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Download Time and Distance Questions for SSC CHSL and MTS<\/a><\/p>\n Enroll to 15 SSC CHSL 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>A boat takes 2 hours to travel from point A to B in still water .To find out it\u2019s speed up-stream ,which of the following information is needed. a)\u00a0All are required<\/p>\n b)\u00a0Even these we cannot found the answer<\/p>\n c)\u00a0Only i,iii, and either ii or iv<\/p>\n d)\u00a0Only i and iii<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Time taken by boat\u00a0to travel from point A to B in still water\u00a0= 2 hours<\/p>\n To find the upstream speed, we definitely need the speed of stream, thus statement (iii) is mandatory.<\/p>\n Also, the distance between points A and B or the speed of boat in still water is needed.<\/p>\n Thus, statements (i) and (iii) are required to find the upstream speed of the boat.<\/p>\n => Ans – (D)<\/p>\n Question 2:\u00a0<\/b>A bus covers a distance of 2,924 km,in 43 hours .what is the bus speed?<\/p>\n a)\u00a072 km\/hr<\/p>\n b)\u00a060 km\/hr<\/p>\n c)\u00a068 km\/hr<\/p>\n d)\u00a0cannot determined<\/p>\n e)\u00a0none of these<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of bus = $s$ km\/hr<\/p>\n Distance covered = 2924 km<\/p>\n Time taken = 43 hours<\/p>\n Using speed = distance\/time<\/p>\n => $s=\\frac{2924}{43}=68$ km\/hr<\/p>\n => Ans – (C)<\/p>\n Question 3:\u00a0<\/b>A man takes 2.2 times as long to row a distance upstream as to row the same distance downstream. If he can row 55 km downstream in 2 hours 30 minutes, what is the speed of the boat in still water? (in km\/h)<\/p>\n a)\u00a040<\/p>\n b)\u00a08<\/p>\n c)\u00a016<\/p>\n d)\u00a024<\/p>\n e)\u00a032<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of boat in still water = $x$ km\/hr<\/p>\n => Speed of current = $y$ km\/hr<\/p>\n Let distance travelled = $d$ km<\/p>\n Acc. to ques, => $2.2 (\\frac{d}{x + y}) = \\frac{d}{x – y}$<\/p>\n => $2.2x – 2.2y = x + y$<\/p>\n => $2.2x – x = y + 2.2y$<\/p>\n => $3x = 8y$ ————-(i)<\/p>\n Also, the man takes 2 hrs 30 mins in travelling 55 km downstream.<\/p>\n => $\\frac{55}{x + y} = 2 + \\frac{1}{2}$<\/p>\n => $\\frac{55}{x + y} = \\frac{5}{2}$<\/p>\n => $x + y = 22$<\/p>\n Multiplying both sides by 8, and using eqn(i), we get\u00a0:<\/p>\n => $8x + 3x = 22 \\times 8$<\/p>\n => $x = \\frac{22 \\times 8}{11} = 16$ km\/hr<\/p>\n Question 4:\u00a0<\/b>A boat takes a total time of twelve hours to travel 105 kms upstream and the same distance downstream. The speed of the boat in still water is six times of the speed of the current. What is the speed of the boat in still water? (in km\/hr)<\/p>\n a)\u00a012<\/p>\n b)\u00a030<\/p>\n c)\u00a018<\/p>\n d)\u00a024<\/p>\n e)\u00a036<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of current = $x$ km\/hr<\/p>\n => Speed of boat in still water = $6x$ km\/hr<\/p>\n Acc. to ques, => $\\frac{105}{7x} + \\frac{105}{5x} = 12$<\/p>\n => $\\frac{15}{x} + \\frac{21}{x} = 12$<\/p>\n => $\\frac{36}{x} = 12$<\/p>\n => $x = \\frac{36}{12} = 3$<\/p>\n $\\therefore$ Speed of boat in still water = $6 \\times 3 = 18$ km\/hr<\/p>\n Question 5:\u00a0<\/b>A boat takes a total time of eight hours to travel 63 kms upstream and the same distance downstream. The speed of the current is ${1 \\over 8}$th of the speed of the boat in still water. What is the speed of the boat in still water? (in km\/hr)<\/p>\n a)\u00a032<\/p>\n b)\u00a024<\/p>\n c)\u00a016<\/p>\n d)\u00a08<\/p>\n e)\u00a038<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of current = $x$ km\/hr<\/p>\n => Speed of boat in still water = $8x$ km\/hr<\/p>\n Acc. to ques, => $\\frac{63}{9x} + \\frac{63}{7x} = 8$<\/p>\n => $\\frac{7}{x} + \\frac{9}{x} = 8$<\/p>\n => $\\frac{16}{x} = 8$<\/p>\n => $x = \\frac{16}{8} = 2$<\/p>\n $\\therefore$ Speed of boat in still water = $8 \\times 2 = 16$ km\/hr<\/p>\n Take a free SSC CHSL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>A boat takes six hours to travel a certain distance downstream and five hours to travel a certain distance upstream. The distance travelled upstream is half of the travelled downstream. If the speed of the current is 4 km\/hr, what is the speed of the boat in still water? (in km\/hr)<\/p>\n a)\u00a016<\/p>\n b)\u00a020<\/p>\n c)\u00a024<\/p>\n d)\u00a010<\/p>\n e)\u00a018<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of boat in still water = $x$ km\/hr<\/p>\n Let distance travelled downstream = $2d$ km<\/p>\n => Distance travelled upstream = $d$ km<\/p>\n Using, $time = \\frac{distance}{speed}$<\/p>\n => $6 = \\frac{2d}{x + 4}$ ———(i)<\/p>\n and $5 = \\frac{d}{x – 4}$ ———(ii)<\/p>\n Dividing eqn(i) from (ii), we get :<\/p>\n => $\\frac{6}{5} = \\frac{\\frac{2d}{x + 4}}{\\frac{d}{x – 4}}$<\/p>\n => $\\frac{6}{5} = \\frac{2 (x – 4)}{x + 4}$<\/p>\n => $6x + 24 = 10x – 40$<\/p>\n => $10x – 6x = 24 + 40 = 64$<\/p>\n => $x = \\frac{64}{4} = 16$ km\/hr<\/p>\n Question 7:\u00a0<\/b>A bus covers a distance of 2,924 kms. in 43 hours. What is the speed of the bus?<\/p>\n a)\u00a072 kmph<\/p>\n b)\u00a060 kmph<\/p>\n c)\u00a068 kmph<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of bus = $\\frac{Distance}{Time}$<\/p>\n = $\\frac{2924}{43}$<\/p>\n = 68 kmph<\/p>\n Question 8:\u00a0<\/b>A car runs at the speed of 40 when not serviced and runs at 65 kmph when serviced. After servicing the car covers a certain distance in 5 hours. How much approximate time will the car take to cover the same distance when not serviced ?<\/p>\n a)\u00a010<\/p>\n b)\u00a07<\/p>\n c)\u00a012<\/p>\n d)\u00a08<\/p>\n e)\u00a06<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n After servicing, the distance covered by car in 5 hours = 65 * 5 = 325 km<\/p>\n Without servicing, speed of car = 40 kmph<\/p>\n => Required time = $\\frac{Distance}{Speed}$<\/p>\n = $\\frac{325}{40}$ = 8.125 hr<\/p>\n $\\approx$ 8 hours<\/p>\n Question 9:\u00a0<\/b>A car runs at the speed of 50 kms per hour when not serviced and runs at 60 km.\/hr. when serviced. After servicing the car covers a certain distance in 6 hours. How much time will the car take to cover the same distance when not serviced?<\/p>\n a)\u00a08.2 hours<\/p>\n b)\u00a06.5 hours<\/p>\n c)\u00a08 hours<\/p>\n d)\u00a07.2 hours<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n After servicing, the distance covered by car in 6 hours = 60 * 6 = 360 km<\/p>\n Without servicing, speed of car = 50 kmph<\/p>\n => Required time = $\\frac{Distance}{Speed}$<\/p>\n = $\\frac{360}{50}$ = 7.2 hr<\/p>\n Question 10:\u00a0<\/b>Yesterday Priti type an essay of 5000 words at the speed of 60 words per minute: Today she type the same essay faster and her speed was 15% more than yesterday. What is the approximate <\/strong>difference in the time she took to type yesterday and the time she took to type today?<\/p>\n a)\u00a020 minutes<\/p>\n b)\u00a030 minutes<\/p>\n c)\u00a010 minutes<\/p>\n d)\u00a040 minutes<\/p>\n e)\u00a01 hour<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Priti’s second day typing speed = $\\frac{115}{100} \\times 60$ = 69 wpm<\/p>\n Required difference in time<\/p>\n = $(\\frac{5000}{60} – \\frac{5000}{69})$<\/p>\n = $5000(\\frac{69 – 60}{60 \\times 69})$<\/p>\n $\\approx$ 10 minutes<\/p>\n Question 11:\u00a0<\/b>The average speed of a train is 3 times the average speed of a car. The car covers a distance of 520 kms in 8 hours. How much distance will the train cover in 13 hours?<\/p>\n a)\u00a02553 kms<\/p>\n b)\u00a02585 kms<\/p>\n c)\u00a02355 kms<\/p>\n d)\u00a02535 kms<\/p>\n e)\u00a0None of these<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of car = $\\frac{Distance}{Time}$<\/p>\n = $\\frac{520}{8} = 65$ kmph<\/p>\n => Speed of train = $65 \\times 3 = 195$ kmph<\/p>\n $\\therefore$ Distance covered by train in 13 hours<\/p>\n = $13 \\times 195 = 2535$ km<\/p>\n Question 12:\u00a0<\/b>If a man runs at 6 kmph from his house, he misses the train at the station by 8 minutes. If he runs at 10 kmph, he reaches the station 7 minutes earlier than the departure of the train. What is the distance of station from his house? (in km)<\/p>\n a)\u00a0$4\\frac{3}{4}$<\/p>\n b)\u00a0$3\\frac{1}{2}$<\/p>\n c)\u00a0$4\\frac{1}{4}$<\/p>\n d)\u00a0$3\\frac{3}{4}$<\/p>\n e)\u00a0$4\\frac{1}{2}$<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the distance of the station from the house of the person = $x$ km<\/p>\n Difference of time = 8 + 7 = 15 minutes = $\\frac{1}{4}$ hr<\/p>\n Since, Time =\u00a0$\\frac{Distance}{Speed}$<\/p>\n $\\therefore \\frac{x}{6} – \\frac{x}{10} = \\frac{1}{4}$<\/p>\n => $\\frac{10x – 6x}{60} = \\frac{1}{4}$<\/p>\n => $\\frac{2x}{30} = \\frac{1}{4}$<\/p>\n => $x = \\frac{15}{4} = 3\\frac{3}{4}$ km<\/p>\n Question 13:\u00a0<\/b>The distance between Shaurya\u2019s house and Pratyusha\u2019s house is 18 km. Shaurya\u2019s speed is 3\/7th of that Pratyusha. Shaurya takes one hour in going to Pratyusha\u2019s house. What is the speed of Pratyusha ?<\/p>\n a)\u00a018 kmph<\/p>\n b)\u00a024 kmph<\/p>\n c)\u00a030 kmph<\/p>\n d)\u00a032 kmph<\/p>\n e)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Shaurya’s speed = $\\frac{18}{1}$ = 18 kmph<\/p>\n => Speed of Pratyusha = $18 \\times \\frac{7}{3}$<\/p>\n = 42 kmph<\/p>\n Question 14:\u00a0<\/b>Two cars, A and B started moving from the same point at the same time towards opposite direction (A toward North and B towards South). If the speed of car A is 34.5 kmph and that of car B is 41.5 kmph. after how much time will they be 684 km apart? (in hours)<\/p>\n a)\u00a07<\/p>\n b)\u00a011<\/p>\n c)\u00a012<\/p>\n d)\u00a08<\/p>\n e)\u00a09<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\because$ The cars are moving in opposite direction<\/p>\n $\\therefore$ Relative speeds = 34.5 + 41.5 = 76 km\/hr<\/p>\n => Required time \u00a0= $\\frac{684}{76} = 9$ hrs<\/p>\n Question 15:\u00a0<\/b>A water tank has one inlet, A and one outlet, B. A takes 5 hours to fill the empty tank, when B is not open and B takes 8 hours to empty the full tank. If the tank is three fifth full, how much time will it take to fill the tank completely when both A and B are opened simultaneously ? (in hours)<\/p>\n a)\u00a0$6\\frac{1}{3}$<\/p>\n b)\u00a0$4\\frac{1}{3}$<\/p>\n c)\u00a0$8\\frac{1}{3}$<\/p>\n d)\u00a0$5\\frac{1}{3}$<\/p>\n e)\u00a0$7\\frac{1}{3}$<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Part of tank filled by A and B in 1 hour<\/p>\n = $\\frac{1}{5} – \\frac{1}{8} = \\frac{3}{40}$<\/p>\n => Time taken to fill the tank completely = $\\frac{40}{3}$ hrs<\/p>\n $\\therefore$ Time taken to fill two-fifth part of tank<\/p>\n = $\\frac{2}{5} \\times \\frac{40}{3}$<\/p>\n = $\\frac{16}{3} = 5\\frac{1}{3}$ hrs<\/p>\n Question 16:\u00a0<\/b>A boat travels from A to B upstream and then from B to C downstream taking the same time. The respective ratio between the distance from A to B and the distance from B to C is 5 : 7. If the boat takes 2 hours 30 min to travel a distance of 35 km downstream, what is the speed of the stream ? (in km\/h)<\/p>\n a)\u00a02 km\/h<\/p>\n b)\u00a01.5 km\/h<\/p>\n c)\u00a02.5 km\/h<\/p>\n d)\u00a02.2 km\/h<\/p>\n e)\u00a0None of these<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let distance between AB = $5x$ and BC = $7x$<\/p>\n Downstream speed = $\\frac{35}{2.5} = 14$ km\/h<\/p>\n Let upstream speed = $y$ km\/h<\/p>\n Now, $time = \\frac{distance}{speed}$<\/p>\n => $\\frac{5x}{y} = \\frac{7x}{14}$<\/p>\n => $\\frac{5}{y} = \\frac{1}{2}$<\/p>\n => $y = 5 \\times 2 = 10$ km\/h<\/p>\n Now, speed of stream = $\\frac{1}{2}$ (downstream – upstream)<\/p>\n = $\\frac{1}{2} (14 – 10)$<\/p>\n = $\\frac{4}{2} = 2$ km\/h<\/p>\n Question 17:\u00a0<\/b>The speeds of John and Max are 30 km\/h and 40 km\/h respectively. Initially, Max is at a place L and John is at a place M. The distance between L and M is 650 km. John started his journey 3 hours earlier than Max to meet each other. If they meet each other at a place P some where between L and M, then the distance between P and M is :<\/p>\n a)\u00a0220 km<\/p>\n b)\u00a0250 km,<\/p>\n c)\u00a0330 km<\/p>\n d)\u00a0320 km<\/p>\n e)\u00a0None of these<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Distance between LM = 650 km<\/p>\n Distance travelled by John in 3 hours = 30 * 3 = 90 km<\/p>\n => Remaining distance = 650 – 90 = 560 km<\/p>\n Since, they are moving in opposite direction, => Relative speed = 40 + 30 = 70 km\/h<\/p>\n => Time taken by John and Max to meet\u00a0at P = $\\frac{560}{70}$ = 8 hrs<\/p>\n => Time taken by John to reach P from beginning = 8 + 3 = 11 hrs<\/p>\n $\\therefore$ Distance travelled by John in 11 hrs = PM = 11 * 30<\/p>\n = 330 km<\/p>\n Question 18:\u00a0<\/b>A tap can empty a tank in one hour. A second tap can empty it in 30 min If both the taps operate simultaneously, how much time is needed to empty the tank ?<\/p>\n a)\u00a020 min<\/p>\n b)\u00a030 min<\/p>\n c)\u00a040 min<\/p>\n d)\u00a045 min<\/p>\n e)\u00a0None of these<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 1 minute work of tap A = $\\frac{1}{60}$<\/p>\n 1 minute work of tap B = $\\frac{1}{30}$<\/p>\n => (A + B)’s 1 minute work = $\\frac{1}{60} + \\frac{1}{30}$<\/p>\n = $\\frac{1 + 2}{60} = \\frac{3}{60}$<\/p>\n = $\\frac{1}{20}$<\/p>\n $\\therefore$ Time taken by (A + B) to empty the tank = 20 min<\/p>\n Question 19:\u00a0<\/b>Train A, 220m long, can cross a platform 340 m long in 32 sec. If the respective ratio of speed of trains A and B is 7 : 9 and the length of train B is 270 m, how much time (in sec) would train B take to cross an electric pole ?<\/p>\n a)\u00a018<\/p>\n b)\u00a012<\/p>\n c)\u00a020<\/p>\n d)\u00a022<\/p>\n e)\u00a014<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Length of train\u00a0A = 220 m<\/p>\n => Speed of train A = $\\frac{220 + 340}{32}$<\/p>\n = $\\frac{560}{32} = 17.5$ m\/s<\/p>\n Speed of train A\u00a0: Speed of train B = 7\u00a0: 9<\/p>\n => Speed of train B = $\\frac{9}{7} \\times 17.5$<\/p>\n = $9 \\times 2.5 = 22.5$ m\/s<\/p>\n Length of train B = 270 m<\/p>\n $\\therefore$ Time taken by train B to cross the pole = $\\frac{270}{22.5}$<\/p>\n = 12 sec<\/p>\n Question 20:\u00a0<\/b>The distance between two points is 36 km. A boat rows in still water at 6 kmph. It takes 8 hours less to cover this distance in downstream in comparison to that in upstream. The rate of stream is<\/p>\n a)\u00a03 kmph<\/p>\n b)\u00a02 kmph<\/p>\n c)\u00a02.5 kmph<\/p>\n d)\u00a04 kmph<\/p>\n e)\u00a0None of these<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of stream = $x$ kmph<\/p>\n => Speed of boat downstream = $(6 + x)$ kmph<\/p>\n Speed upstream = $(6 – x)$ kmph<\/p>\n Using, $time = \\frac{distance}{speed}$<\/p>\n => $\\frac{36}{6 – x} – \\frac{36}{6 + x} = 8$<\/p>\n => $\\frac{1}{6 – x} – \\frac{1}{6 + x} = \\frac{8}{36}$<\/p>\n => $\\frac{(6 + x) – (6 – x)}{(6 + x) (6 – x)} = \\frac{2}{9}$<\/p>\n => $\\frac{2x}{36 – x^2} = \\frac{2}{9}$<\/p>\n => $\\frac{x}{36 – x^2} = \\frac{1}{9}$<\/p>\n => $36 – x^2 = 9x$<\/p>\n => $x^2 + 9x – 36 = 0$<\/p>\n => $x^2 + 12x – 3x – 36 = 0$<\/p>\n => $x (x + 12) – 3 (x + 12) = 0$<\/p>\n => $x = 3 , -12$<\/p>\n $\\because$ Speed cant be negative, => $x = 3$ kmph<\/p>\n Question 21:\u00a0<\/b>Three typists P, Q and R have to type 368 pages. P types one page in 8 minutes, Q in 18 minutes and R in 24 minutes. In what time will these pages be typed if they work together?<\/p>\n a)\u00a025 hours<\/p>\n b)\u00a027.6 hours<\/p>\n c)\u00a027 hours<\/p>\n d)\u00a028 hours<\/p>\n e)\u00a0None of these<\/p>\n 21)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Pages printed by P, Q and R in one minute<\/p>\n = $\\frac{1}{8} + \\frac{1}{18} + \\frac{1}{24}$<\/p>\n = $\\frac{9 + 4 + 3}{72} = \\frac{16}{72}$<\/p>\n = $\\frac{2}{9}$<\/p>\n Number of pages to be printed = 368<\/p>\n $\\therefore$ Time taken = $\\frac{9}{2} \\times 368$<\/p>\n = $1656$ minutes<\/p>\n = $\\frac{1656}{60} = 27.6$ hours<\/p>\n Question 22:\u00a0<\/b>Nitin rides a bicycle at the speed of 15 kms\/hr, but stops for 10 minutes to take rest every 20 kms. How much time will he take to cover a distance of 150 kms, ?<\/p>\n a)\u00a012 hours 30 minutes<\/p>\n b)\u00a011 hours 10 minutes<\/p>\n c)\u00a010 hours 20 minutes<\/p>\n d)\u00a012 hours 10 minutes<\/p>\n e)\u00a0None of these<\/p>\n 22)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Nitin takes rest for 10 minutes after every 20 km<\/p>\n => For a period of 150 km, he will take rest = $\\frac{150}{20} = 7.5$ = 7 times<\/p>\n => Rest period = 7 * 10 = 70 minutes<\/p>\n Time taken to cover 150 km at a speed of 15 kmph<\/p>\n = $\\frac{150}{15} = 10$ hrs<\/p>\n $\\therefore$ Total time taken including rest period = 10 hrs + 70 minutes<\/p>\n = 11 hrs 10 minutes<\/p>\n Question 23:\u00a0<\/b>The respective ratio between the time taken by a boat to travel the same distance downstream in stream A and that in stream B is 8 : 7. The speed of the boat is 12 km\/h and the speed of stream A is half the speed of stream B. What is the speed of stream B ? (in km\/h)<\/p>\n a)\u00a05<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n e)\u00a06<\/p>\n 23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of stream B = $2x$ km\/h<\/p>\n and speed of stream A = $x$ km\/h<\/p>\n Speed of stream = $12$ km\/h<\/p>\n Acc to ques,<\/p>\n => $\\frac{\\frac{1}{12 + x}}{\\frac{1}{12 + 2x}} = \\frac{8}{7}$<\/p>\n => $\\frac{12 + 2x}{12 + x} = \\frac{8}{7}$<\/p>\n => $84 + 14x = 96 + 8x$<\/p>\n => $14x – 8x = 96 – 84$<\/p>\n => $x = \\frac{12}{6} = 2$<\/p>\n $\\therefore$ Speed of stream B = $2 \\times 2 = 4$ km\/h<\/p>\n Question 24:\u00a0<\/b>A boat covers a distance of 2.75 km upstream in 11 minutes. The ratio between speed of current and that of boat downstream is 1 : 7 respectively. The boat covers distance between A and B downstream in 52 minutes. What is the distance between point A and point B ?<\/p>\n a)\u00a019.2 km.<\/p>\n b)\u00a017.2 km.<\/p>\n c)\u00a018.2 km.<\/p>\n d)\u00a016.5 km.<\/p>\n e)\u00a0None of these<\/p>\n 24)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let speed of boat in still water = x kmph<\/p>\n Speed of current = y kmph<\/p>\n => Rate upstream = $(x – y)$ kmph<\/p>\n and Rate downstream = $(x + y)$ kmph<\/p>\n => $\\frac{y}{x + y} = \\frac{1}{7}$<\/p>\n => $7y = x + y$<\/p>\n => $x = 6y$<\/p>\n Again, $\\frac{2.75}{x – y} = \\frac{11}{60}$<\/p>\n => $x – y = \\frac{60 * 2.75}{11}$<\/p>\n => $6y – y = 15$<\/p>\n => $y = \\frac{15}{5} = 3$kmph<\/p>\n and $x = 6 * 3 = 18$ kmph<\/p>\n Thus, rate downstream = 18 + 3 = 21 kmph<\/p>\n $\\therefore$ Distance between points A & B = rate downstream * time<\/p>\n = $21 \\times \\frac{52}{60} = 18.2$ km<\/p>\n Question 25:\u00a0<\/b>Two stations, A and B are 850 km apart from each other. One train starts from station A at 5 am and travels towards station at 62 kmph. Another train starts from station B at 7 am and travels towards station A at 59 kmph. At what time will they meet<\/p>\n a)\u00a01 pm<\/p>\n b)\u00a011 : 45 am<\/p>\n c)\u00a012 : 30 pm<\/p>\n d)\u00a01 : 30 pm<\/p>\n e)\u00a0None of these<\/p>\n 25)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n At 7am the distance between the two trains will be 726kms.
\ni. Distance between point A and B
\nIi.Time taken to travel down stream from B to A
\niii. Speed of the stream of the water
\niv. Effective speed of Boat while traveling Downstream from B to A<\/p>\n
\nSince the trains are moving in opposite directions, the relative speed is the sum of individual speeds.
\nThe relative speed is 121kmph. and the distance is 726 kms.
\nThe time taken to cover 726kms at a speed of 121kmph is nothing but the time taken by the two trains to meet each other.
\nTime taken to meet = 726\/121 = 6 hours
\nHence the trains would meet at 1:00 pm.
\nHence Option A is the correct answer.<\/p>\n