TOPPER SAMPLE PAPER 3 Summative Assessment-II MATHEMATICS CLASS X
TOPPER SAMPLE PAPER 3 Summative Assessment-II MATHEMATICS CLASS X M.M: 80 TIME : 3-3 1 Hrs. 2 GENERAL INSTRUCTIONS : 1. 2. 3. 4. All questions are compulsory. 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) 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. 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 which is 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 common tangent 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 centre of a circle with 5 cm radius is (a) Q5. 2 3 cm 1 12 (c) 3 cm (d) 5 3 cm (b) 20 m (c) 30 m (d) 40 m (b) 1 2 (c) 1 4 (d) 1 6 The value of x, for which the points (x,-1), (2,1) and (4,5) lie on a line is (a) 0 Q8. cm The probability that a randomly chosen number from one to twelve is a divisor of twelve is (a) Q7. 3 2 An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from his eyes is 45 0. The height of the tower is (a) 10 m Q6. (b) (b) 1 (c) 2 (d) 3 2 20 The ratio in which the point R divides the join of , 7 7 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 arc 8.8cm 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 cm respectively are melted to recast a single cube. The surface area of 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 are A(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 7 and 125 respectively, find the 32nd term. Q13. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) A two digit number (ii) a perfect square number Q14. Find two consecutive positive integers, sum of whose squares is 25. Q15. Prove that the tangents at the extremities of any chord make equal angles with the chord. Q16. In given figure, find the area of the circle not included in the rectangle 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. Prove that AB + CD = BC + DA SECTION C Q19. If a b c, prove that the points (a,a2), (b,b2) and (c,c2) can never be collinear. Q20. 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? 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 and the point at which they intersect at an angle of 45°. Also write the steps of construction. Q22. 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 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 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? Q24. In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle. (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 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? Q26. 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 =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 AB respectively of a ∆ABC, whose vertices are A(-4,1), B(6,7) and C(2,-9), then prove that, Ar. ∆DEF = 1 (Ar ∆ABC). 4 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 of the first and the third term exceeds the second term by 29, find the AP. Q30. 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 p tan tan tan tan Q31. 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 been 180. Find the marks obtained in the two subjects separately. Q32. Prove that the lengths of tangents drawn from an external point to a circle are equal. Q33. 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. Q34. 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. SOLUTIONS SECTION A Ans1. Option (c) 2x2- 5x +1=0 D= b2-4ac = 5 2 -4(2)(1) = 5-8 = -3 Since D < 0, therefore the given quadratic equation has no real roots. 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 are equal. 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 centre of a circle with a 5 cm radius is 24 2 2 x2.5 x2.5 360 30 3 The length of the arc is Ans5. Option (c) 2 3 cm. 1 mark 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= AB EB AB 28.5 (as EB = DC = 28.5 m ) AB = 28.5 m Height of the tower = h = AB + BC = AB+DE = 28.5 + 1.5 = 30m Ans6. Option (b) Total number of outcomes = 12 1 mark Numbers which are divisors of 12 are 1,2,3,4,6,12. Number of favourable outcomes = 6 P(divisor of 12) = 6 1 12 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 Ans8. Option (a) 2 20 Let R divides PQ in the ratio K:1 , 7 7 2K 2 4K 2 2 20 K 1 , K 1 7 , 7 2K 2 2 4K 2 20 and K 1 7 K 1 7 1 mark 14K-14 = -2K-2 16K = 12 K= 12 3 16 4 Thus R divides PQ in the ratio 3:4. 1 mark Ans9. Option (c) Length of arc (l) = 2r 360 8.8 360 7 = 30 2 22 r r= 8.8 6 7 cm = 16.8 cm 22 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 x x2 x3 y1 y2 y3 by 1 , i.e. 3 3 a b c b c a , 3 3 1 mark Also, given that centroid is at origin i.e. its coordinates are (0,0). ab c So, 0 a b c 0. 3 1 mark OR Area of quadrilateral ABCD = area of triangle ABC + area of Area of triangle ABC = triangle ACD 1 [1(-3-2) +7(2-1)+12(1+3)] 2 1 (-5+7+48) = 25 sq. units 1 mark 2 1 Area of triangle ACD = [1(2-21) + 12(21-1)+7(1-2)] 2 = 1 (-19+240-7) = 107 sq. units 2 Area of quadrilateral ABCD = 25 + 107 1 mark 2 = 1 mark 2 = 132 sq. Units 1 mark 2 Ans12.Here, n = 60, a = 7 and t60 = 125 7+59d = 125 d=2 1 mark 2 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,.............90 Number of favourable outcomes = 81 81 9 P(two digit number) = 90 10 (ii) Perfect square numbers from 1 to 90 are 1,4,9,16,25,36,49,64,81 9 1 P(perfect square number) = 90 10 1 mark 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) = 0 X = -4 or x = 3 Reject x = -4 1 mark 2 Therefore the given consecutive positive integers are X = 3 and x + 1 = 3 + 1 = 4 1 mark 2 Ans15. 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. 1 mark 2 We have to prove that PAC= PBC 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 PAC = PBC Ans16. ( by SAS ) ( by cpct ) 1 mark 1 mark 2 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 = 10 = 5cm 2 Area of the circle = r2 = 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 1 mark 2 1 mark 2 1 mark 2 = 78.50 – 48 = 30.50 cm2 1 mark 2 Ans17. Let OAB be the given sector. Then, Perimeter of sector OAB = 16.4 cm 1 mark 2 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 = Ans18. 1 1 lr = 6 5.2 = 15.6 cm2 2 2 1 mark 2 1 mark Since tangents drawn from an external point to a circle are equal in 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 x1 y2 y3 x2 y3 y1 x3 y1 y2 2 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 = 1 a b2 c2 b c2 a2 c a2 b2 2 1 mark 1 mark 2 1 mark It is given that a b c Area of the triangle can never be 0 Ans20. The AP is 8, 24, 40, ….., and the sum is 2048. We have to determine n. 1 mark 2 1 mark 2 Here a = 8, d = 16 Since, Sn = 2048, we have 2048 = n [2x8+(n-1)16] 2 1 mark 2048 = n[8+8n-8] 2048 = 8n2 n2 = 256 i.e. n = 16 negative) (as n being number of terms, it can’t be 1 mark Hence, the body will take 16 seconds to fall 2048 metres. 1 mark 2 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 the perpendiculars intersect at point P. (v) Join OP. PA and PB are the required tangents, which make an angle of 45 with OP. 1 mark 2 marks Ans22. 1 mark 2 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 From right ∆PBQ, (as tan 45 =1) 1 mark PQ = tan60 = BQ 3 or h h 100 3 or h = 3 (h-100) 1 mark Therefore, h = 100 3 3 1 = 50(3+ 3 ) i.e. the height of the balloon is 50(3+ 3 ) m. 1 mark 2 OR 1 mark 2 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 m AD = b m (a>b) In right ∆CAB, tanx = AB h tanx = ………………….(i) AC a 1 mark In right ∆DAB, Tan(90-x) = AB AD cotx = h ……………(ii) b 1 mark 2 From (i) and (ii), tanx.cotx = 1= h2 ab h2 ab h2 = ab or h = Thus, the height of the tower is ab metres. 1 mark 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) = x 12 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 Therefore p2=P(Getting a white ball) = x6 18 It is given that p2 = 2p1 x6 x 2 18 12 x6 x 18 6 6(x+6) = 18x 6x+36 = 18x 12x = 36 1 mark 2 1 mark 2 1 mark 2 x= 36 =3 12 1 mark 2 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 6 1 P(same day) = 1 mark 36 6 (ii) Number of different days = 36-6 = 30 Number of favourable outcomes = 30 30 5 P(two different days) = 1 mark 36 6 (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) = Ans24. 10 5 36 18 1 mark 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 1 mark 2 The centre O of the inscribed circle will coincide with the centroid of ∆ABC Therefore OD= AD 3 (as centroid divides each median in the ratio 2:1) 1 mark 2 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 cm 3 3 Area of the incircle = (OD)2 1 mark 2 1 mark 2 2 22 22 = 4 3 cm2 48 cm2 150.85 cm2 7 7 Area of the triangle ABC = = 3 (Side)2 4 3 (24)2 = 249.4 cm2 4 1 mark 2 Therefore, area of the remaining portion of the triangle = (249.4-150.85)cm2 1 mark 2 = 98.55 cm2 OR Area of one design = = 2 60 28 – area of ∆OAB 360 2 28 6 3 28 4 2 1 mark = 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 the canal x depth of the canal x speed of the canal water = 3x1.2x20x1000m3 = 72000m3. In 20 minutes the volume of water = 1 mark 72000 20 m3 60 = 24000 m3. 1 mark Area irrigated in 20 minutes, if 8cm, i.e. 0.08 m standing water 24000 is required = = 300000 m2 = 30 hectares. 1 mark 0.08 Ans26. Radius of the cylinder = 0.7cm Height of the cylinder = 2.4 cm Surface area of the cylinder = 2rh+r2 = [2 0.7 2.4+(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 1 mark 2 surface area of interior conical portion = rl = x0.7 2.5 = 1.75 1 mark 2 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= = = 2 a2 b2 4abab 2ab (a2 +b2 ) ± a4 b4 2a2b2 2ab (a2 +b2 ) ± (a2 +b2 ) ± a2 b2 1 mark 2 1 mark 2 2ab Therefore, x = 1 mark a b or b a 1 mark Ans28. Let A = (x1,y1) = (-4,1), B = (x2,y2) = (6,7) and C = (x3,y3)= (2,-9) 1 [x1 (y2 - y3 )+x2 ( y3- y1) +x3 (y1 - y2)] 2 Area of ∆ABC = Area of ∆ABC = 1 [ -4(7+9) + 6(-9-1) + 2(1-7)] 2 Area of ∆ABC = 1 [ -64-60-12] = 68 sq. Units. 2 1 mark Coordinates of D, E and F are 6 2 7 9 4 2 1 9 4 6 1 7 D , , , , E and F 2 2 2 2 2 2 i.e. D(4,-1), E(-1,-4) and F(1,4) Area of ∆DEF = 1 mark 2 1 [ 4(-4-4) -1(4+1) + 1(-1+4)] 2 1 [ -32-5+3] = 17 sq. Units. 2 1 1 (Area of ∆ABC) = (68) = 17 = Area of ∆DEF 4 4 Area of ∆DEF = 1 mark 1 mark 2 SECTION D Ans29. Here, a =-11, d =-7-(-11) =4,an = 49 We have an = a+(n-1)d 1 mark 2 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 which are 16 16 th and [ 1] th 2 2 i.e., 8th term and the 9th term. 1 mark a8 = a+7d = -11+7 4 =17 1 mark 2 a9 = a+8d = -11+8 4 = 21 1 mark 2 So, the values of the two middle most terms are 17 and 21, 1 respectively. mark 2 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 i.e., 121-d2 = 11+29 i.e., d2 = 81 1 mark 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 the points of observation. AB = p metres and let BC = x metres. In right ∆ACQ, CQ = tan AC ………………(i) h = (p+ x)tan or x = h -p tan 1 mark In right ∆BCQ, CQ = tanβ h = xtanβ BC …………………………(ii) or x = h tan 1 mark From (i) and (ii), h h -p= tan tan 1 1 P=h tan tan tan tan P =h tan tan or h = p tan t an tan tan 1 mark Hence proved. Ans31.Let usual speed = x km / hr and increased speed = (x+5) km/hr According to question, 300 300 2 x x 5 (Because Distance/Speed = time) 1 mark 300 x 5 x 2 x 5 x 1 mark 2 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. 1 mark 2 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 – 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 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 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 are equal) Thus, it is proved that the lengths of the two tangents drawn from 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 But AR = AQ Therefore AQ = 18 cm Also BQ = AB-AQ = 29-18 = 11 cm But BP = BQ Therefore BP = 11 cm 1 mark 2 1 mark 2 1 mark 2 1 mark 2 Also 1 mark 2 OQB = OPB = 900 [Because OQ and OP are perpendiculars to AB and BC respectively.] In quadrilateral BPOQ, 1 mark 2 QOP+ OPB+ OQB+ B = 3600 QOP = 3600-(900+900+900)= 900 1 mark 2 Hence, OQBP is a square. BQ = OQ = OP = BP = 11 cm 1 mark 2 Hence, radius of circle is 11 cm. Ans34. Volume of water in cylinder when full = (3)2 6 = 54 cm3 Volume of solid = 1 mark 2 1 (2)3 + (2)2 4 3 3 = 32 cm3 3 1 mark Volume of water in the cylinder when solid is immersed in it = 54 - 32/3 1 mark cm3 1 mark =136.19
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