Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>The time taken by a boat to travel 13 km downstream is the same as time taken by it to travel 7 km upstream. If the speed of the stream is 3 km\/h, then how much time (in hours) will it take to travel a distance of 44.8 km in still water?<\/p>\n a)\u00a0$4\\frac{12}{25}$<\/p>\n b)\u00a0$5\\frac{3}{5}$<\/p>\n c)\u00a0$5\\frac{2}{5}$<\/p>\n d)\u00a0$4\\frac{13}{25}$<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the stream = 3 km\/h<\/p>\n Let the speed of the boat be v.<\/p>\n Speed in downstream = 3 + v<\/p>\n Speed in upstream = v – 3<\/p>\n Time = distance\/speed<\/p>\n According to the question,<\/p>\n The time taken by a boat to travel 13 km downstream = time taken by boat to travel 7 km upstream<\/p>\n 13\/(3 + v) = 7\/(v – 3)<\/p>\n 13v – 39 = 21 + 7v<\/p>\n 6v = 60<\/p>\n v = 10 km\/h<\/p>\n Speed of boat = 10 km\/h<\/p>\n The time taken by a boat to travel\u00a0a distance of 44.8 km in still water = 44.8\/10 = 112\/25 = $ 4 \\frac{12}{25}$ hr<\/p>\n Question 2:\u00a0<\/b>A train can travel 40% faster then a car.Both the train and the car start from point A at the same time and reach point B, which is 70km away point from A,at the same time.On the way, however,the train lost about 15 minutes while stopping at stations. The speed of the car in km\/h is:<\/p>\n a)\u00a0120<\/p>\n b)\u00a080<\/p>\n c)\u00a090<\/p>\n d)\u00a0100<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the speed of car be x.<\/p>\n then the speed of train =\u00a0$x\\left(1+\\frac{40}{100}\\right)=1.4x$<\/p>\n Time taken by car to cover 70km =\u00a0$\\frac{70}{x}$<\/p>\n Time taken\u00a0 by train to cover 70km =\u00a0$\\frac{70}{1.4x}=\\frac{50}{x}$<\/p>\n According to question,<\/p>\n $\\therefore\\ \\frac{70}{x}-\\frac{50}{x}=\\frac{15}{60}$<\/p>\n $\\therefore\\ \\frac{20}{x}=\\frac{1}{4}$<\/p>\n so, Speed of car =\u00a0x = 80 km\/hr<\/p>\n Hence, Option B is correct<\/p>\n Question 3:\u00a0<\/b>A car can cover a distance of 144 km in 1.8 hours. In what time(in hours) will it cover double the distance when its speed is increased by 20% ?<\/p>\n a)\u00a03<\/p>\n b)\u00a02.5<\/p>\n c)\u00a02<\/p>\n d)\u00a03.2<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the car = $\\frac{144}{1.8}$ = 80 km\/hr<\/p>\n Speed of the car when increased by 20% = $\\frac{120}{100}\\times$80 = 96 km\/hr<\/p>\n Required time = $\\frac{288}{96}$<\/p>\n = 3 hours<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 4:\u00a0<\/b>The speed of a train is 220% of the speed of a car. The car covers a distance of 950 km in 19 hours. How much distance will the train cover in $3 \\frac{1}{2}$ hours?<\/p>\n a)\u00a0380 km<\/p>\n b)\u00a0385 km<\/p>\n c)\u00a0375 km<\/p>\n d)\u00a0285 km<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the car =\u00a0$\\frac{950}{19}$ = 50 km\/h<\/p>\n Given, speed of the train is 220% of the speed of the car<\/p>\n $\\Rightarrow$\u00a0 Speed of the train =\u00a0$\\frac{220}{100}\\times50$ =\u00a0 110 km\/h<\/p>\n $\\therefore\\ $Distance covered by the train in\u00a0$3 \\frac{1}{2}$ hours =\u00a0$110\\times3\\frac{1}{2}$ =\u00a0$110\\times\\frac{7}{2}$ = 385 km<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 5:\u00a0<\/b>A car covered 150 km in 5 hours. If it travels at one-third its usual speed, then how much more time will it take to cover the same distance?<\/p>\n a)\u00a012 hours<\/p>\n b)\u00a014 hours<\/p>\n c)\u00a010 hours<\/p>\n d)\u00a08 hours<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, the car covered\u00a0150 km in 5 hours<\/p>\n Speed of the car =\u00a0$\\frac{150}{5}$ = 30 km\/h<\/p>\n One third of the speed =\u00a0$\\frac{1}{3}\\times30$ = 10 km\/h<\/p>\n Time required for the car to cover 150 km with one third speed =\u00a0$\\frac{\\text{Distance}\\\\ }{\\text{Speed}}$ =\u00a0$\\frac{150}{10}$ = 15 hours<\/p>\n $\\therefore\\ $Extra time required to cover the distance with one third speed = 15 – 5 = 10 hours<\/p>\n Hence, the correct answer is Option C<\/p>\n Take a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>Rahul and Mithun travel a distance of 30 km. The sum of their speeds is 70 km\/h and the total time taken by both to travel the distance is 2 hours 6 minutes. The difference between their speeds is:<\/p>\n a)\u00a035 km\/h<\/p>\n b)\u00a020 km\/h<\/p>\n c)\u00a025 km\/h<\/p>\n d)\u00a030 km\/h<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of Rahul = s<\/p>\n $\\Rightarrow$ Speed of Mithun = 70 – s<\/p>\n Time taken by Rahul to cover 30 km distance =\u00a0$\\frac{30}{s}$<\/p>\n Time taken by Mithun to cover 30 km distance = $\\frac{30}{70-s}$<\/p>\n Given, total time = 2 hours 6 minutes = 2 + $\\frac{6}{60}$ hours = 2 + $\\frac{1}{10}$ hours =\u00a0$\\frac{21}{10}$ hours<\/p>\n $\\Rightarrow$ \u00a0$\\frac{30}{s}+\\frac{30}{70-s}=\\frac{21}{10}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{s}+\\frac{1}{70-s}=\\frac{7}{100}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{70-s+s}{s\\left(70-s\\right)}=\\frac{7}{100}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{70}{70s-s^2}=\\frac{7}{100}$<\/p>\n $\\Rightarrow$ \u00a0$70s-s^2=1000$<\/p>\n $\\Rightarrow$ \u00a0$s^2-70s+1000=0$<\/p>\n $\\Rightarrow$ \u00a0$s^2-50s-20s+1000=0$<\/p>\n $\\Rightarrow$ \u00a0$s\\left(s-50\\right)-20\\left(s-50\\right)=0$<\/p>\n $\\Rightarrow$ \u00a0$\\left(s-50\\right)\\left(s-20\\right)=0$<\/p>\n $\\Rightarrow$ \u00a0$s-50=0$ \u00a0 or \u00a0 $s-20=0$<\/p>\n $\\Rightarrow$\u00a0 s = 50 km\/h or \u00a0 s = 20 km\/h<\/p>\n When speed of Rahul = 50 km\/h, speed of mithun = 20 km\/h<\/p>\n When speed of Rahul = 20 km\/h, speed of mithun = 50 km\/h<\/p>\n $\\therefore\\ $Difference between their speeds = 30 km\/h<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 7:\u00a0<\/b>Mohan travels three equal distances at speeds of 12 km\/h, 18 km\/h and 24 km\/h.If he takes a total of 13 hours, then what is the total distance covered?<\/p>\n a)\u00a0214 km<\/p>\n b)\u00a0212 km<\/p>\n c)\u00a0216 km<\/p>\n d)\u00a0218 km<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total distance be 3d<\/p>\n Time taken to cover the distance at speed 12 km\/h =\u00a0$\\frac{d}{12}$ hour<\/p>\n Time taken to cover the distance at speed 18 km\/h = $\\frac{d}{18}$ hour<\/p>\n Time taken to cover the distance at speed 24 km\/h = $\\frac{d}{24}$ hour<\/p>\n Given, total time = 13 hours<\/p>\n $\\Rightarrow$ \u00a0$\\frac{d}{12}+\\frac{d}{18}+\\frac{d}{24}=13$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{6d+4d+3d}{72}=13$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{13d}{72}=13$<\/p>\n $\\Rightarrow$\u00a0\u00a0 d = 72 km<\/p>\n $\\therefore\\ $Total distance = 3d = 3 x 72 = 216 km<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>Mohan finishes a journey by scooter in 5 hours. He travels the first half of the journey at 30 km\/h and the second half of the journey at 20 km\/h. The distance covered by him is:<\/p>\n a)\u00a0130 km<\/p>\n b)\u00a0120 km<\/p>\n c)\u00a0140 km<\/p>\n d)\u00a0100 km<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total distance covered by Mohan = d<\/p>\n Time taken by Mohan to cover the first half =\u00a0$\\frac{\\frac{d}{2}}{30}$ =\u00a0$\\frac{d}{60}$ hours<\/p>\n Time taken by Mohan to cover the second half =\u00a0$\\frac{\\frac{d}{2}}{20}$ =\u00a0$\\frac{d}{40}$ hours<\/p>\n Total time for the journey = 5 hours<\/p>\n $\\Rightarrow$ $\\frac{d}{60}+\\frac{d}{40}=5$<\/p>\n $\\Rightarrow$ $\\frac{2d+3d}{120}=5$<\/p>\n $\\Rightarrow$ $\\frac{5d}{120}=5$<\/p>\n $\\Rightarrow$\u00a0 d = 120 km<\/p>\n $\\therefore\\ $Distance covered by Mohan = 120 km<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 9:\u00a0<\/b>A man travelled a distance of 35 km in 5 hours. He travelled partly on foot at the rate of 4 km\/h and the rest on bicycle at the rate of 9 km\/h. The distance travelled on foot is:<\/p>\n a)\u00a08 km<\/p>\n b)\u00a010 km<\/p>\n c)\u00a015 km<\/p>\n d)\u00a012 km<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the distance travelled on foot = d<\/p>\n $\\Rightarrow$\u00a0 Distance travelled on bicycle = 35 – d<\/p>\n Speed of the man on foot = 4 km\/h<\/p>\n Time taken by the man on foot =\u00a0$\\frac{d}{4}$<\/p>\n Speed of the man on bicycle = 9 km\/h<\/p>\n Time taken by the man on bicycle = $\\frac{35-d}{9}$<\/p>\n Total time taken by the man = 5 hours<\/p>\n $\\Rightarrow$ \u00a0$\\frac{d}{4}+\\frac{35-d}{9}=5$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{d}{4}+\\frac{35}{9}-\\frac{d}{9}=5$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{9d-4d}{36}=5-\\frac{35}{9}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{5d}{36}=\\frac{10}{9}$<\/p>\n $\\Rightarrow$ \u00a0d = 8 km<\/p>\n $\\therefore\\ $Distance travelled on foot = 8 km<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 10:\u00a0<\/b>A person covers 700 m distance in 6 minutes. What is his speed in km\/h?<\/p>\n a)\u00a03.45 km\/h<\/p>\n b)\u00a07 km\/h<\/p>\n c)\u00a06 km\/h<\/p>\n d)\u00a06.23 km\/h<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Distance = 700 m =\u00a0$\\frac{700}{1000}$ km<\/p>\n Time = 6 minutes =\u00a0$\\frac{6}{60}$ hr<\/p>\n Speed = $\\frac{\\text{Distance}}{\\text{Time}}$ = $\\frac{\\frac{700}{1000}}{\\frac{6}{60}}$ = $\\frac{700}{1000}\\times\\frac{60}{6}$ = 7 km\/h<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 11:\u00a0<\/b>Two cars start from the same place at the same time at right angles to each other. Their speeds are 54 km\/hr and 72 km\/hr, respectively. After 20 seconds the distance between them will be:<\/p>\n a)\u00a0720 m<\/p>\n b)\u00a0480 m<\/p>\n c)\u00a0500 m<\/p>\n d)\u00a0540 m<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the cars start at a point A<\/p>\n Speed of first car = 54 km\/hr = \u00a0$54\\times\\frac{5}{18}$ m\/s = 15 m\/s<\/p>\n Distance travelled by first car in 20 seconds =\u00a0$15\\times20$ = 300 m<\/p>\n Speed of second car = 72 km\/hr =\u00a0$72\\times\\frac{5}{18}$ = 20 m\/s<\/p>\n Distance travelled by second car in 20 seconds = $20\\times20$ = 400 m<\/p>\n Two cars move at right angle to each other<\/p>\n Let the position of the cars are B and C after 20 seconds respectively as shown in figure<\/p>\n From the figure,<\/p>\n $AC^2+AB^2=BC^2$<\/p>\n $=$> \u00a0$400^2+300^2=BC^2$<\/p>\n $=$> \u00a0$1600+900=BC^2$<\/p>\n $=$> \u00a0$BC^2=2500$<\/p>\n $=$> \u00a0$BC=500$ m<\/p>\n $\\therefore\\ $Distance between the cars after 20 seconds = BC = 500 m<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 12:\u00a0<\/b>A person walks a distance from point A to B at 15 km\/h, and from point B to A at 30 km\/h. If he takes 3 hours to complete the journey, then what is the distance from point A to B?<\/p>\n a)\u00a015 km<\/p>\n b)\u00a030 km<\/p>\n c)\u00a025 km<\/p>\n d)\u00a010 km<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the distance between point A and B = ‘d’ km<\/p>\n Speed of the person from point A to B = 15 km\/h<\/p>\n Time taken by the person to walk from point A to B =\u00a0$\\frac{d}{15}$ hours<\/p>\n Speed of the person from point B to A = 30 km\/h<\/p>\n Time taken by the person to walk from point B to A = $\\frac{d}{30}$ hours<\/p>\n Total time taken to complete the journey = 3 hours<\/p>\n $=$> \u00a0$\\frac{d}{15}+\\frac{d}{30}=3$<\/p>\n $=$> \u00a0$\\frac{2d+d}{30}=3$<\/p>\n $=$> \u00a0$\\frac{3d}{30}=3$<\/p>\n $=$> \u00a0$d=30$ km<\/p>\n $\\therefore\\ $Distance from point A to B = 30 km<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 13:\u00a0<\/b>The distance covered by a train in $(x + 1)$ hours is $( x^3 + 1)$ km. What is the speed of the train?<\/p>\n a)\u00a0$(x+ 1)$ km\/h<\/p>\n b)\u00a0$(x^2 + x + 1) km\/h$<\/p>\n c)\u00a0$(x^2 – x\u00a0 + 1) km\/h$<\/p>\n d)\u00a0$(x^3 – 1) km\/h$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Distance covered by the train =\u00a0$( x^3 + 1)$ km<\/p>\n Time taken by the train =\u00a0 $(x+1)$ hours<\/p>\n Speed of the train = $\\frac{\\text{Distance covered}}{\\text{Time taken}}$<\/p>\n $=\\frac{\\left(x^3+1\\right)}{\\left(x+1\\right)}$<\/p>\n $=\\frac{\\left(x+1\\right)\\left(x^2-x+1\\right)}{\\left(x+1\\right)}$<\/p>\n $=\\left(x^2-x+1\\right) km\/h$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 14:\u00a0<\/b>A train, 150 m long is running at 90 km\/h. How long (in seconds) will it take to clear a platform that is 300 m long?<\/p>\n a)\u00a06<\/p>\n b)\u00a012<\/p>\n c)\u00a050<\/p>\n d)\u00a018<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Length of the train (L) = 150 m<\/p>\n Length of the platform (l) = 300 m<\/p>\n Speed of the train (s) = 90 km\/h = $\\text{90}\\times\\frac{5}{18}$ m\/s = $\\text{25}$ m\/s<\/p>\n Time taken by the train to cross the platform = $\\frac{L+l}{s}=\\frac{150+300}{25}=\\frac{450}{25}= \\text{18}$ seconds<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 15:\u00a0<\/b>Tap A can fill a tank in 6 hours and tap B can empty the same tank in 10 hours. If both taps are opened together, then how much time (in hours) will be taken to fill the tank?<\/p>\n a)\u00a020<\/p>\n b)\u00a015<\/p>\n c)\u00a018<\/p>\n d)\u00a016<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the Volume of the tank = $V$<\/p>\n Given,<\/p>\n Tap A can fill the tank in 6 hours<\/p>\n $=$>\u00a0 Volume filled by Tap A in 1 hour = $\\frac{V}{6}$<\/p>\n Tap B can empty the tank in 10 hours<\/p>\n $=$>\u00a0 Volume emptied by Tap B in 1 hour = $\\frac{V}{10}$<\/p>\n $=$> \u00a0Volume filled by both the taps in 1 hour = $\\frac{V}{6}-\\frac{V}{10}=\\frac{5V-3V}{30}=\\frac{V}{15}$<\/p>\n $\\therefore\\ $Number of hours required for both the taps to fill the tank =\u00a0$\\frac{V}{\\frac{V}{15}}=15$ hours<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 16:\u00a0<\/b>Mohan covers a distance of 2.5 km by scooter at the rate of 30 km\/h. The time taken by Mohan to cover the given distance in minutes is:<\/p>\n a)\u00a010<\/p>\n b)\u00a06<\/p>\n c)\u00a08<\/p>\n d)\u00a05<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Distance travelled = 2.5 km<\/p>\n Speed = 30 km\/hr<\/p>\n $\\text{Time taken}=\\ \\frac{\\text{Distance travelled}}{\\text{Speed}}=\\frac{2.5}{30}hr=\\frac{25}{30\\times10}hr=\\frac{25}{30\\times10}\\times60\\min=5\\min$<\/p>\n Hence, The correct answer is Option D<\/p>\n Question 17:\u00a0<\/b>Ramu works 4 times as fast as Somu. If Somu can complete a work in 20 days independently, then the number of days in which Ramu and Somu together can complete the work is:<\/p>\n a)\u00a04 days<\/p>\n b)\u00a05 days<\/p>\n c)\u00a03 days<\/p>\n d)\u00a06 days<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of days required for Somu to complete work = 20 days<\/p>\n Ramu works 4 times as fast as Somu<\/p>\n $=$> Number of days required for Ramu to complete work = $\\frac{20}{4}=5$ days<\/p>\n Let the Total Work = W<\/p>\n Work done by Ramu in 1 day =\u00a0$\\frac{W}{5}$<\/p>\n Work done by Somu in 1 day =\u00a0$\\frac{W}{20}$<\/p>\n Work done by Ramu and Somu in 1 day =\u00a0$\\frac{W}{5}+\\frac{W}{20}=\\frac{5W}{20}=\\frac{W}{4}$<\/p>\n Number of days required for both Ramu and Somu together to complete work =\u00a0$\\frac{W}{\\frac{W}{4}}=4$ days<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 18:\u00a0<\/b>A boat takes 80 minutes to row 12 km upstream and 60 minutes to row 15 km downstream. How long will it take to row a distance of 36 km in still water?<\/p>\n a)\u00a02 hours<\/p>\n b)\u00a03 hours<\/p>\n c)\u00a04 hours<\/p>\n d)\u00a02.5 hours<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s assume the speed of the boat in still water and the speed of the stream are B and C respectively.<\/p>\n Speed of boat\u00a0downstream =\u00a0(B+C)<\/p>\n Speed of boat\u00a0upstream = (B-C)<\/p>\n A boat takes 80 minutes to row 12 km upstream and 60 minutes to row 15 km downstream.<\/p>\n $\\frac{12}{B-C}\\ =\\ \\frac{80}{60}$<\/p>\n $\\frac{12}{B-C}\\ =\\ \\frac{4}{3}$<\/p>\n $\\frac{3}{B-C}\\ =\\ \\frac{1}{3}$<\/p>\n (B-C) = 9\u00a0 \u00a0 Eq.(i)<\/p>\n $\\frac{15}{B+C}\\ =\\ \\frac{60}{60}$<\/p>\n (B+C) = 15\u00a0 \u00a0\u00a0Eq.(ii)<\/p>\n Add\u00a0Eq.(i) and\u00a0Eq.(ii).<\/p>\n (B-C)+(B+C) = 9+15<\/p>\n 2B = 24<\/p>\n B = 12 km\/h<\/strong><\/p>\n Time taken by the boat to\u00a0row a distance of 36 km in still water =\u00a0$\\frac{36}{12}$<\/p>\n = 3 hours<\/p>\n Question 19:\u00a0<\/b>Two trains each having a length of 160 meters moving in opposite direction crossed each other in 9 seconds. If one train crossed a 200-metre-long platform in 18 seconds,then the ratio of their speeds is:<\/p>\n a)\u00a02 : 3<\/p>\n b)\u00a09 : 7<\/p>\n c)\u00a05 : 8<\/p>\n d)\u00a03 : 4<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Two trains each having a length of 160 meters moving in the opposite directions crossed each other in 9 seconds.<\/p>\n Let’s assume the speed of trains are ‘a’ and ‘b’ respectively.<\/p>\n $\\frac{\\left(160+160\\right)}{a+b}\\ =\\ 9$<\/p>\n $\\frac{320}{9}\\ =\\ \\left(a+b\\right)$\u00a0 \u00a0\u00a0Eq.(i)<\/p>\n If one train crossed a 200 metre long platform in 18 seconds.<\/p>\n speed of one train = a =\u00a0$\\frac{\\left(160+200\\right)}{18}$<\/p>\n =\u00a0$\\frac{\\left(360\\right)}{18}$<\/p>\n = 20 m\/s\u00a0 \u00a0 Eq.(ii)<\/p>\n Put\u00a0Eq.(ii) in\u00a0Eq.(i).<\/p>\n $\\frac{320}{9}\\ =\\ \\left(20+b\\right)$ $\\frac{320-180}{9}=b$<\/p>\n $\\frac{140}{9} = b$\u00a0 \u00a0\u00a0Eq.(iii)<\/p>\n Ratio of their speeds is:\u00a0Eq.(ii) :\u00a0Eq.(iii)<\/p>\n $20 :\u00a0\\frac{140}{9}$<\/p>\n $1 : \\frac{7}{9}$<\/p>\n 9 : 7<\/p>\n Question 20:\u00a0<\/b>A boat goes a distance of 4 km upstream in 2 hours and the same distance downstream in 20 minutes. How long will it take to go $10\\frac{1}{2}$ km in still water?<\/p>\n a)\u00a0$1\\frac{1}{2}$ hours<\/p>\n b)\u00a048 minutes<\/p>\n c)\u00a0$1\\frac{1}{4}$ hours<\/p>\n d)\u00a01 hour<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s assume the speed of the boat in still water is B and the speed of the stream is C.<\/p>\n A boat goes a distance of 4 km upstream in 2 hours and the same distance downstream in 20 minutes.<\/p>\n $\\frac{4}{B-C}\\ =\\ 2$<\/p>\n B-C = 2\u00a0 \u00a0 Eq.(i)<\/p>\n $\\frac{4}{B+C}\\ =\\ \\frac{20}{60}$<\/p>\n $\\frac{4}{B+C}\\ =\\ \\frac{1}{3}$ Time taken\u00a0to go $10\\frac{1}{2}$ km in still water =\u00a0$\\frac{10.5}{7}$<\/p>\n = 1.5 hours<\/p>\n =\u00a0$1\\frac{1}{2}$ hours<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/349204.png\")
<\/p>\n
\n$\\frac{320}{9} –\u00a020 = b$<\/p>\n
\nB+C = 12\u00a0 \u00a0\u00a0Eq.(ii)
\nAdd\u00a0Eq.(i) and\u00a0Eq.(ii),
\nB-C+B+C = 2+12
\n2B\u00a0 =14
\nB = 7 km\/h<\/p>\n