1292840747 classx math sample paper saii 50

8/3/2019 1292840747 ClassX Math Sample Paper SAII 50

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TOPPER SAMPLE PAPER 5

Sum m at i ve Assessm en t - I I

MATHEMATI CS

CLASS X

M.M: 80 TI ME : 3 -31

2Hrs .

GENERAL I NSTRUCTI ONS :

1. All questions are compulsory.

2. The question paper consists of 34 questions divided into four

sections, namely

Section A : 10 questions (1 mark each)

Section B : 8 questions (2 marks each)

Section C : 10 questions (3 marks each)

Section D : 6 questions (4 marks each)

3. There is no overall choice. However, internal choice has been

provided in 1 question of two marks, 3 questions of three marks

and 2 questions of four marks each.

4. Use of calculators is not allowed.

SECTI ON A

Q1. If x=b is a solution of x2- (a + b) x + p=0, then the value of p is

(a) ab (b) a + b (c) a-b (d)a

b

Q2. The ratio of the volume of a cube to that of a sphere which will exactly fit

inside the cube is

(a) :8 (b) (c) 8: (d) 6:


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Q3. If A (1,2) , B (4,y), c (x,6) and D (3,5) are the vertices of a parallelogramtaken in order then the values of x and y are

(a) 6 and 5

(b) 6 and 3

(c) 2 and 3

(d) 5 and 2

Q4. In a lottery there are 5 prizes and 20 blanks. The probability of getting a prize

is

(a)1

2(b)

1

3(c)

1

4(d)

1

5

Q5. If x silver coins 1.75 cm in diameter and of thickness 2 mm, must be melted

to form a cuboid 11cmx10cmx7cm, then the value of x is

(a) 1200 (b) 1400 (c) 1600 (d) 1800

Q6. PQ is a tangent drawn from a point P to a circle with centre O and QOR is a

diameter of the circle such that POR=1200, then OPQ is

(a) 600 (b) 450 (c) 300 (d) 900

Q7. In given AP 210 is ............. term: 2,6,10........

(a) 50th (b) 52nd (c) 53rd (d) 54th

Q8. A kite is flying, attached to a thread which is 165m long. The thread makes

an angle of 300 with the ground. The height of the kite from the ground,

assuming that there is no slack in the thread is

(a) 80 m (b) 81.5 m (c) 82.5 m (d) 84 m


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Q9. The points (a, b + c), (b, c + a) and (c, a + b) are

(a) Collinear(b) Non-collinear

(c) Concurrent

(d) All of the above

Q10. If the area of a circle is 154 cm2, then its perimeter is

(a) 11cm (b) 22cm (c) 44cm (d) 55cm

SECTI ON B

Q11. Find the area of a square ABCD, whose vertices are A(5,6), B(1,5), C(2,1)

and D(6,2).

Q12. The height of a right triangle is 7 cm less than its base. If the hypotenuse is

13 cm, find the other two sides.

Q13. Prove that the angle between two tangents drawn from an external point to a

circle is supplementary to the angle subtended by the line segments joiningthe points of contact at the centre.

Q14. The minute hand of a clock is 10 cm long. Find the area of the face of the

clock described by the minute hand between 9AM and 9:35 AM.

Q15. Four equal circles are described about the four corners of a square so that

each touches two of the others as shown in figure. Find the area of the square

not included in the circles if each side of the square measures 14cm.

Q16. A number x is selected from the numbers 1, 2, 3 and then a second number y

is randomly selected from the numbers 1, 4, 9. What is the probability that

the product xy of the two numbers will be less than 9?

OR

Three unbiased coins are tossed together. Find the probability of getting

atmost two tails.


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Q17. Determine the 10th term from the end of the AP 4,9,14,........,254.

Q18. Show that the tangents at the end points of a diameter of a circle are parallel.

SECTI ON C

Q19. The area of a triangle is 5. Two of its vertices are (2, 1) and (3,-2) and the

third vertex lies on y=x+3. Find the third vertex.

Q20. If the roots of the equation (a-b)x2+(b-c)x+(c-a)=0 are equal then prove that2a=b+c.

OR

Solve :1 1 1 1

a b x a b x

Q21. Draw a pair of tangents to a circle of radius 5cm which are inclined to each

other at angle of 600. Also, write the steps of construction.

OR

Construct a ABC in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Then,

construct a triangle ABC such thatC A BA BC 2

CA BA BC 3

.

Q22. The m th term of an AP is n and the n th term is m. find the sum of (m + n)

terms.

Q23. Two pillars of equal height are on either side of a road, which is 100m wide.

The angles of elevation of the top of the pillars are 600 and 300 at a point onthe road between the pillars. Find the height of the pillars.

OR


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As observed from the top of a lighthouse, 100 metres high above sea level,the angle of depression of a ship moving directly towards it, changes from 30o

to 60o. Determine the distance travelled by the ship during the period of

observation.

Q24. Find the coordinates of the vertices of a ABC with A (1,-4) and the mid point

of sides through A being (2,-1) and (0,-1).

Q25. One card is drawn from a well shuffled pack of 52 cards. Calculate theprobability of getting

(i) A king or a queen

(ii) Neither a heart nor a red king

Q26. Find the area of the shaded region if PQ=24cm, PR=7cm and O is the centre

of the circle.

Q27. An iron pole consisting of a cylindrical portion 110cm high and of basediameter 12cm is surmounted by a cone 9cm high. Find the mass of the pole,

given that 1cu cm of iron has 8 gm mass (approx.)

Q28. Water flows at the rate of 10m per minute through a cylindrical pipe having

its diameter as 5mm. how much time will it take to fill a conical vessel whosediameter of base is 40cm and depth is 24cm?

SECTI ON D

Q29. Prove that the coordinates of the centroid of a triangle whose vertices are

(x1,y1),(x2,y2)and (x3,y3) is given by 1 2 3 1 2 3x x x y y y,

3 3

.


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Q30. A vertical tower stands on a horizontal plane and is surmounted by a verticalflagstaff of height h. At a point on the plane, the angles of elevation of thebottom and the top of the flagstaff are and respectively. Prove that the

height of the tower is h tan

tan tan.

Q31. By increasing the list price of a book by Rs10, a person can buy 10 less books

for Rs.1200. Find the original list price of the book.

OR

Two years ago a mans age was three times the square of his sons age.

Three years hence his age will be four times his sons age. Find their present

ages.

Q32. Prove that the lengths of tangents drawn from an external point to a circle are

equal.

OR

Prove that the radius of a circle is perpendicular to the tangent at the point of

contact.

Q33. A circle is touching the side BC of ABC at P and touching AB and AC

produced at Q and R respectively. Prove that

AQ =1

2(perimeter of ABC).

Q34. A bucket is in the form of a frustum of a cone of height 30 cm with radii of its

lower and upper ends as 10 cm and 20 cm, respectively. Find the capacity

and surface area of the bucket. Also, find the cost of the milk which cancompletely fill the container, at the rate of Rs 25 per litre (use = 3.14).


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SOLUTI ONS

SECTI ON A

Ans1. Option (a) ab

Given x=b therefore it satisfies the given equation

b2-(a + b) b+p=0 p=ab

Ans2. Option (d) 6:

Let x be the edge of cube. Then, x is also the diameter of the sphere.

Ratio of the volume of the cube to the volume of sphere

= x3 :

34 x 4

1 : 6 :3 8 24

Ans3. Option (b) 6 and 3

We know that the diagonals of a parallelogram bisect each other.

So, coordinates of mid point of AC = coordinates of mid point of BD.

x 1 6 2 3 4 5 y

, ,2 2 2 2

x 1 7 5 y,4 ,

2 2 2

x 1 7 5 yand 4

2 2 2

X=6 and y=3.


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Ans4. Option (d)1

5

5 prizes and 20 blanks

Total number of outcomes = 25

Number of favourable outcomes =5

P(getting a prize) =5 1

25 5

Ans5. Option (c) 1600

Required number of coins =volumeof cuboid

volumeof eachcoin

=

770 7

22 0.875 0.875 .2

= 40x40 =1600

Ans6. Option (c) 300

Ans7. Option (c) 53rd

an=a+(n-1)d

210=2+(p-1)4 4(p-1)=208 P-1=52 P=53Hence 53rd term is 210.

Ans8. Option (c) 82.5 m


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Let A be the position of the kite. Let O be the observer and OA be the thread.

Given, OA =165m, BOA = 300 and let AB = h m. In right OBA,

AB

OA= sin30

h 1

165 2

h =165

2= 82.5

Thus, the height of the kite from the ground is 82.5 m.

Ans9. Option (a) collinear

Area of a triangle =1

2[{a(c+a)+b(a+b)+c(b+C)}-{b(b+c)+c(c+a)+a(a+b)}]

=1

2[ac+a2+ab+b2+bc+c2-b2-bc-c2-ca-a2-ab]

=1

2(0) = 0


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Since, the area of triangle is 0, therefore, the given points are collinear.

Ans10.Option (c) 44cm

Area of a circle = r2

154 =22

7r2

r2=49 or r=7

perimeter = 2r = 2x22

7x 7 = 44cm.

SECTI ON B

Ans11.In a square, all the sides are equal. So, AB=BC=CD=DA

Distance AB= 2 21 5 (5 6) 17 units 1 mark

Area of Square ABCD = AB2

= 2

17 = 17 sq. Units. 1 mark

Ans12.let base = x cm and height = (x 7) cmHypotenuse = 13 cm

By Pythagoras theorem,

X2 + (x 7)2 = 1321

2mark

X2 + 49 + x2 14x = 169

2x2 14x 120 = 0

X2 12x + 5x 60 = 0


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X( x 12) + 5 (x 12) = 01

2mark

(x 12) (x + 5) = 0

x 12 = 0 or x + 5 = 0

x=12 or x=-5 (rejecting)1

2mark

The other two sides are 12 cm and 5 cm.1

2mark

Ans13.

Let PA and PB be two tangents drawn from an external point P to a circle withcentre O. We have to prove that angles AOB and APB are supplementaryi.e. AOB+ APB=1800.

In right triangles OAP and OBP, we have

PA=PB (tangents drawn from an external point to a circle are equal in length)

OA=OB (radii of same circle)

OP=OP (common)

OAP OBP (by SSS)


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OPA= OPB and AOP= BOP (by cpct) 1 mark

APB = 2 OPA and AOB =2 AOPBut, AOP=90- OPA

2 AOP =180-2 OPA AOB =180- APB AOB+ APB=180 1mark

Ans14.Angle described by the minute hand in 1 minute=60

Angle described by the minute hand in 35 minutes =6x35=21001

2mark

Area swept by the minute hand in 35 minutes=210 22

360 7

x (10)2 1 mark

= 183.3 cm2.1

2mark

Ans15.Area of the square = (14)2 = 196 cm21

2

mark

Area of four quadrants = area of one circle = r2

=22

7x (7)2 =154 cm2 1 mark

Required area = 196-154 = 42 cm2.1

2mark

Ans16.Number x can be selected in three ways and corresponding to each such way

there are three ways of selecting number y. therefore,the two numbers can

be selected in 9 ways as listed below:

(1,1),(1,4),(1,9),(2,1),(2,4),(2,9),(3,1),(3,4),(3,9)

Total number of outcomes =9 1 mark


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The product xy will be less than 9, if x and y are chosen in one of the

following ways:

(1,1),(1,4),(2,1),(2,4), (3,1)1

2mark

Number of favourable outcomes =5

P(product less than 9) =5

9.

1

2mark

OR

S=[HHH,HHT,HTH,THH,HTT,THT,TTH,TTT]

Total number of outcomes = 81

2mark

At most two tails = HHH, HHT, HTH, THH, HTT, THT, TTH

Number of favourable outcomes =7 1 mark

P(at most two tails) =7

8

1

2mark

Ans17.L= last term =254 and d=5 12

mark

Nth term from the end = l- (n-1)d1

2mark

tenth term from the end = 254- (10-1)5

=254-45 = 209. 1 mark


14/29

Ans18.

1

2mark

Let AB be a diameter of a given circle, and let PQ and RS be the tangent lines

drawn to the circle at points A and B respectively. Since tangent at a point to

a circle is perpendicular to the radius through the point.

Therefore, AB is perpendicular to both PQ and RS.

PAB=90 and ABS =90 1 mark PAB= ABSBut, these are a pair of alternate interior angles.

Therefore, PQ is parallel to RS.1

2mark

SECTI ON C

Ans19.Let the third vertex be A(x, y). Other two vertices of the triangle are B(2,1)and C(3,-2).

Area of BC = 5 sq. Units


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1

2{x (1+2) + 2 (-2-y) + 3 (y-1)} = 5 1 mark

1

2{3x-4-2y + 3y-3} = 5

{3x+ y-7} = 10

3x + y -17 = 0 (i)

or 3x +y +3 = 0 (ii) 1 mark

Given that, A(x, y) lies on y = x + 3 (iii)

On solving (i) and (iii) we get, x =7

2and y =

13

2and,

On solving (ii) and (iii) we get, x =3

2

and y =3

2. 1 mark

Ans20.Here, (a-b)x2+(b-c)x+(c-a)=0

The given equation will have equal roots, if

(b-c)2-4(a-b)(c-a)=0 1 mark

b2+c2-2bc-4(ac-bc-a2+ab) =0

b2+c2+4a2+2bc-4ab-4ac=0

2(b+c-2a) =0 1 mark

b+c-2a=0


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b+c=2a 1 mark

OR

1 1 1 1

a b x a b x

1 1 1 1

a b a b x x

x (a b x)a b

ab a b x x

a ba b

ab a b x x

1 mark

ax+bx+x2=-ab

x2+ax+bx+ab=0 1 mark

x(x+a)+b(x+b)=0

(x+a)(x+b)=0

x+a=0 or x+b=0

x=-a,-b 1 mark

Ans21.The tangents on given circle as follows:

1. Draw a circle of 5 cm radius and with centre O.2. Take a point P on circumference of this circle. Extend OP to Q such that OP

= PQ.

3. Midpoint of OQ is P. Draw a circle with radius OP with centre as P.

Let it intersect our circle at R and S. Join QR and QS. QR and QS are required

tangents.1 mark

2 marks

OR


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A triangle ABC is to be constructed such thatC A BA BC 2

CA BA BC 3

. This

means that the triangle ABC is similar to the triangle ABC with scale factor as2

.3

ABC is the required triangle.

3 marks

Ans22.Here, a+(m-1)d =n (i)

And, a+(n-1)d =m ..(ii) 1 mark

On solving (i) and (ii), we get,

d = -1 and a= m+n-1 1 mark

Sm+n=m n

2

[2(m+n-1) + (m+n-1)(-1)] =

m n

2

(m+n-1) 1 mark

Ans23.


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1

2mark

Let AB and CD be two pillars of equal heights, say, h metres. Let P be a pointon road such that AP= x m and PC = 100 x. APB= 600 and CPD =300.

In right PAB,

Tan60 =AB

AP

h

3 h 3xx

.. (1) 1 mark

In right PCD,

tan30 =CD

PC

1 h

100 x3 .(2)1

2 mark

From (1) and (2) we get,

3x=100-x or x=25

From (1) , h=25 3 1 mark


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Thus, the height of the pillars is 25 3 metres.

OR

1

2mark

Let A and B be the two positions of the ship. Let d be the distance travelled

by the ship during the period of observation, i.e., AB=d metres.

Suppose that the observer is at the point P. given that PC = 100m.

Let h be the distance (in metres) from B to C.

From right triangle PCA, we have

d hcot30 3

100

d+h = 100 3 ..(i) 1 mark

Again in triangle PCB, we have,

h 1cot 60

100 3


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h =100

3metres.

1

2mark

Putting the value of h in (i) we get,

d= 1001

33

=

200

3= 115.47 (approx.)

Thus, the distance travelled by the ship from A to B is 115.47 m(approx.)1 mark

Ans24.Let the coordinates of A, B and C be (1, -4), (a, b) and (x, y) resp.

Then,1 a 4 b

,2 2

= (2,-1) 1 mark

Therefore, 1+a=4, -4+b= -2

a=3, b=2 1 mark

Also,1 x 4 y

,2 2

= (0,-1)

Therefore, 1+x=0, -4+y= -2

x= -1 , y=2

Therefore, the coordinates of the vertices of ABC are A (1,-4), B (3, 2)

and C (-1, 2) 1 mark

Ans25.Total number of outcomes =52

(i) Number of kings = 4

Number of queens = 4


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P(king or queen) = 4 4 8 2

52 52 13

1

1

2marks

(ii) Number of cards of hearts =13

Number of red kings =2 ( out of these 1 is already in hearts)

Neither a heart nor a red king = 52 (13+2-1) =52-14 =38

P(neither a heart nor a red king) =38 19

52 26 1

1

2marks

Ans26

.

In QPR, by Pythagoras theorem, we have,

QR2=PR2+PQ2 = 72+242=49+576=625

QR=25cm

Diameter of the circle = 25cm

Radius of the circle =25

21 mark

Area of semicircle =2r 22 25 25 11 625

2 2 7 2 2 28

cm2


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Area of PQR = 21 1

PR PQ 7 24 84cm2 2

1 mark

Area of shaded region =11 625 6875 2352 4523

8428 28 28

= 161.54 cm2 1 mark

Ans27.

Volume of the pole = volume of the cylinder + volume of the cone

= 2 21

6 110 6 93

cu cm 1 mark

= [3960+108] cu cm

= 4068 cu cm 1 mark

Mass of the pole = 4068 x3.14x8

= 102188.16 grams = 102.19 kg 1 mark


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Ans28.Amount of water required to fill the conical vessel = volume of the conicalvessel

=2

1 40

3 2

(24) cu cm =3200 cu cm .(i) 1 mark

Amount of water that flows out of the cylindrical pipe in 1 minute

=2

5

20

(10x100) = 62.5 cu cm ..(ii) 1 mark

From (i) and (ii) , we get,

Time taken to fill the vessel =3200

62.5

minutes = 51.2 minutes 1 mark

SECTI ON D

Ans29.

1

2 mark

Let A(x1,y1), B(x2,y2) and C(x3,y3) be the vertices of ABC whose medians are

AD,BE and CF respectively. So, D,E and F are respectively the mid points of

BC,CA and AB.

1 mark


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Coordinates of D are 3 2 3 2x x y y

,2 2

1

2mark

Coordinates of a point G(x, y) dividing AD in the ratio 2 : 1 are1

2mark

2 3 2 31 1

x y2( ) 2( )

2 2( , )2 1 2 1

x yx y

1 mark

1 2 3 1 2 3x + x y + y

( , )3 3

x y 12 mark

Ans30.

1 mark

Let AB be the tower and BC be the flagstaff. Let OA=x metres, AB= y metres

and BC= h metres.

In right OAB,

Tan =

yABy x tan or x

OA tan..(i) 1mark

In right OAC,


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Tan = y hy h

xx tan

.(ii) 1mark

From (i) and (ii),

y hytan tan

y( tan - tan ) = h tan

y = htan

tan tan

Thus, the height of the tower is

htan

tan tan

. 1 mark

Ans31.Let list price of the book = Rs x

And increased price of the book = Rs (x+10)

According to question,

1200 1200

x x 10 =10 1 mark

(1200)1 1

x x 10

=10

(1200)

x 10 x

x x 10=10

1

2mark

1200=x(x+10)

x2+10x-1200=0 1 mark

(x+40)(x-30)=0X=-40

X=30 1 mark

But x is the cost of the book and hence cant be negative.

Therefore x=30

List price of book=Rs 301

2mark

OR

Two years ago, let sons age = x years


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Therefore mans age = 3x2

Three years hence, sons age = x+2+3=(x+5) years and mans age = 3x2+5

By the given condition

3x2+5=4(x+5) 1 mark

3x2-4x+5-20=0

3x2-4x-15=0 1 mark

3x2-9x+5x-15=0

3x(x-3)+5(x-3)=0

(x-3)(3x+5)=0

x=3

[Because 3x+5=0 means x =5

3

, not possible) 1 mark

Therefore sons present age = x+2=3+2=5 years

Mans present age = 3x2+2=3(3)2+2

= 27+2=29 years 1 mark

Ans32.Given: A circle with centre O; PA and PB are two tangents to the circle drawn

from an external point P.To prove: PA = PB

Construction: Join OA, OB, and OP.

2 marks

It is known that a tangent at any point of a circle is perpendicular to theradius through the point of contact. OA PA and OB PB ... (1)In OPA and OPB:

OAP = OBP (Using (1))OA = OB (Radii of the same circle)

OP = PO (Common side)Therefore, OPA OPB (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 externalpoint to a circle are equal. 2 marks

OR

Given: A circle C (O, r) and a tangent AB at a point P.

To Prove: OP is perpendicular to AB.


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Construction: Take any point Q, other than P, on the tangent AB. Join OQ.

2 marksSince, Q is a point on the tangent AB, other than the point of contact P, so Q

will be outside the circle.Let OQ intersect the circle at R.Then, OQ=OR+RQ

OQ>OR OQ>OP (OR=OP=radius)Thus, OP


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CP=CR (From C) ...............(ii)

And, AQ=AR (From A) ..........(iii) 1 markFrom (iii) we have,

AQ=AR

AB+BQ=AC+CR

AB+BP=AC+CP ...........(iv) 1 mark

Perimeter of ABC = AB+BC+CA= AB+(BP+PC)+AC

= (AB+BP) + (AC+PC)

= 2(AB+BP)

= 2(AB+BQ)

= 2AQ

AQ =1

2(perimeter of ABC). 1 mark

Ans34.Capacity (or volume) of the bucket =h

3

[r21 + r

22 + r1r2]

Here, h = 30 cm, r1 = 20 cm and r2 = 10 cm1

2mark

So, the capacity of bucket = 3.14 x30

3[202 + 102 + 20 x 10] cm3

= 21980 cu cm = 21.980 litres. 1 mark

Cost of 1 litre of milk = Rs 25

Cost of 21980 litres of milk = Rs 21.980 x 25 = Rs 549.501

2mark

Surface area of the bucket


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= curved surface area of the bucket + surface area of the bottom

= l(r1 + r2) + r22 where l=

2 21 2h (r r )

1

2mark

Now, l = 900 100 cm = 31.62 cm1

2mark

Therefore, surface area of the bucket = 3.14 x 31.62 (20+10) + 3.14x 210

= 3.14 x 1048.6 cm2

= 3292.6 cm2 (approx.) 1 mark

1292840747 classx math sample paper saii 50

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