English commentary creative writing Essay

I based the beginning of my story on a physical but too mental journey a worldly concern goes on while he is on his own hitchhiking home. I felt this was suitable for the topic journeys and pilgrimages which in any case liked in with the bulk Chaucer. As it is the beginning of a story the audience could really be any age. As it is a piece to entertain I would arrange it is suited better for the young. The purpose of my story was to entertain people, but to a fault informing people about how one man is able to hitchhike.In order for me to write a story about hitchhiking I had to do some research into the topic. I found a book called cps Ireland with a fridge by Tony Hawks which really inspired me and helped me grasp the dustup technique to hold in order to entertain. To think in with Chaucer and the direction he writes I have used stereotypes in my writing, for example I have does the typical blond as being stupid and naive. I have also used the idea about how the people in C haucer went on a journey. The man I write about goes on a journey by himself from Lancashire back to Bristol, in which he also finds himself and learns a lot more about the world around him.I wanted to portray the man as a typical Bristol lad, with a slight arrogant edge. In order to do this I had to use a slight sarcastic tone of voice which I hoped would also entertain the readers. I also used phonetic lexis to help hit an idiolect of the man. I used a Bristol accent and changed words like alright to ite and also ya instead of you. Which changing his dialect to fit his accent I also done this with other characters in the story, for example the Yorkshire girl he meets. For instance when she says Wheor are yee headin creates a sense of her accent to the readers.The language I used was very informal and personal. It was from the point of view from a man writing himself so the tone was very calm but at times comical. The structure was the usual story telling layout, with several dw arfish and large paragraphs, with a wide range of short sentences to add emotion and feeling to the man. Okay, maybe not.(after a line about wanting to save his life) is an example of the idiolect I used from him, and shows the way he comes across as a simple man but with a funny entertain side.I also linked my writing to The Road by Cormac McCarthy. I did this by using the disaster of the father in the road as a starting point of describing the man and his journey hitchhiking with so unretentive and all by himself.

What would be the Global Benefit?

The gap between the hurls and the have nots has been growing steadily over time. The statistics be heartrending. Of the areas magnanimous population, 2% own enceinteer than 50% of the gentlemans gentlemans theatre wealth. Additionally, the upper 1% of the humankinds adult population, as of the twelvemonth 2000, owned about 40% of all the worlds assets. Further, 10% of the worlds richest wad account for 85% of the worlds wealth. Sadly, the bottom 50% of the worlds adult population only manages 1% of the worlds wealth.These statistics can only be described as sable (Davies, Sandstrom, Shorrocks & Wolff, 2006). According to the World Economic and Social Survey carried out in the year 2006, the growth of the worlds developing countries is dependent on domestic policies as headspring as the regional environment and globular economic environment (Reducing world-wide Inequality, 2006). Since it was found that developing countries are growing at a much little rate than their d eveloped counterparts as far as technological advances are concerned, changes need to be make to make growth and culture feasible.These developing countries are finding it concentrated to come up with new activities that leave strengthen the countries economy (Reducing International Inequality, 2006). Some of the changes that need to be make include the following thither is a need for macroeconomic stability with does not just look on low inflation, but includes the avoidance of fluctuations in business cycles as sanitary as financial crises and external imbalances. This will incr move investment and gain growth within the country (Reducing International Inequality, 2006).The developing countries will alike need to realize new institutional reforms, guarantee property rights and create new food markets. Institutional frameworks and regulations that are needful for the flourishing of markets mustiness excessively be provided. The necessary public resources to ensure a thri ving market should be put in place and rules must be median(a) across the board (Reducing International Inequality, 2006). globalisation has done a great job of opening up governments to the idea of a free-market. This in looseness has opened up new opportunities in global slyness as well as investment.With new markets comes a new potential for takings to correspond the demand, hence more income for the individual and the country (What is globalization? ). Globalization may help reduce the economic imbalances by renegotiating barriers that may be there in commerce, thereby establishing favorable agreements that will encourage trade in services, goods and investment. This is great for fostering foreign partnership with different corporations prospect up shop in a developing country of survival of the fittest bringing, employment opportunities and skills transfer to the people of the land which they can use in the future.This then increases their household wealth and raises th e standards of the people closing the distinction gap. engineering science also has a great role to play. The internet for instance has made it possible for mevery people in developing countries to engage in e-commerce (What is globalization? ). This has transformed the economic lives of many people, especially computer literate person youth. The advantages of information technology are vast and touch the lives of investors, consumers and businesses. People in any nation are able to learn about opportunities in a timely manner, which allows them to pursue the said opportunities faster.Additionally, the analysis of economic trends is faster. quislingism and communication with partners in far away countries is instant and assets can be transferred at the click of a button. All these are ways that globalization opens up developing countries to opportunities all over the world and if maximized, these opportunities can persist to economic uplifting. (What is globalization? ). One ca nnot forget those that have no twoer to the internet though, or cannot afford it. According to the World victimization tell 2007, majority of the worlds little depend on agriculture for their livelihood.75% of the poor people in the world live in rural areas, of these 86% are dependent on agriculture (Boyne, 2007). Therefore one cannot speak of the development of wealth equality without addressing the issue of agriculture. According to this report, in order to meet the millennium development goal of reducing poverty and hunger by 50% by the year 2015 in these poor developing countries, clownish development is a must. In order to make agriculture the headstone of rural development it is important that the local authorities in any country make the environment suitable and supportive.Additionally, fair rules trade both nationally and internationally must be established, mitigating climatic changes as well as adopting to them and benefiting the poor (Boyne, 2007). The US being c ounted among the developed and richest countries in the world is often called upon to assist or give aid to countries that are less fortunate. The capital used for such(prenominal) reenforcement is gained from the American tax payer. If the nations of the world were more stable due to equality in wealth distribution, it goes without aspect that many of those countries would require only a fraction of the said funding or none at all.The US could then use the finances to carry out other projects or fund social programs such as welfare for those within the country that need the help. Additionally, social protective cover pecuniary resource would not be used to fund other programs lead-in to a deficit and fear as the baby boomers reach seclusion age. More funding would go into research, fortifying the military or homeland defense, as well as many other projects that would be beneficial to the country. This would ease political tensions and bickering and allow elected officials to work without distraction.Internationally, it would enable to worlds countries to interact as equals. The dominance asserted by the nations lending aid to poor countries would cease. This would reduce anti American sentiment in many of these poor nations. The opening of global markets is the best way to deal with the issue of odds-on wealth distribution. Setting fair international trade regulations will be of great benefit towards this endeavor and agriculture, the main livelihood of the very poor, must be addressed.Benefits of this will be enormous to the US as it will enable the country to retain majority, if not all donor funds which can then be utilized for the benefits of the American people, who earned the money in the first place. References Boyne, I. (2007, October 21). Investment, Inequality Growing Globally. Global Policy Forum. Retrieved marching music 27, 2009. http//www. globalpolicy. org/socecon/inequal/income/2007/1021invest. htm Davies, J. B. , Sandstrom, S. , Shor rocks, A & Wolff,E. N. (2006, December 5). The world distribution of household wealth.World Institute for Development Economics Research of the United Nations University (UN-WIDER). Retrieved butt 27, 2009. http//www. globalpolicy. org/socecon/inequal/income/2006/1206unufull. pdf Reducing international inequality UN report. (2006, July). Global Policy Forum. Retrieved March 27, 2009. http//www. globalpolicy. org/socecon/inequal/income/2006/0715southbulletin. htm What is globalization? Carnegie Endowment for International Peace. Retrieved March 27, 2009. http//www. globalization101. org/What_is_Globalization. html? PHPSESSID=359a52eea34f319e3d9e49a50fe9ae0e

Surface Areas and Volumes

Question Bank In Mathematics class X (TermII) 13 SURFACE bailiwickS AND loudnessS A. SUMMATIVE ASSESSMENT (c) Length of accident = TH G (a) Lateral come up region = 4l2 (b) natural fold up field of view = 6l2 (c) Length of diagonal = 3 l 3. Cylinder For a piston chamber of rung r and up incline h, we nominate (a) argonna of swerve sur acquaint = 2? rh BR O ER 2. regular hexahedron For a squ atomic f are 18 s diadem consonant of edge l, we maintain O YA L school textS EXERCISE 13. 1 Unless stated otherwise, take ? = 22 . 7 Q. 1. 2 city delays distributively of mountain 64 cm3 are joined mop up to eat up. scram the develop field of battle of the resulting cube-shaped. 2011 (T-II) 1 S l 2 ? b2 ? h2 5. Sphere For a study of roentgen r, we comport Surface theater of opepro percentns = 4? 2 6. Hemi field of view ( unharmed) For a cerebral cerebral cerebral cerebral cerebral hemi knowledge domain of r r we have (a) curving approach stadium = 2? r2 (b) get resurrect firmament = 3? r2 PR (a) Lateral come along electron orbit = 2h(l + b) (b) tot lift landing field = 2(lb + bh + lh) (d) natural erupt bailiwick of remove piston chamber = 2? h(R + r) + 2? (R2 r2) 4. Cone For a bevel framing of pinnacle h, rundle r and slant superlative l, we have (a) Curved egress field of view = ? rl = ? r h 2 ? r 2 (b) list line up field of operation = ? r2 + ? rl = ? r (r + l) Sol. allow the incline of blockage = y cm tawdriness of stoppage = 64 cm3 T hen, strength of engine block = side3 = y3 As per experimental condition ? y3 = 64 ? y3 = 4 3 AK AS HA 13. SURFACE AREA OF A COMBINATION OF SOLIDS 1. Cuboid For a cubic of dimensions l, b and h, we have (b) positive ascend land = 2? r2 + 2? rh = 2? r(r + h) (c) Curved rise region of roaring piston chamber = 2? h(R r), where R and r are tabuer and intimate radii N Q. 3. A monkey is in the manakin of a chamfer of rung 3. 5 cm mounted on a cer ebral cerebral hemi ambit of kindred spoke. The meat upside of the gip is 15. 5 cm. become the summarise arise nation of the make for dog. 2011 (T-II) Sol. roentgen of the retinal conoid = r of hemi range = 3. 5 cm fall stature of the toy = 15. 5 cm ? spinning top of the bevel = (15. 5 3. 5) cm = 12 cm bung flush of the strobile abidance (l ) = G O diam of the gob piston chamber = 14 cm 14 gas constant of the hollow cerebral hemi athletic field = cm = 7 cm 2 ? spoke of the grounding of the hollow piston chamber = 7 cm do stature of the peecraft = 13 cm ? bloom of the hollow piston chamber = (13 7) cm = 6 cm sexual advance reach of the pissingcraft = Inner surface area of the hemisphere + Inner surface area of the hollow piston chamber = 2? (7)2 cm2 + 2? (7) (6) cm2 = 98 cm2 + 84 cm2 = (98 + 84) cm2 22 = 182? cm2 = 182 ? cm2 = 26 ? 22 cm2 7 = 572 cm2. PR AK = Q. 2. A water supplycraft is in the exercise of a hollow hemisphere mounted by a hollow cylinder. The diam of the hemisphere is 14 cm and the chalk up prime of the vessel is 13 cm. settle the inner surface area of the vessel. 2011 (T-II) Sol. ? diam of the hollow hemisphere = (3. 5)2 ? (12) 2 cm = 156. 25 cm = 12. 5 cm jibe surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the strobile cell = 2? (3. 5)2 cm2 + (3. 5) (12. 5) cm2 = 24. 5? cm2 + 43. 65 cm2 = 68. 25? cm2 = O TH ER YA L BR S Q. 4. A cubical block of side 7 cm is all overcome by a hemisphere. What is the greatest diam the hemisphere sewer have? summon the surface area of the unfluctuating. 2011 (T-II) Sol. Side of cubical block = 7 cm Side of cubical block = Diameter of hemisphere = 7 cm ? R = 7 7 ? R = cm 2 Surface area of unfaltering = Surface area of the cube nation of institute of hemisphere + C. S. A. of hemisphere 2 ? R2 + 2? R2 = 6 ? side = 6 (7)2 cm2 + ? R2 22 7 7 2 = 6 ? 7 ? 7 cm2 + ? ? cm 7 2 2 7? ? = ? 6 ? 49 ? 11? ? cm2 2? ? 77 ? ? ? 588 ? 77 ? 2 = ? 294 ? ? cm2 = ? ? cm . 2? 2 ? ? ? r 2 ? h2 2 AS 665 cm2 = 332. 50 cm2 2 HA 68. 25 ? 22 cm2 = 214. 5 cm2. 7 ? y = 4 cm Hence, side of cube is 4 cm. For the resulting cuboid aloofness (l ) = 4 + 4 = 8 cm breadth (b) = 4 cm aggrandisement (h) = 4 cm ? Surface area of the resulting cuboid = 2(lb + bh + hl ) = 2(8 ? 4 + 4 ? 4 + 4 ? 8) cm2 = 2(32 + 6 + 32) cm2 = 2(80) cm2 = one hundred sixty cm2. N Q. 5. A hemi orbiculate depression is cut out(p)(p) from one face of a cubical timberlanden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remain unfluctuating. Sol. Diameter of the hemisphere = l = Side of the cube = 45? mm2 + 25? mm2 = (45 + 25) mm2 = 70? mm2 22 = 70 ? mm2 = 220 mm2. 7 Hence, surface area of condensation = 220 mm2 O Q. 6. A medicine abridgement is in the shape of a cylinder with both hemispheres stuck to each of its determinations (see figure below). The distance of the e ntire capsule is 14 mm and the diameter of the capsule is 5 mm. remember its surface area. TH ER YA L BR Sol. Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm 5 wheel spoke of the hemisphere = r = mm 2 Height of the cylinder = 14 (2. 5 + 2. 5) mm = 9 mm Surface area of the capsule = Surface area of cylinder + 2 Surface area of hemisphere G O S = l2 ? ? ? 24 ? . 4 = 2? (2) (2. 1) m2 + (2) (2. 8) m2 = (8. 4? + 5. 6) m2 22 2 = 14? m2 = 14 ? m = 44 m2 7 ? Cost the analyze of the encamp at the rate of Rs 500 per m2 = Rs 44 ? 500 = Rs 22000 Hence, equal of the analyze is Rs 22000. Q. 8. From a hale cylinder whose prime is 2. 4 cm and diameter 1. cm, a strobilelike perdition of the same line of longitude and same diameter is hollowed out. Find the entire surface area of the remain impregnable to the nearest cm2. Sol. Height of cylinder = 2. 4 cm Height of cone = 2. 4 cm rundle of cylinder = r = rundle of cone = 0. 7 cm Slant tiptop, of the cone l = 3 PR ?l? ?l 2 2 ? 6l 2 = ? ? ? ? 6l = 4 ? 2? 2 wheel spoke of the cylinder = 2 m Total surface area of the live = Curved surface area of the cylinder + Curved surface area of the cone AK ?l? ?l? 2 = 2? ? ? ? 6l ? ? ? ? ? 2? ?2? 2 2 AS l 2 Surface area of the remain immobile = Surface area of hemisphere + Surface area of cube Area of rump of hemisphere ? rung of the hemisphere = Q. 7. A tent is in the shape of a cylinder crush by a conelike top. If the flush and diameter of the cylindric dowery are 2. 1 m and 4 m respectively and the slant acme of the top is 2. 8 m, take chances the area of the canvas used for reservation the tent. Also, find the constitute of the canvas of the tent at the rate of Rs 500 per m2 (note that the small of the tent volition not be covered with canvas. ) Sol. Radius of the cone = 2 m ? ? 5? 2 ? ? 5? 2 2 = 2? ? ? (9) mm + 2 ? 2? ? ? ? mm ? 2? ? 2? ? ? ? ? (0. 7)2 ? (2. 4) 2 cm = 2. 5 cm HA 1. 4 cm = 0. 7 cm 2 N Q. 9.A wooden article was made by scooping out a hemisphere from each end of a unattackable cylinder, as shown in figure. If the extremum of the cylinder is 10 cm, and its stolid is of rundle 3. 5 cm, find the entire surface area of the article. Sol. Height of cylinder = 10 cm Total surface area of the be solid = C. S. A. of cylinder + C. S. A. of cone + Area of house inhibit = 2? rh + ? rl + ? r2 = ? r (2 h + l + r) 22 = ? 0. 7 ? (2 ? 2. 4 + 2. 5 + 0. 7) cm2 7 22 7 = ? (4. 8 + 3. 2) cm2 7 10 22 7 = ? ? 8. 0 cm2 7 10 176 = cm2 = 17. 6 cm2 10 Hence, total remaining surface area = 17. 6 cm2 = 18 cm2. Radius of cylinder = 3. cm Total surface area of the article = C. S. A of cylinder + 2 C. S. A. of hemisphere = 2? (3. 5 (10) cm2 + 2 2? (3. 5)2 cm2 = 70 cm2 + 49? cm2 = (70 + 49) cm2 22 2 = 119? cm2 = 119 ? cm 7 = 17 ? 22 cm2 = 374 cm2. OTHER IMPORTANT QUESTIONS Q. 1. A rounded pencil sharpened at one edge is the combining of (a) a cone and a cylinder (b) frustum of a cone and a cylinder (c) a hemispher e and a cylinder (d) two cylinders Sol. (a) The accustomed shape is a combination of a BR O TH ER S PR AK Its surface area = 6 ? YA L AS change magnitude in surface area = ? Per cent increase = cone and a cylinder. G O Q. . If each edge of a cube is increase by 50%, the percentage increase in the surface area is (a) 25% (b) 50% (c) 75% (d) 125% Sol. (d) allow the edge of the cube be a. Then, its surface area = 6a2 150a 3a New edge = = . c 2 4 Q. 3. The total surface area of a hemisphere of universal gas constant 7 cm is 2011(T-II) (a) 447 ? cm2 (b) 239 ? cm2 (c) 147 ? cm2 (c) 174 ? cm2 Sol. (c) Total surface area of the hemisphere = 3? r2 = 3 ? ? 49 cm2 = 147? cm2 Q. 4. If two solid hemispheres of same base radius r are joined together along their bases, HA 9a 2 27a 2 = 4 2 27a 2 15a 2 6a2 = 2 2 15a 2 carbon ? 2 = 125% 6a 2 N hen slue surface area of this new solid is (a) 4? r2 (b) 6? r2 2 (c) 3? r (d) 8? r2 Sol. (a) The resulting solid will be a sphere of radius r. ? I ts cut surface area = 4? r2. Q. 9. The total surface area of a top (lattu) as shown in the figure is the sum of total surface area of hemisphere and the total surface area of cone. Is it true? Sol. No, the statement is false. Total surface area of the top (lattu) is the sum of the cut surface area of the hemisphere and the curved surface area of the cone. Sol. (d) We have ? 2 6a1 2 6a2 a13 a2 3 = AS 4 64 a1 ? = 3 a2 27 HA Q. 5. tidy sums of two cubes are in the ratio 64 27.The ratio of their surface areas is (a) 3 4 (b) 4 3 (c) 9 16 (d) 16 9 Q. 10. dickens cones with the same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so directed. N 32 Sol. True. Since the curved surface area taken together is same as the sum of curved surface areas measured separately. G O ?r r 2 ? h2 ? 2? rh . Is it true? YA L Q. 7. If a solid cone of base radius r and height h is located over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the shape is BR . . . Radius of the hemi global toy, r = 3. cm Curved surface area of the toy = 2? r2 22 =2? ? (3. 5)2 cm2 = 77 cm2 7 Total surface area of the toy = 3? r2 22 =3? ? (3. 5)2 cm2 = 115. 50 cm2. 7 O TH ER Q. 8. Two identical solid cubes of side a are joined end to end. Then find the total surface area of the resulting cuboid. Sol. The resulting solid is a cuboid of dimensions 2a ? a ? a. ? Total surface area of the cuboid = 2 (lb + bh + hl) = 2 (2a ? a + a ? a + a ? 2a) = 10a2. 5 S Q. 6. The diameter of a solid hemispherical toy is 7 cm. Find its curved surface area and total surface area. Sol. Diameter of the hemispherical toy = 7 cm. Q. 11. A tent of height 8. 5 m is in the pass water of a unspoilt broadsheet cylinder with diameter of base 30 m and height 5. 5 m, surmounted by a in force(p) airman cone of the same base. Find the cost of the canvas of the tent at the rate of Rs 45 per m2. Sol. PR 22 ? 8 ? 17 cm2 7 = 854. 85 cm2 = 855 cm2 (approx. ) = 2 (? rl) = 2 ? Height of the tent = 8. 25 m. Height of the rounded erupt = 5. 5 m . . . Height of the conelike vocalization = (8. 25 5. 5) m = 2. 75 m. 30 Base radius of the tent = m = 15 m. 2 . . . Slant height of the conic discontinue (15)2 + (2. 75)2 m = = 15. 25 m. = AK = 42 16 = = 16 9 9 Sol. Slant height of each cone = 82 ? 152 cm 64 ? 225 cm = 17 cm. ? Surface area of the resulting shape 225 + 7. 5625 m Curved surface area of the tent = curved surface area of the cylindric part + curved surface area of the conical part = 2? rh + ? rl = ? r (2h + l) 22 = ? 15 (2 ? 5. 5 + 15. 25) m2 7 ? 22 ? = ? ? 15 ? 26. 25? m 2 ? 7 ? = 1237. 50 m2. Rate of the canvas = Rs 45 per m2 . . . Cost of the canvas = Rs (1237. 50 ? 45) = Rs 55687. 50. Sol. Slant height of the cone = = = AS and height of the cone = 14 cm BR = 22 ? 7 ( 7 5 + 7) cm2 7 O TH = 7 ? 14 cm = 245 cm = 7 5 cm. Total surface area of the cone = ? rl + ? r2 = ? r (l + r) 2 2 ER Slant height of the cone = r 2 ? h2 = 154 ( 5 + 1) cm2 Surface area of the cube = 6 ? 142 cm2 = 1176 cm2 ? Surface area of the remaining solid left out after the cone is carved out = surface area of the cube area of base of the cone + curved surface area of the cone 22 2 ? ? = ? 1176 ? ? 7 ? 154 5 ? cm2 7 ? ? YA L = ? 1022 ? 154 5 ? cm2. ? ? Q. 13. A toy is in the form of a cone mounted on a hemisphere of super C base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy. 2007, 2011 (T-II) 6 G O S Q. 14. A solid is in the form of a proficient circular cylinder with hemispherical ends.The total height of the solid is 58 cm and the diameter of the cylinder is 28 cm. Find the total surface area of the solid. 2006 Sol. Q. 15. A toy is in the shape of a rightfield circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the PR Q. 12. A cone of maximum size is carved out from a cube of edge 14 cm. Find th e surface area of the cone and of the remaining solid left out after the cone carved out. Sol. Diameter of the cone = 14 cm = 625 cm = 25 cm ? Total surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the cone = 2? r2 + ? rl = ? r (2r + l) =Radius of the each hemisphere = base radius of the cylinder = 14 cm Total height of the toy = 58 cm ? Height of the cylinder = 58 (14 + 14) cm = 30 cm ? Total surface area of the solid = 2? r2 + 2? rh + 2? r2 = 2? r (2r + h) 22 =2? ? 14 (2 ? 14 + 30) cm2 7 = 88 ? 58 cm2 = 5104 cm2. AK 22 ? 7 (14 + 25) cm = 858 cm2. 7 HA N Height of the toy = 31 cm Base radius of the cone = radius of the hemisphere = 7 cm ? Height of the cone = (31 7) cm = 24 cm r 2 ? h2 72 ? 242 cm 49 ? 576 cm cylindrical part are 5 cm and 13 cm respectively. The radii of the hemisphercial and conical parts are the same as that of the cylindrical part.Find the surface area of the toy if the total height of the toy is 30 cm. 2002 Sol. = 2? r2 + 2? rh + ? rl = ? r (2r + 2h + l ) = = 22 ? 5 (2 ? 5 + 2 ? 13 + 13) cm2 7 22 ? 5 ? 49 cm2 = 770 cm2. 7 TH PRACTICE EXERCISE 13. 1 A choose the correct resource (Q 1 7) 1. A funnel is the combination of (a) a cone and a cylinder (b) frustrum of a cone and a cylinder (c) a hemisphere and a cylinder (d) a hemisphere and a cone. 2. A plumbline (shahul) is the combination of (a) a cone and a cylinder (b) a hemisphere and a cone (c) frustrum of a cone and a cylinder (d) a sphere and a cylinderO ER = 144 ? 25 cm = 13 cm. Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cylinder + curved surface area of the cone BR 3. A shuttle cock used for playacting badminton has the shape of the combination of 2011 (T-II) (a) a cylinder and a cone (b) a cylinder and a hemisphere (c) a sphere and a cone (d) frustrum of a cone and a hemisphere 4. The height of a conical tent is 14 m and its floor area is 346. 5 m2. The continuance of 1. 1 m wide 7 G O YA L S canvas required to built the tent is (a) 490 m (b) 525 m (c) 665 m (d) 860 m 5.The ratio of the total surface area to the asquint surface area of a cylinder with base diameter 160 cm and height 20 cm is (a) 1 2 (b) 2 1 (c) 3 1 (d) 5 1 6. The radius of the base of a cone is 5 cm and its height is 12 cm. Its curved surface area is (a) 30? cm2 (b) 65? cm2 2 (c) 80? cm (d) none of these 7. A right circular cylinder of radius r cm and height h cm (h 2r) just encloses a sphere of diameter (a) r cm (b) 2r cm (c) h cm (d) 2h cm 8. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6? r2. Is it true? PR Slant height of the cone = 122 ? 52 cm = 22 ? 612. 75 cm2 = 1925. 78 cm2. 7 ? mandatory cost of painting = Rs 5. 25 ? 1925. 78 = Rs 1010. 38. AK Radius of the cone = Radius of the cylinder = radius of the hemisphere = 5 cm. Total height of the toy = 30 cm Height of the cylinder h = 13 cm ? Height of the cone = 30 (13 + 5) cm = 12 cm. Internal radius (r) of the vessel = 12 cm Total surface area of the vessel = 2? R2 + 2? r2 + ? (R2 r2) = 2 ? (12. 5)2 + 2 ? 122 + (12. 52 122) cm 2 = 312. 5 + 288 + 12. 25 cm 2 AS HA Q. 16. The internal and orthogonal diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively.If the cost of painting 1 cm2 of the surface area is Rs 5. 25, find the total cost of painting the vessel all over. 2001 Sol. External radius (R) of the vessel = 12. 5 cm N ER 16. A rocket is in the form of a cone of height 28 cm, surmounted over a right circular cylinder of height 112 cm. The radius of the bases of cone and cylinder are equal, each being 21 cm. Find the total surface area of the rocket. ? ? = ? ? ? ? 7? 22 G 13. 2 VOLUME OF A COMBINATION OF SOLIDS 1. meretriciousness of a cuboid of dimensions l, b and h = l ? b ? h. 2. mountain of a cube of edge l = l3. 3. hoi polloi of a cylinder of base radius r and height h = ? 2h. O YA L BR 4. hoi polloi of a cone of base radius r and height 1 h = ? r2h. 3 4 3 5. mickle of a sphere of radius r = ? r . 3 2 6. deal of a hemisphere of radius r = ? r3. 3 textbookS EXERCISE 13. 2 22 . 7 O TH Unless stated otherwise, take ? = Q. 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the vividness of the solid in cost of . 8 S PR Sol. AK 9. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape organize is 4? h + 4? r2. Is it true? 10. A solid clod is exactly fitted inside the cubical box of side a. Surface area of the evening gown is 4? a2. Is it true? 11. From a solid cylinder whose height is 2. 4 cm and diameter 1. 4 cm, a conical fossa of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. 12. A de corative block shown below, is made of two solids a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter 4. 2 cm. Find the total surface area of the block. 22 ? ? = ? . ? 7? 2011 (T-II) 3. A tent of height 3. 3 m is in the form of a right circular cylinder of diameter 12 m and height 2. 2 m, surmounted by a right circular cone of the same diameter. Find the cost of canvas of the tent at the rate of Rs 500 per m2. 15. A solid is composed of a cylinder with hemispherical ends. If the entirely distance of the solid is 108 cm and the diameter of hemispherical ends is 36 cm, find the cost of embellish the surface at the rate of 7 paise per cm2. AS HA 14. cardinal cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid. N O YA L BR O 1 ? 3 ? 2 cm = ? cm3. 3 3 ? Q. 2.Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its t wo ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the stack of look contained in the model that Rachel made. (Assume the satellite and inner dimensions of the model to be nearly the same. ) Sol. = ? ?+ ? TH For conical portion Radius of the base (r) = G Height of cone (h1) = 2 cm 3 cm = 1. 5 cm 2 1 2 ? r h 3 9 We know that, loudness of cone = ER 22 3 cm = 66 cm3 7 Hence, the multitude of the air contained in the model that Rachel made is 66 cm3. 21 ? S Q. 3. A gulab jamun, contained sugar syrup up to about 30% of its pot. Find most how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2. 8 cm (see figure). 2011 (T-II) Sol. Gulab jamun is in the shape of cylinder with two hemispherical ends. Diameter of cylinder = 2. 8 cm ? Radius of cylinder = 1. 4 cm Height of cylindrical part = (5 1. 4 1. 4) cm = (5 2. 8) cm = 2. 2 cm PR AK AS Radius of the hemisphere = Radius of cone = 1 cm Height of cone = h = 1 cm 2 2 ? mess of hemisphere = ? r3 = ? (1)3 cm3 3 3 2 = ? m3 .. (i) 3 1 1 ? ledger of cone = ? r2h = ? (1)2 (1) cm3 3 3 1 = ? cm3 .. (ii) 3 intensity of the solid = Volume of the hemisphere + Volume of cone Volume of cone OAB = = 1 2 ? r h1 3 1 (1. 5)2 (2) ? cm3 = 1. 5? cm3 (i) 3 1 Volume of cone A? B? O? = ? r2h1 3 1 = (1. 5)2 ? (2) ? cm3 = 1. 5? cm3 (ii) 3 For cylindrical portion Radius of the base (r) = 1. 5 cm Height of cylinder h2= 12 cm (2 + 2) cm = 8 cm ? Volume of cylinder = ? r2h2 = ? (1. 5)2 (8) cm3 = 18? cm3 .. (iii) Adding equations (i), (ii) and (iii), we have Total people of the model = saturation of the two cones + masses of the cylinder. = 1. 5? cm3 + 1. ? cm3 + 18? cm3 = 21? cm3 HA N Volume of a gulab jamun 2 2 = ? (1. 4)3 cm3 + ? (1. 4)2 (2. 2) cm3 + ? (1. 4)3cm3 3 3 = = 1 22 14 3 ? ? 0. 25 ? cm 3 7 10 ER 4 = ? (1. 4)3 cm3 + ? (1. 4)2 (2. 2)cm3 3 ? 4 ? 1. 4 ? ? 2. 2 ? cm3 = ? (1. 4)2 ? 3 ? ? ? 5. 6 ? 6. 6 ? = ? (1. 96) ? ? cm3 3 ? ? ? (1. 96) (12. 2) = cm3 3 ? Volume of 45 gulab jamuns ? (1. 96) (12. 2) = 45 ? cm3 3 = 15? (1. 96) (12. 2) cm3 22 ? 1. 96 ? 12. 2 cm3 = 15 ? 7 = 15 ? 22 ? 0. 28 ? 12. 2 = 1127. 28 cm3 30 ? Volume of syrup = 1127. 28 ? cm3 100 = 338. 184 = 338 cm3 (approximately) 11 cm3 30 ? Volume of quadruplet conical depressions 11 3 22 3 cm = cm = 1. 7 cm3 30 15 ? Volume of the wood in the pen stand = (525 1. 47) cm3 = 523. 53 cm3. =4? S PR Q. 5. A vessel is in the form of an upside-down cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is modify with water up to the brim. When consume shots, each of which is a sphere of radius 0. 5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of poke out story shots dropped in the vessel. Sol. Radius of cone = 5 cm Height of cone = 8 cm Volume of cone = = AK = = O YA L BR Q. 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens.The dimensions of the cuboid are 15 cm by 10 cm by 3. 5 cm. The radius of each of the depressions is 0. 5 cm and the skill is 1. 4 cm. Find the pile in the entire stands. (See figure). TH O Radius of spherical lead shot, r1 = 0. 5 cm ? Volume of a spherical lead shot G Sol. Length of cuboid, l = 15 cm comprehensiveness of cuboid, b = 10 cm Height of cuboid, h = 3. 5 cm Volume of the cuboid = 15 ? 10 ? 3. 5 cm3 = 525 cm3 Volume of a conical depression = 4 3 4 ? 3 ? r = ? (0. 5)3 cm3 = cm 3 1 3 6 ? Volume of water that flows out = 1 ? (0. 5)2 (1. 4) cm3 3 10 AS 1 ? strength of the cone 4 1 ? 200? ? 50? cm3 ? ? = 4? 3 ? 3 HA 2 1 ? r h = ? (5)2 8 cm3 3 3 200 ? cm3 3 N let the number of lead shots dropped in the vessel be n. Volume of n lead shots = As per condition, ? n? cm3 6 n? 50? = 6 3 = 31680? cm3 + 3840 cm3 = 35520 cm3 = 35520 ? 3. 14 cm3 = 111532. 8 cm3 ? Mass of the pole = 111532. 8 ? 8 g = 892262. 4 g = 892. 26 kg Hence, the mass of the pole is 892. 26 kg (approximately). BR O TH ER S Sol. Diameter of cylinder ABCD = 24 cm 24 cm3 2 = 12 cm Height of cylinder ABCD (h) = 220 cm ? Volume of cylinderABCD = ? r2h = (12)2 (220)cm3 = 31680? cm3 Base radius of cylinder A? B? C? D? , R = 8 cm Height of cylinder A? B? C?D? (H) = 60 cm ? Volume of cylinder A? B? C? D? = ? R2h = (8)2 (60) cm3 = 3840? cm3 ? Volume of solid push pole = Volume of the cylinder ABCD + Volume of the cylinder A? B? C? D? Base radius of cylinder ABCD, r = YA L PR Q. 6. A solid push pole consist of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use ? = 3. 14) Radius of the cone OAB (r) = 60 cm Height of cone OAB (h1) = 120 cm ? Volume of cone OAB 1 2 1 ? r h1 = ? (60)2 (120) cm3 3 3 = 144000? m3 Radius of the hemisphere (r) = 60 cm = ? Volume of hemisphere = = = Radiu s of the cylinder (r) = Height of cylinder (h2) = ? Volume of cylinder = 11 G O AK AS 50? 6 ? ? n = 3 ? ? n = 100 Hence, the number of lead shots dropped in the vessel is 100. Q. 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is one hundred eighty cm. Sol. HA N 2 3 ? r 3 2 ? (60)3 cm3 3 144000? m3 60 cm 180 cm ? r2h2 So, r = OTHER IMPORTANT QUESTIONS Q. 1. Volume of the largest right circular cone that can be cut out from a cube of edge 4. 2 cm is (a) 9. 7 cm3 (b) 77. 6 cm3 3 (c) 58. 2 cm (d) 19. 4 cm3 O TH YA L BR O Sol. (d) Radius of the cone = 4. 2 cm = 2. 1 cm. 2 ER 8. 5 cm 2 S Sol. Diameter of sphere = 8. 5 cm 4 ? 3. 14 ? 4. 25 ? 4. 25 ? 4. 25 cm3 + 8 ? 3. 14 cm3 3 = 321. 39 cm3 + 25. 12 cm3 = 346. 51 cm3 = Henc e, she is correct. The correct volume is 346. 51 cm3. remains unfilled. Then the number of stains that the cube can accommodate is (a) 142296 (b) 142396 (c) 142496 (d) 142596 Sol. a) Volume of the cube = 223 cm3 = 10648 cm3 space which remains unfilled G Height of the cone = 4. 2 cm. 1 ? Volume of the cone= ? r2h 3 = PR Q. 8. A spherical frosting vessel has a cylindrical neck 8 cm long, 2 cm in diameter the diameter of the spherical part is 8. 5 cm. By measuring the kernel of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, winning the above as the inside measurements, and ? = 3. 14. Amount of water it holds = 4 ? 8. 5 ? ? ? ? cm3 + ? 12 (8) cm3 3 ? 2 ? 10648 cm3 = 1331 cm3 8 Remaining space = (10648 1331) cm3 = = 9317 cm3 1 22 ? ? 2. 1 ? 2. 1 ? . 2 cm3 = 19. 404 cm3. 3 7 Q. 2. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0. 5 cm and it is assumed that 1 space of the cube 8 12 4 ? (0. 25)3 cm3 3 Let n m arbles can be accommodated. Volume of 1 marble = Then, n ? AK 3 4 22 ? ? (0. 25)3 = 9317 3 7 AS HA = ? (60)2 (180) cm3 = 648000? cm3 ? Volume of water left in the cylinder = Volume of the cylinder Volume of the cone + Volume of the hemisphere = 648000? cm3 144000? + 144000? cm3 = 648000 cm3 288000? cm3 = 360000 cm3 360000? = m3 = 0. 36? m3 100 ? 100 ? 100 22 3 = 0. 36 ? m = 1. 131 m3 (approx. 7 Radius of cylindrical neck = 1 cm Height of cylindrical neck = 8 cm N ?n= 9317 ? 3 ? 7 4 ? 22 ? (0. 25) 3 = 142296. Q. 3. A medicine capsule is in the shape of a cylinder of diameter 0. 5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The energy of the capsule is (a) 0. 36 cm3 (c) 0. 34 cm3 Sol. (a) (b) 0. 35 cm3 (d) 0. 33 cm3 Q. 5. The volume of a sphere (in cu. cm) is equal to its surface area (in sq. cm). The diameter of the sphere (in cm) is 2011 (T-II) (a) 3 (b) 6 (c) 2 (d) 4 4 3 ? r = 4? r 2 3 ? r = 3 ? d = 2r = 2 ? 3 = 6 cm Sol. (b) BR = 22 ? ? ? (0. 25)2 ? ? 0. 25 ? 1. 5? cm3 3 7 ? ? O TH Height of the cylindrical part = (2 0. 5) cm = 1. 5 cm Radius of each hemispherical part = Radius of the cylindrical part = 0. 25 cm. ? ability of the capsule 4 ? 4 ? = ? r3 + ? r2h = ? r2 ? r ? h ? 3 ? 3 ? Q. 7. The ratio in the midst of the radius of the base and the height of the cylinder is 2 3. If its volume is 1617 cm3, the total surface area of the cylinder is 2011 (T-II) (a) 208 cm2 (b) 77 cm2 (c) 707 cm2 (d) 770 cm2 Sol. (d) Let the radius and height of the cylinder be 2x and 3x respectively. Then, volume of the cylinder = ? r2h 22 ? 1617 = ? 2x)2 ? 3x 7 YA L = 22 ? 5. 5 ? ? (0. 25)2 ? ? cm3 = 0. 36 cm3 7 ? 3 ? ER Q. 4. A solid piece of iron in the form of a cuboid of dimensions 49 cm ? 33 cm ? 24 cm is moulded to form a solid sphere. The radius of the sphere is 2011 (T-II) (a) 25 cm (b) 21 cm (c) 19 cm (d) 23 cm Sol. (b) Volume of sphere = Volume of cuboid S PR 4 3 ? r1 r 8 2 3 = ? 1 = 4 3 27 r2 3 ? r 3 2 ? Ratio between surface areas = 4 9 1617 ? 7 343 = 22 ? 4 ? 3 8 ? x = 3. 5 cm. ? Total surface area of the cylinder = 2? r (h + r) ? x3 = G O AK ? 4 3 ? r = (49 ? 33 ? 24) cm3 = 38808 cm3 3 38808 ? 3 ? 7 cm 3 = 9261 cm 3 4 ? 22 ? r3 = r = 21 cm Q. 8. On increasing each of the radius of the base and the height of a cone by 20%, its volume will be increased by (a) 25% (b) 40% (c) 50% (d) 72. 8% 13 AS Q. 6. The ratio of the volumes of two spheres is 8 27. The ratio between their surface areas is 2011 (T-II) (a) 2 3 (b) 4 27 (c) 8 9 (d) 4 9 Sol. (d ) 22 ? 7 (10. 5 + 7) cm2 7 = 44 ? 17. 5 cm2 = 770 cm2. =2? HA N Sol. (d) Volume of the original cone = New radius New height 1 2 ? r h 3 = 6r 120r = = 5 100 6h 120h = = 5 100 2 4 3 3 2 3 ? = = 3 2? 2? 2? 6 ? 2? 3 ? = 6 ? Hence, ratio of the volume of sphere to that of cube = cm. Then, volume of the metallic solid cylinder of 91 2 ? r h. 375 ? Per cent increase in volume = AK ? 216 ? 125 ? 2 = ? ? ? r h ? 375 ? height 10 = BR Q. 9. A sphere a nd a cube have the same surface. fancy that the ratio of the volume of sphere to that of the cube is 6 ? O 91? 100 ? 3 = 72. 8%. 375 TH ER = 91 2 100 ? r h ? 1 2 375 ? r h 3 2 cm. 3 = Volume of the metal in the spherical shell 32 4 2 = ? 53 ? 33 ? r ? 3 3 32 2 4 r = (125 ? 27) ? 3 3 3 4 ? ? 98 ? r2 = 32 3 49 7 ? r = cm ? r2 = 4 2 Hence, the diameter of the base of the cylinder AS ( Increase in volume = 72 2 1 ? r h ? r2h 3 125 2011 (T-II) Sol. Let the radius of the sphere be r and the edge YA L O of the cube be x. Whole surface area of sphere = 4? r2 and whole surface area of cube = 62. According to question, ? S Q. 11. A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is 4 3 ? a . Is it true? 3 PR = 7 cm. Sol. Diameter of the ball = side of the cube ? Radius of the ball = ? Volume of the ball = G 4? r2 = 62. r2 x 2 = 6 3 r = ? = 4? 2? x 3 2? 4 3 ? r Volume of sphere 3 Now, = Volume of cube x3 = Hence, the statement is false. 4 ? r? 4 ? r? r ? = ? ? 3 ? x? 3 ? x? x 3 2 Q. 12.From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid. 14 HA ) a 2 1 ? 6r ? 6h New volume = ? ? ? ? 3 ? 5 ? 5 72 2 = ? r h. 125 Q. 10. The internal and external radii of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted to form a solid 2 cylinder of height 10 cm, find the diameter of 3 the cylinder. 2011 (T-II) Sol. Let the radius of the base of the cylinder be 4 a3 ? a3 = 3 8 6 N Sol. Volume of the cube = 73 cm3 = 343 cm3 Sol. 1 ? ? 32 ? 7 cm3 3 = 66 cm3 ? Volume of the remaining solid = (343 66) cm3 Volume of the cone = = 277 cm3.AK = = Q. 13. The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm2. If the volume of metal used in do cylinder is 176 cm3, find the outer and inner diameters of the cylinder. 2010 Sol. Let the inner and outer radii of the cylinder be r cm and R cm respectively. Then, the height of the cylinder = 14 cm. Inner surface of the cylinder = 2? r ? 14 cm2 = 28? r cm2 Outer surface of the cylinder = 2? R ? 14 cm2 = 28? R cm2 Difference of the two surfaces = (28? R 28? r) ? 88 = 28? (R r) ? AS Radius of the hemispherical portion = 5 cm = radius of the cone. Height of the conical portion = (10 5) cm = 5 cm. Capacity of the shape = PR TH (R r) = 88 ? 7 =1 28 ? 22 ER 1 2 ? r (2r + h) 3 1 22 = ? ? 5 ? 5 (2 ? 5 + 5) cm3 3 7 2750 22 ? 25 = ? 15 cm3 = cm3. 7 21 ? Rr= 1 (i) Volume of the metal used in do the cylinder = ? (R2 r2) ? 14 cm3 . .. 176 = ? (R + r) (R r) ? 14 BR O S 1 2750 ? cm3. 6 7 ? Required volume of the frosting-skating rink cream Space which remains unfilled = ? 2750 2750 ? ? = ? ? cm3 6? 7 ? ? 7 2750 5 ? cm3 = 327. 4 cm3. 7 6 ? ? (R + r) = YA L 176 ? 7 =4 22 ? 1 ? 14 (ii) R = 2. 5 cm and G Solving (i) and (ii), we have r = 1. cm Hence, inner and outer diameters of the cylinder are 3 cm and 5 cm respectively. Q. 14. An ice cream cone, full of ice cream is having radius 5 cm and height 10 cm as shown. Calculate the volume of ice cream provided that its 1 part is left unfilled with ice cream. 6 O R+r= 4 Q. 15. A solid toy is in the form of a hemisphere surmounted by a right-circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy. Sol.Volume of the toy = Volume of the cone + Volume of the hemisphere = 1 2 2 1 ? r h + ? r3 = ? r2 (h + 2r) 3 3 3 15 HA 2 3 1 2 ? r + ? r h 3 3 N = 1 22 1408 ? ? 4 ? 4 (4 + 8) cm3 = cm3. 3 7 7 Sol. Capacity of the box = 16 ? 8 ? 8 cm3 = 1024 cm3 Volume of the 16 glass spheres 4 = 16 ? ?r3 3 4 22 = 16 ? ? ? 2 ? 2 ? 2 cm3 3 7 11264 = cm3 21 Volume of water filled in the box 11264 ? ? 10240 = ? 1024 ? cm3 ? cm3 = 21 ? ? 21 A cube circumscribes this toy, hence edge of the cube = 8 cm. Volume of the cube = 83 cm3 = 512 cm3 ? Required difference in the volumes of the toy and the cube = 487. 61 cm3. 1408 ? ? = ? 512 ? ? cm3 7 ? ? 2176 cm3 = 310. 6 cm3. 7 Total surface area of the toy = curved curface area of the cone + curved surface area of the hemisphere = 2 2 2 = ? r h ? r ? 2? r 2 ? 2 ? = ? r ? h + r + 2 r ? ? ? = YA L 22 ? 4 ? 16 ? 16 ? 2 ? 4 ? cm2 ? ? 7 BR O TH ER diameter of the dome is equal to its total height above the floor, find the height of the structure. 2001 Sol. Let the internal height of the cylindrical part be h and the internal radius be r. Then, total height of the mental synthesis =h+r Also, 2r = h + r ? h = r. Now, volume of the building = Volume of the cylindrical part + Volume of the hemispherical part ? ? ? ? S PR and contains 41 O 22 ? 4 ? ? 4 2 ? 8 ? cm2 = ? 7 88 ? 4 = 7 ? 2 ? 2 cm2 ? G 88 ? 4 = ? 3. 41 cm2 = 171. 47 cm2. 7 Q. 16. 16 glass spheres each of radius 2 cm are packed into a cubical box of internal dimensions 16 cm ? 8 cm ? 8 cm and then the box is filled with water. Find the volume of water filled in the box. 16 880 ? 3 ? 7 =8 21? 5 ? 22 ? r =2 Hence, height of the building = h + r r3 = = (2 + 2) m = 4 m. AK 41 Q. 17. A building is in the form of a cylinder surmounted by a hemispherical valuted dome 19 m3 of air. If the internal 21 2 880 = ? r3 + ? r3 ? r = h 3 21 5? r 3 880 = 21 3 AS 2 19 = ? r2h + ? r3 3 21 HA N Q. 18. A godown building is in the form as shown in the figure.The good cross section parallel to the width side of the building is a rectangle 7 m ? 3 m, mounted by a semicircle of radius 3. 5 m. The inner measurements of the cuboidal portion of the building are 10 m ? 7 m ? 3 m. Find the volume of the godown and the total interior surface area excluding the floor 22 ? ? (base). ? ? = ? . ? 7 ? ? 1 2? = 2 ? ?r ? = ? r2 ? 2 ? 22 ? (3. 5) 2 m2 = 38. 5 m2 7 Total interior surface area excluding the base floor = area of the four walls = = 250. 5 m2. Sol. The godown building consists of cuboid at the bottom and the top of the building is in the form of half of the cylinder.Length of the cuboid = 10 m, Breadth of the cuboid = 7 m Height of the cuboid = 3 m Volume of the cuboid = lbh = 10 ? 7 ? 3 m3 = 210 m3. Radius of the cylinder = 3. 5 m Length of the cylinder = 10 m 1 2 Volume of the half of the cylinder = ? r h 2 1 22 = ? ? (3. 5)2 ? 10 m3 2 7 = 192. 5 m3 Volume of the godown = volume of the cuboid + volume of the half cylinder = (210 + 192. 5) m3 = 402. 5 m3 inner(a) surface area of the cuboid = Area of four walls = 2 (l + b) h = 2(10 + 7) 3 m2 = 102 m2 Interior curved surface area of half of the cylinder 22 = ? rh = ? 3. 5 ? 10 m2 = one hundred ten m2 7 YA L BR O TH ER Q. 19.A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2. 1 m and 4 m respectively and the slant height of the top is 2. 8 m, find the area of canvas used for making the tent. Find the cost of the canvas of the tent at the rate of Rs 550 per m2. Also, find the volume of air enwrap in the tent. 2008C Sol. O S G PR Height of the cone, H = AK ? 2. 8 ? 2 ? 22 = 7. 84 ? 4 m = 1. 95 m Area of canvas required for making the tent = Curved surface area of the tent = Curved surface area of the cylindrical part + curved surface area of the conical part = 2? rh + ? l = ? r (2h + l ) = Interior area of two semicircles 17 22 ? 2 (2 ? 2. 1 + 2. 8) m2 7 AS m HA 1 (curved surface area of the cylinder) 2 + 2 (area of the semicircle) = (102 + 110 + 38. 5) m2 + N 44 ? 7 m2 = 44 m2. 7 Cost of canvas = Rs 500 ? 44 = Rs 22000. Volume of the air enclosed in the tent = Volume of the cylindrical part + Volume of the conical part = = ? r2h + = = 88 8. 25 3 ? m = 34. 57 m3. 7 3 ER Q. 20. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two places of decimal.Also find the total surface area of th e remaining solid. (Take ? = 3. 14) 2008, 2011 (T-II) Q. 21. A juice seller serves his customers using a glass as shown in the figure. The inner diamater of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raise which reduces the depicted object of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use ? = 3. 14) 2009 Sol. Radius of the cylindrical glass r = 2. 5 cm Radius of the cylinder = radius of the cone = 6 cm. Height of the cylinder = height of the cone = 8 cm. Volume of the remaining solid 1 2 = ? 2h ? r2h = ? r2h 3 3 2 = ? 3. 1416 ? 36 ? 8 cm3 3 = 603. 19 cm3 Slant height of the cone, l O YA L BR O TH Sol. G S Q. 22. A cylindrical vessel with internal diamater 10 cm and height 10. 5 cm is full of water. A solid cone of the diameter 7 cm and height of 6 cm is completely immersed in water. Find the volume of (i) water displaced out of the cylindrical vessel. (ii) water left in the cylindrical vessel. Take ? = 18 PR Height of the glass = 10 cm unvarnished capacity of the glass = ? r2h = 3. 14 ? 2. 5 ? 2. 5 ? 10 cm3 = 196. 25 cm3 Volume of the hemispherical portion 2 2 = ? r3 = ? 3. 14 ? 2. 5 ? 2. 5 ? 2. 5 cm3 3 3 = 32. 71 cm3 ?Actual capacity of the glass = (196. 25 32. 71) cm3 = 163. 54 cm3. AK AS 22 7 HA 1. 95 ? 22 ? ? 22 ? 2. 1 ? m3 3 ? 7 ? ? N H? 1 2 ? ?r H = ? r2 ? ? h ? ? 3? 3 ? = 36 ? 64 cm = 10 cm Total surface area of the remaining solid = curved surface area of the cylinder + area of top + curved surface area of the cone = 2? rh + ? r2 + ? rl = ? r (2h + r + l) = 3. 14 ? 6 (16 + 6 + 10) cm2 = 18. 84 ? 32 cm2 = 602. 88 cm2. = r 2 ? h2 2009 Sol. Radius of the cylinder, r = 5 cm Height of the cylinder, h = 10. 5 cm Capacity of the vessel = ? r2h 22 = ? 5 ? 5 ? 10. 5 cm3 = 825 cm3 7 1 Volume of the cone = ? r2h 3 1 22 = ? ? 3. 5 ? 3. 5 ? 6 cm3 = 77 cm3. 7 (i) Water displaced out of the cylinder = Volume of the cone = 77 cm3 (ii) Water left in the cylindrical vessel = Capacity of the vessel Volume of the cone = (825 77) cm3 = 748 cm3. 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0. 5 cm and abstrusity is 2. 1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand. Sol. Volume of a cuboid = 10 ? 5 ? 4 cm3 = 200 cm3. Volume of the conical depression Choose the correct option (Q 1 5) 1. The surface area of a sphere is 154 cm2. The volume of the sphere is 2 1 (a) 179 cm3 (b) 359 cm3 3 2 2 3 1 (c) 1215 cm (d) 1374 cm3 3 3 2.The ratio of the volumes of two spheres is 8 27. The ratio between their surface areas is (a) 2 3 (b) 4 27 (c) 8 9 (d) 4 9 3. The curved surface area of a cylinder is 264 m2 and its volume is 924 m3. The height of the cylinder is (a) 3 m (b) 4 m (c) 6 m (d) 8 m 4. The radii of the base of a cylinder and a cone of same height are in the ratio 3 4. The ratio between their volumes is (a) 9 8 (b) 9 4 (c) 3 1 (d) 27 16 TH ER PR ACTICE EXERCISE 13. 2A 5. The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the figure is (a) ? 2h (b) ? r 2 ? 3h ? 2r ? 3 ? r 2 ? 3h ? 2r ? (c) 3 YA L BR O S 6. Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is 2 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder. G O 7. Marbles of diameter 1. 4 cm are dropped into a cylindrical beaker of diameter 7 cm containing 19 PR Q. 23. A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold pens and pins respectively. The dimensions of the cuboid are 4 22 ? ? (0. 5)2 ? 2. cm3 3 7 = 2. 2 cm3 Volume of cubical depression = 33 cm3 = 27 cm3. ? Volume of wood in the entire stand = 200 (2. 2 + 27) cm3 = 170. 8 cm3. = (d) ?r 3 (3h + 4r ) 3 AK AS HA 1 2 1 22 ? r h = ? ? (0. 5)2 ? 2. 1 cm3 3 3 7 Volume of 4 con ical depressions = N 11. An ice cream cone consists of a right circular cone of height 14 cm and the diameter of the circular top is 5 cm. It has a hemispherical scoop of ice cream on the top with the same diameter as of the circular top of the cone. Find the volume of ice cream in the cone. 12. A solid toy is in the form of a hemisphere surmounted by a right circular cone.Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover. 2011 (T-II) 13. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and gum olibanum the level of water is raised by 6. 75 cm. What is the radius of the ball? 13. 3 CONVERSION OF SOLID FROM ONE SHAPE TO ANOTHER TEXTBOOKS EXERCISE 13. 3 22 , unless stated otherwise. 7 Q. 1. A metallic sphere of radius 4. 2 cm is melted and redo into the shape of a cylinder of Take ? = 20 G O YA L BRO TH ER S 16. A heap of rice is in the form of a cone of diameter 9 m and height 3. 5 m. Find the volume of the rice. How much canvas material is required to just cover the heap? 17. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0. 04 m3. 18. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm respectively.If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket. (Take ? = 3. 14) radius 6 cm. Find the height of the cylinder. Sol. Radius of sphere = 4. 2 cm ? Volume of sphere = PR some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5. 6 cm. 8. A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3. 5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water.If the radius of the cylinder is 5 cm and height 10. 5 cm, find the volume of water left in the cylindrical tub. 9. The largest possible sphere is carved out from a solid cube of side 7 cm. Find the volume of the sphere. 10. A cylindrical boiler, 2 m high, is 3. 5 m in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered 22 ? ? by the lid. ? ? = ? ? 7 ? 14. From a solid cylinder of height 12 cm and base diameter 10 cm, a conical cavity with the same height and diameter is carved out. Find the volume of the remaining solid. 15.A building is in the form of a cylinder surmounted by a hemispherical dome as shown in the figure. The base diameter of the dome is equal 2 of the total height of the building. Find the 3 height of the building, if it contains 67 1 m3 of 27 to AK AS air. HA N 2011 (T-II) 4 3 4 ? r = ? (4. 2)3 cm3 3 3 Volume of cylinder = ? R2H = ? (6)2H cm3 As per condition, Volume of the sphere = Volume of the cylinder 4 ? ? (4. 2)3 = ? (6)2H 3 ? ? Radius (r) = 7 m 2 2 deepness (h) = 20 m Volume of sphere of radius 6 cm 4 = ? (6)3 cm3 3 Volume of sphere of radius 8 cm ? (i) Hence, the height of the platform is 2. m. = As per condition, G ? ? 4 3 4 4 4 ? R = ? (6)3 + ? (8)3 + ? (10)3 3 3 3 3 3 = (6)3 + (8)3 + (10)3 R R3 = 1728 O YA L 4 3 3 ? R cm 3 BR 4 ? (10)3 cm3 (iii) 3 Let the radius of the resulting sphere be R cm. Then volume of the resulting sphere = TH ER 4 ? (8)3 cm3 3 Volume of sphere of radius 10 cm = (ii) Q. 4. A well of diameter 3 m dug 14 m deep. The dry land taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. 2011 (T-II) Sol. For well S PR O (iv) 3 m 2 Depth of well (h) = 14 m ?Volume o f earth taken out = ? r2h Radius of well (r) = AK H = Sol. We know that, volume of the sphere = 4 3 ? r 3 AS Q. 2. aluminiferous spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere. 245? 245 ? 22 ? H= ? H= 2. 5 308 308 ? 7 Diameter = 3 m 63 ? 3? = ? ? ? (14) m3 = ? m3 2 ? 2? Width of the embankment = 4 m Let the height of the embankment be H m. ? Radius of the well with embankment, R ? R = 3 1728 ? R = 12 Hence, the radius of the resulting sphere is 12 cm. Q. 3. A 20 m deep well with diameter 7 m is dug and the earth from digging is

Multiple Choice Questions

CH 1 Answers bigeminal plectrum Questions 1. E 8. E 15. E 2. E 9. E 16. A 3. E 10. C 17. D 4. E 11. E 18. E 5. D 12. A 6. B 13. D 7. E 14. E CH 2 Answers Multiple election Questions 1. C 11. B 21. E 2. C 12. E 22. A 3. B 13. E 23. C 4. B 14. E 24. D 5. D 15. B 25. C 6. E 16. D 26. A 7. A 17. D 27. D 8. D 18. E 28. B 9. A 19. C 10. D 20. E CH 3 Answers Multiple filling Questions 1 B 9. B 17. D 2. C 10. A 18. E 3. B 11. A 19. C 4. D 12. B 20. D 5. E 13. D 21. C 6. A 14. A 7. C 15. D 8. E 16. A CH 4 Answers Multiple filling Questions 1 B 9. B 17. D 2. C 10. A 18. E 3. B 11. A 19. C 4. D 12. B 20. D 5. E 13. D 21. C 6. A 14. A 7.C 15. D 8. E 16. A 9. D 18. B CH 5 Answers Multiple prize Questions 1. E 11. B 21. C 2. B 12 B 22. A 3. C 13. B 23 B 4. C 14. E 24 A 5. B 15. E 25. C 6. B 16. C 26. E 7. D 17. D 27. B 8. A 18. A 28. A 9. B 19. D 29. D 10. B 20. A antecedents 16. Solution occasion equality (5-4) (. 32 . 30)/. 30 x (360/180) = 13. 3% 17. Solution use equation (5-4) (. 30 . 32)/. 32 x (360/180) = -12. 5% 18. Solution cross order . 28/. 86 = . 3256 19. Solution cross rate DM. 31/FF FF1/DM. 31 = FF3. 23/DM FF3. 23/$. 35 = FF9. 228/$ 20. Solution use Equation (5-1) (. 0045 . 0035)/. 0035 = 29% 21. Solution use Equation (5-8) (. 864 S)/S) x (360/90) = . 10 . 04 S = . 3807 22. Solution use Equation (5-6). Remember that Cr$3342. 63 = $0. 0002991. new vary rate = $0. 0002991(1 + . 05)/(1 + . 90) = $. 0001652/Cr$ or Cr$1/$. 0001653 = Cr$6053. 27/$ 23. Solution accustom Equation (5-7) tokenish rate = real rate + inflation rate. nominal rate = 5% + 4% = 9% Solution point in the U. S. $10,000 x 1. 01 = $10,100 invest in the U. K. and cover in the forward market. get pounds at the present spot rate $10,000/1. 8 = ? 5,555 invest in the U. K ? 5,555 x 1. 015 = ? 5,638 sell pounds forward ? 5,638 x 1. 8 = $10,036 The investor would piddle $64 more by investing in the United States preferably of the United Kingdom. 25. Solution use Equation (5-8) and solve for the forward rate (F 1. 800)/1. 800 x (360/90) = 0. 04 0. 06 F = ? 1. 809 26. Solution use Equation (5-8) and solve for the U. K. reside rate. (1. 780 1. 800)/1. 800 x (360/90) = 0. 04 if if = 0. 084 27. Solution function Equation (5-1). % Change = (0. 68 0. 64)/0. 64 = 0. 0625 or 6. 25% 28. Solution Converting the above example into indirect quotations, the Swiss franc interpolates from 1. 5625 francs to 1. 4706 francs. usance Equation (5-2) to solve this problem. % Change = (1. 5625 1. 4706)/1. 4706 = 6. 25% 29. Solution practise Equation (5-3). Spread = (0. 68 0. 64)/0. 64 = 0. 0625 or 6. 25% CH 6 Answers Multiple resource Questions 1. E 10. E 19. B 2. E 11. B 20. E 3. E 12. E 21. E 4. B 13. A 22. C 5. C 14. A 23. D 6. E 15. D 24. A 7. C 16. C 25. B 8. B 17. D 9. B 18. A Solutions 16. $ think of = $0. 50 x DM10,000,000 = $5,000,000 17. Investment =? 62,500 x $1. 65 x 0. 02 = $2,062. 50 Profit = ? 62,500 ($1. 67 $1. 65) = 1,250 Rate of return = (1,250/2,062. 50) x (12/6) = 121% 18. Potential attain = $1. 65 $1. 62 = $0. 3 19. Potential way out = $1. 62 $1. 65 = -$0. 03 20. Buy call pickaxs on March 19 -$0. 04 Exercises the selection on September 19 -$0. 80 Sell the pounds on September 19 +$0. 92 meshwork profit as of September 19 +$0. 08 Net profit for three contracts = pile$150,000 x $0. 08= $12,000 21. Total loss = Can$150,000 x $0. 04 = $6,000 22. Intrinsic value = $0. 16 $0. 15 = $0. 01 23. Breakeven point = $1. 75 + $0. 07 = $1. 82 24. Total receipts = FF10,000,000 x $0. 20 = $2,000,000 total subvention = FF10,000,000 x $0. 05 =$ 500,000 send away receipts = $1,500,000 25. Breakeven point = $0. 70 $0. 05 = $0. 65 CH 7Answers Multiple Choice Questions 1. E 10. C 19. C 2. E 11. B 20. B 3. E 12. B 21. D 4. C 13. A 22. E 5. E 14. B 23. A 6. A 15. B 24. C 7. E 16. C 25. D 8. A 17. E 26. A 9. C 18. A 27. D Solutions 20. $7,500,000 x (0. 082 0. 08) = $15,000. 21. $15,000 x the annuity discount factor of $1 for 5 years at 8 pe rcent = $15,000 x 3. 993 = $59,895. 22. You depart receive a total of $30,000 for the first two years $7,500,000 x (0. 082 0. 080) x 2. The new floating rate that you will receive 8. 2% -1. 5% = 6. 7%. You will pay a total of $292,500 for the digest three years $7,500,000 x (0. 67 0. 08) x 3 years. Thus, your net payment over the five years will be -$262,500 ($30,000 $292,500). 23. $500,000 x SFr1. 4 = SFr700,000. 24. $500,000 x 0. 09 = $45,000. 25. $500,000 x (0. 09 0. 08) = $5,000. 26. SFr700,000 (1. 08) = SFr756,000. 27. $500,000 (1. 09) = $545,000. CH 8 Answers Multiple Choice Questions 1. E 9. E 17. D 2. C 10. A 18. D 3. E 11. D 19. A 4. D 12. E 20. C 5. E 13. C 21. E 6. D 14. E 22. C 7. E 15. D 23. C 8. B 16. C 24. E Solutions 18. call Equation (8-1) % Change = ( 0. 70 0. 65 ) / 0. 65 = 7. 69% 19. Use Equation (8-2) % Change = ( 0. 65 0. 0 ) / 0. 70 = -7. 14% 20. Use Equation (8-1) % Change = ( 0. 44 0. 40 ) / 0. 40 = 10% 21. Use Equation (8-3) Predicted Rate = $0. 4 x ( 1 + 0. 05 ) / ( 1 + 0. 03 ) = $0. 4078 22. ( 0. 4400 0. 4078 ) / 0. 4078 = 7. 9% 23. Use Equation (8-5) Predicted Rate = $0. 50 x ( 1 + 0. 12 ) 5 / ( 1 + 0. 08 ) 5 = $0. 5997 CH 9 Answers Multiple Choice Questions 1. B 10. D 19. C 2. E 11. D 20. D 3. D 12. A 21. A 4. C 13. A 22. A 5. E 14. E 23. C 6. E 15. E 24. D 7. C 16. A 8. E 17. C 9. D 18. B Solutions 12. Call option = ? 50,000 x $1. 7 = $85,000 Spot transaction = ? 50,000 x $1. = $90,000 Thus, the U. S. company should sour the option. 16. Net Exposure = Ps300 gazillion Ps200 million = Ps100 million Gain or loss = $0. 0001 x (-Ps100 million)= $10,000 17. (Expected amount) $15 million (actual amount) $14 million = exchange loss of $1 million 18. (Profit after taxes) ? 50 million + (depreciation) ? 10 million = (cash flows from operation) ? 60 million Exchange gain or loss = ? 60 million x $0. 02 = $1. 2 million 19. Gain or loss = $15 million x 3 = $45 million 20. ( 1. 8090 1. 800 ) / 1. 8000 x 360 / 90 = 0. 08 0. 1 0. 02 = 0. 02 21. irect loan credit swap 50,000y + (250,000y 500,000) = 50,000y + 50,000 y = 2. 2 22. $value = $2. 02 x ? 10,000 = $20,200 23. 1) borrow ? 9,709 (10,000/1. 03) 2) buy $19,515 (? 9709 x $2. 01) 3) invest $19,515 in the U. S. at 2% 4) receive $19,905 ($19,515 x 1. 02) 24. Call option = ? 50,000 x $1. 7 = $85,000 CH 10 Answers Multiple Choice Questions 1. E 10. D 19. e 2. E 11. E 20. E 3. C 12. E 21. D 4. B 13. D 22. A 5. E 14. D 23. E 6. E 15. E 7. D 16. E 8. E 17. B 9. E 18. C CH 11 Answers Multiple Choice Questions 1. B 9. A 17. B 2. B 10. B 18. E 3. E 11. D 19. B 4. E 12. D 20. C 5. B 13. A 21.D 6. D 14. E 22. E 7. B 15. A 8. D 16. C CH 12 Answers Multiple Choice Questions 1. E 10. B 19. C 2. A 11. A 20. D 3. E 12. C 21. D 4. E 13. C 22. B 5. B 14. B 23. D 6. D 15. E 24. C 7. D 16. E 25. C 8. E 17. E 9. E 18. A CH 13 Answers Multiple Choice Questions 1. C 8. C 15. E 2. A 9. D 16. A 3. D 10. E 17. D 4. E 11. C 18. B 5. D 12. B 19. E 6. D 13. B 20. B 7. D 14. D 21. D CH 14 Answers Multiple Choice Questions 1. E 11. E 21. D 2. D 12. E 22. E 3. E 13. B 23. C 4. E 14. A 24. B 5. A 15. B 25. A 6. A 16. C 26. D 7. B 17. B 27. D 8. B 18. E 28. C 9. E 19. C 10. D 20. A Solutions 25. Solution U. S. nvestment earns 1 percent. Percentage change in mark = ($0. 40 $0. 50)/$0. 50 = -20%. German investment loses 18. 8 percent (1 + 0. 015)(1 + (- 0. 20) 1 = -18. 8%. 26. Solution Convert DM100,000 to $50,000 at $0. 50 rate. Invest $50,000 in the U. S. at 11 percent. ($50,000 x 1. 11 = $55,500) reconvert dollars to marks. ($55,500/$0. 46 = DM120,652) Yield = (DM120,652 DM100,000)/DM100,000 = 20. 65%. 27. Solution Use Equation (14-1). 0. 10 = (1 + 0. 13)(1 + ie) 1 solve the equating for ie (percentage depreciation). ie = (1 + 0. 10/(1 + 0. 13) 1 = -2. 65%. 28. SolutionUse Equation (14-1). 0. 09 = (1 + 0. 60)(1 + ie) 1 solve the equation for ie (percentage depreciation). ie = (1 + 0. 09)/(1 + 0. 60) 1 = -31. 88%. CH 15 Answers Multiple Choice Questio ns 1. B 9. A 17. D 2. E 10. E 18. A 3. D 11. B 19. E 4. E 12. D 20. A 5. A 13. B 21. A 6. E 14. B 22. B 7. A 15. E 23. E 8. B 16. C 24. A Solution 20. Use Equation (15-2) R = . 07 + (. 15 . 07) 1. 4 = 18. 2% 21. Use Equation (15-2) R = . 05 + (. 11 . 05) 1. 2 = 12. 2% 20% 22. Use Equation (15-4) Rp = (. 4) (. 12) + (. 6) (. 20) = 16. 8%. 23. Use Equation (15-4) 0. 17 = (0. 60) (Rus) + (0. 40) (0. 20). Rus = 15%. 4. Average price = (40 + 50 + 60) / 3 = $50. Use Equation (15-1) for the standard deviation Standard deviation = ( 40 50 ) 2 + ( 50 50 ) 2 + ( 60 50 ) 2 / ( 3 1 ) ? =$10. 1. The coefficient of variation = 10 / 50 = 0. 20. CH 16 Answers Multiple Choice Questions 1. E 11. A 21. C 2. B 12. C 22. E 3. B 13. D 23. C 4. E 14. E 24. B 5. A 15. B 25. B 6. C 16. D 26. B 7. C 17. B 27. A 8. B 18. E 28. E 9. D 19. C 10. C 20. A CH 17 Answers Multiple Choice Questions 1. A 11. D 2. D 12. E 3. B 13. E 4. B 14. E 5. D 15. A 6. E 16. B 7. C 17. A 8. C 18. C 9. B 19. A 10. A Solut ions 15. ayback period = 1 + (15,000 8,000)/9,000 = 1. 8 years. 16. NPV = $8,000/(1. 12) + $9,000/(1. 12)2 + $10,000/(1. 12)3 + $10,000/(1. 12)4 -$ 15,000 = $13,433. 17. NPV = $8,000/(1. 20) + $9,000/(1. 20)2 + $10,000/(1. 20)3 + $10,000/(1. 20)4 -$ 15,000 = $9,002. 18. category 1 DM12,000,000 x $0. 60 = $ 7,200,000 Year 2 DM30,000,000 x $0. 60 = $18,000,000 Net present value = $7,200,000/(1. 18) + $18,000,000/(1. 18)2 $8,000,000 = $11,029,015. 19. NPV = $900 (0. 75)/(1. 06) + $1,000(0. 55) /(1. 06)2 + $1,400(0. 35)/(1. 06)3 -$ 1,400 = $138. CH 18 Answers Multiple Choice Questions 1. D 10.C 19. E 2. C 11. C 20. B 3. D 12. E 21. E 4. E 13. B 22. C 5. B 14. B 23. B 6. C 15. E 24. E 7. B 16. A 25. D 8. D 17. A 26. E 9. B 18. E 27. A 28. D Solutions 21. Use Equation (18-2) Cost of coarse stock = 4 / 54 + . 09 = 16. 4% 22. Use Equation (18-5) Cost of bond = . 124 (1 . 40) = 7. 4% 23. Use Equation (18-1) Cost of capital = (120,000/200,000). 164 + (80,000/200,000). 074 = 12. 8% 24. Us e Equation (18-3) Cost of common stock = 0. 06 + (0. 08 0. 06) 1. 2 = 8. 4%. 25. Use Equation (18-6) The before-tax cost of debt = 0. 30 x 0. 85 0. 15 = 0. 105. After-tax cost of debt = 0. 105 (1 0. 35) = 6. 3% 26. Use Equation (18-4) The cost of common stock = 1 / 25 = 4%. 27. If you rearrange Equation (18-2) for the market price of equity, you will have market price = dividend / (cost of equity annual dividend growth rate) = $1. 2 / (0. 20 0. 04) = $7. 50. 28. lap up Equation (18-2) for the market price of equity Because the dividend per share is $2. 40 ($4. 00 x 0. 60), market price of the stock = $2. 4 / (0. 12 0. 05) = $34. 29. CH 19 Answers Multiple Choice Questions 1. E 11. E 2. D 12. A 3. D 13. E 4. A 14. A 5. D 15. D 6. A 16. E 7. E 17. D 8. E 18. C 9. C 19. C 10. D 20. A

Community based tourism

On the one hand, al topical anesthetic anesthetic level, opportunities have been observed to modernize plans for the elaboration of naked actors and strategies, and for the selectment of civil corporations in the discipline process. tourism is the travel for recreational, leisure, family or business drives, usually of a limited duration. touristry is refers to travel to another location within the like country and as well as trans-national travel. The World tourism Organization defines tourists as pack traveling to and staying In places outside their usual environment for not much than one consecutive year for leisure, business and other purposes.The concept of Community-based tourism (CB) squeeze out be found in the work of Murphy (1985), where aspects concerning tourism and developing topical anesthetic communities atomic offspring 18 analyzed, and in a further cultivation by the same author in 2004 (Murphy and Murphy, 2004). Along with these two studies, on that point ar several other research papers analyzing the relationship mingled with tourism and topical anaesthetic communities ( much(prenominal) as Richards and Hall, 2000). This concept paves the way for untried lines of probe and for the possibility of tourism development together with other alternatives such as Pro-PoorTourism (PPTP) Community Benefit touring car Initiatives (Cobalt) (Simpson, 2008) or Community-Based Enterprises (Cubes) (Mammary and Jones, 2007). To summaries, all these Initiatives agree that the termination fri reverseship should be involved In the tourism planning and way decision-making process, owed to three mall reasons It considers them to be part of the tourist product, topical anesthetic communities adjust to changes easily, and it helps to open their minds. Several projects based on CB Kibosh, 2008), Asia (Enplane et al. 2006 Kaki, 2008 Kate, 2010), Oceania (Dyer et al. 2003), and in antithetic countries of Latin America such as Brazil (Grouc hier, 2007), Ecuador (Uric teal. , 2008), Mexico (Brings and Israel, 2004) and Peru (Zorn and Farthing, 2007). CB is based on the active participation of the local club. This is why the creation of partnership events which may favors this type of tourism, part at the same while helping to create a relationship between the local partnership and visitors, is so important.To facilitate this, different macrocosm administrations, Non- Governmental Organizations (Nags), private institutions and the local confederation itself should get heterogeneous and work together. According to Enplane et al. (2006), the main limitations local communities have to pose when implementing tourism projects are the following lack of fiscal re characters, infrastructure or know-how limitations of a heathenish kind and potence conflicts between the different public administrations.At the same metre, the following factors are described as being passing important for CB implementation (Kibosh, 200 8) the inclusion of stakeholders, the evaluation of individual and collective earns, the circumstance of objectives, And analysis f decisions to be implemented. The main benefits of lodge tourism are the direct economic impact on families, socioeconomic improvements, and sustainable variegation of lifestyles (Mammary and Jones, 2007 Raster, 2010).CB is certainly an effective way of implementing policy coordination, avoiding conflicts between different actors in tourism, and obtaining synergies based on the exchange of knowledge, analysis and ability among all members of the union. unitary of the most controversial aspects in scientific literature is determining the tote up and type of tourists. Thus, Enplane et al. 2006) highschoollight the importance of receiving a small human activity of tourists, which means more interlocutor with local culture and society. This way, the risk of tourists invading private aspects of the local culture is avoided.However, at the same time, his limited number of visitors reduces the economic resources generated by tourism Having taken this element into consideration, the creation of cooperatives allowing the union to manage its own tourist resources is seen a fundamental element (Leap, 2007 Groans and Kaufmann, 2010). However, there are around negative aspects in the development of the tourist product, including he potential link between a high number of tourists, provoke and alcohol, and the loss of cultural identity and the hypothetical degradation of lifelike resources (They et al. 2002). In this fibre, residents have five alternatives to minimize this negative impact on the community resistance, retreat, boundary maintenance, revitalization and adoption (Dugan, 1989). We must remember that tourism fecal matter change (or event destroy) the local culture if it is being treated as Just another tourist attraction and only aims for a quick development of the area (Dyer et al. , 2003). CB is based on the reacti on of tourist products characterized by community participation in their development.CB emerged as a possible event to the negative effects of mass tourism in developing countries, and was, and the same time, a strategy for community organization in order to describe better living conditions. Its core idea is the integration of hotel management, aliment and beverages, complementary color services and tourism management, but in any case accommodates other subsystems (infrastructure, health, education and environment) as main characteristics, thus presenting a sustainable between the local community and visitors as a key element in the development of a aorist product (Choice et al. 2007). CB is protected and emboldened by different international organizations, such as the World Tourism Organization (2002), and has several objectives, among which community sanction and ownership, social and economic development, conservation of natural and cultural resources, and a high quality visitor have intercourse, should be noted. Community participation in tourism development has been originally unquestionable and implemented in the so-called developed world, where tourism supply emerged.In these regions, special computer programmemers have supported locally set development. An example of this is the Elses LEADER programmers focusing on development of sustainable countryfied areas. It is now claimed that developing countries could avoid many of the problems that have plagued past tourism development by involving diverse social groups from the popular sectors of local communities in decision making. They should become convinced of the benefits of tourism and thus support its development or at least acquiesce to it with resignation rather than impassibility.From the compendious reflected above it becomes clear that a community-based approach requires interchangeable elaboration and co-operation, conceptualized as a process of Joint-decision-making among auton omous keys take holders of an inter-organizational, community tourism domain. It should be resolving planning problems and managing issues related to development of that domain. And it is exactly this that is missing in most developing countries. Now we come to the discussion of preserve or develop that is never ending, especially when focusing on developing countries.Community participation has been use to attempt readjusting the balance of mogul and reassert local community views against those of the developers or local authorities. Furthermore, though we think that community participation is a superpowerful tool to educate the community in rights, laws and political in effect(p) sense, the question should be asked to what extent local people, with sometimes limited pinch of wider issues, push aside accurately decide what is best frothier long term well-being.Also, community participation should not only focus on this political dimension, ignoring financial and economic consi derations which are often the primary drivers at local level. divinatory FRAMEWORK Community-Based Tourism (CB) enables tourists to discover local habitats and lilied, celebrates and respects conventional cultures, rituals, and wisdom. The community leave be aware of the commercial and social value fixed on their natural and cultural heritage through tourism, and this will further community based conservation of these resources (www. Communicativenesss. Info). There are many require related on CB. Kaki (2008) developed the model of CB integrating the concepts the turn tail of participation, power redistribution, collaboration processes, and social capital. The study demonstrated that the model displace be used to assess the actual participation level in a study site. Bothersome, Byrd (2007) argued that there is not a definable genius generic interest for the waiter community. The interests will be community can either assist in keeping an individual in a community or add their chance to leave the community.Current tourism and tourism development in the community will also influence their support for future endeavors and their interactions with visitors. The support and interactions will in turn influence the general success of the tourism development. Moreover, Baggier and Console (2009) elaborated the meaning of community in tourism. They argued that customers may interact with companies and other customers and may achieve more knowledge that allows them to reduce their revealing asymmetry and become more empowered than they were in a pre-let era.Numerous recent researches report that online reviews and comments do influence individuals purchase decisions, bear on the evolution of demand within communities. In this respect, virtual communities play a pivotal role in boosting tourist product innovation by leveraging learning from customer relationships. Customers may become a source of innovation since they gain an economic benefit from innovat ion which boosts their creativity, and show gig expertise which may be transferred to firms and among communities.Virtual communities allow people who interact to revenge their own needs and to share purpose such as an interest, need, information exchange, or service that trys a reason for the community. Firms may leverage these communities by providing a suitable context where customers may share social conventions, language, and protocols. Bothersome, Bananas and Jasmines (2011) examined the potential of CB in Villain district municipality, as well as to put up measures to thread this activity feasible.The study demonstrates that communities considered their participation in ours development inform a number of difficulties they have confronted with. The problems were mentioned most often disagreement with the local government lack of communal land lack of finance apathy and lack of initiative amongst local residents lack of sociality and solidarity. Moreover, Lappers (2010) assesses the potential contribution of Community-Based Tourism Enterprises (Cubes) to poverty alleviation and empowerment.It shows that tourism income captured locally improves rural households livelihoods and generates linkages in the local economy. On the Job learning, training sessions and extensive support by non-governmental organizations and donors are further shown to empower rural actors and unlock socioeconomic opportunities for the future. In this context, Cubes can be characterized as pro-poor initiatives. However, this study provides counter evidence that the sustainability of such community tourism ventures is to be questioned.First, mainstreaming these projects within the competitive tourism commodity scope proves extremely challenging and existly second, communities institutional and managerial capacity is unclouded and thus Cubes viability is limited finally, inadequate support by donors and non-governmental organizations fails to tackle challenges aced by commun ity tourism ventures. Bothersome, Mammary and Jones (2007) evaluate the potential of Community-Based Enterprise (CUBE) as avenues of poverty alleviation in Kenya and the challenges facing them.The key factors that could influence local community attitudes towards CUBE fall into two categories motivational factors and community factors. A sense of ownership was seen as critical so that local communities were adequately empowered and involved. CUBE initiatives had to be also have to be addressed. Furthermore, Foreleg and chick (2012) studied Community- based customize. It aims at environmental conservation but it is also a way to empower communities, by allowing them a degree of sustain over tourism projects and their impacts.Foreleg and Boo explored the varying degrees of empowerment of host communities provided by community-based customize through a meta study analysis of six case studies of tourism projects. Not all contemporary tourism projects take local populations into consid eration thus the six case studies are nonrandom selections for the purpose of representing the concept embodied in the thesis and showing its appropriateness with the new tourists expectations. Furthermore, Salary (2012), studied community-based tourism using long-term anthropological fieldwork in Tanzania.The study critically analyzes how well generally accepted community-based tourism discourses fall upon with the reality on the ground. It focuses on how local moderates dish out their role as ambassadors of communal cultural heritage and how community members react to their narratives and practices. It pays special upkeep to the time-limited, project-based development method, the need for an effective exit strategy, for quality control, tour guide training and long-term our guide retention.Findings reveal fivefold complex issues of power and resistance that illustrate many community-based tourism conflicts. The encounter with the Other is shown to be central and that the role of professional intermediaries in facilitating this consume of cultural contact is crucial. Tour guides are often the only locals with whom tourists spend considerable time they have considerable agency in the image-building process of the peoples and places visited, (re)shaping tourist end images and indirectly influencing the self-image of those visited too.Community Based Tourism subject matter Building Capacity building is important in preparing the community as a host as it will educate and prepare all community members to provide the best services. Through capacity building, acquirements are begind. Training is given to increase the skill level and to strengthen the capacity of individuals and organizations to develop and support the program of CBT as a tourist activity. Capacity building for Kampung Sinaran Baru, Kempas homestay can be done based on study tour, where the community is learning to be better in Malaysia or abroad and the training program or course of commun ication are provided.Such courses are languages, computer skills training, finance and marketing courses, and homestay management. Part of the program for capacity building for the local community consists of eight modules described below and it is designed to educate and develop the skills of local communities. The modules are Capacity Building -APEC TOURISM WORKING GROUP (TWG02/2008A) both the participants attain the course to learn how to behave in order to acquire good behavioral attributes. Every homestay has a first aid kit. nutrient for thought safety, quality and knowledge is essential to the participants.Good level of communication with the tourists is highly encouraged because communication is very important to enhance learning and mutual understanding. Body language and signs aid high understanding and help participants to handle the tourists. The ministry of culture and tourism has a program in place, which is aimed at selecting a hand-full of participants and to trai n them on Basic English language skills. Many of them end up as better people by operating a public cookies kitchen which is part of the village tour by women or Teaching basket weaving and making handicrafts to the touristsStyle of leadership The success of CBT projects is essentially dependent on leadership and organisation. It is common for some members of a community to have more advanced skills or areas of possess than others. These people can champion their skills, show leadership in their skill area, and share their knowledge base with others in the community for everyones benefit (Effective Community Based Tourism Sally Asker, Louise Boronyak . ) In 2006, a female person called Mrs. Norbi Binti Ahmad, is an Elder of the Kampung Sinaran Baru, Kempas homestay.As she teaches Quran, she had gained so much respect from the local community, she had the power to fight for the local communitys rights, and she is hardworking. Since 2006 she has been the only champion. Every 3 years they have an election to select local champion. The last time they had an election, competing with another nominee, she gained 40 out of 48 votes. The final decision is do by the local champion herself. Product development and packaging A tourism product in Malaysia that is becoming increasingly popular among local and foreign tourists is the homestay programme.Under the homestay programme, tourists live with a local family in a kampung (village) house and learn about close-knit family relationships, enjoy sumptuous home-cooked food and discover the simple lifestyle in the country. It flings a chance for visitors to experience the daily life outside of bustling capital cities and puts tourists closer to our natural assets, the culture and friendly people of Malaysia. Each homestay village has something unique to passing game tourists and organizes its own special activities for their guests such as jungle trekking, fishing, rubber tapping, harvest picking, handicraft-making, etc .From just RM75 a day, a guest will be hosted by a family and share their meals and enjoy the daily routines of a kampung life. modification is usually in traditional wooden houses on stilts set amidst a pretty landscaped garden. Basic amenities include bed and bath (either en suite or commonly shared). The homestay packages are priced between RM75 and RM295 ( per person) which include on-board bunk beds and meals, an entertainment coach with karaoke facilities for groups of 40 and above, as well as a full range of authentic traditional and cultural activities at the chosen homestay.The main objective of the programme is to promote the authentic experience of culture, tradition and lifestyle of the homestay communities. Sinaran Baru, Kempas homestay The experience derived from Kampung Sinaran Baru, Kempas homestay could be described as an astound way of tradition life. Imaging been woke up first thing in the morning by the singing of birds and cockerels. Then feeling like sleepi ng more but the sounds of the birds continues to remind you to get up from the bed and prospect the beauty of the day.As you rise from the bed, open up your windows, plenty of smart natural air salutes you. This unpolluted air brings in a new sense of life and energy for the day activities. Savory local food follows remarkable having lunch in the mushroom farm and the participation in a host of delightful activities throughout the day begins, such activities include harvesting local crops, visiting a fish farm, learning how to make traditional delicacies and playing traditional games.The unique experience which continues when the villagers try to involve tourists or rather engages them in the participation of cultural performances by inviting them to jump to the tune of their traditional music. This natural local enjoyment continues as winning spectrum of cultures and traditions in a Malay village of Kampung Sinaran Baru, Kempas indulge tourist in various village pastimes, from enjoyable indoor games to exciting outdoor activities. Furthermore, Tourist can acquire some basic farming skills for example, they can learn how to tap a rubber tree and learn how to weave a basket.They could visit the plantations, acquire some skills in harvesting, rubber tapping and fruits picking. Products and Activities Lets relax and take your time to enjoy the beautiful scenery while mingling with the locals. You can experience the following at Sinaran Baru safe tapping, Homestay and pluck farming. There are activities like traditional dance, plantation visiting, mushroom farm, traditional game and handicrafts. 1. Mushroom farm (visit the farm and have lunch) especially for agribusiness students. agro-tourism) Azmi give tongue to the company was started in 1986 with a shed has now gravid to 10 barns that could accommodate as many as 60,000 blocks of mushroom. He verbalize the response the company by the name of Mushroom Biotech Enterprise is go one step further by intro ducing the concept of agro-tourism since February2007. He said the company is now able to attract tourists, especially from Singapore. Through this company, they offer tour packages are preparing breakfast of nasi lemak, two traditional cakes and drinks and fried mushrooms cost RM8 per person. . Typical Malay kampung house & local way of life (lifestyle) Visitors will experience the full spectrum of village life. Home- cookery lessons and meal times will for sure be a memorable part of your stay. The wide variety and extravagance of painstakingly prepared dishes can sometimes be enough to dishearten most people. Guests will learn how to prepare the food. This might include gathering, cooking and washing it as well. Experience the charm of Malay traditions that have been maintain throughout the ages, which remain very much alive.Live with your hosts, eat with them, be a part of their family. And when you return home, take with you a memorable cultural experience and the friendshi ps of a family happy to welcome you whenever you return to Malaysia. 3. Traditional Malay dance & music (Zapin dance , kompang music) Zapin Zapin is a popular dance in the state of Johor. Introduced by Muslim missionaries from the Middle East, the original dance was performed to Muslim devotional chanting to spread knowledge about the history of the Muslim civilisation. KompangArguably the most popular Malay traditional instrument, the kompang is widely used in a variety of social occasions such as the National Day parades, official functions and weddings. Similar to the tambourine but without the jingling admixture discs, this hand drum is most commonly played in gigantic kompang ensembles, where various rhythmic composite patterns are produced by overlapping multiple layers of different rhythms. 4. Local traditional game (sepak raga,congkak,batu seremban) Tourists could experience playing Malay community traditional games such as sepak raga and congkak. 5. Fruit farm 6. Rubber & oil palm plantations

The Rise of Christianity – Did Jesus Want to Start a New Religion?

If the Gospels ar accurate in reporting that divinity fudge sent messiah to farming, what did deliveryman amaze to earth to do? What was his mission? Lets briefly consider a few of Jesus knowledge statements almost wherefore he came. How better to discover his mission than to render what he himself had to say about it? Do non think that I befuddle come to abolish the honor or the Prophets, he said, thereof endor fumbleg the program line of the entire ageing Testament. I have non come to abolish them, that to fulfill them (Matthew 517). Jesus came to fulfill the Old Testament, in part, by living in complete obedience to God.Although he faced the same temptations that stalk us, it is written that he never sinned once. Thats why he could become the perfect sacrifice for every nations sins, as the script tells us God made him who had no sin Jesus to be sin for us, so that in him we might become the responsibility of God (2 Corinthians 521). The Christian message is no t merely about eternal life some day, after(prenominal) we die, in some far heaven. It is about living life to the maximum proper(a) here and right now. There really is no other way to take Jesus wrangling The thief comes only to steal and kill and destroy I have come that they may have life (John 1010).In the era that Jesus walked the Earth in, we can imagine the papistic Empire. The Roman government safe syncretism, accepting that all religious dogmas, philosophical teachings, and government systems are finally compatible, or a reflection of, a larger system the Roman system. They practiced one of the first one country, devil systems policies pronouncing that all masses had religious freedom, political freedom, and freedom of thought, yet maintaining strict control. The Jews held much intuition and often hatred for the Roman Empire they were unwilling subjects.At the time of Jesus birth, the local Roman ruler, King Herod had initiated a massacre of all Judaic baby b oys born at the time. Such actions added more reasons for Jewish fury of the foreign Roman government. The Jews understood the world to be divided into two types of people Jewish and Gentile (non-Jew). The Jews worked hard to disassociate themselves from the Gentiles. King Herod apply many an(prenominal) laborers by commissioning many public works (e. g. expression temple in Jerusalem, palaces, ports, fortresses, stadiums, ornate stone carvings, etc. ) There was a very large disparity between rich and poor.Jesus approached the scene with the promulgation that he had come to dispatch the difference between rich or poor, Gentile or Jew, and sinner or saint. While accumulating followers, others inappropriate his ministry with the belief that he was gaining fame all for the sake of wanting to be seen as God Almighty. All the same, Jesus never stated or claimed to be God. He did ,nonetheless ,claim to be the son of God. It is true that Jesus never said, I am God askly. However, mo st of His teachings were done through parables, and not by direct statements. His disciples asked Him why He taught this way.His answer was To you it has been granted to know the mysteries of the kingdom of heaven, scarce to them it has not been granted. For whoever has, to him shall more be given, and he shall have an abundance but whoever does not have, even what he has shall be taken away from him. so I speak to them in parables because while seeing they do not see, and while hearing they do not hear, nor do they understand. (Matthew 1311-13) on with witnessed miracles performed, his technique and approach reeled in crowds of all kinds of denominations.Some may hold the whim that he hoodwinked desperate believers by using magic or tricks to charm them into his ministry. Others were awfully intimidated by Jesus and his prophecies , thus forth came his crucifixion. nevertheless after death, there was still more to come. Dont be alarmed, he said, You are looking for Jesus the Nazarene, who was crucified. He has risen He is not here. take hold of the place where they laid him. (Mark 166) The book from the evangelist Mark records that after tercet days, Jesus Christ arose from his grave and ascended back to Heaven to reunify with his suffer, God Almighty.After death and acclaimed resurrection, many still testified to spiritual and physical encounters with Jesus. unrivaled person who can testify to this is the apostle capital of Minnesota. Apart from Jesus, Paul is the most evidentiary figure in the development of early Christianity. He has profoundly influenced such(prenominal) monumental figures as Augustine, Luther, and Calvin. Of the 27 books of the New Testament, thirteen are attributed to Paul. Luke tells us in Acts 9 that Paul was on the road to Damascus when he saw a light from heaven and heard a role saying, Saul, Saul, why do you persecute me?Paul responds, Who are you, Lord? and the say came, I am Jesus whom you are persecuting. Paul w as blind for three days, and thus his site was restored when Ananias laid hands on him. This story is not in Pauls own writing, although Paul does tell us something of his conversion draw (See Gal. 1. 13. -16 1 Cor. 15. 3-9). In one way or another, it seems that Paul believes that he received a revelation of the risen Christ. whatever happened, it completely changed Pauls life, and he went from being a persecutor of Christians to the greatest missionary of the early church.Pauls influence of early Christianity was one of great measure. Paul was the author of many writings that were written during his imprisonment of an estimated six to seven years in Rome. During this period of time Paul authored books in earn to all churches holding different denominations with instruction and inspiration to motivate and rector to his readers about following the teachings of Christ and repenting their sins to God through prayer and worship. The Bible does not tell us how or when the apostle Paul died, and history does not provide us with any information.The only thing we have to go on is Christian tradition, which has Paul being behead in Rome, most the mid 60s A. D. , during the reign of Nero. His role in Christianity has resulted in giving him the deed of conveyance The Founder of Christianity. Just as Paul was captivated by his experience with Jesus many others also became believers because of their experiences. Recorded in the book agree to the evangelist Luke, there was a woman with the inability to stand uncoiled up (a condition referred to as stooped ) who meets Jesus on a Sabbath Day (A Saturday supposedly) in one of the synagogues of the city he was then in.Although the Law of Moses states that the Sabbath is a day of rest Jesus heals this woman of her condition and she is then enabled to stand upright without unease and strain. Jesus is said to have opposed many rules and laws through his miracles, but nonetheless continued to heal and lecture to many wh ether the situation and timing was convenient or not gaining many followers in the process of doing so.In response to the question of whether or not Jesus was setting out to form his own religion it is my spirit that that is a misconception to say so. Jesus, in my perspective, was set on saving lives and teaching sinners to follow God and repent their sins so that they could join he and his father in Heaven. It will most likely always be souls opinion that Jesus had other motives behind his healing and ministries, merely the revealing of the truth is yet to arrive to us. Was he our messiah or our magician? You decide.

For four years the guilt still haunted my Mum

As we sit down down for dinner or tea as it has come to be c each(prenominal)ed in our sm each two up-two down house, 174, Lonsdale Street, at that place was that same ominous feeling in the air that all similarly often filled me with dread and a sinking sensation in the pit of my stomach. That would set the tone for the remainder of the evening and most in all likelihood con canfulue late on in to the night and early hours.The starve that at one time eluded my on the whole body had earlier been forced to become a headway over matter issue as I struggled unproductively with my maths spotwork, hardly now, tea was ready. Unfortunately though, my hunger was no much satisfied even after I had ravenously finished kill every last bite of my half yes my whole half of the epicure dish of Heinz beans and sausage, so expertly cooked or rather opened from the tin by Paul. This was a nonher tell-tale sign of what was to come in the next few hours. As per usual, the rather blan d evening meal was complemented by authentic happenings in our house that even I struggled to deal with as I maybe would have done in the past.By now, we had grown utilize to, and even begun to accept, the fact that Dad was no longer around to spiel a sense of buoyancy and a friendly, smiling face to our home and our lives. Maths homework was left unfinished and certain D.I.Y jobs, such as the burst water pipe in the small dingy spurside at the end of the landing, were left leaking. But we all understand that sustenance must go on and we must stay strong if not for our sake, for mutes. This was something we had done very well since Dad left. Just the two of us, until she set up Paul.As soon as we finished our tea, Mum set active busily clearing the table as I put the tympanic membrane on, offering Paul a drink and anxiously awaiting his most-probable violent reaction. a good deal to my disbelief, he simply just declined the drink, only if left the kitchen hurriedly, in a similar fashion to the one in which he had entered, irritably maunder something ab off Janet and how he had had to put up with this for too long now and how he is not putting up with it for much longer.With this, my thoughts and fears were confirmed and my smell pounded formerly again. It had been plainly obvious from the siemens I awoke that morning that they had been argue the night beforehand and it would once again be Mum who came out worse off. The atmosphere was tense all too often in our house. It was impossible to tell when things would boil over, but it was obvious that the tension was ever so at a light simmer, biding its time and appearing to await the complete(a) moment. It was for exactly this reason that I never invited any of my friends round to eternal rest or even to watch a film or for tea. The whole environment and aura is too aggressive, too heated for others to understand. In fact, everything that happens in our house with regards to this stake is k ept to myself.Just as the kettle poached my mind jolted back into reality and I poured water in to each of the three mugs. I heard the front door loaded with a bang and the letter box swung backwards and fore, probably reprieve off even more than it had before.We shant be seein im again tonight, verbalize Mum softly in her ever-calming voice.Hell be back, I replied tentatively.Oh, hell be back all right. Drunk as a skunk, no doubt. She even said this in a subtle tone, as if she had seen it all before and was now conclusion the whole scenario increasingly monotonous.I decided that the best reply, was no reply.We sat in the lounge in silence. No sound apart from Matthew Kelly on Stars in their Eyes gaily welcoming some middle recovered house-wife from Wales as Christina Aguilera. I glanced across at Mum, still glaring at the tele plenty intently, steel-eyed. She did not even look up, not a second thought. Nothing. Perhaps all recent happenings had hardened her to such petty cond uctWhen Gillian had finished her piece as Christina, I drained the remainder of my mug, said good-night to Mum and proceeded up the stairs that creaked more and more with each processional step higher, and to my fork outroom. The unaccompanied place of sanctity I could find.After having only a wash due to the broken shower, I wearily clambered into bed and took out my football magazine from my bedside drawer. The last thing I remembered before dozing off to sleep was groggily reading an article about the demise of Spain on the World stage.The lines upon lines of text slowly transformed into horizontal lines in my mind and became wood, lots of wood all lined up neatly, now vertical, forming walls. turn over panels. My eyes came into focus as I gazed around what slowly materialized into a beautiful open space, lined with borders and rockeries immediately before the mahogany slats, containing amazingly intensityful trees, plants and grasses that thrived in the gorgeous weather. No w my mind caught up with my vision and I recognised the setting as our old house. It was peaceful. Perfect. Birds chirruped in the trees and a plethora of insects busily went about their duties. Bees buzzed in and out of open petals and buds all around the garden.Ladybirds landed gracefully on emerald green leaves and butterflies parading stunning spectrums of colour fluttered aimlessly in the gentle breeze. Beyond a finely pruned turn off towards the opposite end of the garden was a ray of light that glared bright in my eyes. It drew me towards it with an almost hypnotic supremacy as I felt the heat of the beating sun on my neck. I stood over the pool and watched tiny insignificant creatures skate on the waters surface, backwards and forwards with no evident purpose.Gradually, large elegant petals floated down from an overhanging tree, landing on the surface of the water charmingly. One by one they came down in to the water. At first with grace, causing minuscule disturbances on the glistening mirror-like surface, but with each descent came a larger and more portentous landing until the pool was littered with petals that gave the water a mystical purple-red glow.At once or so it seemed the garden was silent. Insects and birds still seemed to buzz, sing and twitter contentedly in and around the garden, but no sound. Silence. I turned back to the pool to see its emptiness. No buds, no petals, not even a single solitary blade of grass.My eye lids opened with surprising exempt to focus on the blankness of the familiar surroundings. Despite my disorientation, there were recognisable sounds from beyond my immediate surroundings. It took another moment for it to make sense. Mum?There was a specious, ear piercing smash, followed by a thud then an frantic cry of pain and torment I was right. All my thoughts and fears were so damningly correct, but what could I do? Was I to go down? My mind wavered. Could I help her, or was it not worth it? Would it only worsen the situation? Could it only worsen the situation? It did on the previous occasion.Another loud scream later, followed by a familiar bang and it was over. He was home. Her haven. She whimpered in his arms, but he was oblivious. Immensely intoxicated by drink, he had locomote backwards onto the empty bottle on the table. Well, that was our version. Who would doubt it though? Her many breaks and bruises were plain for all to see.For four years the guilt still stalk my Mum. And Chloe, the result of that fateful night.