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Download Time Distance and Speed Questions for TISSNET<\/a><\/p>\n Enroll to TISSNET 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>A person crosses a 900-metre long street in 6 minutes.The speed of the person in km\/h is:<\/p>\n a)\u00a09<\/p>\n b)\u00a010<\/p>\n c)\u00a015<\/p>\n d)\u00a012<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Distance travelled = 900 m =\u00a0$\\frac{900}{1000}=0.9\\ km$<\/p>\n Time taken = 6 minutes =\u00a0$\\frac{6}{60}=\\frac{1}{10\\ }=0.1\\ hr$<\/p>\n Speed of person (km\/hr) =\u00a0$\\frac{Dis\\tan ce}{time\\ }=\\frac{0.9}{0.1}$<\/p>\n = 9 km\/hr<\/p>\n Hence, Option A is correct.<\/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>The speed of a motor boat in still water is 20 km\/h.It travels 150 km downstream and then returns to the starting point. If the round trip takes a total of 16 hours, what is the speed (in km\/h) of the flow of river?<\/p>\n a)\u00a08<\/p>\n b)\u00a06<\/p>\n c)\u00a05<\/p>\n d)\u00a04<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the flow of the river = s<\/p>\n The speed of motor boat in still water(m)\u00a0= 20 km\/h<\/p>\n Downstream speed = 20 + s<\/p>\n Upstream speed = 20 – s<\/p>\n Time taken for boat to travel 150 km downstream = $\\frac{150}{20+s}$<\/p>\n Time taken for boat to travel 150 km upstream = $\\frac{150}{20-s}$<\/p>\n Total time taken = 16 hours<\/p>\n $\\frac{150}{20+s}$ +\u00a0$\\frac{150}{20-s}$ = 16<\/p>\n 150[$\\frac{20-s+20+s}{(20+s)(20-s)}$] = 16<\/p>\n $\\frac{150\\times40}{16}$ =\u00a0(20+s)(20-s)<\/p>\n 375 = 400 – s$^2$<\/p>\n s$^2$ = 25<\/p>\n s = 5 km\/h<\/p>\n The speed of the flow of the river = s = 5 km\/h<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 4:\u00a0<\/b>A train is to cover 370 km at a uniform speed. After running 100 km,the train could run at a speed 5 km\/h less than its normal speed due to some technical fault. The train got delayed by 36 minutes. What is the normal speed of the train, in km\/h?<\/p>\n a)\u00a048<\/p>\n b)\u00a045<\/p>\n c)\u00a040<\/p>\n d)\u00a050<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the normal speed of the train = s km\/h<\/p>\n According to the problem,<\/p>\n $\\frac{100}{s}+\\frac{270}{s-5}=\\frac{370}{s}+\\frac{36}{60}$<\/p>\n $\\frac{270}{s-5}-\\frac{270}{s}=\\frac{3}{5}$<\/p>\n $90\\left[\\frac{s-s+5}{\\left(s-5\\right)s}=\\frac{1}{5}\\right]$<\/p>\n $s^2-5s-2250=0$<\/p>\n s = 50 or s = -45<\/p>\n ‘s’ cannot be negative.<\/p>\n So s = 50 km\/h<\/p>\n Normal speed of the train = s = 50 km\/h<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 5:\u00a0<\/b>A train running at $40\\frac{1}{2}$ km\/h takes 24 seconds to cross a pole. How much time (in seconds) will it take to pass a 450 m long bridge?<\/p>\n a)\u00a056<\/p>\n b)\u00a052<\/p>\n c)\u00a060<\/p>\n d)\u00a064<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the train = L<\/p>\n Speed of the train =\u00a0$40\\frac{1}{2}$ km\/h =\u00a0$\\frac{81}{2}$ km\/h =\u00a0$\\frac{81}{2}\\times\\frac{5}{18}$ m\/sec = $\\frac{45}{4}$ m\/sec<\/p>\n Train crosses a pole in 24 seconds.<\/p>\n $\\frac{L}{\\frac{45}{4}}$ = 24<\/p>\n $\\frac{4L}{45}$ = 24<\/p>\n L = 270 m<\/p>\n Length of the train = L = 270 m<\/p>\n Time required for train to pass\u00a0a 450 m long bridge = $\\frac{L+450}{\\frac{45}{4}}$<\/p>\n =\u00a0$\\frac{270+450}{\\frac{45}{4}}$<\/p>\n =\u00a0$\\frac{720\\times4}{45}$<\/p>\n = 64 sec<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 6:\u00a0<\/b>A train covers 450 km at a uniform speed. If the speed had been 5 km\/h more,\u00a0it would have taken 1 hour less to cover the same distance. How much time will it take to cover 315 km at its usual speed?<\/p>\n a)\u00a07h 52m<\/p>\n b)\u00a06h 30m<\/p>\n c)\u00a06h 18m<\/p>\n d)\u00a07h<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the uniform speed of train = s<\/p>\n According to the problem,<\/p>\n $\\frac{450}{s+5}=\\frac{450}{s}-1$<\/p>\n $\\frac{450}{s}-\\frac{450}{s+5}=1$<\/p>\n $450\\left(\\frac{s+5-s}{s\\left(s+5\\right)}\\right)=1$<\/p>\n $450\\left(\\frac{5}{s^2+5s}\\right)=1$<\/p>\n $s^2+5s-2250=0$<\/p>\n $\\left(s+50\\right)\\left(s-45\\right)=0$<\/p>\n s = -50 or s = 45<\/p>\n s cannot be negative, so s = 45 km\/h<\/p>\n The uniform speed of train =\u00a045 km\/h<\/p>\n Time taken by train to cover 315 km at its usual speed =\u00a0$\\frac{315}{45}$ = 7 hours<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 7:\u00a0<\/b>A train running at 48 km\/h crosses a man going with the speed of 12 km\/h, in the same direction, in 18 seconds and passes a woman coming from the opposite direction in 12 seconds. The speed (in km\/h) of the woman is :<\/p>\n a)\u00a06<\/p>\n b)\u00a08<\/p>\n c)\u00a09<\/p>\n d)\u00a010<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the train = L<\/p>\n Relative speed between train and man = 48 – 12 = 36 km\/h<\/p>\n = 36 x $\\frac{5}{18}$ m\/sec<\/p>\n = 10 m\/sec<\/p>\n Time taken by train to cross the man = 18 seconds<\/p>\n $\\frac{L}{10}$ = 18<\/p>\n L = 180 m<\/p>\n Length of the train = 180 m<\/p>\n Let the speed of the woman = s km\/h<\/p>\n Relative\u00a0speed between train and woman = (48 + s) km\/h<\/p>\n =\u00a0(48 + s)\u00a0x $\\frac{5}{18}$ m\/sec<\/p>\n Time taken by train to cross the woman = 12 seconds<\/p>\n $\\frac{L}{\\left(48+s\\right)\\times\\frac{5}{18}}=12$<\/p>\n $\\frac{180}{\\left(48+s\\right)\\times\\frac{5}{18}}=12$<\/p>\n $\\frac{180\\times18}{\\left(48+s\\right)5}=12$<\/p>\n 48 + s = 54<\/p>\n s = 6<\/p>\n Speed of the woman = 6 km\/h<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 8:\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 8)\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 9:\u00a0<\/b>A boat goes 30 km upstream in 3 hours and downstream in 1 hour. How much time (in hours) will this boat take to cover 60 km in still water?<\/p>\n a)\u00a06<\/p>\n b)\u00a03<\/p>\n c)\u00a02<\/p>\n d)\u00a05<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat in still water = m<\/p>\n Speed of the stream = s<\/p>\n Upstream speed = m – s<\/p>\n $\\frac{30}{3}$ = m – s<\/p>\n m – s = 10…………(1)<\/p>\n Downstream speed = m + s<\/p>\n $\\frac{30}{1}$ = m + s<\/p>\n m + s = 30………..(2)<\/p>\n Adding (1) and (2),<\/p>\n 2m = 40<\/p>\n m = 20<\/p>\n Speed of the boat in still water = 20 km\/h<\/p>\n Time required for the\u00a0boat to cover 60 km in still water = $\\frac{60}{20}$ = 3 hours<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 10:\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 10)\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 11:\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 11)\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 Question 12:\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 12)\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\\ $Difference between their speeds = 30 km\/h<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 13:\u00a0<\/b>Mohan travels three equal distances at speeds of 12 km\/h, 18 km\/h and 24 km\/h.If he takes a total of 13 hours, then what is the total distance covered?<\/p>\n a)\u00a0214 km<\/p>\n b)\u00a0212 km<\/p>\n c)\u00a0216 km<\/p>\n d)\u00a0218 km<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total distance be 3d<\/p>\n Time taken to cover the distance at speed 12 km\/h =\u00a0$\\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\\ $Total distance = 3d = 3 x 72 = 216 km<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 14:\u00a0<\/b>The diameter of a wheel is 49 cm. The number of revolutions in which it will have to cover a distance of 770 m, is:<\/p>\n a)\u00a0400<\/p>\n b)\u00a0500<\/p>\n c)\u00a0700<\/p>\n d)\u00a0600<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, diameter of the wheel = 49 cm<\/p>\n Perimeter of the wheel = 49$\\pi =49\\times\\frac{22}{7}=154$cm<\/p>\n Distance to be covered = 770 m = 770$\\times$100 cm<\/p>\n $\\therefore\\ $The number of revolutions in which it will have to cover a distance of 770 m = $\\frac{770\\times100}{154}=500$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 15:\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 15)\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\\ $Distance covered by Mohan = 120 km<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 16:\u00a0<\/b>A car covers 15 km, 20 km, 30 km and 12 km at speeds of 20 km\/h, 30 km\/h, 40 km\/h and 30 km\/h,respectively. The average speed of the car for the total journey is :<\/p>\n a)\u00a088 km\/h<\/p>\n b)\u00a060 km\/h<\/p>\n c)\u00a030 km\/h<\/p>\n d)\u00a040 km\/h<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Time taken by the car to cover 15 km distance at speed of 20 km\/h =\u00a0$\\frac{15}{20}$ h<\/p>\n Time taken by the car to cover 20 km distance at speed of 30 km\/h = $\\frac{20}{30}$ h<\/p>\n Time taken by the car to cover 30 km distance at speed of 40 km\/h = $\\frac{30}{40}$ h<\/p>\n Time taken by the car to cover 12 km distance at speed of 30 km\/h = $\\frac{12}{30}$ h<\/p>\n Total Time =\u00a0$\\frac{15}{20}+\\frac{20}{30}+\\frac{30}{40}+\\frac{12}{30}$ =\u00a0$\\frac{90+80+90+48}{120}$ =\u00a0$\\frac{308}{120}$ =\u00a0$\\frac{77}{30}$ h<\/p>\n $\\therefore\\ $Average speed of the car for the total journey = $\\frac{\\text{Total distance}}{\\text{Total Time}}$ = $\\frac{15+20+3+12}{\\frac{77}{30}}$ = $\\frac{77\\times30}{77}$ = $30$ km\/h<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 17:\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 17)\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 18:\u00a0<\/b>The wheel of a car has 210 cm diameter. How many revolutions per minute must the wheel make so that the speed of the car is kept at 120 km\/h ?<\/p>\n a)\u00a0245<\/p>\n b)\u00a0289<\/p>\n c)\u00a0326.42<\/p>\n d)\u00a0303.03<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the car = 120 km\/h =\u00a0$120\\times\\frac{1000\\times100}{60}$ cm\/min = 2,00,000 cm\/min<\/p>\n $\\Rightarrow$ Car has to travel 2,00,000 cm in 1 minute<\/p>\n Diameter of the wheel of the car (d) = 210 cm<\/p>\n Perimeter of the wheel of the car = $\\pi\\ d$ = $\\frac{22}{7}\\times210$ = 660 cm<\/p>\n For one revolution, distance travelled by the car = 660 cm<\/p>\n Number of revolutions made by the car to cover 2,00,000 cm =\u00a0$\\frac{200000}{660}$ = 303.03<\/p>\n $\\therefore\\ $Number of revolutions made in a minute = 303.03<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 19:\u00a0<\/b>A person covers 700 m distance in 6 minutes. What is his speed in km\/h?<\/p>\n a)\u00a03.45 km\/h<\/p>\n b)\u00a07 km\/h<\/p>\n c)\u00a06 km\/h<\/p>\n d)\u00a06.23 km\/h<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Distance = 700 m =\u00a0$\\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 20:\u00a0<\/b>Ravi starts for his school from his house on his cycle at 8:20 a.m. If he runs his cycle at a speed of 10 km\/h, he reaches his school 8 minutes late, and if he drives the cycle at a speed of 16 km\/h, he reaches his school 10 minutes early. The school starts at:<\/p>\n a)\u00a09:40 a.m.<\/p>\n b)\u00a08:40 a.m.<\/p>\n c)\u00a08:50 a.m.<\/p>\n d)\u00a09:00 a.m.<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the distance between Ravi’s house and his school = $d$<\/p>\n Time taken by him when he runs his cycle at a speed of 10km\/h =\u00a0$\\frac{d}{10}$<\/p>\n Time taken by him when he drives his cycle at a speed of 16km\/h = $\\frac{d}{16}$<\/p>\n According to the problem,<\/p>\n $\\frac{d}{10}-\\frac{d}{16}=\\frac{18}{60}$ hours<\/p>\n $=$> \u00a0$\\frac{8d-5d}{80}=\\frac{18}{60}$<\/p>\n $=$> \u00a0$\\frac{3d}{80}=\\frac{18}{60}$<\/p>\n $=$> \u00a0$d=8$ km<\/p>\n Time taken by him when he runs his cycle at a speed of 10km\/h = $\\frac{8}{10}$ hour =\u00a0$\\frac{8}{10}\\times60$ minutes = 48 minutes<\/p>\n The time at which school starts = 8:20 a.m+ 48 minutes – 8 minutes = 8:20 + 40 minutes = 9:00 a.m<\/p>\n Hence, the correct answer is Option D<\/p>\n