53 + Interview Questions in Surface Area And Volumes in Mathematics

1.	What is the total surface area of a cylinder with a radius of 7 m and a height of 8 m?(a) 609.4 m^2(b) 659.4 m^2(c) 650.4 m^2(d) 689.4 m^2This question was addressed to me in exam.My question is based upon Surface Area and Volume of Different Solid Shapes topic in division Surface Area and Volumes of Mathematics – Class 10
Answer» Right option is (b) 659.4 m^2 The BEST explanation: Total surface AREA of a cylinder = 2πr(h + R) = 2 × \(\frac {22}{7}\) × 7 × (8 + 7) = 659.4 m^2

Discussion

2.	A funnel is in the shape of a right circular cone with a base radius of 3 cm and a height of 4 cm. Find the slant height of the funnel.(a) 4 cm(b) 5 cm(c) 7 cm(d) 7.5 cmThe question was asked during an interview for a job.Query is from Surface Area and Volume of Different Solid Shapes in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» The correct CHOICE is (B) 5 cm The explanation is: SLANT HEIGHT = \(\sqrt {h^2+r^2}\) = \(\sqrt {4^2+3^2}\) = \(\sqrt {16+9}\) = √25 = 5 cm

Discussion

3.	The difference between the radii is 12 cm and slant height is given as 15 cm. Find the height of the frustum.(a) 5.4 cm(b) 1.2 cm(c) 3 cm(d) 9 cmThe question was posed to me during a job interview.This intriguing question comes from Surface Area & Volumes in section Surface Area and Volumes of Mathematics – Class 10
Answer» The correct CHOICE is (d) 9 cm The explanation is: GIVEN R – r = 12 cm and slant height = 15 cm The slant height of a cone s^2 = h^2 + (R – r)^2 225 = h^2 + 144 h^2 = 225 – 144 h^2 = 81 h = 9 cm

Discussion

4.	How many coins of 1 cm in diameter and thickness of 1.2 cm need to be melted to form a cuboid with dimensions of 5 cm × 10 cm × 4 cm?(a) 193(b) 213(c) 184(d) 282This question was addressed to me during an online exam.This interesting question is from Conversion of Solid from One Shape to Another in division Surface Area and Volumes of Mathematics – Class 10
Answer» Correct choice is (B) 213 Explanation: VOLUME of the CUBOID = NUMBER of coins(volume of a coin) lbh = number of coins(πr^2h) 5 × 10 × 4 = number of coins(3.14 × 0.5^2 × 1.2) 200 = number of coins(0.94) number of coins = 213

Discussion

5.	What is the skeletal formula to find the T.S.A of the tank consisting of a circular cylinder with a hemisphere attached on either end?(a) 2πrh + 2(2πr^3)(b) 2πrh + 2(πr^2)(c) 2πrh + 2(\(\frac {2}{3}\)πr^2)(d) 2πrh + 2(2πr^2)The question was asked by my school principal while I was bunking the class.I need to ask this question from Surface Area and Volume of Combination of Solids in portion Surface Area and Volumes of Mathematics – Class 10
Answer» Correct answer is (d) 2πrh + 2(2πr^2) The best I can EXPLAIN: T.S.A of the TANK = C.S.A of the CYLINDER + 2(C.S.A of the hemisphere) = 2πrh + 2(2πr^2)

Discussion

6.	Find the volume of the right prism with an area of base 121 m^2 and a height of 23 m.(a) 4793 m^3(b) 2763 m^3(c) 2783 m^3(d) 4783 m^3The question was asked during an online interview.This question is from Surface Area and Volume of Different Solid Shapes in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» Correct answer is (c) 2783 m^3 The explanation is: The VOLUME of the RIGHT PRISM = Area of BASE × height = 121 × 23 = 2783 m^3

Discussion

7.	Find the slant height if the Frustum of height 22 cm has its diameter as 34 and 50cm.(a) 83.40 cm(b) 43.90 cm(c) 23.40 cm(d) 73.48 cmI got this question during an interview.This key question is from Surface Area & Volumes in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» Right ANSWER is (c) 23.40 cm To elaborate: h = 22 cm, R = 25 cm, r = 17 cm The slant height of a CONE (s^2) = h^2 + (R – r)^2 s^2 = 484 + (25 – 17)^2 s^2 =548 s = 23.40 cm

Discussion

8.	A metallic sphere whose radius is 4 cm is melted and cast into the shape of a right circular cone of radius 7 cm. Find the height of the cylinder?(a) 14.48 cm(b) 22.36 cm(c) 16.40 cm(d) 20.32 cmThis question was posed to me in an online interview.This interesting question is from Conversion of Solid from One Shape to Another in division Surface Area and Volumes of Mathematics – Class 10
Answer» RIGHT choice is (c) 16.40 cm Explanation: Volume of the SPHERE = volume of the CONE \(\FRAC {4}{3}\) πr^3 = \(\frac {1}{3}\) πr^2h \(\frac {4}{3}\) × 3.14 × 4^3 = 3.14 × 7^2 × h h = \(\frac {4 \times 3.14 \times 3.14 \times 64}{3.14 \times 49}\) h = 16.40 cm

Discussion

9.	What is the length of the resulting solid if two identical cubes of side 7 cm are joined end to end?(a) 26 cm(b) 16 cm(c) 21 cm(d) 14 cmI have been asked this question during an interview.I need to ask this question from Surface Area and Volume of Combination of Solids topic in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» The correct CHOICE is (d) 14 CM Explanation: Length of resulting cuboid = 2 × side of the CUBE = 2 × 7 cm = 14 cm

Discussion

10.	A solid is in the form of a cone mounted on a hemisphere. The radius and height of the cone are 3 m and 4 m. Find the surface area of the given solid.(a) 114.4 m^2(b) 103.62 m^2(c) 70 m^2(d) 72.5 m^2This question was addressed to me in exam.Asked question is from Surface Area and Volume of Combination of Solids in portion Surface Area and Volumes of Mathematics – Class 10
Answer» The correct option is (B) 103.62 m^2 The explanation: Slant height = \(\sqrt {h^2 + r^2}\) = \(\sqrt {4^2 + 3^2}\) = √25 = 5 m The surface area of the TOY = C.S.A of the cone + C.S.A of the sphere = πrl + 2πr^2 = (3.14 × 3 × 5) + (2 × 3.14 × 3^2) = 47.1 + 56.52 = 103.62 m^2

Discussion

11.	Find the lateral surface area of a rectangular room with a height of 22 m, length 27 m and a breadth of 23 m.(a) 4793 m^2(b) 2263 m^2(c) 2200 m^2(d) 4783 m^2The question was posed to me by my school teacher while I was bunking the class.My enquiry is from Surface Area and Volume of Different Solid Shapes in division Surface Area and Volumes of Mathematics – Class 10
Answer» Right OPTION is (c) 2200 m^2 Best explanation: The LATERAL surface AREA of a cuboid = 2h(L + b) = 2 × 22 (27 + 23) = 2200 m^2

Discussion

12.	Find the total surface area of a square-shaped box having a length of 2 m for its side.(a) 47 m^2(b) 22 m^2(c) 24 m^2(d) 48 m^2This question was addressed to me during an interview.I would like to ask this question from Surface Area and Volume of Different Solid Shapes topic in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» The correct ANSWER is (C) 24 m^2 For EXPLANATION I would say: The TOTAL surface area of a cube = 6a^2 = 6 × 2^2 = 24 m^2

Discussion

13.	Find the slant height of the right circular cone if the base diameter of the right circular cone is 14 cm and the height is 24 cm.(a) 11 cm(b) 13 cm(c) 28 cm(d) 25 cmThis question was posed to me in final exam.Query is from Surface Area and Volume of Different Solid Shapes topic in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» RIGHT OPTION is (d) 25 cm The EXPLANATION: Radius (r) = \(\frac {diameter}{2} = \frac {14}{2}\) = 7 cm Slant height = \(\SQRT {h^2+r^2}\) = \(\sqrt {24^2+7^2}\) = \(\sqrt {576+49}\) = 25 cm

Discussion

14.	What is the total surface area of an iron sphere having a radius of 11 cm?(a) 1529.76 cm^2(b) 1514.76 cm^2(c) 1519.76 cm^2(d) 1419.76 cm^2This question was posed to me in an online interview.This is a very interesting question from Surface Area and Volume of Different Solid Shapes in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» Correct CHOICE is (c) 1519.76 cm^2 The EXPLANATION: The total surface area of an iron SPHERE = 4πr^2 = 4 × \(\FRAC {22}{7}\) × 11^2 = 1519.76 cm^2

Discussion

15.	Find the volume of the largest right circular cone that can be cut out of cube having 5 cm as its length of the side.(a) 32.72 cm^3(b) 15 cm^3(c) 25.4 cm^3(d) 37.2 cm^3I had been asked this question during an interview for a job.Asked question is from Surface Area and Volume of Combination of Solids in division Surface Area and Volumes of Mathematics – Class 10
Answer» The correct option is (a) 32.72 cm^3 The best I can explain: Length of the side of the CUBE = height of the CONE = 5 cm The RADIUS of the BASE of the cone = \(\frac {5}{2}\) cm = 2.5 cm The volume of the cone = \(\frac {1}{3}\)πr^2h = \(\frac {1}{3}\) × 3.14 × 2.5^2 × 5 = 32.72 cm^3

Discussion

16.	The frustum of a cone has its circular ends radius as 5 cm and 15 cm where the height is given as 8 cm. Find the volume of the frustum.(a) 2152.76 cm^3(b) 1254 cm^3(c) 2721.33 cm^3(d) 1421.76 cm^3This question was addressed to me in unit test.Question is from Surface Area & Volumes in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» The correct choice is (c) 2721.33 cm^3 The best I can explain: R = 5 cm, R = 15 cm and h = 8 cm The volume of a Frustum = \(\FRAC {1}{3}\) πh(r^2 + rR + r^2) = \(\frac {1}{3}\) × 3.14 × 8(25 + 75 + 225) = 2721.33 cm^3

Discussion

17.	A medicine capsule is in the form of a cylinder with two hemispheres joined together at the ends. Find the surface area if the length of the capsule is 14 mm and the width is 5 mm.(a) 219.8 mm^2(b) 105 mm^2(c) 115.4 mm^2(d) 317.2 mm^2This question was posed to me in final exam.My doubt is from Surface Area and Volume of Combination of Solids in division Surface Area and Volumes of Mathematics – Class 10
Answer» Right choice is (a) 219.8 MM^2 To explain: RADIUS of the common base (r) = 2.5 mm because the width of the capsule is EQUAL to the diameter of the CYLINDER. Length of the cylinder (h) = length of the capsule – 2(radius of the hemisphere) = 14 – 2(2.5) = 9 mm The surface area of the capsule = C.S.A of the cylinder + 2(C.S.A of the hemisphere) = 2πrh + 2(2πr^2) = (2 × 3.14 × 2.5 × 9) + 2(2 × 3.14 × 2.5^2) = 219.8 mm^2

Discussion

18.	The lateral surface area of a right circular cone is 2020 cm^2 and its radius is 10 cm. What is its slant height?(a) 28.3 cm(b) 25.6 cm(c) 68.23 cm(d) 64.33 cmThe question was posed to me during an interview.My query is from Surface Area and Volume of Different Solid Shapes in division Surface Area and Volumes of Mathematics – Class 10
Answer» RIGHT ANSWER is (d) 64.33 CM Explanation: The lateral surface area of a right circular CONE is 2020 cm^2. πrl = 2020 3.14 × 10 × L = 2020 l = \(\frac {2020}{3.14 \times 10}\) = 64.33 cm

Discussion

19.	Find the Lateral surface area of the frustum, if the slant height is 12 cm and having radii as 15cm and 35 cm.(a) 1984 cm^2(b) 1804 cm^2(c) 1984 cm^2(d) 1884 cm^2I had been asked this question in homework.Enquiry is from Surface Area & Volumes topic in division Surface Area and Volumes of Mathematics – Class 10
Answer» The correct choice is (d) 1884 cm^2 To EXPLAIN I would say: s = 12 cm, R = 35 cm, r = 15 cm Lateral Surface area of FRUSTUM = π(r + R)s L = 3.14 × (15 + 35) × 12 = 1884 cm^2

Discussion

20.	Find the total surface area ofa frustum whose slant height is 10 cm and having radii as 10 cm and 20 cm.(a) 1923.21 m^3(b) 2512 cm^2(c) 2184.21 m^3(d) 2842.21 m^3I had been asked this question at a job interview.This is a very interesting question from Surface Area & Volumes topic in portion Surface Area and Volumes of Mathematics – Class 10
Answer» Correct OPTION is (B) 2512 CM^2 The best I can explain: s = 10 cm, R = 20 cm, r = 10 cm Total surface area of a CONE = π(R + r)s + πR^2 + πr^2 = 3.14(20 + 10)10 + (3.14 × 20^2) + (3.14 × 10^2) = 942 + 1256 + 314 = 2512 cm^2

Discussion

21.	What is the volume of an article which is made by digging out a hemisphere from each end of a solid cylinder where the radius, height of the cylinder is 5 cm, 8 cm respectively and the radius of the hemisphere is 5 cm?(a) 104.40 cm^3(b) 205.6 cm^3(c) 168.23 cm^3(d) 604 cm^3I got this question in a national level competition.This intriguing question comes from Surface Area and Volume of Combination of Solids topic in section Surface Area and Volumes of Mathematics – Class 10
Answer» RIGHT ANSWER is (a) 104.40 cm^3 To elaborate: Volume of the ARTICLE = volume of the CYLINDER – 2(volume of the hemisphere) = πr^2h – 2(\(\frac {2}{3}\)πr^3) = (3.14 × 5^2 × 8) – 2(\(\frac {2}{3}\) × 3.14 × 5^3) = 104.40 cm^3

Discussion

22.	A sphere of radius 14 cm is melted and cast into a number of tiny cones of radius 2.33 cm each and height 6 cm. Find the number of cones that will be formed?(a) 726(b) 816(c) 721(d) 821This question was posed to me in an international level competition.Query is from Conversion of Solid from One Shape to Another topic in portion Surface Area and Volumes of Mathematics – Class 10
Answer» The CORRECT OPTION is (d) 821 Easy EXPLANATION: Number of cones(VOLUME of a cone) = volume of sphere number of cones(\(\frac {1}{3}\)πr^2h) = \(\frac {4}{3}\)πr^3 number of cones(\(\frac {1}{3}\) × 3.14 × 2.33^2 × 6) = \(\frac {4}{3}\) × 3.14 × 14^3 number of cones(14) = 11488.21 number of cones = 821

Discussion

23.	Find the difference between the radii of a Frustum, if the slant height and height are 10 cm and 8 cm.(a) 4 cm(b) 1 cm(c) 2 cm(d) 6 cmI have been asked this question in homework.The query is from Surface Area & Volumes topic in section Surface Area and Volumes of Mathematics – Class 10
Answer» CORRECT choice is (d) 6 cm Easiest explanation: The slant height of a CONE = h^2 + (R – r)^2 10^2 = 8^2 + (R – r)^2 (R – r)^2 = 100 – 64 (R – r) = 6 cm

Discussion

24.	Find the slant height if the frustum of height 10cm and having its diameter as 20 cm and 40 cm.(a) 10.4 cm(b) 14.14 cm(c) 15.50 cm(d) 89.4 cmThis question was posed to me in class test.The above asked question is from Surface Area & Volumes topic in division Surface Area and Volumes of Mathematics – Class 10
Answer» Correct option is (b) 14.14 CM Best explanation: h = 10 cm, R = 20 cm, r = 10 cm The SLANT HEIGHT of a cone s^2 = h^2 + (R – r)^2 s^2 = √(100 + (20 – 10)^2) s = 14.14 cm

Discussion

25.	A well of depth 25 m with a radius 4 m is dug from the earth forming a platform of length 28 m and a breadth of 16 m. Find the height of the platform.(a) 2.8 m^3(b) 5 m^3(c) 5.4 m^3(d) 7.2 m^3I have been asked this question in homework.My question is taken from Conversion of Solid from One Shape to Another in portion Surface Area and Volumes of Mathematics – Class 10
Answer» Correct OPTION is (a) 2.8 m^3 Easy explanation: VOLUME of the well = volume of the platform πr^2h = lbh 3.14 × 4^2 × 25 = 28 × 16 × h h = 2.8 m^3

Discussion

26.	What is the C.S.A of resulting solid if two identical cubes are joined end to end together with the length of the sides of the cube is 4 m?(a) 160 cm^2(b) 205.6 cm^2(c) 168.23 cm^2(d) 604 cm^2I have been asked this question in class test.My question is based upon Surface Area and Volume of Combination of Solids in portion Surface Area and Volumes of Mathematics – Class 10
Answer» Right option is (a) 160 cm^2 Easy explanation: Cuboid is the resulting solid when two identical CUBES are JOINED END to end together. Length of the cuboid (L) = 4 + 4 = 16 cm Height of the cuboid (h) = 4 cm Breadth of the cuboid (b) = 4 cm The curved surface AREA of cuboid = 2h(l + b) = (2 × 4)(16 + 4) = 160 cm^2

Discussion

27.	A cylindrical hole of depth 20 m with a radius 5 m is dug from the earth forming a platform of length 14 m and a breadth of 12 m. Find the height of the platform.(a) 8.34 m^3(b) 11.84 m^3(c) 7.64 m^3(d) 9.34 m^3I had been asked this question by my school principal while I was bunking the class.This key question is from Conversion of Solid from One Shape to Another in portion Surface Area and Volumes of Mathematics – Class 10
Answer» RIGHT choice is (d) 9.34 m^3 To ELABORATE: Volume of the cylindrical hole = volume of the platform πr^2h = lbh 3.14 × 5^2 × 20 = 14 × 12 × h h = 9.34 m^3

Discussion

28.	What is the formula required to find the height of a solid in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end?(a) 4793 m^3(b) 2763 m^3(c) 2783 m^3(d) 4783 m^3I had been asked this question during an interview.The doubt is from Surface Area and Volume of Combination of Solids in section Surface Area and Volumes of Mathematics – Class 10
Answer» The correct CHOICE is (c) 2783 m^3 The best I can explain: To find the HEIGHT of the solid we REQUIRE the height of the cone, height of the cylinder and the RADIUS of the hemisphere. Height of the solid = height of the cone + height of the cylinder + radius of the hemisphere

Discussion

29.	A sphere having a radius of 3 cm is melted and elongated into a wire having a circular cross-section of radius 0.1 cm. Find the length of the wire?(a) 2400 cm(b) 3100 cm(c) 1200 cm(d) 3600 cmI have been asked this question during a job interview.My question comes from Conversion of Solid from One Shape to Another in section Surface Area and Volumes of Mathematics – Class 10
Answer» Correct CHOICE is (d) 3600 cm The EXPLANATION is: Volume of the wire = volume of the sphere πr^2h = \(\frac {4}{3}\)πr^3 3.14 × 0.1^2 × h = \(\frac {4}{3}\) × 3.14 × 27 h = 3600 cm

Discussion

30.	How many cylinders having 2.1 cm of radius and 1.4 cm of height can be made out of a cuboid metal box having dimensions 33 cm, 21 cm, 10.5 cm?(a) 152(b) 154(c) 844(d) 841This question was posed to me by my school principal while I was bunking the class.The doubt is from Conversion of Solid from One Shape to Another topic in section Surface Area and Volumes of Mathematics – Class 10
Answer» The correct ANSWER is (c) 844 The explanation: Number of CYLINDERS(volume of a cylinder) = volume of a cuboid Number of cylinders (πr^2h) = lbh Number of cylinders (3.14 × 1.4^2 × 1.4) = 33 × 21 × 10.5 Number of cylinders = 844

Discussion

31.	The length, breadth and height of the cuboid is 8 cm, 4 cm and 4 cm. Find the volume of the cuboid?(a) 152.76 cm^3(b) 154 cm^3(c) 128 cm^3(d) 141.76 cm^3The question was asked in class test.I need to ask this question from Surface Area and Volume of Combination of Solids in division Surface Area and Volumes of Mathematics – Class 10
Answer» The CORRECT CHOICE is (c) 128 cm^3 Easiest EXPLANATION: The volume of the cuboid = lbh = 8 × 4 × 4 = 128 cm^3

Discussion

32.	A wooden box is in the shape of a cuboid with three conical depressions. 7 cm × 3 cm × 4 cm are the dimensions of the cuboid and the radius and depth of the conical depressions are 0.5 cm and 1.2 cm. Find the volume of the entire wooden box?(a) 109.4 cm^3(b) 80.05 cm^3(c) 150 cm^3(d) 89.4 cm^3I got this question by my college professor while I was bunking the class.My question comes from Surface Area and Volume of Combination of Solids topic in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» Correct choice is (b) 80.05 cm^3 Explanation: VOLUME of the WOODEN box = volume of the cuboid – the volume of 3 CONICAL depressions = lbh – 3(\(\frac {1}{3}\)πr^2h) = (7 × 3 × 4) – 3(\(\frac {1}{3}\) × 3.14 × 0.5^2 × 1.2) = 80.05 cm^3

Discussion

33.	A solid is in the form of a cone mounted on a hemisphere. The radius and height of the cone are 3 m and 4 m. Find the volume of the given solid?(a) 93.21 m^3(b) 94.21 m^3(c) 84.21 m^3(d) 82.21 m^3This question was addressed to me during an interview for a job.The question is from Surface Area and Volume of Combination of Solids in section Surface Area and Volumes of Mathematics – Class 10
Answer» The correct OPTION is (b) 94.21 m^3 Easiest explanation: Volume of the SOLID = volume of the cone + volume of the hemisphere = \(\frac {1}{3}\)πr^2h + \(\frac {2}{3}\)πr^3 = (\(\frac {1}{3}\) × 3.14 × 3^2 × 4) + (\(\frac {2}{3}\) × 3.14 × 3^3) = 94.21 m^3

Discussion

34.	Find the Lateral surface area of the frustum, if the slant height is 15cm and having radii as 14 cm and 21 cm.(a) 1648.5 cm^2(b) 1115 cm^2(c) 2105.4 cm^2(d) 1237.2 cm^2The question was posed to me in examination.I'm obligated to ask this question of Surface Area & Volumes in portion Surface Area and Volumes of Mathematics – Class 10
Answer» RIGHT answer is (a) 1648.5 cm^2 The best explanation: s = 15 cm, R = 42 cm, r = 28 cm Lateral Surface AREA of FRUSTUM (L) = π(R + r)s L = 3.14(21 + 14 )15 = 1648.5 cm^2

Discussion

35.	Three metallic spheres of radius 3 cm, 6 cm, 9 cm are melted into a single sphere. Find the radius of the resulting sphere.(a) 109.4 cm^3(b) 4071.48 cm^3(c) 1520 cm^3(d) 1869.4 cm^3I got this question in quiz.This intriguing question originated from Conversion of Solid from One Shape to Another in section Surface Area and Volumes of Mathematics – Class 10
Answer» Correct OPTION is (B) 4071.48 cm^3 To elaborate: Volume of the RESULTING SPHERE = sum of the volumes of all three spheres = \(\frac {4}{3}\)π3^3 + \(\frac {4}{3}\)π6^3 + \(\frac {4}{3}\)π9^3 = 113.09 + 904.77 + 3053.62 = 4071.48 cm^3

Discussion

36.	A toy is in the form of a cone mounted on a hemisphere and a cylinder. The radius and height of the cone are 3 m and 4 m. Find the volume of the given solid?(a) 193.21 m^3(b) 207.30 m^3(c) 184.21 m^3(d) 282.21 m^3The question was asked at a job interview.Question is taken from Surface Area and Volume of Combination of Solids topic in portion Surface Area and Volumes of Mathematics – Class 10
Answer» The correct CHOICE is (b) 207.30 m^3 Explanation: Volume of the TOY = volume of the cone + volume of the hemisphere + volume of the CYLINDER = \(\frac {1}{3}\)πr^2h + \(\frac {2}{3}\)πr^3 + πr^2h = (\(\frac {1}{3}\) × 3.14 × 3^2 × 4) + (\(\frac {2}{3}\) × 3.14 × 3^3) + (3.14 × 3^2 × 4) = 207.30 m^3

Discussion

37.	What is the formula to find the rise in the water level when ‘x’ spherical balls are dropped into a cylindrical beaker?(a) \(\frac {Volume \, of \, ‘x’ \, spherical \, balls}{Base \, area \, of \, cylinder}\)(b) \(\frac {Volume \, of \, ‘x’ \, spherical \, balls}{2(Base \, area \, of \, cylinder)}\)(c) The volume of the cylinder + volume of the beaker(d) The volume of the cylinder – 2(volume of the beaker)This question was addressed to me at a job interview.Origin of the question is Conversion of Solid from One Shape to Another topic in portion Surface Area and Volumes of Mathematics – Class 10
Answer» Correct option is (a) \(\frac {VOLUME \, of \, ‘x’ \, spherical \, BALLS}{Base \, AREA \, of \, cylinder}\) Explanation: To find the rise in the water LEVEL we require the volume of ‘x’ spherical balls and the base area of the cylinder Rise in the water level = \(\frac {Volume \, of \, ‘x’ \, spherical \, balls}{Base \, area \, of \, cylinder}\)

Discussion

38.	What is the formula to find the height of the tank consisting of a circular cylinder with a hemisphere attached on either end?(a) Radius of the cylinder + 2(height of the hemisphere)(b) Height of the cylinder + 2(height of the hemisphere)(c) Radius of the cylinder + 2(radius of the hemisphere)(d) Height of the cylinder + 2(radius of the hemisphere)The question was asked in an online quiz.This is a very interesting question from Surface Area and Volume of Combination of Solids in section Surface Area and Volumes of Mathematics – Class 10
Answer» Right answer is (d) Height of the cylinder + 2(radius of the HEMISPHERE) The explanation: To find the height of the tank consisting of a CIRCULAR cylinder with a hemisphere attached on either end REQUIRES the height of the cylinder and radius of the hemisphere. Height of the tank = height of the cylinder + 2(radius of the hemisphere)

Discussion

39.	What is the volume of a hemisphere if the radius of the hemisphere is 3 m?(a) 135.4 m^3(b) 56.52 m^3(c) 120.23 m^3(d) 105.5 m^3I had been asked this question in a job interview.I need to ask this question from Surface Area and Volume of Different Solid Shapes in division Surface Area and Volumes of Mathematics – Class 10
Answer» The correct CHOICE is (B) 56.52 m^3 The best I can explain: RADIUS of the hemisphere = \(\FRAC {2}{3}\)πr^3 = \(\frac {2}{3}\) × 3.14 × 3^3 = 56.52 m^3

Discussion

40.	What is the volume of an article which is made by digging out a hemisphere from each end of a solid cylinder?(a) πr^2h + 2(2πr^3)(b) 2πrh – 2(πr^2)(c) 2πrh + 2(\(\frac {2}{3}\) πr^2)(d) πr^2h – 2(\(\frac {2}{3}\)πr^3)I have been asked this question during an interview.This intriguing question originated from Surface Area and Volume of Combination of Solids in chapter Surface Area and Volumes of Mathematics – Class 10
Answer» CORRECT choice is (d) πr^2h – 2(\(\FRAC {2}{3}\)πr^3) For explanation: VOLUME of the article = volume of the CYLINDER – 2(volume of the hemisphere) = πr^2h – 2(\(\frac {2}{3}\)πr^3)

Discussion

41.	Find the surface area of the given solid which is in the form of a cone mounted on a hemisphere. The radius and height of the cone are 5cm and 12cm.(a) 214.4 cm^2(b) 279.53 cm^2(c) 70 cm^2(d) 72.5 cm^2The question was asked in an online interview.Asked question is from Surface Area and Volume of Combination of Solids in section Surface Area and Volumes of Mathematics – Class 10
Answer» The correct answer is (B) 279.53 cm^2 Easiest EXPLANATION: Slant height = \(\sqrt {h^2+r^2}\) = \(\sqrt {12^2+5^2}\) = √169 = 13cm The surface AREA of the TOY = C.S.A of the cone + C.S.A of the sphere = πrl + 2πr^2 = (3.14 × 3 × 13) + (2 × 3.14 × 5^2) = 47.1 + 56.52 = 279.53cm^2

Discussion

42.	A metallic sphere whose radius is 5 cm is melted and cast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.(a) 21.4 cm(b) 43.63 cm(c) 70 cm(d) 72.5 cm^2This question was addressed to me by my school principal while I was bunking the class.The origin of the question is Conversion of Solid from One Shape to Another topic in portion Surface Area and Volumes of Mathematics – Class 10
Answer» The CORRECT choice is (b) 43.63 cm For explanation I WOULD say: Volume of the sphere = volume of the CYLINDER \(\frac {4}{3}\)πr^3 = πr^2h \(\frac {4}{3}\) × 3.14 × 5^3 = 3.14 × 6^2 × h h = \(\frac {4 \times 3.14 \times 3.14 \times 125}{3.14 \times 36}\) h = 43.63 cm

Discussion

43.	Two cubes each of volume 27 cm^3 are joined together. Find the surface area of the resulting solid?(a) 109.4 cm^2(b) 126 cm^2(c) 150 cm^2(d) 189.4 cm^2I have been asked this question in examination.The question is from Surface Area and Volume of Combination of Solids in division Surface Area and Volumes of Mathematics – Class 10
Answer» Correct choice is (B) 126 cm^2 The explanation: Volume of cube = a^3 = 27 a = 3 cm Joining 2 cubes RESULTS in a cuboid. Length of the cuboid (l) = 3 + 3 = 9 cm Height of the cuboid (h) = 3 cm Breadth of the cuboid (b) = 3 cm Surface area of cuboid = 2(lb + BH + hl) = 2(27 + 9 + 27) = 126 cm^2

Discussion

44.	What is the length of the resulting solid if two identical cubes of side 8 cm are joined end to end?(a) 26 cm(b) 16 cm(c) 21 cm(d) 14 cmI got this question in an interview for job.This intriguing question originated from Surface Area and Volume of Combination of Solids topic in division Surface Area and Volumes of Mathematics – Class 10
Answer» Right OPTION is (b) 16 cm The explanation: Length of RESULTING CUBOID = 2 × SIDE of the cube = 2 × 8 cm = 16 cm

Discussion

45.	Find the total surface area of a frustum whose slant height is 18 cm and having radii as 8 cm and 16 cm.(a) 2361.28 cm^2(b) 2205.6 cm^2(c) 1628.23 cm^2(d) 2604 cm^2This question was addressed to me in an interview for job.Question is taken from Surface Area & Volumes in division Surface Area and Volumes of Mathematics – Class 10
Answer» Right OPTION is (a) 2361.28 cm^2 The best explanation: s = 18cm, R = 16 cm, r = 8 cm Total SURFACE AREA of a cone = π(R + r)s + πR^2 + πr^2 = 3.14(16 + 8)18 + (3.14 × 256) + (3.14 × 64) = 1356.48 + 803.84 + 200.96 = 2361.28 cm^2

Discussion

46.	Three metallic spheres of radius 2 cm, 4 cm, 8 cm are melted into a single sphere. Find the radius of the resulting sphere.(a) 1009.4 cm^3(b) 2446.25 cm^3(c) 1520 cm^3(d) 2869.4 cm^3I got this question during a job interview.This interesting question is from Conversion of Solid from One Shape to Another topic in section Surface Area and Volumes of Mathematics – Class 10
Answer» Right answer is (b) 2446.25 cm^3 Explanation: Volume of the resulting sphere = SUM of the volumes of all three spheres = \(\frac {4}{3}\)π2^3 + \(\frac {4}{3}\)π4^3 + \(\frac {4}{3}\)π8^3 = \(\frac {4}{3}\) × 3.14(2^3 + 4^3 + 8^3) = 2446.25 cm^3

Discussion

47.	What is the formula required to use for T.S.A of an article which is made by digging out a hemisphere from each end of a solid cylinder?(a) C.S.A of the cylinder – 2(C.S.A of the hemisphere)(b) C.S.A of the cylinder + C.S.A of the hemisphere(c) C.S.A of the cylinder + 2(C.S.A of the hemisphere)(d) C.S.A of the cylinder – C.S.A of the hemisphereThe question was posed to me in an interview for job.Origin of the question is Surface Area and Volume of Combination of Solids topic in section Surface Area and Volumes of Mathematics – Class 10
Answer» The correct choice is (c) C.S.A of the CYLINDER + 2(C.S.A of the hemisphere) The best I can explain: To FIND the T.S.A of an article which is made by digging out a hemisphere from each END of a solid cylinder, we NEED C.S.A of the cylinder and C.S.A of the hemisphere. T.S.A of the article = C.S.A of the cylinder + 2(C.S.A of the hemisphere).

Discussion

48.	What is the formula to find the height of an iron pillar consisting of a cylinder and a cone mounted on it?(a) The radius of the cylinder + 2(height of the cone)(b) Height of the cylinder + 2(height of the cone)(c) The radius of the cylinder + radius of the cone(d) Height of the cylinder + height of the coneI have been asked this question in an international level competition.The query is from Surface Area and Volume of Combination of Solids topic in section Surface Area and Volumes of Mathematics – Class 10
Answer» The CORRECT CHOICE is (d) HEIGHT of the cylinder + height of the cone Easy explanation: To find the height of an iron pillar consisting of a cylinder and a cone MOUNTED on it, we require heights of both cone and cylinder. Height of the pillar = height of the cylinder + height of the cone.

Discussion

49.	The frustum of a cone has its circular ends radius as 15 cm and 30 cm where the height is 10cm. Find the volume of the frustum.(a) 21544.4 cm^2(b) 16485 cm^2(c) 16570 cm^2(d) 16572.5 cm^2The question was asked in quiz.This is a very interesting question from Surface Area & Volumes in division Surface Area and Volumes of Mathematics – Class 10
Answer» Correct answer is (b) 16485 cm^2 For explanation: r = 15 cm, R = 30 cm, h = 10 cm The volume of a Frustum = \(\FRAC {1}{3}\)πh(r^2 + rR + r^2) = \(\frac {1}{3}\) × 3.14 × 10 (225 + 450 + 900) = 16485 cm^3

Discussion

50.	Two cubes each of volume 64 cm^3 are joined together. Find the volume of the resulting solid?(a) 152.76 cm^3(b) 154 cm^3(c) 256 cm^3(d) 141.76 cm^3This question was addressed to me in an international level competition.This intriguing question originated from Surface Area and Volume of Combination of Solids in portion Surface Area and Volumes of Mathematics – Class 10
Answer» Correct answer is (c) 256 cm^3 Explanation: Volume of cube = a^3 = 64 a = 4 cm is the side of each cube. Joining 2 cubes results in a CUBOID. LENGTH of the cuboid (l) = 4 + 4 = 16 cm Height of the cuboid (H) = 4 cm Breadth of the cuboid (b) = 4 cm The volume of the cuboid = LBH = 16 × 4 × 4 = 256 cm^3

Discussion

Explore topic-wise InterviewSolutions in .

Find the slant height if the Frustum of height 22 cm has its diameter as 34 and 50cm.(a) 83.40 cm(b) 43.90 cm(c) 23.40 cm(d) 73.48 cmI got this question during an interview.This key question is from Surface Area & Volumes in chapter Surface Area and Volumes of Mathematics – Class 10

Find the difference between the radii of a Frustum, if the slant height and height are 10 cm and 8 cm.(a) 4 cm(b) 1 cm(c) 2 cm(d) 6 cmI have been asked this question in homework.The query is from Surface Area & Volumes topic in section Surface Area and Volumes of Mathematics – Class 10