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- 1. Numerical Reasoning
- 2. Chain RuleDirect Proportion : A B A BIndirect Proportion: A B A B
- 3. Type - 1Direct Proportion: Price Quantity 100 15 150 xPrice increase no. of quantity will also increaseDirect proportion means Cross Multiply100*x = 150*15 ;X = 150*15/100 = 22.5
- 4. Type - 2Indirect Proportion:Men Days 15 50 x 10To complete the work in less number of days, they need to deploy more number of people.15*50 = x*20;X = 15*50/20 = 75
- 5. TipsIn given problem, if you have more than 2 variables, you need to,1. Compare data you need to find with other data which is given.2. While comparing based on the proportion, do either cross or direct multiply.
- 6. Problem - 1If 6 men and 8 boys can do a piece of work in 10days and 26 men and 48 boys can do the samework in 2 days, what would be the time taken by15 men and 20 boys to do the same type of work?
- 7. Solution6 men + 8 Boys = 10 days, 26 men and 48 Boys = 2 days60 men + 80 Boys = 52 Men + 96 Boys8 M = 16 B1 Man = 2 Boys15 M + 20 B = 30 B + 20 B = 50 Boys6 men + 8 boys = 12 Boys + 8 Boys = 20 Boys = 10 DaysBoy Days20 1050 x50x = 20* 10x = 20*10/50 = 4 days.
- 8. Problem - 220 men can complete one third of a work in 20days. How many more men should be employedto finish the rest of the work in 25 more days?
- 9. SolutionMen Work Days 20 1/3 20 x 2/3 25x * 1/3 * 25 = 20*20*2/3x = 20*20*2 /25 = 32Already we have 20 men, so 12 more menshould be deployed
- 10. Problem - 395 men have provision for 200 days. After 5days, 30 men die due to an epidemic. For howmany days will the remaining food last?
- 11. SolutionRemaining days = 200 – 5 = 95Remaining men = 95 – 30 = 65Men Days 95 195 65 x65x = 195*95x = 195*95/65 = 285 days
- 12. Problem - 4A certain number of men completed a piece ofwork in 60 days. If there were 8 more men, thework could be finished in 10 days less. Howmany men were originally there?
- 13. SolutionDays Men 60 x 50 x+860x = 50*x+860x/50 = x + 86x = 5x + 40x = 40
- 14. Problem - 5A contractor employs 60 people to complete atask in 20 days. After 10 days, he finds that only30% of the work is completed. In order toachieve the target of completion of the task in 20days, how many more people should he employnow?
- 15. SolutionPeople Days Work 60 10 30 x 10 7030*x*10 = 60*70*10x = 60*70*10/30*10x = 140Already there are 60 men, so 80 more men are needed.
- 16. Problem - 6If 6 engines consume 15 metric tons of coal,when each is running 9 hours a day, how muchcoal will be required for 8 engines each running12 hours a day with the fact that 3 engines of theformer type consume as much as 4 engines of thelatter type?
- 17. SolutionThree engines of former type consume = 1 unitOne engine of former type consume = 1/3Four engines of latter type consume = 11 engine of latter type consume = ¼Engine Hours Consumption Coal 6 9 1/3 15 8 12 ¼ xx *6*9*1/3 = 8*15*12*1/4x = 20
- 18. Problem - 78 men working for 9 hours a day complete apiece of work in 20 days. In how many days can7 men working for 10 hours a day complete thesame piece of work?
- 19. SolutionMen Hours Days 8 9 20 7 10 x7*x*10 = 8*20*9x = 20 4/7 days
- 20. Problem - 835 cattle can graze a piece of land for 56 days.How many cattle will graze a field three times aslarge in 35 days?
- 21. SolutionCattle Graze Days35 1 56 x 3 35x*35 = 35*3 *56x = 56*35*3/35 = 168
- 22. Problem - 930 men can do a work in 50 days. After 20 days10 men leave the work. The whole work will bedone in?
- 23. SolutionMen Days 30 30 20 xX = 30*30/20 = 45 days
- 24. Problem - 10The work done by a man, a woman and a boy arein the ratio 3 : 2 : 1. There are in a factory 24men, 20 women and 16 boys, whose weeklywages amount to Rs. 224. What will be theyearly wages of 27 men, 40 women, 15 boys?
- 25. Solution1 man = 3 boys, 1 woman = 2 boys24 M + 20 W + 16 B = 24*3 + 20*2 + 16*1 = 128 boys27 M + 40 W + 15 B = 27*3 + 40*2 + 15*1 = 176 boysBoys Duration Wages 128 1 224 176 52 xx*128 = 52*224*176x = 52*224*176/128 = Rs.16,016
- 26. Problem - 11A contractor undertakes to dig a canal 12 kmlong in 350 days and employs 45 men. He findsafter 200 days of working that only 4½ km ofcanal has been dug. How many extra men mustbe employed to finish the work on time?
- 27. SolutionMen Days KM 45 200 4½ x 150 7½200*45*15/2 = 150*x*9/2x= 100Extra men = 100 – 45 = 55 men
- 28. Problem - 128 men can build a foundation 18 m long, 2mbroad and 12m high in 10 days, working 9 hoursa day. Find how many men will be able to builda foundation 32m long, 3m broad and 9m high in8 days, working 6 hours a day.
- 29. SolutionMen Metre Breadth Height Day Hour 8 18 2 12 10 9 x 32 3 9 8 6x*18*2*12*8*6 = 32*8*3*9*10*9x = 30
- 30. Time and Work
- 31. Type - 1A can do work in 10 daysB can do work in 15 daysThey can complete the work by working together?A’s 1 day work = 1/10B’s 1 day work = 1/15A + B = 1/10 + 1/15 = 25/150 = 1/66 days they will complete their work.
- 32. Type - 2A and B can complete the work in 10 daysB and C can complete the work in 15 daysC and A can complete the work in 20 daysHow many days they will take to complete the work by working all together.A + B 1 day work = 1/10B + C 1 day work = 1/15C + A 1 day work = 1/20A+B+B+A+C+A = 1/10 + 1/15 + 1/202(A+B+C) = 6/60 + 4/60 + 3/60 = 13/60A+ B + C = 13*2/60 = 26/60 = 13/30 = 30 /13 = 2 4/13 days
- 33. Types - 3A and B can complete the work in 10 daysB alone can complete the work in 30 daysHow many days A will take to complete the work?A+B = 1/10B = 1/30A = A+B – B = 1/10 – 1/30 = 3/30 – 1/30 = 2/30A = 1/15 = 15 days
- 34. Type - 4A can complete the work in 30 daysB can complete the work in 50 daysC can complete the work in 40 daysA is working continuously from the starting, B and C supporting A alternatively.Day 1 A+B = 1/30 + 1/50 = 8/150Day 2 A+ C = 1/30 + 1/40 = 7/1202 days work = 8/150+ 7/120 = 67/600 Cont…
- 35. Type - 42* 8 days = 67*8/600 = 67/75In 16 days they will complete 67/75 of the workRemaining work = 1- 67/75 = 8/7517th day A+ B = 8/150Work left = 8/75 – 8/150 = 8 / 150 = 4/75On 18th day A+C will work they will finish in 120/7 = 120 / 7 * 4/75 = 32 /35Whole work will be done in 17 32/35 days
- 36. Type - 5A is 60% more efficient than B.B can complete the work in 20 daysA will take how many days to complete the work. A : BEff 160 : 100 8 : 5Days 5 : 8 (More efficient person will take less Days) Cont…
- 37. Types - 5Use chain rule method Effi Days 5 20 8 xMore efficient person takes less days8x = 20*5X = 20*5/8 = 12.5 days
- 38. Problem – 18 men and 12 children can do a work in 9 days.A child takes double the time to do a work thanthe man. In how many days 12 men cancomplete double the work?
- 39. Solution8 men and 12 children in 9 days1 man = 2 children6 men = 12 children8 man + 6 man (12 Children) = 14 menMen Days Work14 9 112 men x 212x = 14*9*2x = 14*9*2 /12 = 21 days
- 40. Problem - 2A is three times more efficient than B, and istherefore able to complete a work in 60 daysearlier than A. How long A and B together willtake to complete the work?
- 41. SolutionA is three times more efficient than B A B Effi 3 1 Days 1 3If A takes 10 days A B 10 30 20 60 30 90 (60 days difference)So, A takes 30 days and B takes 90 days.A and B together = 1/30 + 1/90 = 4/90 = 90/4 =21 ½ days
- 42. Problem – 3A can do ½ of the work in 5 days, B can do 3/5of the same work in 9 days and C can do 2/5 ofthat work in 8 days. In how many days can threeof them together do the work?
- 43. SolutionThe number of Days taken by A, B and C to completeworkA = ½ work in 5 days = 1 work = 10 daysB = 3/5 of the work in 9 days = 1 work = 9*5/3= 15 daysC = 2/3 of the work in 8 days = 1 work = 8*3/2= 12 daysA + B + C = 1/10 + 1/15 + 1/12 = 6+4+5/60 = 15/60=¼A, B and C together take 4 days to complete the work
- 44. Problem - 4A, B and C can do a piece of work in 6, 8 and 10days respectively. They begin the work together.A continues to work till it is completed. B leavesoff 1day before and C leaves off ½ day beforethe work is completed. In what time is the workcompleted?
- 45. SolutionA’s x days work = x/6B’s (x-1) day work = x – 1/8C’s (x-1/2) day work = 2x-1/2*10 = 2x-1/20x/6 + x-1/8 + 2x -1/20 = 120x +15x – 15+12x – 6/120 = 120x + 15x – 15+12x – 6 = 12047x = 120 + 21x = 141/47 = 3 days
- 46. Problem - 5A can do a piece of work in 12 days, while Balone can do it in 15 days. With the help of Cthey can finish it in 5 days. If all the three werepaid Rs. 960 for the whole work, what is theshare of A?
- 47. SolutionA + B = 1/12 + 1/15 = 9/60 = 3/ 20A + B +C = 1/5(A+B+C) – (A+B) = 1/3 – 3/20 = 4 -3/20 = 1/20C will take 20 days to complete the workA:B:C = 1/12 : 1/15 : 1/20= 5/60 : 4/60 : 3/60 = 5 : 4 : 3A’s Share = 960 * 5/12 = Rs.400
- 48. Problem - 6A and B can do a piece of work in 40 days. Afterworking for 10 days, they are assisted by C andwork is finished in 20 days more. If C does asmuch as work as what B does in 3 days, howmany days could A alone do the same work?
- 49. SolutionA and B’s one day work = 1/40 Day Work 1 1/40 10 xx = 1/40*10 = ¼Remaining work = 1- ¼ = ¾A+B+C took 20 days to complete ¾ of the work. Days Work 20 ¾ 1 xx = ¾ * 1/20 = 3/80
- 50. SolutionC’s one day work = 1 day (A+B+C) - 1 day (A+B) = 3/80 – 1/40 = 1/80C takes 80 days to complete the work C B Effi 1 3 Days 80 240B takes 240 days to complete the workA’s 1 day work = A + B – B = 1/40 – 1/240 = 6 – 1 /240= 5/40 = 1/48A will take 48 days to complete the work.
- 51. Problem - 7To complete a work A takes 50% more time thanB. If together they take 18 days to complete thework, how much time should B take to do it?
- 52. SolutionA + B = 18 daysB takes = x daysA takes = x +x/2 days = 3x/21/x + 2/3x = 18 days5/3x = 1/183x = 18*5x = 18*5 /3 = 30 days
- 53. Problem - 8A and B can complete a piece of work in 12 and18 days respectively. If A begins to do the workand they work alternatively one at a time for oneday each, then in how many days will the wholework be completed?
- 54. Solution1 day’s work of A = 1/121 day’s work of B = 1/182 days’ work of A and B = 1/12 + 1/18 = 5/367 times repeating 2 days work7*2 = 5*7/36Amount of work completed after 14 days = 35/36Remaining work = 1 – 35/36 = 1/3615th day A is working= 1/36 *12 = 12/36 = 1/3 days14 1/3 days to complete the work.
- 55. Problem - 912 men can complete a piece of work in 36 days.18 women can complete the same piece of workin 60 days. 8 men and 20 women work togetherfor 20 days. If only women were to complete theremaining piece of work in 4 days, how manywomen would be required?
- 56. Solution12 men take 36 days and 18 women take 60 days to complete a work.Ratio of efficiency = 12*36 = 18*60, 432 = 1080, 2 : 52 men = 5 women, 8 men = 20 womenWomen Days 18 60 40 x40*x = 18*60 = x = 18*60/4 = 27 daysThey worked for only 20 days, 1 day’s work = 1/27, so 20 days’ work = 20/27Remaining work = 7/27Women Days Work 40 27 1 x 7 7/274x = 40*27*7/27 = 40*7/4 = 70 Women
- 57. Problem - 10A, B and C are employed to do a piece of workfor Rs.529. A and B together are supposed to do19/23 of the work and B and C together 8/23 ofthe work. What is the share of A?
- 58. SolutionA + B = 19/23; remaining work by C = 1 – 19/23= 4/23B + C = 8/23, B = 8/23 – C = 8/23 – 4/23 = 4/23A = A+B – B = 19/23 – B = 19/23 – 4/23= 15/23A : B : C = 15:4:4A’s share = 529*15/23 = Rs. 345
- 59. Problem - 1112 children take 16 days to complete a workwhich can be completed by 8 adults in 12 days.16 adults started working and after 3 days 10adults left and 4 children joined them. Howmany days will they take to complete theremaining work?
- 60. Solution12 children take 16 days, 8 Adults take 12 days to complete work16 adults would complete the work in 6 days but they worked for 3 days, sothey completed ½ work. Remaining ½ work is yet to be completed.12 children’s 1 day work = 12*16 = 1928 adults’ 1 day work = 12 * 8 = 96192 : 96 = 2 : 12 Children = 1 Adult4 children are added instead of 10 adults4 children = 2 Adults2A +6A = 8 adults are going to complete ½ work8 adults took 12 days to complete the whole work.They will require 6 days to complete the remaining half work.
- 61. Problem - 12Anu can complete a work in 10 days. Manu is25% more efficient than Anu and Sonu is 60%more efficient than Manu. Working together,how long would they take to finish the job?
- 62. SolutionAnu takes 10 days to complete work;Manu takes 10*100/160 = 8 days tocomplete work;Sonu takes 8*100/160 = 5 days tocomplete work.1/10 + 1/8 + 1/5 = 4 /40 + 5/40+8/40 =17/40Number of Days = 40/17 = 2 6/17
- 63. Pipes and Cisterns
- 64. Pipes and CisternPipes and Cistern and Time and Work problems and method of solving are same.Only difference, instead of Man and Days, in pipes and cistern, they will give pipes / taps and cistern / tank.
- 65. Pipes and Cisterns• P1 and P2 both working simultaneously which fills in x hrs and empties in y hrs resp ( y>x) then net part filled is 1/x – 1/y• P1 can fill a tank in X hours and P2 can empty the full tank in y hours( where x>y), then on opening both pipes, the net part empties in hour 1/y -1/x
- 66. Problem 1Taps A and B fills a bucket in 12 and 15 minutesrespectively. If both are opened and A is closedafter 3 minutes, how much further time would ittake for B to fill the bucket?
- 67. Solution1 minute = A + B = 1/12 + 1/15 = 9 /603 minutes = 3*9/60 = 9 /20Remaining Bucket = 1 – 9/20 = 11/20Pipe B time Bucket 15 1 x 11/20x = 11/20 *15 = 33/4Time required to fill remaining bucket by Balone = 8 ¼ minutes.
- 68. Problem 2Three pipes A, B and C working together can fill acistern in 6 hours. After working together for 2 hours Cis closed and A & B fill it in 7 hours more. How manyhours will C alone take to fill the cistern?
- 69. SolutionA + B = 7 hoursA + B + C = 6 hoursHours Tank 6 1 2 x6x = 2x = 1/3 tank is full, remaining tank = 1 -1/3 = 2/32/3 *1/7 = 2/21A+B+C – A+B = 1/6 – 2/21 = 21 – 12 /126 = 9/126 = 1/14Time taken by C alone to fill the cistern = 14 Hours
- 70. Problem - 3A swimming pool is filled with three pipes withuniform flow. The first two pipes operatingsimultaneously fill the pool in the same timeduring which the pool is filled by the third pipealone. The second pipe fills the pool five hoursfaster than first pipe and 4 hours slower than thethird pipe. What is the time required by eachpipe to fill the pool separately?
- 71. A+B=C SolutionLet the first pipe be XX+5=X=X–4A + B = C ------ 1/(x + 5) + 1/x = 1/(x – 4)1/(x+5) + 1/x = x/x(X+5) + (x+5)/x(x+5)= 2x +5/x(x+5)Time taken = x(x+5) /2x+5x(x+5)/2x+5 = (x-4)x2 – 8x – 20 = 0(x+10) (x – 2)x = 10, -2x = 10A = 10 +5 = 15 hoursB = 10 HoursC = 10 – 4 = 6 hours
- 72. Problem - 4A ship 55 km from the shore springs a leakadmits 2 tons of water in 6 minutes. The ship canmanage 80 tons of water from sinking. Thepumps can throw out 12 tons an hour. Find theaverage rate of sailing such that the ship mayreach the shore just to avoid sinking.
- 73. SolutionRate of admission of water = 2/6 = 1/3Rate of pumping out = 12/60 = 1/5Rate of accumulation of water = 1/3 – 1/5 = 2/15Time to accumulate 80 tons of water = 80/2/15= 80*15/2 = 600 min = 10 hoursAverage rate of sailing = 55/10 = 5.5 km /hr
- 74. Problem - 5Two pipes A and B can fill a tank in 12 and 16minutes respectively. Both pipes are openedtogether, but 4 minutes before the tank is full,one pipe A is closed. How much time will theytake to fill the tank?
- 75. SolutionB’s 4 minutes’ work = 1/16*4 = ¼Remaining tank = 1 – ¼ = ¾A + B = 1/12 + 1/16 = 48/7 to fill 1 tankTank Time 1 48/7 ¾ xx = 48*4/7*3 = 64/7 = 9 1/7 minutes
- 76. Problem - 6An electric pump can fill a tank in 3 hours,because of leak in the tank it took 3½ hours tofill the tank. If the tank is full, how much timewill the tank take to empty it?
- 77. Solution1/3 – 1/x = x – 3 / 3xTime Taken = 3x /x – 33x / x – 3 = 7/26x = 7x – 21x = 21 hours
- 78. Problem - 7Two pipes A and B can fill a tank in 24 minutesand 32 min respectively. If both pipes are openedsimultaneously after how much time, B shouldbe closed so that the tank is full in 18 minutes?
- 79. SolutionPart filled by A + B in x min + part filled by A in 18-xmin = 1x/24 + x/32 + (18-x)1/24 =17x/96 + (18-x)/24 = 17x – 4x – 72/96 = 17x – 4x – 72 = 963x = 96 -72x = 8 min
- 80. Problem - 8Two pips can fill a cistern in 14 hours and 16hours respectively. The pipes are openedsimultaneously and it is found that due toleakage in the bottom, 32 minutes extra are takenfor the cistern to be filled up. If the cistern is full,in what time will the leak empty it?
- 81. SolutionA + B = 1/14 + 1/0 = 15/112 = 112 /15 = 7 hours 28 minutes32 min extra due to leakage, so total time takento fill the tank = 7.28 + 32 min = 8 hours15/112 – 1/8 = 15/112 – 14/112 = 1/112 = 112 hours
- 82. Problem - 9Two pipes A and B fill a cistern in 10 minutesand 15 minutes respectively, and a tap C canempty the full cistern in 20 minutes. All the threewere opened for 1 minute and then the emptyingtap was closed. How many minutes more wouldit take for the cistern to be filled?
- 83. Solution1 minute work of A+B –C = 1/10 + 1/15 – 1/20 = 7/60Remaining tank = 1 – 7/60 = 53/60A + B = 1/10 + 1/15 = 6 minA+B take 6 minutes to fill the tankTime Cistern6 1x 53/60x = 53/60*6 = 5 3/10
- 84. Problem - 10A pipe can fill a bath in 20 minutes and anothercan fill it in 30 minutes. A person opens both thepipes alternatively for 1 minute each. When thebath was full, he finds that the waste pipe wasopen. he then closes the waste pipe and in 3more minutes the bath is full. In what time wouldthe waste pipe empty it?
- 85. Solution1st min A = 1/202nd min B = 1/30 2 minutes A + B = 1/20 + 1/30 = 5/60 = 1/122*12 = 12/12 so 24 minutes 1 bath will be filled2 minutes = 1/123rd minutes = 1/20 = 1/12 + 1/20 = 5+3/60 = 8/60 = 2/152/15 of bath in 24 minutesEmpty the full bath in = 24*15/2 = 180 = 3 hours
- 86. Problem - 11A tap can fill a tank in 6 hours. After half thetank is filled, three more similar taps are opened.What is the total time taken to fill the tankcompletely?
- 87. Solution1 tank 6 hours: half tank = 3 hours4 taps = 1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3To fill one tank = 3/2 = 1 ½ hoursTo fill half tank 45 minutesTotal time taken to fill the tank = 3 hours 45 minutes
- 88. Problem - 12A tank 10m by 6m by 3m has an inlet pumpwhich pours water at the rate of ¼ cubic metresin a minute. An emptying pipe can empty thetank when it is full in 4 hours. If the tank is fulland both pipes are open, how long will it take toempty it?
- 89. SolutionCapacity of the tank = 10*6*3 = 180 cu.mThe emptying pipe can remove 180 cu.m in 4 hoursIn 1 minute it can remove 180/4*60 cu.m (i.e) 3/4 cu.mThe inner pipe pours ¼ cu.m in 1 minute¾ - ¼ = 2/4 = ½ cu.m of water will be removed in 1 min.Empty the tank = 10*6*3/ ½ = 360 minutes = 6 hours
- 90. Problems on Trains
- 91. Problems on TrainsImportant Formula:1. Distance = Speed * Time2. Speed = Distance / Time3. Time = Distance / SpeedTime taken to Cross Each Other:1. Opposite Direction (Moving Object) = S1 + S22. Same Direction (Moving Object) = S1 – S2 Cont…
- 92. Problems on Trains3. Opposite Direction (both object with Length) = L1 + L2 / S1 + S24. Same Direction (both object with Length) = L1 + L2 / S1- S25. Both object with Length / One non moving Object = L1 + L2 / S16. One object with Length / both object are moving (opp.dire) = L1 / S1 + S27. One object with Length / both object are moving (same.dire) = L1 / S1- S2
- 93. Problems on Trains• Speed from km/hr to m/sec - ( * 5/18).• Speed from m/sec to km/h, - ( * 18/5).• Average Speed:- Average speed = Total distance traveled Total time taken
- 94. Problem - 1Two trains start from stations A and B andtravels towards each other at speeds of 50 km/hrand 60 km/hr respectively. At the time of theirmeeting the second train has traveled 120 kmmore than the first train. What is the distancebetween Stations A and B?
- 95. SolutionSpeed difference in 1 hr = 10 km120 km difference will come after 12 hrs oftraveling. So both the trains have traveled 12hours each.Distance covered in 12 hrs,Train 1 = 12*50 = 600Train 2 = 12*60 = 720Distance between stations A and B is = 600 +720 = 1320
- 96. Problem - 2Trains are running with speeds of 30 km/hr and58 km/hr in the same direction. A man in theslower train passes by the faster train in 18seconds. What is the length of the faster train inmeters?
- 97. SolutionSpeed of the train (same direction) = 58 – 30 = 28 Km/hr = 28*5/18 = 70/9 m/secDistance (Length of the train) = S * time = 70/9*18 = 140 metres
- 98. Problem - 3A train passes two bridges of lengths 800m and400m in 100 sec and 60 sec respectively. What isthe length of the train?
- 99. SolutionLet the length of the train be x.Speed = Distance / Time(x + 800)/100 = x+400/60x = 200 m
- 100. Problem - 4A train does a journey without stoppage in 8hours. If it had traveled 5 km an hour faster, itwould have done the journey in 6 hours 40 min.What is its slower speed?
- 101. SolutionLet the Speed be xSpeed Time x 8 x+5 6 hr 40 minSp Time x 480 x+5 400480x = 400(x+5)480x – 400x = 200080x = 2000x = 25Slower Speed = 25
- 102. Problem - 5Without stoppage a train travels with an averagespeed of 80 km/hr and with stoppages, it coversthe same distance with an average speed of 60km/hr. How many minutes per hour the trainstops?
- 103. SolutionStoppage time / hour = = Difference in speed / Faster Speed = 80 – 60/80 = 20/80 =1/4 = 15 minutes
- 104. Problem - 6Two trains of lengths 190m and 210mrespectively start from a station in oppositedirections on parallel tracks. If their speeds are40 km/hr and 32 km/hr respectively, in whattime will they cross each other?
- 105. SolutionTime = Distance / SpeedOpp. Direction = S1 + S2 = 190 +210/40+32 = 400/72*5/18 = 400/20 = 20 Sec
- 106. Problem - 7A train 100 m long takes 6 sec to cross a manwalking at 5 km/hr in a direction opposite to thatof the train. Find the speed of the train.
- 107. SolutionSpeed = Distance / Time 100/(x+5)*5/18= 6 = 30 (x+5) = 1800X + 5 = 1800/30 = 60x = 60 – 5 = 55 km/hr
- 108. Problem - 8Two trains running in opposite directions cross aman standing on the platform in 27 sec and 17sec respectively and they cross each other in 23seconds. What is the ratio of their speeds?
- 109. SolutionSp. of Slower train Sp. of Faster train 17 27 23(cross Each other) 27-23 = 4 23-17 = 6 2 : 3Ratio of Faster and Slower train = 3:2
- 110. Problem - 9Two trains of equal lengths takes 10 seconds and15 sec respectively to cross a telegraphic post. Ifthe length of each train be 120m, in what timewill they cross each other traveling in oppositedirection?
- 111. SolutionS1 = 120/10 = 12S2 = 120/15 = 8Opposite Direction = S1 + S2 = 12 +8 = 20Time taken to cross each other = D/S120 + 120/20 = 12 Sec
- 112. Problem - 10A train consists of 12 bogies, each bogie is 15metres long. The train crosses the telegraphicpost in 18 sec. Due to some problem, two bogieswere detached. In what time will the train crossthe Telegraph post now?
- 113. Solution12 bogies each 15 m longTotal length = 12*15 = 18012 bogies 180 m long10 bogies 150 m longLength Time180 18150 x180x = 18*150x = 18*150/180x = 15 sec
- 114. Problem - 11A train 300 m long overtook a man walkingalong the line in the same direction at the rate 4km/hr and passed him in 30 sec. The trainreached the station in 15 min after it had passedthe man. In what time will the man reach thestation?
- 115. SolutionL = 300 mSpeed of the man = 4 km/hrTime took to pass = 30 sec15 min to reach station after crossing a manRelative Speed in same direction = x – 4*5/18m/sec = 300/(x-4)*5/18 = 30 = x – 4 = 180/5, x – 4 = 36, x = 40 km/hr
- 116. SolutionSpeed of the Train = 40 km/hourDistance Time 40 60 x 1560x = 40*15; x = 10 km = Distance to reach the StationSpeed Time40 km/hr 154 km/hr X = 4x = 40*15x = 40*15/4 = 150 min = 2 ½ hrs
- 117. Problem - 12A train starts from station X at the rate of 80km/hr and reaches station Y in 48 minutes. If thespeed is reduced by 8 km/hr, how much moretime will the train take to return from station Yto station X?
- 118. SolutionSpeed Time 80 48 72 x72x = 80*48 = x = 80*48/72 = 53 1/348 – 53 1/3 = 5 1/3 = 5 min 20 sec
- 119. Time and Distance
- 120. Time and DistanceSpeed:- Distance covered per unit time is called speed.Speed = distance/time (or)Distance = speed*time (or)Time = distance/speed
- 121. Problem - 1In 1760m race, A can beat B by 44m while in a1320m race, B can beat C by 30m. By whatdistance will A beat C in a 880m race?
- 122. SolutionIn 1760 m race, A = 1760m B = 1760 – 44 = 1716mIn 1320 m race, B = 1320, C = 1290m B C 1716 x 1320 1290C = 1716*1290/1320 = 1677mA = 1760, B = 1716, C = 1677In 1760 m race, A beats C by 83 mIn 880 m race, A will beat C by 41 ½ m
- 123. Problem - 2Two trains traveling in opposite direction startfrom a station at the same time with the speedsof 40 km/hr and 120 km/hr respectively. What isthe distance between them after 15 minutes?
- 124. SolutionIn 1 hour Both trains covers A + B = 40 + 120 =160 kmDistance covered in 15 minutes = 160*15/60 =16*15/6 = 40 km
- 125. Problem - 3A hare preceded by a greyhound is 87.5m beforehim. While the hare takes 4 leaps, the greyhoundtakes 3 leaps. If the greyhound and the hare go2¾ m and 1¾ m respectively in one leap, in howmany leaps will the greyhound overtake thehare?
- 126. SolutionDis. Covered by Hare = 4* 1¾ = 4*7/4 = 7Dis. Covered by Greyhound = 3*11/4 = 33 / 4 = 8.25Dis. gained by greyhound in 3 leaps= 8.25 – 7 = 1.25 m Dis. Leaps 1.25 3 87.5 xx = 87.5 * 3 / 1.25 = 210 leaps
- 127. Problem - 4A monkey is climbing up a greased pole ascends5m and slips down 2m in alternate minute. If thepole is 35 m high then when will the monkeytouch the top?
- 128. Solution1st min = 5m up2nd min = 2m down = 2 minutes = 3m up2*11 = 3*11 m22 min = 33 mRemaining = 35 -33 = 2mMeter Min5m 60 sec2m xx = 2*60/5 = 24 secTotal time taken = 11 minutes 24 seconds
- 129. Problem - 5Walking 6/7 of his usual rate a man is 25minutes late. Find his usual time?
- 130. SolutionUsual time = Nr * Late= 6*25 = 150 Min = 2 hr 30 Min
- 131. Problem - 6A hare runs at 5 m/sec and a tortoise runs at 0.5m/sec. They start from the same point in thesame direction. How far ahead will the hare beafter 30 min?
- 132. Solution30*60 = 1800 SecPer sec difference = 5 – 0.5 = 4.5 m / sec30 min = 4.5*30*60 / 1000 = 8.1 km
- 133. Problem - 7Two cars A and B are running towards eachother from two different places 88 km apart. Ifthe ratio of the speeds of the cars A and B is 5 : 6and the speed of the car B is 90 km / hr. At whattime will the two meet each other?
- 134. SolutionRatio A : B=5 : 6A : B = x : 90x*6/11 = 90 x = 165(Total Speed ) 165 – (B’s Speed) 90 = 75Speed of car A = 75Time to meet each other = 88/75+90 = 88/165hrs88/165*60 = 32 min
- 135. Problem - 8A train travelling at the rate of 60 km/hr, whileinside a tunnel meets another train of half itslength traveling at 90 km/hr and passescompletely in 4 ½ sec. Find the length of thetunnel if the first train passes completely throughit in 4 minutes 37 ½ sec?
- 136. SolutionRelative Speed = 60 + 90 = 150Dis. = 150*5/18*9/2 = 375/2 mLength of the 1st train is = xLength of the 2nd train is = x/2X+ x/2 = 375/23x = 375X = 375/3 = 125Time taken by the first train to cross the tunnel, Cont….
- 137. SolutionSpeed = 60*5/18 = 50/3 m /secTime = 4 min 75/2 sec = 4*60 +75/2 = 555/2Distance traveled by 1st train is,Sec Metre 1 50/3 555/2 x50/3*555/2 = 4625Length of the Tunnel = Total Dis. – Length of the Train = 4625 – 125 = 4500m or 4.5 km
- 138. Problem - 9A man standing on a 170 m long platformwatches that a train takes 7 ½ sec to pass himand 21 seconds to cross the platform. Find theLength of the train and its Speed?
- 139. SolutionSp. to cross a man = x/15/2 = 2x/15Sp. To cross a platform = 170+x/21170 +x/21 = 2x/152550 + 15x = 42xx = 850/9 mSpeed = 2* 850/15*9 = 12 16/27 m/sec
- 140. Problem - 10A and B starts at the same time with speeds of 40km/hr and 50 km/hr respectively. If in coveringthe journey, A takes 15 minutes longer than B.Find the Total distance of the Journey?
- 141. SolutionA (1 hour) covers = 40 kmB (I hour ) covers = 50 kmA took 15 minutes longer than BDistance cover in 15 minutes = 40*1/4 = 10 kmTotal Distance = 50 km
- 142. Problem - 11A train 110 m in length passes a man at the rateof 6 km/hr against it in 6 sec. How much time itwill take to pass another man walking at thesame speed in same direction?
- 143. SolutionOpp. Dir = x + 6 km /hr x+6 *5/18*6 = 110 m x = 60Time taken to pass 2nd man = 110/60-6 = 7 1/3 sec
- 144. Problem - 12A train running at the speed of 20 m/sec crossesa pole in 24 seconds less than the time it requiredto cross a platform thrice its length at the samespeed. What is the length of the train?
- 145. SolutionDistance = Speed * TimeLength of the train be xTime = x/20 secTo cross the platform 3x+x/20 = 4x/20To cross the Pole = x/20Diff. in time to cross the platform and pole, 4x/20 – x/20 = 24 3x/20 = 24 3x = 20*24; x = 160 m
- 146. Boats and Streams
- 147. Boats and Streams• Up stream – against the stream• Down stream – along the stream• u = speed of the boat in still water• v = speed of stream• Down stream speed (a)= u+v km / hr• Up stream speed (b)=u-v km / hr• u = ½(a+b) km/hr• V = ½(a-b) km / hr
- 148. Problem - 1The speed of a boat in still water in 8 km/hr. If itcan travel 20 km downstream at the same time asit can travel 12 km upstream, what is the rate ofstream (in km/hr)?
- 149. SolutionRate of stream = x km/hrSp. Of boat in still water = 8 km/hrUpstream speed = 8 – x km/hrDownstream Speed = 8+x km/hrGiven, 20/8+x = 12/8-x= 160 – 20x = 96+12x32x = 64xX = 64/32 = 2 km/hr
- 150. Problem - 2A boat’s crew rowed down a stream from A to Band up again in 7 ½ hours. If the stream flows at3km/hr and speed of boat in still water is 5 km/hr.find the distance from A to B ?
- 151. SolutionSolution:Down Stream = Sp. of the boat + Sp. of the stream = 5 +3 =8Up Stream = Sp. of the boat – Sp. of the stream = 5-3 = 2Let distance be xDistance/Speed = Time X/8 + x/2 = 7 ½ X/8 +4x/8 = 15/2 5x / 8 = 15/2 5x = 15/2 * 8 x =12
- 152. Problem - 3A boat goes 40 km upstream in 8 hours and 36km downstream in 6 hours. Find the speed of theboat in still water in km/hr?
- 153. SolutionSolution: Speed of the boat in upstream = 40/8 = 5 km/hr Speed of the boat in downstream = 36/6 = 6km/hr Speed of the boat in still water = 5+6/2 = 5.5km/hr
- 154. Problem - 4A man rows to place 48km distant and back in14 hours. He finds that he can row 4 km with thestream in the same time as 3 km against thestream. Find the rate of the stream?
- 155. SolutionSolution:Down stream 4km in x hours. Then,Speed of Downstream = 4/x km/hr,Speed of Upstream = 3/x km/hr48/ (4/x) + 48/(3/x) = 14 , x = 1/2Speed of Down stream = 8,Speed of upstream = 6Rate of the stream = ½ (8-6) km/hr = 1 km/hr
- 156. Problem - 5In a stream running at 2 km/h a motor boat goes10 km upstream and back again to the startingpoint in 55 minutes. Find the speed of the motorboat in still water.
- 157. SolutionSolution:Let the speed be x km/h 10/(x+2) + 10/(x-2) = 55/60Solving the above equation, x = 22 km/h
- 158. Problem - 6A swimmer can swim a certain distance in thedirection of current in 5 hours and return thesame distance in 7 hours. If the stream flows atthe rate of 1 km/h, find the speed of the swimmerin still water.
- 159. SolutionSolution:Let the speed of the swimmer in still water be xDistance = speed x timeDistance covered at Downstream = (x+1) * 5Distance covered at Upstream = (x-1) * 7 5(x+1) = 7(x-1)Solving the equation, x = 6 Speed of the swimmer = 6 km/h
- 160. Problem - 7A person can row a boat D km upstream and thesame distance downstream in 5 hours 15minutes. Also, he can row the boat 2D kmupstream in 7 hours. How long will it take to rowthe same distance 2D km in downstream?
- 161. SolutionSpeed of boat and stream be x and y km/hr respectively.Rate of Downstream = (x+y) km/hrRate of Upstream = (x – y) km/hrGiven,D / x+y + D / x-y = 21/4 hrs ---------1And, 2D/x-y = 7; D / x-y = 7/2------2Subtracting equation 2 from 1D/x+y = 21/4 – 7/2; D/x+y = 7/42D/x+y = 7/2 hrs
- 162. Problem - 8A boat takes 19 hours for traveling downstreamfrom point A to point B and coming back to apoint C, midway between A and B. If thevelocity of the stream is 4 km/hr and the speed ofthe boat in still water is 14 km/hr. What is thedistance between A and B?
- 163. SolutionSpeed of Downstream = 14 + 4 = 18 km/hrSpeed of Upstream = 14 – 4 = 10 km/hrx/18 + x/2/10 = 19x/18 + x/20 = 1910x+9x/180 = 1919*180/19 = 180 km
- 164. Problem - 9Speed of a boat in standing water is 9 km/hr andthe speed of the stream is 1.5 km/hr. A man rowto a distance of 105 km and comes back to thestarting point. Find the total time taken by him?
- 165. SolutionSp. of upstream = 7.5 km/hrSp. Of Downstream = 10.5 km/hrTotal time taken = 105/7.5 + 105/10.5 = 24 hrs
- 166. Problem - 10A man can row 40 km upstream and 55 kmdownstream in 13 hours. Also, he can row 30 kmupstream and 44 km downstream in 10 hours.Find the speed of the man in still water and thespeed of the current?
- 167. SolutionLet the rate of upstream = x km/hr andrate of downstream = y km/hr40/x + 55/y = 13-------130/x + 44/y = 10-------21*3 = 120/x + 165/y = 392*4 = 120/x + 176/y = 40Subtracting 1 and 2, we get y = 1140/x + 55/y = 1340/x + 55/11 = 13; x = 5 km/hrRate in still water = ½(11+5) = 8 km/hrRate of Current = ½(11 – 5) = 3 km/hr
- 168. Problem - 11A man can row 7 ½ km/hr in still water. If in ariver running at 1.5 km an hour, it takes him 50minutes to row to a place and back, how far offis the place?
- 169. SolutionSp. of Downstream = 7.5 + 1.5 = 9 km/hrSp. of Upstream = 7.5 – 1.5 = 6 km/hrLet required distance = x kmx/9 + x/6 = 50/60 ; 5x/18 = 5/6;x = 5/6*18/5 = 3 km
- 170. Problem - 12The speed of a boat in still water is 15 km/hr andthe rate of current is 3 km/hr. Find the distancetraveled downstream in 15 min?
- 171. SolutionDistance = Speed * Time = 15 + 13 *15/60 = 28*1/4 = 7 km
- 172. Simple Interest and Compound Interest
- 173. Simple / Compound InterestSimple Interest = PNR / 100Amount A = P + PNR / 100When Interest is Compound annually:Amount = P (1 + R / 100)nC.I = A-P
- 174. Simple / Compound Interest• Half-yearly C.I.: Amount = P (1+(R/2)/100)2n• Quarterly C.I. : Amount = P (1+(R/4)/100)4n
- 175. Simple /Compound InterestDifference between C.I and S.I for 2 years = P*(R/100)2.Difference between C.I and S.I for 3 years = P{(R/100)3+ 3(R/100)2 }
- 176. Problem - 1A sum of money amounts to Rs. 6690 after 3years and to Rs. 10035 after 6 years on C.I. Findthe sum.
- 177. SolutionSolution:P(1 + R/100)3 = 6690 -----------1P(1 + R/100)6 = 10035 -----------2Dividing (1 + R/100)3 = 10035/6690 = 3/2Substitute in equation 1 then P(3/2)=6690 P = 6690 * 2/3 = 4460 The sum is Rs.4460
- 178. Problem - 2What will be the difference between S.I and C.Ion a sum of Rs. 4500 put for 2 years at 5% perannum?
- 179. SolutionC.I – S.I = P (R/100)2Difference = Rs. 11.25
- 180. Problem - 3The simple interest on a certain sum is 16 / 25of the sum. Find the rate percent and time ifboth are numerically equal.
- 181. SolutionR = N and I = 16/25 PR = I * 100 / P*NR*N = 64 where R = NR=8 The rate of interest is 8%
- 182. Problem - 4The difference between the compound andsimple interest on a certain sum for 2 years at therate of 8% per annum is Rs.80,What is the sum?
- 183. SolutionC.I – S.I = P(R/100)2 80 = P*(8*8/100*100) The sum is Rs.12,500
- 184. Problem - 5If a sum of money compounded annuallyamounts of thrice itself in 3 years, in how manyyears will it become 9 times itself?
- 185. Solution A = 3P and find n when A = 9PThe number of years required = 6 years
- 186. Problem - 6What will be the C.I on Rs. 15625 for 2½ yearsat 4% per annum?
- 187. SolutionA = P(1+R/100)nA = 15625 ((1+4/100)2 (1+4*1/2/100)) = 17238CI = A –P = 17238 - 15625 Compound interest = Rs. 1613
- 188. Problem - 7What is the S.I. on Rs. 3000 at 18% perannum for the period from 4th Feb 1995 to18th April 1995
- 189. SolutionTime = 24+ 31+17 = 73 days = 73/365 = 1/5P = 3000; R = 18%;= PNR/100 = 3000*1*18/100*5The simple interest is Rs. 108
- 190. Problem - 8Raja borrowed a certain money at a certainrate of S.I. After 5 years, he had to pay backtwice the amount that he had borrowed. Whatwas the rate of interest?
- 191. SolutionA = 2PA = P + PNR/1002P = P(1+NR/100)2 = (1+5*R/100)1 = R/20The rate of interest is 20%
- 192. Problem - 9In simple interest what sum amounts to Rs. 1120in 4 years and Rs. 1200 in 5 years? CTS Question
- 193. SolutionInterest for 1 year = 1200 – 1120 = 80Interest for 4 year = 80*4 = 320A = 1120P = A – P = 1120 – 320The Principal is Rs. 800
- 194. Problem - 10A simple interest amount for Rs. 5000 for 6months is Rs. 200. What is the annual rate ofinterest? CTS Question
- 195. SolutionP = 5000; N = 6/12 = ½I = 200R = I *100 / P*N =200*100*2/5000*1 = 40/5 = 8% The annual rate of interest is 8%
- 196. Problem - 11A man earns Rs. 450 as an interest in 2 years ona certain sum invested with a company at the rateof 12% per annum. Find the sum invested.
- 197. SolutionP = I*100/R*N = 450*100/12*2Principal = Rs. 1875
- 198. Problem - 12If Rs. 85 amounts to Rs. 95 in 3 years, whatRs. 102 will amount in 5 years at the same ratepercent?
- 199. SolutionLet P = Rs. 85; A = Rs. 95; I = 10/3 in 1 yearRate = I*100/P*N = 4% ( app)Amount = P+PNR/100 = 102+20 =122Hence the amount in 5 years = Rs. 122

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