7.5…………Let in nth week her weekly saving be Rs 19.75Here, a = 4 and d = 5.75 – 4 = 1.75 and Tn = 19.75 346We know, Tn = a +(n – 1)d

19.75 = 4 + (n – 1) 1.75 15.75 = (n – 1) 1.75

n = 10

In 10th week her weekly saving be Rs 19.75.

32]

Volume of water in cylinder when full = p(3)2 6 = 54p cm3

Volume of solid = p(2)3 + p (2)2 4

= cm3

Volume of water in the cylinder when solid is immersed in it = 54p - 32p/3

=136.19 cm3

33] Total number of balls = 5 + 7 + 4 + 2 = 18(i) Favourable outcomes = 5 + 2 = 7

\\Required probability = (ii) Favourable outcomes = 7 + 4 = 11

\\ Required probability = (iii) Number of balls which are not white = 18 – 5 = 13

\\Required probability = (iv) Number of balls which are neither white nor black = Number of balls which are red or blue = 7 + 2 = 9

\\ Required probability = =

34]