# Time and Distance

Two trains are running with speed 30 km/hr. and 58 km/hr. in the same directions. A man in the slower train passes the faster train in 18 seconds. The length (in metres) of the faster trains is:

**Answer**: (d)

The relative speed = 58 - 30 = 28 Km/hr. = \(\frac{140}{18}\) m/s.

Therefore in 18 secs the faster train travels = \(\left ( \frac{140}{18} \right ) \times 18 = 140\)

Two trains travel in the same direction at the speeds of 56 km/hr. and 29 km/hr. respectively. The faster train passes a man in the slower trains in 10 seconds. The length of the faster train (in metres) is

**Answer**: (c)

The relative speed is 56 − 29 Km/hr = \(\left ( \frac{27}{18} \right ) \times 5\) = 7.5 m/s

The faster train travels only the length of it’s own = 7.5 × 10 = 75 m

A bus moving at a speed of 45 km/hr. overtakes a truck 150 metres ahead going in the same directions in 30 seconds. The speed of the truck is

**Answer**: (a)

The lag of 150 m has been covered by the bus in 30 seconds.

Therefore, the relative speed of the bus and truck is = \(\frac{5 \times 18}{5} =18\) km/h

The speed of truck is 45 − 18 = 27 km/hr.

Two trains of equal length are running on parallel lines in the same directions at 46 km/hr. and 36 km/hr. The faster trains pass the slower train in 36 seconds. The length of each train is:

**Answer**: (b)

Let the length of each train be *x* metres.

Then, distance covered = 2*x* metres.

Relative speed = 46 − 36 km/hr

= \(\frac{10 \times 5}{18}\) m/sec

= \(\frac{25}{9}\) m/s

∴ \(\frac{2x}{36} = \frac{25}{9}\)

⇒ 2*x* = 100

⇒ *x* = 50

Two trains start from a certain place on two parallel tracks in the same directions. The speed of the trains are 45km/hr. and 40km/hr. respectively. The distance between the two trains after 45 minutes will be

**Answer**: (c)

In one hour the distance between the trains is 5 km.

In 45 mins = ¾ hour the distance would be equal to \(\frac{3}{4} \times 50\)

= 3.75 km = 3 km 750 m.

The number of degrees that the hour hand of a clock moves through between noon and 2.30 in the afternoon of the same day is

**Answer**: (c)

The hour hand moves from pointing to 12 to pointing to half way between 2 and 3.

The angle covered between each hour marking on the clock is \(\frac{360}{12} = 30\).

Since the hand has covered 2.5 of these divisions the angle moved through is 75.

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?

**Answer**: (a)

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr

Distance covered in 6 minutes = \(\left ( \frac{1}{60} \times 6 \right )\) km = \(\frac{1}{10}\) km = 100 m

∴ Distance between the thief and policeman = (200 – 100) m = 100 m.

The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m and travels towards A at 75 Km/hr. At what time do they meet?

**Answer**: (c)

Suppose they meet *x* hrs after 8 a.m.

then,

[Distance moved by first in *x* hrs] + [Distance moved by second in (*x* − 1) hrs] = 330.

Therefore, 60*x* + 75(*x* − 1) = 330.

⇒ *x* = 3.

So, they meet at (8 + 3) i.e, 11 a.m.

A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. The length of the bridge (in metres) is

**Answer**: (d)

\(Speed = \left ( 5 \times \frac{5}{18} \right )\) m/sec

= \(\frac{25}{18}\) m/sec.

Distance covered in 15 minutes = \(\left ( \frac{25}{18} \times 15 \times 60 \right )\) m

= 1250 m.

The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?

**Answer**: (c)

Total distance travelled in 12 hours = (35 + 37 + 39 + ..... upto 12 terms)

This is an A.P with first term, a = 35,

number of terms, n = 12 , d = 2.

Required distance = \(\frac{12}{2}\left [ 2 \times 35 + \left ( 12 - 1 \right ) \times 2 \right ]\) = 6(70 + 23)

= 552 kms.

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