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

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