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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
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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