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

A → ___160________Meeting____140_____← B
Ratio of their distance covered = $$\frac{{160}}{{140}}$$ = 8 : 7
Then,
Ratio of their speeds = 8 : 7 [When time is constant Speed is proportional to Distance covered]

$$\eqalign{ & \text{Let }x\text{ be the distance, then} \cr & \frac{x}{5} - \frac{x}{8} = \frac{3}{2} \cr & \Rightarrow \frac{8x - 5x}{40} = \frac{3}{2} \cr & \Rightarrow \frac{3x}{40} = \frac{3}{2} \cr & \therefore x = 20\,{\text{km}} \cr} $$

In train he travels = 4 × 45 = 180 km = 180000 m
Thus, the no. of poles,
= $$\frac{{180000}}{{50}}$$   = 3600 [Poles are at the distance of 50 m.]

Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute.
Hence,
If it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min/hour.

Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute. Hence, if it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min per hour.
Time taken to travel 300 km with stoppage,
$$ = \frac{{300}}{{40}} = 7.5\,\,{\text{hours}}$$
Time taken for stoppage as it stops for 20 min per hour
= 7 × 20 + 10
= 140 + 10
= 150 minutes.
= $$\frac{{150}}{{60}}$$  hours = 2.5 hours.

By increasing his speed by 25%, he will reduce his time by 20%. (This corresponds to a 6 minutes drop in his time.)
Hence, his time originally must have been 30 minutes.
Thus required distance = 20 kmph × 0.5 hours = 10 km.

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,
$$\eqalign{ & {\frac{X}{{30}}} + {\frac{X}{{40}}} = 17.5 \cr & {\frac{{ {40X + 30X} }}{{1200}}} = 17.5 \cr & 70X = 17.5 \times 1200 \cr & X = 300\,km \cr & {\text{Total}}\,{\text{distance}},\,2x \cr & = 600\,km \cr} $$

He proceeds at $$\frac{4}{5}$$ S where S is his usual speed means $$\frac{1}{5}$$ decrease in speed which will lead to $$\frac{1}{4}$$ increase in time.
Now the main difference comes in those 12km (30 - 18) and the change in difference of time = (45 - 36) min = 9 min

Thus,
$$\frac{1}{4}$$ × T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km
T = 36 min = $$\frac{36}{60}$$ hours = 0.6 hours.
Speed of the car = $$\frac{12}{0.6}$$ = 20 kmph.

In the case of increasing gap between two objects, we have
$$\frac{{{\text{Speed }}\,{\text{of }}\,{\text{sound}}}}{{{\text{Speed }}\,{\text{of }}\,{\text{tiger}}}} = $$     $$\frac{{{\text{Time }}\,{\text{Utilized}}}}{{{\text{Difference }}\,{\text{in }}\,{\text{time}}}}$$
$$\frac{{1195.2}}{{\text{S}}} = \frac{{83}}{7}$$
S = 100.8 km/h.

Downstream (steamer) = 40 min
Downstream (boat) = 60 min
Upstream (steamer) = 60 min
Upstream (boat) = 90 min
Required time = 40 + 30 + 45 = 115 minutes