Speed Time And Distance - Aptitude MCQ Questions and Solutions with Explanations

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.