Speed Time And Distance — Part 1 MCQs with Explanations

Let Speed of the man is x kmph. Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph. Or, $$20 \times \frac{{10}}{{60}} = \frac{8}{60}\times\left(20+x\right) $$ Or, 200 = 160 + 8x Or, 8x = 40 Hence, x = 5kmph. Detailed Explanation: A _____________M_______________B A = Bus Terminal. B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well. Distance Covered by Bus in 10 min = AB = $$\frac{{20}}{{60}} \times 10$$ = $$\frac{{10}}{3}$$ km. Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes. Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each other. Let Speed of the man = x kmph. Relative speed = 20 + x To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 + x) kmph = distance covered by bus in 10 minuted with speed 20 kmph.

1 st method: $$\frac{4}{3}$$ of usual time = Usual time + 16 minutes; Hence, $$\frac{1}{3}{\text{rd}}$$ of usual time = 16 minutes; Thus, Usual time = 16 × 3 = 48 minutes. 2 nd method: When speed goes down to $$\frac{3}{4}{\text{th}}$$ (i.e. 75%) time will go up to $$\frac{4}{3}{\text{rd}}$$ (or 133.33%) of the original time. Since, the extra time required is 16 minutes; it should be equated to $$\frac{1}{3}{\text{rd}}$$ of the normal time. Hence, the usual time required will be 48 minutes.

Difference in time of departure between two trains = 45 min. = $$\frac{{45}}{{60}}$$ hour = $$\frac{{3}}{{4}}$$ hour. Let the distance be x km from Delhi where the two trains will be together. Time taken to cover x km with speed 136 kmph be t hour and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be $$ {{\text{t}} + \frac{3}{4}} $$ = $$ {\frac{{4{\text{t}} + 3}}{4}} $$ Now, 100 × $$ {\frac{{4{\text{t}} + 3}}{4}} $$ = 136t Or, 25(4t + 3) = 136t Or, 100t + 75 = 136t Or, 36t = 75 Or, t = $$\frac{{75}}{{36}}$$ = 2.083 hours Then, distance x km = 136 × 2.083 ≈ 283.33 km.

Time taken in walking both the ways = 7 hours 45 minutes -------- (i) Time taken in walking one way and riding back = 6 hours 15 minutes ----------- (ii) By the equation (ii) × 2 - (i), we have, Time taken by the man in riding both ways, = 12 hours 30 minutes - 7 hours 45 minutes = 4 hours 45 minutes.

$$\eqalign{ & {\text{Let the total distance be 100 km}}. \cr & {\text{Average speed}} \cr & = \frac{{{\text{total}}\,{\text{distance}}\,{\text{covered}}}}{{{\text{time}}\,{\text{taken}}}} \cr & = \frac{{100}}{{ { {\frac{{30}}{{20}}} + {\frac{{60}}{{40}}} + {\frac{{10}}{{10}}} } }} \cr & = \frac{{100}}{{ { {\frac{3}{2}} + {\frac{3}{2}} + 1 } }} \cr & = \frac{{100}}{{ {\frac{{ {3 + 3 + 2} }}{2}} }} \cr & = \frac{{ {100 \times 2} }}{8} \cr & = 25\,\text{kmph} \cr} $$ Alternate 10% of journey's = 40 km Then, total journey = 400 kms $$\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr} $$

A →_______60Km_________← B Let the speed of A = x kmph and that of B = y kmph According to the question; x × 6 + y × 6 = 60 Or, x + y = 10 --------- (i) And, $$\left( {\frac{{2{\text{x}}}}{3} \times 5} \right) + \left( {2{\text{y}} \times 5} \right) = 60$$ Or, 10x + 30y = 180 Or, x + 3y = 18 ---------- (ii) From equation (i) × 3 - (ii) 3x + 3y - x - 3y = 30 - 18 Or, 2x = 12 Hence, x = 6 kmph Alternate $$\because $$ They meet after 6 hours if they walk towards each other i.e., their speed will be added. So, their relative speed in opposite direction $$ = \frac{{{\text{Distance }}}}{{{\text{Time }}}} = \frac{{60}}{6}$$ Relative speed in opposite direction : $$\left( \rightleftharpoons \right) = 10{\text{ km/h}}.....{\text{(i)}}$$ According to the question, $$\eqalign{ & \Rightarrow \frac{2}{3}A + 2B = \frac{{60}}{5} \cr & \Rightarrow \frac{2}{3}A + 2B = 12 \cr & \Rightarrow A + 3B = 18 \cr & \Rightarrow B's{\text{ Speed = }}\frac{{18 - A}}{3} \cr & \Rightarrow A + B = 10 \cr & \Rightarrow A + \frac{{18 - A}}{3} = 10 \cr & \Rightarrow 3A + 18 - A = 30 \cr & \Rightarrow 2A = 12 \cr & \Rightarrow A{\text{'s speed = 6 km/h}} \cr} $$

Time taken to complete the revolution: A → $$\frac{{12}}{{4}}$$ = 3 hours B → $$\frac{{12}}{{3}}$$ = 4 hours C → 12 × $$\frac{{2}}{{3}}$$ = 8 hours Required time,= LCM of 3, 4, 8. = 24 hours.

Ajay → (x + 4) kmph. A ________ 60 km _________ B Ravi → x kmph. Let the speed of Ravi be x kmph; Hence, Ajay's speed = (x + 4) kmph; Distance covered by Ajay = 60 + 12 = 72 km; Distance covered by Ravi = 60 - 12 = 48 km. According to question, $$\eqalign{ & \frac{{72}}{{x + 4}} = \frac{{48}}{x} \cr & {\text{or,}}\,\frac{3}{{x + 4}} = \frac{2}{x} \cr & {\text{or,}}3x = 2x + 8 \cr & {\text{or,}}x = 8\,{\text{kmph}} \cr} $$

Ratio of speed = 3 : 4 Ratio of time taken = 4 : 3 (As Speed ∝ $$\frac{1}{{{\text{Time}}}},$$ When distance remains constant.) Let time taken by A and B be 4x and 3x hour respectively. Then, 4x - 3x = $$\frac{{20}}{{60}}$$ Or, x = $$\frac{{1}}{{3}}$$ Hence, time taken by A = 4x hours = 4 × $$\frac{1}{3}$$ = $$1\frac{1}{3}$$ hours.

$$\eqalign{ & {\text{Average}}\,{\text{speed}} \cr & = \frac{{2xy}}{{ {x + y} }} \cr & = \frac{{2 \times 6 \times 3}}{{ {6 + 3} }} \cr & = \frac{{36}}{9} \cr & = 4\,{\text{kmph}} \cr} $$

Difference of time = 6 min - 5 mins. 52 secs. = 8 secs. Distance covered by man in 5 mins. 52 secs. = Distance covered by sound in 8 secs. = 330 × 8 = 2640 m. ∴ Speed of man $$\eqalign{ & = \frac{2640\text{ m}}{\text{5 min. 52 secs.}} \cr & = \frac{2640}{352}\text{ m/secs} \cr & = \frac{2640}{352}\times\frac{18}{5}\text{ kmph} \cr & = 27 \text{ kmph} }$$

When the athlete walks at $$\frac{5}{4}$$ of his usual speed then he takes $$\frac{4}{5}$$ of his usual time $$\left( {{\text{As}}\,\,{\text{Speed}}\, \propto \frac{1}{{{\text{Time}}}}} \right)$$ and he saves 5 minutes. → Usual time - $$\frac{4}{5}$$ × usual time = 5 minutes → $$\frac{1}{5}$$ × Usual time = 5 minutes Usual time = 25 minutes.

When A runs 1000m, B runs 900m. Hence, when A runs 500m, B runs 450m. Again, when B runs 400m, C runs 360m. And, when B runs 450m, C runs = 360 × $$\frac{{450}}{{400}}$$ = 405m. Required distance = 500 - 405 = 95 meter. That means when A runs 500 meter then B can run 450m then C runs 405m.

Ratio of the speed of A, B and C = 6 : 3 : 1 Then, ratio of time taken; = $$\frac{{1}}{{6}}$$ : $$\frac{{1}}{{3}}$$ : 1 = 1 : 2 : 6 Hence, time taken by A = $$\frac{{72}}{{6}}$$ = 12 minutes.

1 hour 45 minutes = $$1 + \frac{{45}}{{60}}$$ = $$\frac{7}{4}$$ hours. Speed of the motorboat up-stream, $$\eqalign{ & = \frac{{{\text{Distance}}}}{{{\text{Time}}\,\,{\text{Taken}}}} \cr & = \frac{{56\,{\text{km}}}}{{\frac{7}{4}{\text{hours}}}} \cr & = \frac{{56 \times 4}}{7} \cr & = 32\,{\text{kmph}} \cr} $$ Let the speed of the current = x kmph Hence, 36 - x = 32 Or, x = 36 - 32 = 4 kmph Speed of boat down the stream = 36 + 4 = 40 kmph. ∴ Time taken to cover 56 km at 40 kmph = $$\frac{{56}}{{40}}$$ = $$\frac{7}{5}$$ hours or 1 hours 24 minutes.

$$\eqalign{ & {\text{Speed of athlete}}, \cr & = \frac{{200\,{\text{m}}}}{{24\,{\text{secs}}}} \cr & = \frac{{200 \times 18}}{{5 \times 24}} \cr & = 30\,{\text{km/hour}} \cr} $$

Let the distance between school and home be x km. The difference of time when A goes school to school with these two different speed is 10 min $$\eqalign{ & = \frac{{10}}{{60}}\,{\text{hour}} \cr & {\text{Now,}} \cr & {\frac{x}{3}} - {\frac{x}{4}} = \frac{{10}}{{60}} \cr & {\text{Or}},\,\frac{x}{{12}} = \frac{1}{6} \cr & {\text{Or}},\,x = \frac{{12}}{6} \cr & = 2\,{\text{km}} \cr} $$

1 st Method: Man climbs 1m and slips down 50 cm (0.5m) in one minute i.e. he climbs (1 - 0.5 = 0.5 m) in one minute. But in the last minute he will be climbing 1m as he gets on the top so no slip. Time taken to climb 11 meter = $$ {\frac{{10}}{{0.5}} + 1} $$ = 21 minutes. He climbs the wall at 5:21 pm 2 nd Method (short-cut): $$ = \frac{{{\text{hight}}\,\,{\text{of}}\,\,{\text{pole}} - {\text{slipped}}\,\,{\text{distance}}}}{{{\text{climbed}}\,\,{\text{distance}} - {\text{slipped}}\,\,{\text{distance}}}}$$ $$\eqalign{ & = {\frac{{x - z}}{{y - z}}} \times t \cr & = {\frac{{11 - 0.5}}{{1 - 0.5}}} \times 1 \cr & = 21\,\,{\text{minutes}} \cr} $$

Monkey climbs 6 m in 1 st minute and slips down 3 m in next minute i.e. Monkey climbs 3 meter in 2 minute then he climbs in one minute, = $$\frac{3}{2}$$ m. But in the last minute he climbs 6 m as he gets on the top so there is no slip. Time required = $$2 \times \frac{{54}}{3} + \frac{6}{6}$$ = 37 minutes.

Since it climbs 7 meter in every 2 hr. At the end of 14 hours it would have climbed up to 49 mt. In 15th hour it will reach 61 mt. Then drops back to 56 mt in the 16th hour. It will take another 35 min to travel remaining 7 mt since it will be ascending in the 17th hour. Thus, 16 hr and 35 min. Detail Explanation: Given, Ant ascends 12 meters in first hours and slips down 5 meters in second hour. This is repeated in every alternate hour. It means in every two hours Ant ascends (12 - 5 = 7 meters) Therefore, in ends of 14 hours, Ant ascends = $$\frac{{14 \times 7}}{2}$$ = 49 meters. So, In next hour i.e 15 hour, Ant will ascends another 12 meter. Till end of 15 th hour, Ant climbed = 49 + 12 = 61 meters. But in 16 th hour, Ant will slips down 5 meters (61 - 5 = 56 meters). Remaining pole to climb, = 63 - 56 = 7

Let the speed of train = x km/ hr Length of train = 300 metres Their relative speed in same direction = (x - 3) km/hr According to the question, $$\frac{{\left( {300 + 0} \right)m}}{{\left( {x - 3} \right) \times \frac{5}{{18}}m/s}} = 33$$ [Here man's length is 0 metre] $$\eqalign{ & \frac{{100 \times 18}}{{5x - 15}} = 11 \cr & 1800 = 55x - 165 \cr & 55x = 1965 \cr} $$ ∴ Speed of the train = $$\frac{{1965}}{{55}}$$ = $$35\frac{8}{{11}}$$ kmph

A →____________________← B Ratio of the speed of trains is given by; $$\eqalign{ & \frac{{{\text{Speed}}\,{\text{of}}\,{\text{train}}\,A}}{{{\text{Speed}}\,{\text{train}}\,{\text{of}}\,B}} = \frac{{\sqrt B }}{{\sqrt A }} \cr & = \frac{{\sqrt 9 }}{{\sqrt 4 }} \cr & = \frac{3}{2} \cr & = 3:2 \cr} $$

Let length of each train be x meter. Then, speed of 1 st train = $$\frac{x}{{18}}$$ m/sec Speed of 2 nd train = $$\frac{x}{{12}}$$ m/sec Now, When both trains cross each other, time taken $$\eqalign{ & = {\frac{{2x}}{{ { {\frac{x}{{18}}} + {\frac{x}{{12}}} } }}} \cr & = \frac{{2x}}{{ {\frac{{ {2x + 3x} }}{{36}}} }} \cr & = \frac{{2x \times 36}}{{5x}} \cr & = \frac{{72}}{5} \cr & = 14.4\,\,{\text{seconds}} \cr} $$

Let the length of the train traveling at 48 kmph be 2x meters. And length of the platform is y meters. $$\eqalign{ & {\text{Relative speed of train}} \cr & = \left( {48 + 42} \right)\,{\text{kmph}} \cr & = {\frac{{90 \times 5}}{{18}}} \cr & = 25\,m/\sec \cr & {\text{And}}\,{\text{48}}\,{\text{kmph}} \cr & = \frac{{48 \times 5}}{{18}} \cr & = \frac{{40}}{3}\,m/\sec \cr & \cr & {\text{According to the question}}, \cr & \frac{{ {2x + x} }}{{25}} = 12; \cr & Or,\,3x = 12 \times 25 = 300 \cr & Or,\,x = \frac{{300}}{3} = 100m \cr & {\text{Then, length of the train}} \cr & = 2x \cr & = 100 \times 2 = 200m \cr & \frac{{200 + y}}{{ {\frac{{40}}{3}} }} = 45 \cr & 600 + 3y = 40 \times 45 \cr & {\text{Or}},\,3y = 1800 - 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200 \cr & {\text{Or}},\,y = \frac{{1200}}{3} = 400m \cr & {\text{Length of the platform}} \cr & = 400m \cr} $$

Length of the 1 st train = 105 m Length of the 2 nd train = 90 m Relative speed of the trains,= 45 + 72 = 117 kmph= $$\frac{{117 \times 5}}{{18}}$$ = 32.5 m/sec Time taken to cross each other,= $$\frac{{{\text{Length}}\,{\text{of}}\,{1^{{\text{st}}}}\,{\text{train}} + {\text{length of}}\,{2^{{\text{nd}}}}\,{\text{train}}}}{{{\text{relative }}\,{\text{speed }}\,{\text{of}}\,{\text{the }}\,{\text{trains}}}}$$ ∴ Time taken = $$\frac{{195}}{{32.5}}$$ = 6 secs.

1 st method: Let the speed of the train be S. And length of the train be x. When a train crosses a man, its travels its own distance. $$\eqalign{ & {\text{According to question}}; \cr & \frac{x}{{ {\left( {s - 3} \right) \times {\frac{5}{{18}}} } }} = 10 \cr & {\text{or}},\,18x = 50 \times s - 150.....({\text{i}}) \cr & {\text{and}} \cr & \frac{x}{{ {\left( {x - 5} \right) \times {\frac{5}{{18}}} } }} = 11 \cr & 18x = 55 \times s - 275......({\text{ii}}) \cr & {\text{Equating equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 50 \times s - 150 = 55 \times s - 275 \cr & {\text{or}},\,5 \times s = 125 \cr & {\text{or}},\,s = 25\,{\text{kmph}} \cr} $$

1 st Method: Let length of the train be x m and speed of the train is s kmph. Speed, s = $$\frac{{x + 800}}{{100}}$$ . . . . . (i) Speed, s = $$\frac{{x + 400}}{{60}}$$ . . . . . (ii) Equating equation (i) and (ii), we get, Or, $$\frac{{x + 800}}{{100}}$$ = $$\frac{{x + 400}}{{60}}$$ Or, 5x + 200 = 3x + 2400 Or, 2x = 400 Or, x = 200m 2 nd Method: As in both cases, the speed of the train is constant, and then we have; Time α distance $$\eqalign{ & \frac{{100}}{{60}} = \frac{{x + 800}}{{x + 400}} \cr & {\text{or,}}\,\,x = 200{\text{m}} \cr} $$

Let speed of the 2 nd train is S m/sec. And, 60 kmph $$ = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3}$$ m/sec. As trains are traveling in same distance, Then Relative distance, $$\eqalign{ & = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3} \cr & \frac{{50}}{3} - S = \frac{{120}}{{18}} \cr & \Rightarrow \frac{{50 - 3S}}{3} = \frac{{20}}{3} \cr & \Rightarrow 50 - 3S = 20 \cr & \Rightarrow 3S = 50 - 20 \cr & \Rightarrow 3S = 30 \cr & \therefore S = 10\,\,{\text{m/sec}} \cr} $$ Or, Speed of the 2 nd train = $${\text{10}} \times \frac{{18}}{5}$$ = 36 kmph.

Let the distance of the destination from the starting point = x km. Speed downstream = (10 + 4) = 14 kmph Speed upstream = (10 - 4) = 6 kmph According to the question, Total time taken = 5 hours $$\eqalign{ & {\frac{x}{{14}}} + {\frac{x}{6}} = 5 \cr & \frac{{ {3x + 7x} }}{{42}} = 5 \cr & {\text{or}},\,10x = 42 \times 5 \cr & {\text{or}},\,x = \frac{{42 \times 5}}{{10}} \cr & {\text{So}},\,x = 21\,{\text{km}} \cr} $$

Let the distance covered be x km and speed of stream = y kmph. Speed downstream = $$\frac{{15}}{2} + y$$ kmph Speed upstream = $$\frac{{15}}{2} - y$$ kmph $$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\frac{{2x}}{{ { {\frac{{15}}{2}} + y} }}} = {\frac{x}{{ { {\frac{{15}}{2}} - y} }}} \cr & or,\,15 - 2y = {\frac{{15}}{2}} + y \cr & or,3y = 15 - {\frac{{15}}{2}} = \frac{{15}}{2} \cr & or,y = \frac{{15}}{6} = 2.5\,{\text{kmph}} \cr} $$

Let the speed of the stream be x kmph and both the boats meet after t hour. According to the question, (12 + x) × t + (15 - x) × t = 108 Or, 12t + 15t = 108 Or, 27t = 108 ∴ t = $$\frac{{108}}{{27}}$$ = 4 hours

Let the speed of the motor boat = 36x kmph and speed of current = 5x kmph. The boat goes along with the current in 5 hours 10 minutes = $$\frac{{31}}{6}$$ hour $$\eqalign{ & {\text{Hence, Distance}} \cr & = {\frac{{31}}{6}} \times \left( {36x + 5x} \right) \cr & = \frac{{41x + 31}}{6}\,km \cr & {\text{Speed of boat upstream}} \cr & = 36x + 5x = 31x\,{\text{kmph}} \cr & {\text{Hence, time taken to come back}} \cr & = {\frac{{ {41x \times {\frac{{31}}{6}} } }}{{31x}}} \cr & = \frac{{41}}{6}\,{\text{hours}} \cr & = 6\,{\text{hours}}\,\,50\,{\text{minutes}} \cr} $$

$$\eqalign{ & {\text{Time taken by Kamal}} \cr & = \frac{{100}}{{ {\frac{{18 \times 5}}{{18}}} }} \cr & = 20\,{\text{seconds}} \cr & {\text{Hence, }} \cr & {\text{Time taken by Bimal}} \cr & 20 + 5 = 25\,{\text{seconds}} \cr & {\text{So, Bimal's speed}} \cr & = \frac{{100}}{{25}} \cr & = 4\ \cr & = \frac{{4 \times 18}}{5} \cr & = 14.4\,{\text{kmph}} \cr} $$

According to question, When B runs 200 m meters, A runs 190 meters; Hence, when B runs 180 meters, A runs = $$\frac{{190 \times 180}}{{200}}$$ = 171 meters; When C runs 200m, B runs 180 meters. Hence, C will give a start to A by = 200 - 171= 29 meters

When A runs 100 meters, B runs 970 meters. Hence, when A runs 100 meter, b runs 97 meter. When B runs 500 meter, C runs 475 meter. When B runs 97 meter, C runs, $$\frac{{475 \times 97}}{{500}}$$ = 92.15 meter. Hence, A will beat C by (100 - 92.15) = 7.85 meters.

By increasing the speed by 33.33%, it would be able to reduce the time taken for traveling by 25%. But since this is able to overcome the time delay of 30 minutes, 30 minutes must be equivalent to 25% of the time originally taken. Hence, the original time must have been 2 hours and the original speed would be 750 kmph. Hence, the new speed would be 1000 kmph.

In 6 minutes, the car goes ahead by 0.6 km. Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph. Hence, Net speed of the car is 8 kmph.

Distance covered in one complete revolution $$\eqalign{ & = 2\pi r \cr & = \frac{{2 \times 100 \times 22}}{7} \cr & \approx 628\,m \cr & {\text{Average}}\,{\text{speed}} \cr & = \frac{{628}}{2} \cr & = 314\,m/\min \cr} $$

It will come 11 times together.

Distance travelled by man-train in 11 minute = distance traveled by sound in (11 min 45 sec - 11 min) = 45 sec. Let the speed of train be x kmph. Now, Or, $$x \times \frac{{11}}{{60}} = \frac{{45}}{{3600}} \times 330 \times \frac{{18}}{5}$$ Or, x = 81 kmph.