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- 1. Questions of boat & speed for Aptitude tests by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india FOR – CSE & PGPSE STUDENTS (CSE & PGPSE are free online programmes open for all, free for all) mobile : 91+9414430763
- 2. My words..... My purpose here is to give a few questions, which are often asked in aptitude tests and competitive examinations. Please prepare well for your examinations. Please pass this presentation to all those who might need it. Let us spread knowledge as widely as possible. I welcome your suggestions. I also request you to help me in spreading social entrepreneurship across the globe – for which I need support of you people – not of any VIP. With your help, I can spread the ideas – for which we stand....
- 3. In one hour a boat goes 11 km long the stream and 5 km against the stream. The speed of the boat in still water ? Solution : Take simple average = (11+5) / 2 = 8 km per hour The speed of water = (11-5)/2 = 3 km per hour. Thus speed of boat with stream = 8+3 = 11 speed of boat against stream=8-3=5 speed of boat on still water=8 km per hour
- 4. A man can row 18 kmph in still water. It takes him thrice as long as row up as to row down the river. find the rate of stream Solution : (18+x) = 3(18-x) x=54-3x-18 4x=36 or X = 9 thus speed of stream = 9 (answer) speed with stream = 27 and against stream=9
- 5. A man can row 7 + 1/2kmph in still water . if in a river running at 1.5 km an hour, if takes him 50 min to row to place and back. how far off is the place? Solution : Speed of the person with stream=9 against stream =6. distance covered = X X/6+X/9 = 50/60 15X = (50*54)/60 = 45 X = 3 answer
- 6. A boat can travel with a speed of 13 kmph in still water. if the speed of stream is 4 kmph, find the time taken by the boat to go 68 km downstream? Solution : speed of boat downstream = 13+4 =17 km p.h. time required = 68/17 = 4 hours. Answer
- 7. A boat takes 90 min less to travel 36 miles downstream then to travel the same distence upstream. if the speed of the boat in still water is 10 mph . the speed of the stream is ? Solution : let the speed of the stream = X 36/ (10-X) - (36/(10+X) ) = 1.5 by simple observation, we know that : 36/8 -36/12=1.5 so X = 2, or speed of stream =2 answer
- 8. Boat travels from point A(2,3), to B (2,1) to C (0,3) & back to A what is the area of this triangle covered by boat? Area = Modal value of (Ax(By-Cy) +Bx(Cy-Ay)+Cx(Ay-by)) / 2 =modal value of (2(1-3)+2(3-3)+0(3-1))/2 =2 (answer )
- 9. There is a road beside a river. two friends started from a place A, moved to a temple situated at another place B and then returned to A again. one of them moves on a cycle at a speed of 12 kmph, while the other sails on a boat at a speed of 10 kmph . if the river flows at the speedof 4 kmph , which of the two friends will return to place A first ? Solution : let us assume the distance = 84 km (I have taken LCM of 14 & 6 ) time taken by boat = 6+14 = 20 hours time by cycle (total of coming and going) = 168/12 =14 hours thus cyle reaches first
- 10. A boat takes 19 hrs for travelling downstream from point A to point B. and coming back to a point C midway between A and B. if the velocity of the sream is 4 kmph . and the speed of the boat in still water is 14 kmph. what is the distence between A and B? Solution: Let us assume Ato C = X 2X/18 + X / 10 = 19 38X =19 * 180 X = 90 A to B = 180 km answer
- 11. At his usual rowing rate, Rahul takes6 hours less for downstream than it takes him to travel the same distence upstream of 12 miles, but if he could double his usual rowing rate for his 24 miles round, the down stream 12 miles would then take only one hour less than the up stream 12 miles. what is the speed of the current in miles per hours? Solution : first situation : 12/(X-Y) – 12/(X+Y) = 6 new situation 12/(2X-y) -12/(2x+Y) = 1 new X and Y = 3 and 4, speed of water = .5 his speed now is 3+.5 = 3.5 earlier his speed was : 3.5/2 = 1.75
- 13. A train 140 m long running at 72 kmph.In how much time will it pass a platform 260m long. Total distance to cover = 140+260 = 400 time = distance / speed speed in meters per second = 72 *5/18 = 20 meters per second time = 400/20 = 20 seconds answer
- 14. A man is standing on a railway bridge which is 180 m.He finds that a train crosses the bridge in 20 seconds but himself in 8 sec. Find the length of the train and its sppeed Let us assume the length of train = X the speed of train = x meters/8 180 meters are covered by train in (20-8) seconds 180/12 = 15 meters per second. Thus the length of train is : 15*8 = 120 meters. Answer
- 15. A train 150m long is running with a speed of 54 Km per hour. In what time will it pass a man who is running at a speed of 9 km ph in the same direction in which the train is going Speed of train = 54 * 5/18 = 15 meters per second speed of mann = 9 * 5/18 = 2.5 meters per second distance to cover = 150 time = 150/(15-2.5) =12 seconds answer
- 16. A train 220m long is running with a speed of 59 k m ph ..In what time will it pass a man who is running at 4 kmph in the direction opposite to that in which train is going. Distance to cover = 220 meters speed = 59+4 = 63 km per hour. In meters per second = 63*5/18 = 17.5 meters per second. Time required : 220/17.5 =12.57 seconds. Answer
- 17. Two trains 137m and 163m in length are running towards each other on parallel lines,one at the rate of 42kmph & another at 48 mph.In wht time will they be clear of each other from the moment they meet. Distance to cover 137+163 = 300 meters speed = 42+48 = 90 km per hour speed in meters per second = 90 * 5/18 = 25 meters per second time required = 300/25 = 12 seconds answer
- 18. A train running at 54 kmph takes 20 sec to pass a platform. Next it takes 12 sec to pass a man walking at 6kmph in the same direction in which the train is going.Find length of the train and length of platform Solution : Train v/s man speed = 54 -6 = 48 km per hour speed in m/s =48 * 5/18 = 13.33 m / s length of train = 12*13.33 = 159.6 meters speed for platform =54*5/18 = 15 m / s length of platform+ train = 20*15 = 300 length of platform = 300 – 159 = 140 meters approx.
- 19. A man sitting in a train which is travelling at 50mph observes that a goods train travelling in opposite direction takes 9 sec to pass him .If the goods train is 150m long find its speed Solution : - Distance travelled = 150 speed =150/9 = 16.66 meters per second or 16.66 * 18/5 = 60 km per hour approx. Thus the speed of goods train is 60-50 = 10 km per hour. Answer
- 20. Two trains are moving in the same direction at 65kmph and 47kmph. The faster train crosses a man in slower train in18sec.the length of the faster train is Solution : = When the trains are going in same direction, we take difference of their speed. 65-47 =18 km per hour or 5 meters per second distance travelled =(time 18 seconds * speed 5 meters per second) = 18 * 5 = 90 meters. Thus the length of faster train is 90 meters. Answer
- 21. A train overtakes two persons who are walking in the same direction in which the train is going at the rate of 2kmph and 4kmph and passes them completely in 9 sec and 10 sec respectively. The length of train is Solution : - let us assume the speed of train to be X. (X-2) * 9/3600 = (X-4) *10/3600 9X – 18 = 10X – 40 X=22 km per hour. thus distance travelled = (22-4) * 5/18 = 5 m/s time=10 seconds so length of train = 5*10 = 50 meters. Answer
- 22. Two stations A & B are 110 km apart on a straight line. One train starts from A at 7am and travels towards B at 20kmph. Another train starts from B at 8am an travels toward A at a speed of 25kmph.At what time will they meet From 7 am to 8 am only A is travelling. It would travel 20 km. Now 90 km is to be covered. 90 / (20+25) =2 hours so at 10 am they will meet.
- 23. A train travelling at 48kmph completely crosses another train having half its length an travelling inopposite direction at 42kmph in12 sec.It also passes a railway platform in 45sec.the length of platform is Distance by 2 trains = (48+42) = 90 or 25 m/s 25 * 12=300 meters. So the length of train is 200 meters. Platform : 48 * 5/18 =13.33 m/ s 45*13.33 = 600 so length of platform = 600-200=400 meters. Answer
- 24. Find the time taken by a train 180m long,running at 72kmph in crossing an electric pole Speed of train = 72 * 5/18 = 20 m / s time required = 180/20 = 6 seconds. Answer
- 25. Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring. Solution: Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 352/7 radius = 8 Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 528/7 radius = 12 thus width of ring = 12-8 = 4 answer
- 26. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is Solution : area of one circle = (22/7) * 7 * 7 =154 square of on one circle = 14*14 = 196 difference of area : 42 one side 42/4 =10.5 we have 4 cirlces, each has 10.5 cm of space enclosed, so total space enclosed is 42 sq. cm. Answer
- 27. A semicircular shaped window has diameter of 63cm. Its perimeter equals Circumference of circle = diameter * 22/7 = 63 * 22/7 = 198 it is semicircle so divide by 2 = 99 add diameter also – to denote one side : 99+63 = 162 cm answer
- 28. The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter. Total area = 5.5 * 3.75 =20.63 multiply it by 800 =Rs.16500
- 29. The no of revolutions a wheel of diameter 49 cm makes in traveling a distance of 176m is Solution : circumference = 22/7 * 49 = 154 17600 / 154 = 114.29 thus the wheel will make 115 revolutions. Answer
- 30. .A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field? Solution Area 22/7 * 14 * 14 = 616 time required = 616/100 = 6.16 so cow will take little over 6 days to completely graze the whole field. Answer
- 31. A man runs round a circular field of radius 49m at the speed of 120 m/hr. What is the time taken by the man to take twenty rounds of the field? Solution : circumference = 2* 22/7 * 49 =308 total distance to be travelled = 308 * 20 = 6160 time required = 6160/120 =51.3 hours.
- 32. .The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec. What is the speed of motorcycle in km/hr? Speed covered in 1 seconds : 4 * 22/7 *70 =880 cm or 8.8 meters. Speed in km per hour : 8.8 * 18/5 = 31.68 km per hour answer
- 33. A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be Solution : its circumference is : 2 * 22/7 *56 = 352 when you make a circle out of it, one side will be : 352/4 = 88 thus its area will be : 88*88 =7744 answer
- 34. The area of the largest triangle that can be inscribed in a semicircle of radius 2 is? Area of triangle = Formula = 1/2 * base * height = ½ * (2+2) *2 =4 answer
- 35. A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed? The total boundary = 2(90+50)= 280 280/5 = 56 so we will need 56 poles
- 36. The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is the length of the plot in meter? Total fencing = 5300/26.5 = 200 2(L+B) = 200 2(B+20+B) = 200 2B+20=100 or B=40 and L=60 so length is 60 meters answer
- 37. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required? Length = X bredth = 20 20X =680, or X = 34 boundary =2(34+20)=108 but we will not cover one side =108-20=88 feet. Answer
- 38. A rectangular paper when folded into two congruent parts had a perimeter of 34cm for each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper? When we fold from mid of length = L+2B=34 When we fold from mid of width = 2L+B=38 add them = 3L+3B=72 L+B=24, where L =14, B=10 so area of paper =14*10 =140 answer
- 39. A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is? Diagonal = 52*15/60=13 ½ of perimeter=68*15/60=17 X+Y=17 -- 1 st X^2+Y^2 =169 -- 2 nd square of 1 st equation:X^2+Y^2+2XY=289 2XY=120 or XY =60 thus area of field is 60 sq.meter.
- 40. The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m Boundary = 10.08/.20 = 50.4 one side is 12.6 area : 158.76 area without pavement : (12.6-6)^2= 43.56 pavement = 115.2 cost = 115.2*.5 = Rs. 57.6 answer
- 41. Aman walked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges? Let us assume that the side of square is 1. had he walked along edges, he would have travelled 2. he walked on diagonal so he walked sqrt(2) = 1.41 thus he has saved (2-1.41) =.59 or we can say that he has saved 29.5% answer
- 42. .A man walking at the speed of 4 km p.h. crosses a square field diagonally in 30 minutes. The area of the field is The diagonal is 2 km or 2000 meters side of square is : 2000/sqrt(2) =1414 area of field =2000000 sq. m. Or 200 hectares
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- 54. http://www.scribd.com/doc/6583303/Reasoning-Afterschoool
- 55. http://www.scribd.com/doc/23393680/REASONING-AFTERSCHOOOL http://www.scribd.com/doc/23393719/reasonning-44 http://www.scribd.com/doc/6583347/DI-and-Reasoning http://www.scribd.com/doc/23407636/10-July-reasoning http://www.scribd.com/doc/6583273/Reasoning-Quiz http://www.scribd.com/doc/23393476/10-July-reasoning http://www.scribd.com/doc/23393716/REASONING-QUIZ http://www.scribd.com/doc/14705025/17-Reasoning-II http://www.scribd.com/doc/23393478/10-July-reasoning-II
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- 57. THANKS.... GIVE YOUR SUGGESTIONS AND JOIN AFTERSCHOOOL NETWORK / START AFTERSCHOOOL SOCIAL ENTREPRENEURSHIP NETWORK IN YOUR CITY / CONDUCT WORKSHOP ON SOCIAL ENTREPRENEURSHIP IN YOUR COLLEGE / SCHOOL / CITY [email_address] JOIN OUR NETWORK TO PROMOTE SOCIAL ENTREPRENEURSHIP

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