Practice MCQ Questions and Answer on Speed Time and Distance
401.
Two trains 105 meters and 90 meters long, run at the speeds of 45 kmph and 72 kmph respectively, in opposite directions on parallel tracks. The time which they take to cross each other, is:
(A) 8 seconds
(B) 6 seconds
(C) 7 seconds
(D) 5 seconds
Solution:
Length of the 1st train = 105 m Length of the 2nd 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.
402.
An ant moved for several seconds and covered 3 mm in the first second and 4 mm more in each successive second than its predecessor. If the ant had covered 1 mm in the first second and 8 mm more in each successive second, then the difference between the path it would cover during the same time and actual path would be more than 6 mm but less than 30 mm. find the time for which the ant moved (in seconds).
(A) 5 s
(B) 4 s
(C) 6 s
(D) 2 s
Solution:
3 + 7 + 11 + 15 + . . . . . (1) 1 + 9 + 17 + 25 + . . . . . (2) The condition is satisfied for the 4 seconds in 2nd journey, ant covered 17 m more than 1st one.
403.
Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in:
(A) 14.4 seconds
(B) 15.5 seconds
(C) 18.8 seconds
(D) 20.2 seconds
Solution:
Let length of each train be x meter. Then, speed of 1st train = $$\frac{x}{{18}}$$ m/sec Speed of 2nd train = $$\frac{x}{{12}}$$ m/sec Now, When both trains cross each other, time taken $$\eqalign{ & = {\frac{{2x}}{{ { {\frac{x}{{18}}} + {\frac{x}{{12}}} } }}} \cr & = \frac{{2x}}{{ {\frac{{ {2x + 3x} }}{{36}}} }} \cr & = \frac{{2x \times 36}}{{5x}} \cr & = \frac{{72}}{5} \cr & = 14.4\,\,{\text{seconds}} \cr} $$
404.
Places A and B are 396 km apart. Train X leaves from A for B and train Y leaves from B for A at the same time on the same day on parallel tracks. Both trains meet after $$5\frac{1}{2}$$ÃÂÃÂ hours. The speed of Y is 10 km/h more than that of X. What is the speed (in km/h) of Y?
(A) 41
(B) 54
(C) 31
(D) 56
Solution:
\[A\xrightarrow{{\,\,\,\,\,\,\,\,\,\,396\,\,\,\,\,\,\,\,\,\,}}B\] $$\eqalign{ & \frac{{396}}{{A + B}} = \frac{{11}}{2} \cr & X + Y = \frac{{396 \times 2}}{{11}} \cr & X + Y = 36 \times 2 \cr & X + Y = 72 \cr & {\text{Let,}} \cr & X = a \cr & Y = a + 10 \cr & a + a + 10 = 72 \cr & 2a = 62 \cr & a = 31{\text{ km/h}} \cr & Y = 31 + 10 = 41 \cr} $$
405.
A train is 250 m long. If the train takes 50 second to cross a tree by the railway line, then the speed of the train on Km /hr is :
Two planes move along a circle of circumference 1.2 km with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction, one plane overtakes the other every 60 seconds. Find the speed of the slower plane.
(A) 0.04 km/s
(B) 0.03 km/s
(C) 0.05 km/s
(D) 0.02 km/s
Solution:
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
407.
A plane left half an hour later than the scheduled time and in order to reach its destination 1500 kilometers away in time, it had to increase its speed by 33.33 percent over its usual speed. Find its increased speed.
(A) 250 kmph
(B) 500 kmph
(C) 750 kmph
(D) 1000 kmph
Solution:
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.
408.
A truck covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minutes. The ratio of their speed is :
If a train runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph, it is late by 5 min only. The correct time for the train to complete its journey is :
(A) 13 min
(B) 15 min
(C) 19 min
(D) 21 min
Solution:
Let the correct time to complete the journey be x min Distance covered in (x + 11) min at 40 kmph = Distance covered in (x + 5) min at 50 kmph $$\eqalign{ & \therefore \frac{{\left( {x + 11} \right)}}{{60}} \times 40 = \frac{{\left( {x + 5} \right)}}{{60}} \times 50 \cr & \Leftrightarrow x = 19{\text{ min}} \cr} $$
410.
A motor car does a journey in 17.5 hours, covering the first half at 30 km/h and the second half at 40 km/h. Find the distance of the journey?
(A) 684 km
(B) 600 km
(C) 624 km
(D) 584 km
Solution:
Let the total distance be 2X. A_____X km_____M_____X km_____B Total time taken in the journey = 17.5 hours Time taken to cover X km at 30 km/h = $$\frac{{\text{X}}}{{30}}$$ Time taken to cover X km at 40 km/h = $$\frac{{\text{X}}}{{40}}$$ Now, $$\eqalign{ & {\frac{X}{{30}}} + {\frac{X}{{40}}} = 17.5 \cr & {\frac{{ {40X + 30X} }}{{1200}}} = 17.5 \cr & 70X = 17.5 \times 1200 \cr & X = 300\,km \cr & {\text{Total}}\,{\text{distance}},\,2x \cr & = 600\,km \cr} $$