Company Act

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This material is a part of our PGPSE programe. Our programme is available for any student after class 12th / graduation. AFTERSCHO☺OL conducts PGPSE, which is available free to all online students. There are no charges. PGPSE is a very rigorous programme, designed to give a comprehensive training in social entrepreneurship / spiritual entrepreneurship. This programme is aimed at those persons, who want to ultimately set up their own business enterprises which can benefit society substantially. PGPSE is a unique programme, as it combines industry consultancy, business solutions and case studies in addition to spirituality and social concerns. You can read the details at www.afterschoool.tk or at www.afterschool.tk

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  • Company Act

    1. 1. BASIC MATHEMATICS AFTERSCHO☺OL – DEVELOPING CHANGE MAKERS CENTRE FOR SOCIAL ENTREPRENEURSHIP PGPSE PROGRAMME – World’ Most Comprehensive programme in social entrepreneurship & spiritual entrepreneurship OPEN FOR ALL FREE FOR ALL www.afterschoool.tk AFTERSCHO☺OL's MATERIAL FOR PGPSE PARTICIPANTS
    2. 2. BASIC MATHEMATICS Dr. T.K. Jain. AFTERSCHO☺OL Centre for social entrepreneurship Bikaner M: 9414430763 [email_address] www.afterschool.tk , www.afterschoool.tk www.afterschoool.tk AFTERSCHO☺OL's MATERIAL FOR PGPSE PARTICIPANTS
    3. 3. A man cycles at the rate of 15.6 kmph. How many metres does he cover in 2 minutes? <ul><li>Convert the speed into meters per seconds. For this multiply it by 5/18 </li></ul><ul><li>15.6*5/18 now multiply it by 120 (for 2 minutes) </li></ul><ul><li>15.6*5/18 *120 =520 meters. Answer </li></ul>
    4. 4. If a boy takes as much time in running 10 m as a car takes in covering 25 m, the distance covered by the boy during the time car covers 1 km is <ul><li>10/25 *1000 = 400 meters answer. </li></ul>
    5. 5. A man takes 5 hours 45 mm in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is: <ul><li>W + R = 5.75 or 2W + 2R = 11.5 </li></ul><ul><li>2 R = 3.75 thus 2W = (11.5-3.75) = 7.75 or </li></ul><ul><li>7 hours and 45 minutes answer. </li></ul>
    6. 6. Prerna rides at the rate of 10 KM per hour but stops for 10 minutes to take rest at the end of every 15 Km. How many hours will she take to cover 100 km <ul><li>100/10 = 10 hours. </li></ul><ul><li>Stoppages: there will be 6 stoppages. Thus she will stop for 6*10 minutes. Thus total she will take 11 hours. Answer. </li></ul>
    7. 7. The ratio between the rates of walking of A and B is 3 : 4. If the time taken by B to cover a certain distance is 48 minutes, the time taken (in minutes) by A to cover that distance is: <ul><li>48 * 4/3 = 64 minutes. </li></ul>
    8. 8. A man walking at the rate of 3 kmph crosses a square field diagonally in 2 minutes. The area of the field (in ares) is <ul><li>The person covers 3000*2/60 = 100 meters. Area of square = ½ * diagonal ^2 </li></ul><ul><li>=1/2 *100 *100 = 5000 sq. meters. </li></ul><ul><li>= 50 Ares ( 1 are = 100 sq.meters). Answer. </li></ul>
    9. 9. A man riding a cycle at 12 km/hr can reach a village in 4.5 hours. If he is delayed by 1.5 hours at the start, then in order to reach his destination in time, he should ride with a speed of: <ul><li>Total distance travelled = 12* 4.5 = 54 KM. </li></ul><ul><li>Now he has 3 hours left (as he is delayed by 1.5 hours). 54/3 = 18 Km per hour. </li></ul><ul><li>Thus he should ride @18 KM per hour to cover the delay. </li></ul>
    10. 10. Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of them. The ratio of their speeds is: <ul><li>11: 9 answer. </li></ul>
    11. 11. Two boys start together to walk to a certain distance, one at 3.75 km an hour and another at 3 km an hour. The former arrives half an hour before the latter. The distance (in km) is: t <ul><li>Let us assume that they are travelling (3 * 3.75)= 11.25 KM. the first will take 3 hours and the second will take 3.75 hours. Thus their gap is 45 minutes (.75). If the gap is 45 minutes, the distance is 11.25, so the distance for 30 minutes should be : 11.25 * 30/45 = 7.5 KM. answer. </li></ul>
    12. 12. A is twice as fast as B and B is thrice as fast as C is. The journey covered by C in 42 minutes wilt be covered by A in: <ul><li>7 minutes answer </li></ul>
    13. 13. A man travels 35 km partly at 4 kmph and partly at 5 kmph. If he covers the former distance at 5 kmph and the later distance at 4 kmph, he could cover 2 km more in the same time. The time taken to cover the whole distance at the original rate (in hours) is: <ul><li>Distance = speed * time. Let us assume that the time are X and Y hours respectively. </li></ul><ul><li>4X + 5 Y = 35 and 5X + 4 Y = 37. Solving this we get : 9Y = (35*5) – (37*4) = 27 or Y = 3. Thus X = 5. Thus total time is 5+3 = 8 hours answer. </li></ul>
    14. 14. Two trains start at the same time from Bikaner and Udaipur and proceed towards each other at 36 kmph and 42 kmph respectively. When they meet, it is found that one train has travelled 48 km more than the other. The distance between the two stations (in km) is ? <ul><li>In one hour, the faster train moves 6 KM more than the slower one. Thus 8 hours were taken by them. (36 + 42) * 8 = 624 KM is the distance between the two stations. </li></ul>
    15. 15. 11% of a number exceeds 7% of the same by 18, then What is the number? <ul><li>As we can see the difference is (11-7) = 4 </li></ul><ul><li>4% is equal to 18 </li></ul><ul><li>Thus 100% should be equal to : </li></ul><ul><li>18/4 * 100 = 450 answer. </li></ul>
    16. 16. WHAT IS THE TOTAL SURFACE AREA OF A CUBE, WHOSE ONE SIDE IS 2 C.M.? <ul><li>FORMULA = 6 * SIDE ^2 </li></ul><ul><li>= 6 * 2^2 = 24 SQ. CM. ANSWER. </li></ul>
    17. 17. A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is <ul><li>Base = 22/7 * r * r (in both cases), so radius is same in both the cases. </li></ul><ul><li>Volume of cone : 1/3 *22/7* r*r*h </li></ul><ul><li>Volume of hemisphere : 2/3 *22/7*r*r*r </li></ul><ul><li>1/3h = 2/3r or H = 2r thus the ratio of their heights is 2:1. answer. </li></ul>
    18. 18. A room is in the shape of a cube, with one side of 3 meters. What is the length of the longest rod that can be put in this room? <ul><li>Diagonal of the cube = sqrt(3) * side </li></ul><ul><li>= 3 * sqrt(3) </li></ul>
    19. 19. If all the edges of a rectangular prism are doubled and its shape is left unchanged, then its volume becomes <ul><li>Volume of prism = area of base * height </li></ul><ul><li>As we can see that each of the edge of the rectangular base has been doubled. Thus we can say : (2L * 2 B ) * 2 h = 8 times the original volume. Answer. </li></ul>
    20. 20. A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2cm. The radius of the third ball is <ul><li>Volume of original ball : 4/3*22/7*3*3*3 </li></ul><ul><li>Volume of the new balls : 4/3*22/7*1.5^3 </li></ul><ul><li>+ 4/3*22/7*2^3 the resultant will be the volume of the third ball. 36 pi – 47.66 = 65.47 </li></ul><ul><li>4/3 *22/7 *r^3 = 65.47 or r^3 = 65.47*21/88 = 15.62 or Radius = 2.5 answer. </li></ul>
    21. 21. A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl? <ul><li>2/3 * 22/7*9^3 divided by 22/7*1.5^2*4 </li></ul><ul><li>1527.4 / 28.28 = 54 bottles answer. </li></ul>
    22. 22. A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is <ul><li>Volume of cone : 1/3*pi*r^2*h </li></ul><ul><li>Volume of hemisphere = 2/3*pi*r^3 </li></ul><ul><li>Volume of cylinder=pi*r^2*h </li></ul><ul><li>Height = radius. Thus we have ratio : 1/3: 2/3: 1 or 2:4:6 OR 1:2:3 ANSWER. </li></ul>
    23. 23. A cylindrical tub of radius 12 cm contains water upto a depth of 20cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. The radius of the ball is <ul><li>Old volume: 22/7*12^2*20 = 9051 </li></ul><ul><li>New volume : 22/7*12^2 *26.75 = 12106 </li></ul><ul><li>The difference = 3054.8 </li></ul><ul><li>Or the radius is : cubic root of (3054/(4/3*22/7)) </li></ul><ul><li>=9 Cm Answer. </li></ul>
    24. 24. The diameter of a copper sphere is 6cm. The sphere is melted and drawn into a long wire of uniform circular cross section. If the length of the wire is 36 cm, the radius of wire is <ul><li>Volume of sphere : 4/3*22/7*3^3= 113.14 </li></ul><ul><li>Radius of wire is :sqrt( 113.14/(22/7*36)) </li></ul><ul><li>= 1 answer. </li></ul>
    25. 25. A right cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is <ul><li>Pi*r^2*h = 1/3 pi*r^2*h </li></ul><ul><li>Thus we can say that the ratio of heights is : 1:3 answer. </li></ul>
    26. 26. The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height will be <ul><li>3:1 answer. </li></ul>
    27. 27. The average marks of a student in 8 subjects are 87. Of these, the highest marks are 2 more than the one next in value. If these two subjects are eliminated, the average marks of the remaining subjects are 85. What are the highest marks obtained by him? <ul><li>Total of 8 = 87*8 = 696 </li></ul><ul><li>Total of 6 = 85*6 = 510 </li></ul><ul><li>The difference is : 186. Thus highest marks are : 94 and second highest is 92 answer. </li></ul>
    28. 28. Amit, Sachin and Veeru join a 2000m race. Sachin beats Veeru by 120 metres and Amit by 250 metres. By how many metres does Veeru beat Amit over the full distance? <ul><li>When Veeru runs 1880 meters, Amit runs only 1750 meters. Thus Amit will run : </li></ul><ul><li>2000*1750/1880 = 1861 thus Veeru will beat Amit by 139 meters. Answer. </li></ul>
    29. 30. If x is 90% of y, what percent of x is y? <ul><li>Let us assume Y to be 100. thus X is 90 </li></ul><ul><li>Y is 100/90*100 = 111% of X. answer. </li></ul>
    30. 31. If the numerator of a fraction is increased by 20% and its denominator is decreased by 10%, the fraction becomes <ul><li>Let us assume the numerator has become : 120 </li></ul><ul><li>Similarly, the denominator has become 90. </li></ul><ul><li>Thus the new fraction is : 4/3 of the previous fraction. Answer. </li></ul>
    31. 32. The average weight of a group of boys and girls is 38 kg. average weight of the boys is 42 kg, and that of the girls is 33kg. If the number of boys is 25, find the number of girls? <ul><li>38 is the mid point, so take difference from both the numbers. We get (42-38) = 4 and (38-33) = 5. thus the ratio of boys and girls is 5: 4. Thus the number of girls is 4/5*25= 20 answer. </li></ul>
    32. 33. The average age of 10 boys is l5years.Of this 3 boys having an average age of 22 years left. What is the average age of the remaining boys? <ul><li>Total age = 10 *15 = 150 </li></ul><ul><li>3 left out : 22*3 = 66 so remaining 150-66 =84 </li></ul><ul><li>The average of remaining =84/7= 12 years answer. </li></ul>
    33. 34. The cost of carpeting a room 15 metres long with a carpet 75 cm wide at 30 paise per metre is Rs. 36. The breadth of the room is <ul><li>Area of room = length * bredth </li></ul><ul><li>Total length of carpet : 36/.3 = 120 meters </li></ul><ul><li>Total area of carpet used :=.75 * 120= 90 sqmeters. </li></ul><ul><li>Thus width of the room = 90/15= 6 meters. Answer. </li></ul><ul><li>(length = 15, width = 6 meters). Answer. </li></ul>
    34. 35. The length of a rectangle is doubled while its breadth is halved. What is the percentage change in area ? <ul><li>Let us assume that length is 20 and width is 10. </li></ul><ul><li>Length is doubled = 40, but wideth is halved = 5. new area is 200. Thus area remain unchanged. </li></ul>
    35. 36. The length of a rectangle is increased by 60%. By what percent would the width have to be decreased to maintain the same area ? <ul><li>Let us assume the length be 100, width be 50. We have new length 160. Thus new width is 5000/160 = 31.25. Thus there is reduction of 37.5% in the width. Answer. </li></ul>
    36. 37. A rectangular plot of land with an area 600 sq.m is fenced, length of fencing being 100 m. Length of the plot is <ul><li>Area = length X bredth </li></ul><ul><li>L + B = 50 (half of total fence, which is 100 meters) and L * B = 600 </li></ul><ul><li>The length and width are 30 and 20 respectively. Answer. </li></ul>
    37. 38. A 5 m wide lawn is cultivated all along the outside of a rectangular plot measuring 90 m X 40 m. The total area of the lawn is <ul><li>We can add lawn on either side of length and width. We have new area 100*50 = 5000. The area of rectangle is 90*40 = 3600. Thus area of lawn is 5000-3600=1400 sq.m. answer. </li></ul>
    38. 39. Around a rectangular garden of length 10 m and width 5 m, a road I m wide is laid. The cost of metalling the road at Rs. 200 per sq. m is <ul><li>Let us add 1 meters on either side of length and width. We get 12*7 = 84. the area of garden is 50. Thus the area of road is 84-50 = 34 sq.meters. The cost is : 34*200 = Rs. 6800 answer. </li></ul>
    39. 40. The area of a rectangular field is 52000 sq m This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field is <ul><li>The length is 3.25*100= 325 meters. The width is : 52000/325= 160 meters. thus the width is 1.6 CM answer. </li></ul>
    40. 41. What is the square root of .4? <ul><li>It is not .4, it is .40 </li></ul><ul><li>Thus square root of .40 is .632 … answer. </li></ul>
    41. 42. The cost of fencing a square garden is Rs. 400 at the rate of Re. I per metre. The area of the garden in square metres is <ul><li>One wall is of 100 meters (because 4 walls are of 400 meters). </li></ul><ul><li>The area of square field is : 100*100 </li></ul><ul><li>= 10000 Sq. meters. </li></ul>
    42. 43. If the side of a square be increased by 50%, the percent increase in area is <ul><li>Suppose the side of square 100 </li></ul><ul><li>Area : 100 * 100 = 10000 </li></ul><ul><li>Now we increase the side by 50% : </li></ul><ul><li>150 * 100 = 15000 </li></ul><ul><li>The increase is by 5000 or 50%. </li></ul><ul><li>If both the sides are increased, the new are will be : 150 * 150 = 22500 the growth is of 125% increase. Answer. </li></ul>
    43. 44. If the area of an equilateral triangle is 24 .5 sq. m, then its perimeter is <ul><li>Area of equilateral triangle </li></ul><ul><li>= sqrt(3)/4 * side * side = 24.5 </li></ul><ul><li>Side *side = 24.5 * 4 / (sqrt (3) </li></ul><ul><li>Side * side = 98 / sqrt (3) = 56 </li></ul><ul><li>Side = 7.52 </li></ul><ul><li>Perimter = 7.52 * 3 = 22.5 answer. </li></ul>
    44. 45. The ratio of the area of a square to that of the square drawn on its diagonal is <ul><li>Let us assume the side of square is 1. its diagonal is sqrt (2) </li></ul><ul><li>Area of square is 1, then area of square on diagonal is 2. Thus the ratio is 1: 2 answer. </li></ul>Sqrt(2) 1 (side)
    45. 46. The base of a right angled triangle is 5 metres and hypotenuse is 13 metres. Its area will be <ul><li>Altitude </li></ul><ul><li>= sqrt (13^2 – 5^2) </li></ul><ul><li>= sqrt (169 – 25) </li></ul><ul><li>=sqrt (144) </li></ul><ul><li>= 12 </li></ul><ul><li>Area = ½ * base * height </li></ul><ul><li>=1/2 * 5 * 12 </li></ul><ul><li>= 30 answer. </li></ul>13 5
    46. 47. The area of an equilateral triangle whose side is 8 cms, is <ul><li>Area of equilateral triangle </li></ul><ul><li>= sqrt(3)/4 * side * side = </li></ul><ul><li>=sqrt (3) / 4 * 8 * 8 </li></ul><ul><li>=27.3 approx. answer. </li></ul>
    47. 48. A room 5.44 m x 3.74 m is to be paved with square tiles. The least number of tiles required to cover the floor is <ul><li>Let us find HCF 544 and 374. for this divide 544 by 374, we get remainder of 170, now divide 374 by 170, we get 34 as remainder, now divide 170 by 34, we get 0 as remainder. Thus one side of tile should be 34 cm. the number of tiles is (544*374)/ (34*34) = 176 answer. </li></ul>
    48. 49. The sides of a triangular board are 13 metres, 14 metres and 15 metres. The cost of painting it at the rate of Rs. 8.75 per sq. m. is <ul><li>(13 + 14 + 15) / 2 = 21 </li></ul><ul><li>= sqrt((21-13) * (21- 14) * (21-15) *21) </li></ul><ul><li>=84 </li></ul><ul><li>Total cost = 84 * 8.75 </li></ul><ul><li>= 735 rupees. Answer. </li></ul>
    49. 50. Area of four walls of a room is 77 square metres. the length and breadth of the room are 7.5 metres and 3.5 metres respectively. The height of the room is <ul><li>2(7.5*height) + 2(3.5*height) = 77 </li></ul><ul><li>15 H + 7 H = 77 </li></ul><ul><li>22 H = 77 or H = 77/22 = 3.5 answer. </li></ul>
    50. 51. The cost of papering the four walls of a room is Rs. 480. Each one of the length, breadth and height of another room is double that of this room. The cost of papring the walls of this new room is <ul><li>The cost will be 4 times, thus cost will be 1920 . </li></ul>
    51. 52. The length of a rope by which a cow must be tethered in order that she may be able to graze an area of 9856 sq.m. is <ul><li>Area of circle : pi * radius * radius </li></ul><ul><li>Radius * radius = 9856 / (22/7) </li></ul><ul><li>Radius = 56 meters. Answer. </li></ul>
    52. 53. The length of minute hand on a wall clock is 7 cms . The area swept by the minute hand in 30 minutes is <ul><li>Area of half circle = ½ * pi * radius * radius </li></ul><ul><li>= ½ * 22/7 * 7 * 7 </li></ul><ul><li>= 77 sq. cm. </li></ul>
    53. 54. A and B start a business with initial investments in the ratio 12: 11 and their annual profits were in the ratio 4: 1. If A invested the money for 11 months, B invested for : <ul><li>(12 * 11 ) / (11*X) = 4/1 </li></ul><ul><li>44X = 12*11 </li></ul><ul><li>X = 12*11/ 44 = 3 </li></ul><ul><li>B invested his capital for 3 months only. </li></ul>
    54. 55. Goti started a business with Rs. 21000 and is joined afterwards by Vinod with Rs. 36000. After how many months did Vinod join, if the profits at the end of the year are divided equally <ul><li>21000*12 = 36000*X </li></ul><ul><li>X = 12*21000/ 36000 </li></ul><ul><li>X = 7 </li></ul><ul><li>Thus Vinod invested his capital after 5 months (12-7) after the start of the business. </li></ul>
    55. 56. A and B entered into partnership investing Rs 12000 and Rs. 16000 respectively. After 3 months, B withdrew Rs. 5000 while A invested Rs. 5000 more. Out of total annual profit of Rs. 16000, the share of A exceeds that of B by : <ul><li>A (12*3 + 17*9) = 189 </li></ul><ul><li>B (16*3 + 11*9) = 147 </li></ul><ul><li>The ratio of A & B = 9:7 </li></ul><ul><li>Thus A will get 9000 & B will get 7000. answer. </li></ul>
    56. 57. A, B, C subscribe Rs. 47000 for a business. If A subscribes Rs. 7000 more than B and B Rs. 5000 more than C, then out of a total profit of Ks. 9400, B receives <ul><li>Let us assume that C invested X. B invested X + 5000. A invested X + 12000 </li></ul><ul><li>3X +17000 = 47000 or X = 10000 </li></ul><ul><li>Ratio of capital : 22:15:10. B will get 15/47*9400 = 3000 answer. </li></ul>
    57. 58. A, B, C start a business. If A invests 3 times as much as B invests and B invests two-third of what C invests, then the ratio of capitals of A, B, C is <ul><li>Let us assume that C invested Rs. 300. B invested 200. A invested 600. </li></ul><ul><li>Thus the ratio is : 6:2:3 answer. </li></ul>
    58. 59. If 6 (A’ capital) = 8 (B’s capital) = 10 (Cs capital), then the ratio of their capitals is: <ul><li>For A = take product of 8 * 10 = 80 </li></ul><ul><li>For B = 6 *10 = 60 </li></ul><ul><li>For C = 6*8 </li></ul><ul><li>Thus ratio : </li></ul><ul><li>80:60: 48 </li></ul><ul><li>Or : 20: 15: 12 answer. </li></ul>
    59. 60. If A’s capital is equal to twice B’s capital and B’s capital is three times C’s capital, then the ratio of their capitals is : <ul><li>Let us assume C’s capital to be 1. B’s capital is 3, and A’s capital is 6. </li></ul><ul><li>Thus their ratio is 6:3:1 answer. </li></ul>
    60. 61. A, B Center into partnership. A invests some money at the beginning. B invests double the amount after 6 months and C invests thrice the amount after 8 months. If the annual profit be Rs. 9000, then Cs share is: <ul><li>Let us assume A invested $ 1. </li></ul><ul><li>1 *12 (for A), B = (2*6) , for C = (3*4) </li></ul><ul><li>Thus their ratio is equal. Each will get Rs. 3000 out of 9000. answer. </li></ul>
    61. 62. Monu, Sonu and kapsa rented a video cassette for one week at a rent of Rs. 350. If they used it for 6 hours, 10 hours and 12 hours respectively, what is the rent to be paid by Kapsa? <ul><li>Ratio is : 6:10:12 or 3:5:6 </li></ul><ul><li>Thus Kapsa will pay : 6/14 *350 </li></ul><ul><li>= 150 Rupees. Answer. </li></ul>
    62. 63. A, B, C start a business with capitals in the ratio 1/3, ¼, 1/5. A withdraws half his capital at the end of 4 months. Out of a total annual profit of Rs. 847, A’s share is: <ul><li>Let us assume that their capitals were : 20, 15 and 12. the profit sharing ratio is : </li></ul><ul><li>(20*4+10*8):15*12: 12*12 = 160:180:144 or </li></ul><ul><li>40:45:36 A’s share : 40/(121) * 847 = 280 answer. </li></ul>
    63. 64. A, B, c hired a taxi for Rs. 520 and used it for 7, 8 and 11 hours respectively. The amount of hire charges paid by B was: <ul><li>B = 8/ (7+8+11) * 520 </li></ul><ul><li>= 160 answer. </li></ul>
    64. 65. A and B entered into partnership with capitals in the ratio 4 and 5. After 3 months, A withdrew 1/4 th of his capital and B withdrew 1/5 th of his capital. The gain at the end of 10 months was Ks. 760. The share of A in this profit is
    65. 66. Solution … <ul><li>Let us assume that A and B invested Rs. 4 and 5. for first 3 months: 12 & 15. Now A and B withdraw Re. 1 each. Thus new amounts are 3 and 4 for 7 months. </li></ul><ul><li>Thus 21 and 28 are the new amounts. </li></ul><ul><li>The ratio is ((12+21): (15+28) = 33:43 </li></ul><ul><li>Thus A will get : 33/ (33+43) * 760 = 330 answer. </li></ul>
    66. 67. Four milkmen rented a pasture. A grazed 18 cows for 4 months, B 25 cows for 2 months, C 28 cows for 5 months and D 21 cows for 3 months. If A’s share of rent in Rs. 360, the total rent of the field (in rupees) is: <ul><li>A = 18*4 = 72, B = 25*2 = 50, C = 28*5 =140, D = 21*3 = 63 </li></ul><ul><li>Thus 325/72 * 360 </li></ul><ul><li>Total amount is = 1625 answer. </li></ul>
    67. 68. A and B enter into partnership. A invests Rs. 16000 for 8 months and B remains in the business for 4 months. Out of a total annual profit, B claims 2/7of the profit. The contribution of B (in rupees) was : <ul><li>Ratio of A : B = 5:2 </li></ul><ul><li>B = 2/5 * 128000 = 51200 now divide it by 4 to get the capital of B. 12800 answer. </li></ul>
    68. 69. A circular pipe is to be designed in such a way that water flowing through it at a velocity of 7 m per min. is collected at its open end at the rate of 11 cubic meter per min. what should be the inner radius of the pipe: <ul><li>Formula : pi * radius * radius * height </li></ul><ul><li>R^2 * 22/7*7 = 11 </li></ul><ul><li>R^2 = 11/22 = 1/2 </li></ul><ul><li>R = 1/ (sqrt(2)) answer. </li></ul>
    69. 70. Two rectangular sheets of paper each 30 cm x 18 cm are made into two right circular cylinders, one by rolling the paper along its length and the other along the breadth. The ratio of the volumes of the two cylinders, thus formed, is <ul><li>Radius of the first : 2*22/7*r = 18 </li></ul><ul><li>R = 18 * 7/44 = 2.86 </li></ul><ul><li>Radius of the second = 2*22/7*r = 30 </li></ul><ul><li>R = 30*7 / 44 = 4.77 </li></ul><ul><li>Volume of the first : 22/7*2.86*2.86*30 and volume of the second is : 22/7*4.77*4.77*18 the ratio is :1 : 1.66 answer. </li></ul>
    70. 71. The altitude of a circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is <ul><li>Formula of lateral surface area : 2*pi * r*h </li></ul><ul><li>Radius and hight will change. Let us assume the area of base has reduced by 1/9. thus r = sqrt (1/9 *pi) thus we can compare them as : (2 pi are common) = r*h : 1/3 r*6h = the new area is 2 times the old area. Answer. </li></ul>
    71. 72. Two cylindrical vessels with radii 15 cm & 10 cm and heights 35 cm & 15 cm respectively are filled with water. If this water is poured into a cylindrical vessel 15 cm in height, then the radius of the vessel is <ul><li>Volume of first : 22/7*15*15*35=24750 </li></ul><ul><li>Volume of 2 nd =22/7*10*10*15 = 4714 </li></ul><ul><li>Add them : 29464 </li></ul><ul><li>22/7*r*r*15 = 29464, thus radius^2 = 625 or r=25 </li></ul>
    72. 73. Two circular cylinders of equal volume have their heights in the ratio 1: 2. Ratio of their radii is <ul><li>Pi * R^2*h = pi*r^2 2h </li></ul><ul><li>R^2 / r^2 = 2/1 </li></ul><ul><li>Or the ratio of the radius is sqrt(2) : 1 </li></ul><ul><li>Answer. </li></ul>
    73. 74. A solid cylinder has a total surface area of 231 cm If its curved surface area is two-third of the total surface area, the volume of the cylinder is? <ul><li>Total base surface area is : 1/3 *231 = 77 </li></ul><ul><li>2 * ( pi *radius * radius) = 77 </li></ul><ul><li>Radius ^2 = 77 *7 / (2*22) = 12.25 or R= 3.5 </li></ul><ul><li>2*22/7*35*h = 154 or h = 154*7 / (2*22*35) =.7 </li></ul><ul><li>Vol. = 22/7 *3.5*3.5* .7 = 26.9 answer </li></ul>
    74. 75. The radius of a wire is decreased to one-third; If volume remains the same, the height will become: <ul><li>New formula : </li></ul><ul><li>22/7 * (1/3 r)^2 * h = 22/7 * r^2 * H </li></ul><ul><li>We can say that 1/9h = H </li></ul><ul><li>Or new height is 9 times the earlier height. </li></ul>
    75. 76. The sum of the radius of the base and the height of a solid cylinder is 37 m. If the total surface area of the cylinder be 1628 m its volume is ? <ul><li>R + h = 37 </li></ul><ul><li>Total surface area = 2pi r * h + 2 pi * r^2 </li></ul><ul><li>2pi r (h+r) = 1628 or r = 1628 *7/ (37**2*22)=7 </li></ul><ul><li>Thus 22/7*7*7*30 = 4620 cubic meters </li></ul>
    76. 77. The radii of the two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is <ul><li>First cylinder : 22/7*2r *2r*5h </li></ul><ul><li>Second cylinder : 22/7*3r*3r * 3h </li></ul><ul><li>=their ratio is : 20:27 answer. </li></ul>
    77. 78. The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is <ul><li>Bigger cylinder=22/7*2.25*2.25*10 </li></ul><ul><li>Coin = 22/7*.75*.75*.2 </li></ul><ul><li>Divide cylinder by coin (volumes) we get : 450 as the answer. </li></ul>
    78. 79. A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just to touch each other. The volume of the cylinder (in cm) is <ul><li>2 * 22/7 * radius = 5 </li></ul><ul><li>Radius = 5/2 * pi </li></ul><ul><li>=22/7 *(5/2 * pi) ^ 12 = 23.86 answer. </li></ul>
    79. 80. A hollow garden roller 63 cm wide with a girth of 440 cm is made of iron 4 cm thick. The volume of the iron used is: <ul><li>Radius of outer cover </li></ul><ul><li>2 * 22/7 * r = 440 or r= 70 </li></ul><ul><li>Inner radius = 66 </li></ul><ul><li>Volume of iron 22/7 * (70 – 66)* (70 +66) * 63 = 107712 answer. (pi *( R^2 – r^2 ) * h ) </li></ul><ul><li>Volume of outer : 22/7*70*70*63 =970200 </li></ul><ul><li>Volume of inner : 22/7*66*66*63 = 862488 </li></ul>
    80. 81. 2.2 cubic dm. of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is <ul><li>Convert this into cm, we get 2200 cubit cm. </li></ul><ul><li>22/7 * .25*.25 * h = 2200 </li></ul><ul><li>H = 2200 * 7 / (22*.25*.25) = 11200 cm ans. </li></ul>
    81. 82. If the area of a circle be equal to that of a square. then the ratio of the side of the square and the radius of the circle is <ul><li>Side ^2 = pi * radius * radius </li></ul><ul><li>Let us assume that side is 1. </li></ul><ul><li>1 = pi * radius * radius </li></ul><ul><li>Thus radius = sqrt(1/pi) </li></ul><ul><li>Thus ratio of side and radius is 1: .56 answer. </li></ul>
    82. 83. The perimeter of a square whose area is equal to that of a circle with perimeter 2 pi x, is <ul><li>Radius of the circle is X. (circle boundary = 2 pi r) </li></ul><ul><li>Area of circle : pi * x^2 </li></ul><ul><li>Area of square = pi x^2 = side square </li></ul><ul><li>Side = sqrt (pi) * x </li></ul><ul><li>Perimeter of square = 4 * side </li></ul><ul><li>Perimeter = 4x sqrt(pi) or 7.09 X answer. </li></ul>
    83. 84. The radius of a circle is increased so that its circumference is increased by 5%. The area of the circle will increase by <ul><li>the circumference has increased by 5%, so radius must have also increased by 5%. Thus new radius is 105% in comparison to earlier radius. Thus the increase in area is : (105)^2 – (100)^2 or 11025-10000 or 10.25% answer. </li></ul>
    84. 85. The ratio of the areas of the incircle and circumcircle of an equilateral triangle <ul><li>The radius of incircle = side of traingle / (2 * sqrt(3)) </li></ul><ul><li>The radius of circumcirlce is = side of triangle / (sqrt(3) </li></ul><ul><li>Area of the first: pi*side square / 12 </li></ul><ul><li>Area of 2 nd : pi * side^2/3 thus the ratio of area is : 1:4 answer </li></ul>
    85. 86. The area of the circle inscribed in an equilateral triangle of side 24 cms, <ul><li>Radius = side / (2 * sqrt(3)) </li></ul><ul><li>Radius = 24/(2 * sqrt(3)) = 4 * sqrt(3) </li></ul><ul><li>Area = pi * radius * radius </li></ul><ul><li>= 22/7 * 4* sqrt(3) * 4 * sqrt(3) </li></ul><ul><li>= 150 sq. cm. answer. </li></ul>
    86. 87. From the following find the best option for FARM ? (First row number & then column no.)
    87. 88. WHICH IS BEST OPTION FOR FARM?
    88. 89. Solution… <ul><li>From the options given, the first option is the best option as it gives the word FARM as the combination of all the letters give you the word FARM. Answer. </li></ul>
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    90. 91. Branches of AFTERSCHO☺OL <ul><li>PGPSE programme is open all over the world as free online programme. </li></ul><ul><li>Those who complete PSPSE have the freedom to start branches of AFTERSCHO☺OL </li></ul><ul><li>A few branches have already started - one such branch is at KOTA (Rajasthan). </li></ul>
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    92. 93. FREE ONLINE PROGRAMME <ul><li>AFTERSCHO☺OL is absolutely free programme available online – any person can join it. The programme has four components : </li></ul><ul><li>1. case studies – writing and analysing – using latest tools of management </li></ul><ul><li>2. articles / reports writing & presentation of them in conferences / seminars </li></ul><ul><li>3. Study material / books / ebooks / audio / audio visual material to support the study </li></ul><ul><li>4. business plan preparation and presentations of those plans in conferences / seminars </li></ul>
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