Speed Time And Distance — Part 2 MCQs with Explanations

Since, the two horses meet after 200 minutes; they cover 0.5% of the distance per minute (combined) or 30% distance per hour. This condition is satisfied only if the slower rider takes 10 hours (thereby covering 10% per hour) and the faster rider takes 5 hours.

A__________300 km_______B Let speed of the 1 st ship is s kmph and speed of 2 nd ship is (s + 10) kmph. Let 2 nd ship takes time t to cover the distance 300 km then 1 st ship takes (t + 8) hours. As distance is constant, we have, $$\frac{{\text{s}}}{{{\text{s}} + 10}} = \frac{{\text{t}}}{{{\text{t}} + 8}}$$ Or, st + 8s = st + 10t Or, 8s = 10t If the slower ship took 20 hours (option D) the faster ship would take 12 hours and their respective speeds would be 15 and 25 kmph.

In the final minute of collision, the two boats travel $$\frac{{5}}{{60}}$$ km and $$\frac{{10}}{{60}}$$ km. As they moves towards each other relative distance covered in 1 minute, $$\eqalign{ & = \frac{1}{{12}} + \frac{1}{6} \cr & = \frac{1}{4}\,\,{\text{km}} \cr} $$

In the 1 st race when Karan runs 100 metre, Arjun runs only 90 metre. Hence, the ratio of speeds of Arjun and Karan is 90 : 100 = 9 : 10 In 2 nd race, Karan has run 110 metre. When he finishes the race, Arjun would have run $$\frac{9}{{10}} \times 110 = 99\,{\text{metre}}$$ ∴ Karan beats Arjun by 1 metre

A _______18miles_____B ___x(let)___C Let the speed of the ship be 1 mile/h and then speed of the plane becomes 10 miles/h. Let time t is taken to cover the distance x from point B to point C by ship. Hence, in time t plane will cover (18 + x) miles. Now, $$\frac{{\text{x}}}{1} = \frac{{18 + {\text{x}}}}{{10}}$$ Or, 10x = 18 + x Or, x = 2 miles. Hence, ship will cover 2 miles in 2 hours and plane will cover 20 miles in same time. Short method: Relative speed = (10 - 1) = 9 miles/h. hence, time taken to catch up the ship is 2 hours so distance covered by the ship = 20 miles.

The sum of speeds would be 0.8 m/s (relative speed in opposite direction). Also if we go by option (B) the speeds will be 0.03 and 0.05 m/s respectively. At this speed the overlapping would occur in every 60 second. Alternate : Let their speeds be x m/sec and y m/sec respectively. Then, $$\eqalign{ & \frac{{1200}}{{x + y}} = 15 \cr & \Rightarrow x + y = 80.....(i) \cr} $$ And, $$\eqalign{ & \frac{{1200}}{{x - y}} = 60 \cr & \Rightarrow x - y = 20.....(ii) \cr} $$ Adding (i) and (ii), we get : 2x = 100 or x = 50 Putting x = 50 in (i), we get : y = 30 Hence, speed of slower plane : = 30 m/sec = 0.03 km/sec

$$\eqalign{ & {\text{Speed of first Jogger}} \cr & = {\frac{{ {x - 1} }}{{42}}} \times 60\,{\text{kmph}} \cr & {\text{Speed of }}\,{2^{nd}}\,{\text{jogger}} \cr & = {\frac{{ {x - 2} }}{{52}}} \times 60\,{\text{kmph}} \cr & {\text{Then}}, \cr & {\frac{{x - 2}}{{{s_b}}}} - {\frac{{x - 1}}{{{s_b}}}} \cr & \cr} $$ Now, we use option checking method which gives us that option (B) is correct.

3 + 7 + 11 + 15 + . . . . . (1) 1 + 9 + 17 + 25 + . . . . . (2) The condition is satisfied for the 4 seconds in 2 nd journey, ant covered 17 m more than 1 st one.

A _______ B ________ C Let both the trains start from point A. At point B third train overtook the first train. To overtake the first train by third train, Third train needs to cover, Distance covered by first train in $$1\frac{1}{2}$$ h = distance covered by third train in 1 h. [as third train has started 30 minutes later] In this situation distance is constant, then from, s × t = d; we get, S α $$\frac{1}{{\text{t}}}$$ Now, $$\eqalign{ & \frac{{{\text{t}} + \frac{1}{2}}}{{\text{t}}} = \frac{{\text{s}}}{{40}} \cr & \frac{{2{\text{t}} + 1}}{{\text{t}}} = \frac{{\text{s}}}{{40}}\,.\,.\,.\,.\,.\,.\,.\,.\left( 1 \right) \cr} $$ From equation (1), It is clear that time is in the ratio 3 : 2 then speed will be in 2 : 3 ratios. Hence, Speed of the Third train will be 60 km/h.

The given conditions are met on a Pythagoras triplet 6, 8, 10. Since, the racetrack only consists of the legs of the right angle triangle the length must be,= 6 + 8 = 14 km.

The distance covered in two hour, = 2 × 25 = 50 km Time taken by first individual = (3h + 2h) = 5h Then, the speed of predecessor, = $$\frac{{50}}{5}$$ = 10 kmph.

Let the number of puffs abe, $$3x \times \left( {3 \times \frac{1}{{150}}} \right) + \left( {13 \times 60 - 3x} \right) \times \frac{1}{{150}} = 6$$ On solving, We get, x = 20 puffs.

For every 15 minutes he takes a rest of 2 minutes. Hence, for 90 minutes of drive he would require 12 minutes of rest. Thus, he will be traveling for 90 - 12 = 78 minutes. In 60 minutes he covers 80 Km. In 1 minute he would cover $$\frac{{80}}{{60}}$$ Km. In 78 minutes he would cover $$\frac{{80}}{{60}} \times 78$$ = 104 Km.

Relative speed of the trains = 40 + 30 = 70 kmph. Time taken by trains to collide, = $$\frac{{560}}{{70}}$$ = 8h. In 8h sparrow will cover, = 8 × 80 = 640 km.

Time taken to collide the trains, = one and half hour = $$\frac{3}{2}$$ h. So, in $$\frac{3}{2}$$ h bird travels, = $$\frac{{3 \times 60}}{2}$$ = 90 km.

In 20 minutes the difference between man and son, = 20 × 20 = 400 m. Distance traveled by dog when he goes towards the child, = $$400 \times \frac{{60}}{{40}}$$ = 600 m and time required is 10 minutes. In 10 min remaining distance between man and child, = 400 - (20 × 10) = 200 m. Time taken by dog to meet the man, = $$\frac{{200}}{{100}}$$ = 2 min (Relative speed of dog with child is 40 m/min and same with man is 100 m/min.) Remaining distance in 2 min,= 200 - (2 × 20),= 160 m. Now, the time taken by dog to meet the child again,= $$\frac{{160}}{{40}}$$ = 4 min. In 4 min he covers = 4 × 60 = 240 m. Now, remaining distance in 4 min = 160 - (4 × 20) = 80 m. Time required by dog to meet the man once again = $$\frac{{80}}{{100}}$$ = 0.8 min. Now remaining distance = 80 - (0.8 × 20) = 64 m.

Speed of tiger = 40 m/min Speed of deer = 20 m/min. Relative speed = 40 - 20 = 20 m/min. Initial difference in distance = 50 × 8 = 400 m Time take to catch = $$\frac{{400}}{{20}}$$ = 20 min. Distance traveled in 20 min,= 20 × 40 = 800 m

(6 - x) = (8 - 1.5x) Or, x = 4 cm. So, It will take 4 hours to burn it in such a way that they will remain equal in lengths.

Speed of man, = 4 kmph = $$\frac{{4 \times 5}}{{18}}$$ m/sec In 3 min (180 sec) man will go = $$\frac{{20 \times 180}}{8}$$ = 200 m. That means the diagonal of the square field = 200 m. Diagonal of square, = Side of Square × $$\sqrt 2 $$ = 200 m. → Side of Square = $$\frac{{200}}{{\sqrt 2 }}$$ Area of Square = Side 2 Area = $$\frac{{200 \times 200}}{2}$$ = 20000 sqm.

Train(T)__________ A_____5k____CAT__________B T<-------x--------------><----12k--------------------------> Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then $$\eqalign{ & \Rightarrow \frac{v}{u} = \frac{x}{{5k}} = \frac{{ {x + 12k} }}{{7k}} \cr & \Rightarrow 7x = 5\left( {x + 12k} \right) \cr & \Rightarrow \frac{x}{k} = \frac{{30}}{1} \cr & {\text{Thus}}, \cr & \Rightarrow \frac{u}{v} = \frac{{30}}{5} = \frac{6}{1} \cr} $$ $${\text{or,}}\,\,6:1$$

Relative Speed, $$\eqalign{ & = \frac{{{\text{Total }}\,{\text{distance}}}}{{{\text{Total time}}}} \cr & = \frac{{60 + 40}}{{20}} \cr & = 5\,\,{\text{m/s}} \cr & = \frac{{5 \times 18}}{5} \cr & = 18\,\,{\text{kmph}} \cr} $$ Relative Speed = (speed of ambulance - speed of school bus) Speed of school bus = speed of ambulance - relative speed. = 30 - 18 = 12 kmph.

Let Usual Speed = x kmph. Now, Speed of bus after increasing the speed = (x + 14)kmph. . . . . . . (1) A ____________1680km ____________B In first case, Time taken to covered the distance 1680 km = $$\frac{{1680}}{x}$$ . . . . . . . (2) In Second Case, Time Taken to covered the distance 1680 km = $$\frac{{1680}}{{x + 14}}$$ Time difference = 6 Hours. So, $$\eqalign{ & \Rightarrow {\frac{{1680}}{x} - \frac{{1680}}{{ {x + 14} }}} = 6 \cr & \Rightarrow 1680 {\frac{{14}}{{ {{x^2} + 14x} }}} = 6 \cr & \Rightarrow 280 \times 14 = {x^2} + 14x \cr & \Rightarrow {x^2} + 70x - 56x - 3920 = 0 \cr & \Rightarrow x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & \Rightarrow x = - 70,56. \cr & {\text{hence,}}\,{\text{speed}}\,{\text{of}}\,{\text{minibus}}\,{\text{is}}\,56\,km/h \cr & {\text{put}}\,x = 56\,{\text{in}}\,{\text{equation}}\,(2) \cr & T = \frac{{1680}}{{56}} \cr & \,\,\,\,\,\, = 30\,{\text{hours}} \cr} $$

Usual time = 9 × 10 = 90 min = $$\frac{3}{2}$$ h Thus, Distance traveled = Speed × time = $$20 \times \frac{3}{2}$$ = 30 km. Alternatively, As the speed decreases from 20 kmph to 18 kmph i.e. 10 % increment in usual time. 10% = 10 min 100% = 100 min. Now, Distance traveled by him,= $$\frac{{100}}{{60}} \times 18$$ = 30 km.

Let the original speed be S 1 and time t 1 and distance be D. Now, $$\eqalign{ & \frac{{ {\frac{D}{2}} }}{{2{t_1}}} = {S_2} \cr & {S_2} = \frac{D}{{4{t_1}}}\,{\text{and}}\,{S_1} = \frac{D}{{{t_1}}} \cr & {\text{Thus}}, \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{4}{1} = 4:1 \cr} $$

The ratio of speeds of A, B, C = 6 : 3 : 1 The ratio of time taken by A, B, C = $$\frac{1}{6}$$ : $$\frac{1}{3}$$ : 1 To simplify it, we will multiply it by LCM of ratio of speeds given. Hence, the ratio of time taken by A, B, C = 1 : 2 : 6 [Speed is inversely proportional to time, means if speed increase time decreases. So, ratio of time would be reciprocal of the ratio of speed given. ] Time taken by C to covered given distance = 78 = 6 × 13 The ratio of time of A and C = 1 : 6 Thus, time taken by A = 13 min.

Since, the speeds of Vinay and Versha are in the ratio 5 : 3 i.e. when Vinay covers 5 rounds, then Versa covers 3 rounds, but first time Vinay and Versha meet when Vinay completes $$\left\{ {2\frac{1}{2} = 2.5} \right\}$$ round and Versha completes $$\frac{1}{2}$$ round. For Vinay to pass Versha seventh time, Vinay would have completed, = 7 × 2.5 rounds Since, each round is 1.5 km, the distance covered by Vinay is, = 7 × 2.5 × 1.5 = 26.25 km.

$$\eqalign{ & \frac{{{{\left( {90 - 30} \right)}^\text{o}}}}{{5.5}} \cr & = \frac{{\left( {60 \times 2} \right)}}{{11}} \cr & = 10\left( {\frac{{10}}{{11}}} \right) \cr & = 10\,\min \,54\,\sec \cr & {\text{Thus}}, \cr & {\text{Required}}\,{\text{time}} \cr & = 3:10:54 \cr} $$

Given, Speed of the faster boat Downstream = 1.5 × speed of the slower boat downstream ----------(1) [Difference in First hour ] Speed of the Faster Boat Downstream = Speed of the slower boat + 8 ------------- (2) Using Equation (1) and (2), we get Speed of the faster Boat Downstream = 16 kmph Now, $$\frac{{{\text{Time taken by the faster Boat}}}}{{{\text{Time taken by the Slower boat Upstream}}}}$$ = $$\frac{1}{2}$$ Hence, Time taken by the faster Boat Upstream = 2 × Time taken by the slower Boat Upstream . . . . . . . (3) And, Faster boat's speed upstream - 8 = Slower boat's speed upstream . . . . . . . . (4) Using (4) and (3), we get Speed of the faster Boat upstream = 8 kmph Thus, Speed of the faster Boat in still water = 20 kmph

He proceeds at $$\frac{3}{4}$$ S where S is his usual speed means $$\frac{1}{4}$$ decrease in the speed which will lead to $$\frac{1}{3}$$ increase in the time. Now the main difference comes in those 24km and the change in difference of time = 35 - 25 m = 10 m. ⇒ $$\frac{1}{3}$$ × T = 10 where T is the time required to cover the distance of (74 - 50) = 24 km. T = 30 min = 0.5 hours. Speed of the dog = $$\frac{{24}}{{0.5}}$$ = 48 kmph.

Let the radius be r, then difference in the distance, πr - 2r = r (π - 2) = r [(3.14) -2] = 60 × 3 2r = 315 m. [πr = semi perimeter, 2r = diameter]

Let the speed of the current be x km/h. Now, according to the question, $$\frac{{15 + x}}{{15 - x}} = \frac{2}{1}$$ 15 + x = 30 - 2x 3x = 30 - 15 3x = 15 x = 5 km/h

Let downstream speed be D and Upstream speed be U. Then $$\eqalign{ & {\frac{{48}}{U} + \frac{{72}}{D}} = 12\,.\,.\,.\,.\,.\,.\,.\left( 1 \right) \cr & {\text{and}} \cr & {\frac{{72}}{U} + \frac{{48}}{D}} = 13\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$ On solving the equation (1) and (2), we get D = 12 kmph and U = 8 kmph Thus, speed of the current $$ = \frac{{D - U}}{2} = \frac{4}{2} = 2\,{\text{kmph}}$$

Downstream speed, = $$\frac{9}{2}$$ = 4.5 kmph Upstream speed,= $$\frac{9}{6}$$ = 1.5 kmph Thus, speed of boat in still water, = $$\frac{{4.5 + 1.5}}{2}$$ = 3 km/h Speed of current = $$\frac{{4.5 - 1.5}}{2}$$ = 1.5 km/h

A ____100 km____ P 1 ________300 km_____ P 2 ____X______ B If speed becomes $$\frac{5}{6}$$ then time taken will be $$\frac{6}{5}$$ of original time. Thus, extra time = 2 hour $$\frac{1}{5}$$ = 2 hour Thus unusual time which is require to complete the journey = 5 × 2 = 10 hours. It means It covers 300 km distance in 5 hours. then speed = $$\frac{{300}}{5}$$ = 60 km/h

. A B Reason (ST=D) Speed 7 4 Given Time 4 7 Since, Speed ∝ $$\frac{1}{{{\text{Time}}}}$$ Distance 4 7 Distance ∝ time Now, 7x - 4x = 300 (A runs 7x m and B runs 4x) x = 100 7x = 7 × 100 = 700 m Winning post is 700 m away, As A runs 700 m to complete the race.

Circumference = One revolution Distance covered in 10 revolutions = 300 * 10 = 3000 cm = 30 m 30 meters are covered in 6 seconds. Speed of the wheel = $$\frac{{30}}{{\text{6}}}$$ = 5 m/s 5 m/s = $$\frac{{5 \times 18}}{5}$$ = 18 km/h. [ To convert m/s to km/h, we use to multiply by factor $$\frac{{18}}{5}$$ ]

Initial speed of the police = 10 m/s Increased speed of the Police = 20 m/s Speed of the thief = 15 m/s Initial difference between speed of thief and police = 250 m After 5 seconds difference between thief and police = 200 + (5 × 10) = 250 m Now, the time required by police to catch the thief = $$\frac{{250}}{5}$$ = 50 s Total time = 50 + 15 = 65 s Total distance = 1000 + (15 × 10) = 1150 m.

Lucknow(X →)_____120 km _____(← Y)Kanpur Relative speed = (25+35) = 60 km/h Time taken to complete 120 km, = $$\frac{{120}}{{60}}$$ = 2 h Thus, They will meet 2 hours later.

When time is constant the distance covered by A and B will be in the ratio of their speeds. i.e. ratio of distance is 3 : 5 as time is constant.

When distance is constant then Speed is inversely proportional to time. ST = D When distance constant, S ∝ $$\frac{1}{{\text{T}}}$$ So, ratio of time taken by A and B = 3 : 2