TOPPER SAMPLE PAPER 3Summative Assessment-II

MATHEMATICSCLASS X

M.M: 80 TIME : 3-3 12

Hrs.

GENERAL INSTRUCTIONS :

1. All questions are compulsory.2. The question paper consists of 34 questions divided into four

sections, namelySection 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 beenprovided in 1 question of two marks, 3 questions of three marksand 2 questions of four marks each.

4. Use of calculators is not allowed.

SECTION A

Q1. The quadratic equation 2x2- 5x +1=0 has

(a) Two distinct real roots(b) two equal real roots(c) no real roots

(d) more than two real roots

Q2. The 9th term of an AP is 449 and 449th term is 9. The term whichis equal to zero is

(a) 501th (b) 502th (c) 458th (d) 459th

Q3. Two circles touch each other externally at C and AB is a commontangent to the circles. Then, ACB=

(a) 600 (b) 450 (c) 300 (d) 90

Q4. The length of an arc that subtends an angle of 24o at the centreof a circle with 5 cm radius is

(a) 23 cm (b) 3

2 cm (c)

3 cm (d) 5

3 cm

Q5. An observer 1.5 m tall is 28.5 m away from a tower. The angleof elevation of the top of the tower from his eyes is 450. Theheight of the tower is

(a) 10 m (b) 20 m (c) 30 m (d) 40 m

Q6. The probability that a randomly chosen number from one totwelve is a divisor of twelve is

(a) 112

(b) 12

(c) 14

(d) 16

Q7. The value of x, for which the points (x,-1), (2,1) and (4,5) lie ona line is

(a) 0 (b) 1 (c) 2 (d) 3

Q8. The ratio in which the point R 2 20,7 7

divides the join of

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

(a) 3:4 (b) 4:3 (c) 2:1 (d) 1:2

Q9. A pendulum swings through an angle of 300 and describes an arc8.8cm in length. The length of the pendulum is

(a) 8.8 cm (b) 15.8 cm (c) 16.8 cm (d) 17 cm

Q10. Three cubes whose edges measure 3cm, 4cm and 5 cmrespectively are melted to recast a single cube. The surface areaof the new cube is

(a) 216 cm2 (b) 200 cm2 (c) 215 cm2 (d) 220 cm2

SECTION B

Q11. 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.

OR

Find the area of the quadrilateral ABCD whose vertices areA(1,1), B(7,-3), C(12,2) and D(7,21) respectively.

Q12. An A.P. consists of 60 terms. If the first and the last term be 7and 125 respectively, find the 32nd term.

Q13. A box contains 90 discs which are numbered from 1 to 90. If onedisc is drawn at random from the box, find the probability that itbears

(i) A two digit number (ii) a perfect square number

Q14. Find two consecutive positive integers, sum of whose squares is25.

Q15. Prove that the tangents at the extremities of any chord makeequal angles with the chord.

Q16. In given figure, find the area of the circle not included in therectangle whose sides are 8cm and 6cm respectively.

Q17. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm,find the area of the sector.

Q18. A circle touches all the four sides of a quadrilateral ABCD. Provethat

AB + CD = BC + DA

SECTION C

Q19. If a b c, prove that the points (a,a2), (b,b2) and (c,c2) cannever be collinear.

Q20. A body falls 8 metres in the first second of its motion, 24 metresin the second, 40 metres in the third second and so on. Howlong will it take to fall 2048 metres?

Q21. Draw a pair of tangents to a circle of any convenient radius,which are inclined to the line joining the centre of the circle andthe point at which they intersect at an angle of 45°. Also writethe steps of construction.

Q22. From a balloon vertically above a straight road, the angles ofdepression 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 adistance a and b metres from the base and in the same straightline with it are complementary. Prove that the height of thetower is ab metres.

Q23. A bag contains 12 balls out of which x are white.

(i) If one ball is drawn at random, what is the probability that itwill be a white ball?

(ii) If 6 more white balls are put in the bag, then the probabilityof 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 onany day of the week. What is the probability that both will visitthe shop on

(i) the same day?(ii) two different days?(iii) consecutive days?

Q24. In an equilateral triangle of side 24 cm, a circle is inscribedtouching its sides. Find the area of the remaining portion of thetriangle. (take 3 =1.732)

OR

A round table cover has six equal designs as shown in the figure.

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.

Q25. A canal is 300cm wide and 120cm deep. The water in the canalis flowing with a speed of 20km/h. How much area will it irrigatein 20 minutes if 8cm of standing water is desired?

Q26. From a solid cylinder whose height is 2.4 cm and diameter1.4cm, a conical cavity of the same height and same diameter ishollowed out. Find the total surface area of the remaining solidto the nearest cm2.(use =3.1416)

Q27. Solve for x, using the quadratic formula:

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

Q28. 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) andC(2,-9), then prove that,

Ar. ∆DEF = 14

(Ar ∆ABC).

SECTION D

Q29. 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 ofthe first and the third term exceeds the second term by 29, findthe AP.

Q30. The angle of elevation of the top of a tower from a point on thesame level as the foot of the tower is . On advancing p metrestowards the foot of the tower, the angle of elevation becomes β.Show that the height h of the tower is given by

p tan tantan tan

Q31. A passenger train takes 2 hour less for a journey of 300 km, ifits speed is increased by 5km/hr from its usual speed. Find itsusual speed.

OR

In a class test, the sum of the marks obtained by a student P inMathematics and Science is 28. Had he got 3 more marks inMathematics and 4 marks less in Science, the product of marksobtained in the two subjects would have been 180. Find themarks obtained in the two subjects separately.

Q32. Prove that the lengths of tangents drawn from an external pointto a circle are equal.

Q33. In given figure, a circle is inscribed in a quadrilateral ABCD inwhich B=900. If AD=23 cm, AB=29 cm, and DS=5 cm, findthe radius r of the circle.

Q34. A solid consisting of a right circular cone, standing on ahemisphere, is placed upright, in a right circular cylinder, full of

water, and touches the bottom. Find the volume of water left inthe cylinder, having been given that the radius of the cylinder is3 cm and its height is 6 cm, the radius of the hemisphere is 2 cmand the height of the cone is 4 cm.

SOLUTIONS

SECTION A

Ans1.Option (c)

2x2- 5x +1=0

D= b2-4ac

= 25 -4(2)(1)

= 5-8 = -3

Since D < 0, therefore the given quadratic equation has no realroots. 1 mark

Ans2.Option (c)

a+8d = 449 and,

a+448d = 9

on solving we get, 440d = -440 i.e. d = -1

therefore, a-8 = 449 i.e. a = 457

let its nth term be zero.

an = a+(n-1)d

0 = 457+(n-1)(-1)

457 = n-1 or n = 458 1 mark

Ans3.Option (d)

Lengths of tangents drawn from an external point to a circle areequal.

PA=PC (from P)

Therefore, PAC= PCA = x (say)

Also, PC=PB (from P)

PBC= PCB=y

In ∆ABC,

ABC+ ACB+ BAC=180

y + (x + y) + x = 180

2(x + y) = 180

x+ y = 90

ACB = 900. 1 mark

Ans 4.

Option (a)

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

24 2x2 .5 x2 .5360 30

23

The length of the arc is 23 cm. 1 mark

Ans5.

Option (c)

Let AC be the tower of height h metres and ED be the observerof height 1.5 m at a distance of DC=28.5 m from the tower AC.

In right ∆AED,

Tan45 = ABEB

1= AB28.5 (as EB = DC = 28.5 m )

AB = 28.5 m

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

= 28.5 + 1.5 = 30m 1 mark

Ans6.Option (b)

Total number of outcomes = 12

Numbers which are divisors of 12 are 1,2,3,4,6,12.

Number of favourable outcomes = 6

P(divisor of 12) = 6 112 2

1 mark

Ans7.Option (b)

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 1 mark

Ans8.Option (a)

Let R 2 20,7 7

divides PQ in the ratio K:1

2K 2 4K 2 2 20, ,K 1 K 1 7 7

2K 2 2K 1 7

and 4K 2 20

K 1 7

14K-14 = -2K-2

16K = 12

K= 12 316 4

Thus R divides PQ in the ratio 3:4. 1 mark

Ans9.Option (c)

Length of arc (l) = 2r360

8.8 360 7 = 30 2 22 r

r= 8.8 6 722 cm = 16.8 cm 1 mark

Ans10. Option (a)

Let x be the edge of the new cube. Then,

Volume of the new cube = sum of the volumes of three cubes

x3 = 33 + 43 + 53

x3 = 216 i.e. x= 6cm

Surface area of the new cube = 6x2 = 6 (6)2 cm2 = 216 cm2

1 mark

SECTION B

Ans11.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 1 2 3 1 2 3x x x y y y,

3 3

i.e.

a b c b c a,3 3

1 mark

Also, given that centroid is at origin i.e. its coordinates are (0,0).

So,

a b c 0 a b c 03

. 1 mark

OR

Area of quadrilateral ABCD

= area of triangle ABC + area of triangle ACD

Area of triangle ABC = 12

[1(-3-2) +7(2-1)+12(1+3)]

= 12

(-5+7+48) = 25 sq. units 1 mark

Area of triangle ACD = 12

[1(2-21) + 12(21-1)+7(1-2)]

= 12

(-19+240-7) = 107 sq. units 12

mark

Area of quadrilateral ABCD = 25 + 107

= 132 sq. Units 12

mark

Ans12.Here, n = 60, a = 7 and 60t = 125 12

mark

7+59d = 125

d=2 12

mark

Therefore, 32nd term (t32) = a + 31d = 7 + 31 x 2 = 69 1 mark

Ans13.Total number of outcomes = 90

(i) Two digit numbers are 10,11,12,13,.............90Number of favourable outcomes = 81

P(two digit number) = 81 990 10

1 mark

(ii) Perfect square numbers from 1 to 90 are1,4,9,16,25,36,49,64,81

P(perfect square number) = 9 190 10

1 mark

Ans14. let the required numbers be x and x+1

Given x2 + (x+1)2 = 25 1 mark

2x2 +2x-24 =0 x2 + x – 12 = 0(x+4) (x-3) = 0X = -4 or x = 3

Reject x = -4 12

mark

Therefore the given consecutive positive integers are

X = 3 and x + 1 = 3 + 1 = 4 12

mark

Ans15.

Let AB be a chord of a circle with centre O, and let AP and BP bethe tangents at A and B respectively. Suppose the tangents meetat P. join OP. suppose OP meets A at C.

We have to prove that PAC= PBC 12

mark

In triangles PCA and PCB, we have

PA = 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 ) 1 mark

PAC = PBC ( by cpct )12

mark

Ans16. Diameter of the circle = diagonal BD of the rectangle ABCDIn right triangle BCD,

BD2 = BC2+BD2 = 62+82 = 100

BD2 = 100 i.e.BD = 10 cm

Radius of the circle =102

= 5cm12

mark

Area of the circle = r2 = 3.14(5)2 = 78.50 cm2 12

mark

Area of rectangle ABCD = 68 = 48 cm2 12

mark

Area of the circle not included in the rectangle

= area of the circle – area of rectangle ABCD

= 78.50 – 48 = 30.50 cm2 12

mark

Ans17.

Let OAB be the given sector. Then,

Perimeter of sector OAB = 16.4 cm12

mark

OA+OB+ arc AB = 16.4 cm

5.2+5.2+arc AB =16.4

Arc AB = 6 cm i.e. l =6cm12

mark

Area of sector = 12

lr = 12

6 5.2 = 15.6 cm2 1 mark

Ans18.

Since tangents drawn from an external point to a circle are equalin length.

AP = AS (From A) ………..(i)

BP = BQ (From B) ………(ii)

CR = CQ (From C) ………(iii)

DR = DS (From D) ………(iv) 1 mark

Adding (i) ,(ii), (iii) and (iv), we get,

(AP+BP)+(CR+DR) = (AS+DS)+(BQ+CQ)

AB+CD = AD+BC. 1 mark

SECTION C

Ans19. If the area of the triangle formed by joining the given points is

zero, then only the points are collinear.

Area of triangle = 1 2 3 2 3 1 3 1 21 x y y x y y x y y2

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

Substituting the values in the formula for area of a triangle, weget

Area of triangle = 2 2 2 2 2 21 a b c b c a c a b

21 mark

12

mark

1 mark

It is given that a b c

Area of the triangle can never be 012

mark

Ans20. The AP is 8, 24, 40, ….., and the sum is 2048.

We have to determine n.

Here a = 8, d = 1612

mark

Since, Sn = 2048, we have

2048 = n2

[2x8+(n-1)16] 1 mark

2048 = n[8+8n-8]

2048 = 8n2

n2 = 256 i.e. n = 16 (as n being number of terms, it can’t benegative) 1 mark

Hence, the body will take 16 seconds to fall 2048 metres.12

mark

Ans21. 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 theperpendiculars intersect at point P.

(v) Join OP.PA and PB are the required tangents, which make an angle of45 with OP.

1 mark

2 marks

Ans22.

12

mark

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

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

From right ∆PBQ,

PQBQ

= tan60 = 3 or h 3h 100

or h = 3 (h-100)

1 mark

Therefore, h = 100 33 1

= 50(3+ 3 )

i.e. the height of the balloon is 50(3+ 3 ) m. 12

mark

OR

12

mark

Let AB be the tower of height h metres and C and D be twopoints at a distance of a and b respectively from the base of thetower.

AC = a m

AD = b m (a>b)

In right ∆CAB,

tanx = ABAC

tanx = ha

………………….(i) 1 mark

In right ∆DAB,

Tan(90-x) = ABAD

cotx = hb

……………(ii) 12

mark

From (i) and (ii),

tanx.cotx =2h

ab

1 =2h

ab h2 = ab or h = ab metres. 1 mark

Thus, the height of the tower is ab metres.

Ans23. 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 = 12

There 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) = x12

1 mark

If 6 more white balls are put in the bag, then

Total number of balls in the bag = 12+6 = 18

Number of white balls in the bags = x + 6 12

mark

Therefore p2=P(Getting a white ball) = x 618 1

2mark

It is given that

p2 = 2p1

x 6 x218 12

12

mark

x 6 x18 6

6(x+6) = 18x

6x+36 = 18x

12x = 36

x = 3612

= 3 12

mark

OR

Total number of equally likely outcomes for visiting shop in thesame 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) = 6 136 6

1 mark

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

P(two different days) = 30 536 6

1 mark

(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) = 10 536 18

1 mark

Ans24.

Let ABC be an equilateral triangle of side 24cm, and let AD be analtitude from A on BC. Since the triangle is equilateral, so ADbisects BC.

Therefore BD=CD=12 cm 12

mark

The centre O of the inscribed circle will coincide with the centroidof ∆ABC

Therefore OD= AD3

(as centroid divides each median in the ratio 2:1)

12

mark

In right ∆ABD, we have

AB2 = AD2+BD2 [Using Pythagoras Theorem]

242 = AD2+122

AD = 12 cm

Therefore OD =

AD 1 12 3 cm 4 3 cm3 3

12

mark

Area of the incircle = (OD)2

=

22 2 222 224 3 cm 48 cm 150.85cm

7 712

mark

Area of the triangle ABC = 34

(Side)2

= 34

(24)2 = 249.4 cm2 12

mark

Therefore, area of the remaining portion of the triangle

= (249.4-150.85)cm2

= 98.55 cm2 12

mark

OR

Area of one design = 260 28360

– area of ∆OAB 1 mark

= 228

6-

22834

= 282(0.5238-0.433)

=282x0.0908

= 71.1872

=71.19 cm2 (approx.) 1 mark

Total cost = 6 71.19x 0.35

= Rs149.50 1 mark

Ans25.Volume of water flows in the canal in one hour = width of thecanal x depth of the canal x speed of the canal water

= 3x1.2x20x1000m3 = 72000m3. 1 mark

In 20 minutes the volume of water = 72000 2060

m3

= 24000 m3. 1 mark

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

is required = 240000.08

= 300000 m2 = 30 hectares. 1 mark

Ans26.

Radius of the cylinder = 0.7cm

Height of the cylinder = 2.4 cm

Surface area of the cylinder = 2rh+r2

= [20.72.4+(0.7)2]

= 3.85 cm2 1 mark

Slant height of the cone is given by

l2 = (0.7)2+(2.4)2 = 6.25 l = 2.5 cm 12

mark

surface area of interior conical portion = rl = x0.72.5

= 1.75 12

mark

Total surface area = 3.85 +1.75 = 5.60 cm2

= 5.6 3.1416 = 17.59 cm2. 1 mark

Ans27.Here, abx2-(a2+b2)x + ab = 0

X =

22 2 2 2(a +b ) ± a b 4abab

2ab1 mark

=

2 2 4 4 2 2(a +b ) ± a b 2a b

2ab12

mark

= 2 2 2 2(a +b ) ± a b

2ab12

mark

Therefore, x = a borb a

1 mark

Ans28. Let A = (x1,y1) = (-4,1), B = (x2,y2) = (6,7) and

C = (x3,y3)= (2,-9)

Area of ∆ABC = 12

[x1 (y2 - y3 )+x2 ( y3- y1) +x3 (y1 - y2)]

Area of ∆ABC = 12

[ -4(7+9) + 6(-9-1) + 2(1-7)]

Area of ∆ABC = 12

[ -64-60-12] = 68 sq. Units. 1 mark

Coordinates of D, E and F are

D 6 2 7 9 4 2 1 9, , E ,

2 2 2 2

and F 4 6 1 7,

2 2

i.e. D(4,-1), E(-1,-4) and F(1,4) 12

mark

Area of ∆DEF =12

[ 4(-4-4) -1(4+1) + 1(-1+4)]

Area of ∆DEF = 12

[ -32-5+3] = 17 sq. Units. 1 mark

14

(Area of ∆ABC) = 14

(68) = 17 = Area of ∆DEF 12

mark

SECTION D

Ans29. Here, a =-11, d =-7-(-11) =4,an = 49

We have an = a+(n-1)d 12

mark

So, 49 = -11+(n-1) 4

i.e., 60 = (n-1) 4

i.e., n = 16 1 mark

As n is an even number, there will be two middle terms whichare

162

th and

16[ 1]2

th

i.e., 8th term and the 9th term. 1 mark

a8 = a+7d = -11+74 =17 12

mark

a9 = a+8d = -11+84 = 21 12

mark

So, the values of the two middle most terms are 17 and 21,

respectively. 12

mark

OR

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

so, a-d+a+a+d=33

or a = 11 1 mark

Also, (a-d)(a+ d) = a+29

i.e., a2-d2 = a+29 1 mark

i.e., 121-d2 = 11+29

i.e., d2 = 81

i.e., d = ± 9 1 mark

So there will be two APs and they are:2, 11, 20,...

And 20, 11, 2, ... 1 mark

Ans30.

1 mark

Let CQ be the tower of height h metres and A and B be thepoints of observation. AB = p metres and let BC = x metres.

In right ∆ACQ,

CQAC

= tan h = (p+ x)tan or x =

htan

- p

………………(i)

1 mark

In right ∆BCQ,

CQBC

= tanβ h = xtanβ or x = htan

…………………………(ii)1 mark

From (i) and (ii),

htan

- p = htan

P = h 1 1tan tan

P =h tan tantan tan

or h = p tan t antan tan

1 mark

Hence proved.

Ans31.Let usual speed = x km / hr and

increased speed = (x+5) km/hr

According to question,

300 300 2x x 5

1 mark

(Because Distance/Speed = time)

x 5 x300x 5 x

2 1

2mark

1500 = 2x(x + 5)

x2+5x = 750

x2+5x-750 = 0 1 mark

(x+30) (x-25) = 0

X = -30 or x = 25 1 mark

But x being speed can’t be negative.

Therefore original speed = x = 25 km/hr 12

mark

OR

Let the marks obtained by P in Mathematics be x.

Therefore marks obtained by P in Science = 28 – x

New marks in Mathematics = x + 3

New marks in science = 28-x-4 = 24-x

ATQ,

(x+3) (24-x) = 180 1 mark

24x+72-x2-3x = 180

-x2+21x = 180-72

x2-21x+108 = 0 1 mark

x2-12x-9x+108 = 0

x(x-12)-9(x-12) = 0

(x-12)(x-9) = 0

x = 12,9 1 mark

Therefore marks in Mathematics = 12

Marks in Science = 28-12 = 16

and marks in Mathematics = 9

Marks in Science = 28-9 = 19 1 markAns32. 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.

2 marksIt is known that a tangent at any point of a circle isperpendicular to the radius 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 areequal)

Thus, it is proved that the lengths of the two tangents drawnfrom an external point to a circle are equal. 2 marks

Ans33. 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 12

mark

But AR = AQ

Therefore AQ = 18 cm 12

mark

Also BQ = AB-AQ = 29-18 = 11 cm 12

mark

But BP = BQ

Therefore BP = 11 cm 12

mark

Also OQB = OPB = 900 12

mark

[Because OQ and OP are perpendiculars to AB and BC

respectively.]

In quadrilateral BPOQ,

QOP+OPB+OQB+B = 3600 12

mark

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

Hence, OQBP is a square. 12

mark

BQ = OQ = OP = BP = 11 cm

Hence, radius of circle is 11 cm. 12

mark

Ans34. Volume of water in cylinder when full = (3)2 6

= 54 cm3 1 mark

Volume of solid = 23(2)3 + 1

3 (2)2 4

= 323 cm3 1 mark

Volume of water in the cylinder when solid is immersed in it

= 54 - 32/3 1 mark

=136.19 cm3 1 mark