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Class 10 Maths Objective Test Solutions

Published by IMAX, 2020-12-01 04:57:55

Description: Class 10 Maths Objective Test Solutions

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10M11+10M13 - Constructions + Surface Areas & Volumes Objective Test - Answers & Solutions - Set A Tes t Code: X-X-11-16 Q No. Ans wer Level S kill C hapter 1. A Constructions 2. C E asy Concept Knowledge Surface Areas and Volumes 3. A Surface Areas and Volumes 4. C Medium Surface Areas and Volumes 5. A Surface Areas and Volumes 6. D E asy Recall Constructions 7. B Surface Areas and Volumes 8. D Medium Concept Knowledge Constructions 9. C Surface Areas and Volumes 10. C Medium Problem Solving Surface Areas and Volumes 11. B Surface Areas and Volumes 12. D E asy Problem Solving Surface Areas and Volumes 13. A Constructions 14. B E asy Problem Solving Surface Areas and Volumes 15. A Constructions E asy Concept Knowledge Medium Problem Solving Medium Concept Knowledge Hard Critical T hinking Hard Problem Solving Medium Concept Knowledge E asy Problem Solving Medium Concept Knowledge Solutions Q1. (A) Q2. (C) Q3. (A) Q4. (C) Q5. (A) In a right circular cone with base radius and slant height and semi vertical angle , CSA of cone Q6. (D ) Required angle Q7. (B) Q8. (D ) For congruent triangles, scale factor . We always use the larger number from the scale factor to divide the acute angle drawn from the base line. Hence, both statements are false. Avanti Learning Centres Pvt Ltd. All rights reserved. Page - 1 / 2

10M11+10M13 - Constructions + Surface Areas & Volumes Objective Test - Answers & Solutions - Set A Q9. (C) i.e., Q10. (C) T he amount of material inside the cylinder will remain the same. Hence, the volume of cylinder must also remain the same. Hence, I is false. If the radius of the new cylinder is and its volume is the same, it means: Hence, III is false. Now with this new radius and height, surface area of cylinder becomes Surface area of original cylinder. Hence, II is Q11. true. Q12. (B) =Area of triangular face height Volume of combined solid= + (D ) Suppose the level of the water in the tank will rise by in , so in hours. Since, the water is flowing at the rate of hours, length of water equals We have, D iameter of cylindrical pipe Radius of cylindrical pipe, Volume of the water flowing through the cylindrical pipe in hours Also, volume of the water that falls into the tank in hours Volume of the water flowing, through the cylindrical pipe in hours = Volume of water that falls in the tank in hours Hence, the level of water in the tank will rise by in hours. (A) Q13. Q14. If and then Since point divides in the ratio , that is into parts, . Hence, . G eneral Case: and For any two positive integers always. T herefore, always. (B) G iven, radius of cylinder and hemisphere Height of cylinder Total surface area of container = Area of base + CSA of cylinder + CSA of hemisphere Q15. Hence, the total surface area of the container is (A) using compass and scale as It is possible to divide a line segment in the ratio As point lies outside the circle [ ], it is possible to draw pair of tangents from this point to the circle. Avanti Learning Centres Pvt Ltd. All rights reserved. Page - 2 / 2


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