## Speed, Time and Distance

#### Speed, Time and Distance

1. In a kilometre race, A beats B by 30 seconds and B beats C by 15 seconds. If A beats C by 180 metres, the time taken by A to run 1 kilometre is
1. 250 seconds
2. 205 seconds
3. 200 seconds
4. 210 seconds

1. A beats B by 30 seconds and B beats C by 15 seconds.
Clearly, A beats C by 45 seconds.
Also, A beats C by 180 metres.
Hence, C covers 180 metres in 45 seconds.

 ∴  Speed of C = 180 = 4 m/sec 45

∴  Time taken by C to cover 1000 m
 = 1000 = 250 sec. 4

∴  Time taken by A to cover 1000 m
= 250–45 = 205 sec.

##### Correct Option: B

A beats B by 30 seconds and B beats C by 15 seconds.
Clearly, A beats C by 45 seconds.
Also, A beats C by 180 metres.
Hence, C covers 180 metres in 45 seconds.

 ∴  Speed of C = 180 = 4 m/sec 45

∴  Time taken by C to cover 1000 m
 = 1000 = 250 sec. 4

∴  Time taken by A to cover 1000 m
= 250–45 = 205 sec.

1. Two cars start at the same time from one point and move along two roads at right angles to each other. Their speeds are 36 km/ hour and 48 km/hour respectively. After 15 seconds the distance between them will be
1. 400 m
2. 150 m
3. 300 m
4. 250 m

1. Let O be the starting point. The car running at 36 kmph is moving along OB and that at 48 kmph moving along OA. Also let they reach at B and A respectively after 15 seconds.

 ∴   OA = 48 × 5 × 15 = 200 m 18

 ∴   and OB = 36 × 5 × 15 = 150 m 18

∴  Required distance = AB
= √(200)2 + (150)2
W (By Pythagoras theorem)
= √40000 + 22500
= √62500
= 250 m

##### Correct Option: D

Let O be the starting point. The car running at 36 kmph is moving along OB and that at 48 kmph moving along OA. Also let they reach at B and A respectively after 15 seconds.

 ∴   OA = 48 × 5 × 15 = 200 m 18

 ∴   and OB = 36 × 5 × 15 = 150 m 18

∴  Required distance = AB
= √(200)2 + (150)2
W (By Pythagoras theorem)
= √40000 + 22500
= √62500
= 250 m

1. A and B run a kilometre and A wins by 25 sec. A and C run a kilometre and A wins by 275 m. When B and C run the same distance, B wins by 30 sec. The time taken by A to run a kilometre is
1. 2 min 25 sec
2. 2 min 50 sec
3. 3 min 20 sec
4. 3 min 30 sec

1. If A covers the distance of 1 km in x seconds, B covers the distance of 1 km in (x + 25) seconds. If A covers the distance of 1 km, then in the same time C covers only 725 metres.
If B covers 1 km in (x + 25) seconds, then C covers 1 km in (x + 55) seconds.
Thus in x seconds, C covers the distance of 725 m.

 ∴ x × 1000 = x + 55 725

⇒  x = 145
∴   A covers the distance of 1 km in 2 minutes 25 seconds.

##### Correct Option: A

If A covers the distance of 1 km in x seconds, B covers the distance of 1 km in (x + 25) seconds. If A covers the distance of 1 km, then in the same time C covers only 725 metres.
If B covers 1 km in (x + 25) seconds, then C covers 1 km in (x + 55) seconds.
Thus in x seconds, C covers the distance of 725 m.

 ∴ x × 1000 = x + 55 725

⇒  x = 145
∴   A covers the distance of 1 km in 2 minutes 25 seconds.

1. A car completes a journey in 10 hours. If it covers half of the journey at 40 kmph and the remaining half at 60 kmph, the distance covered by car is
1. 400 km
2. 480 km
3. 380 km
4. 300 km

1. Suppose distance be x km.

 x + x = 10 2 × 40 2 × 60

 ⇒ x + x = 10 80 120

 ⇒ 3x + 2x = 10 240

 ⇒ 5x = 10 240

x = 480 km

##### Correct Option: B

Suppose distance be x km.

 x + x = 10 2 × 40 2 × 60

 ⇒ x + x = 10 80 120

 ⇒ 3x + 2x = 10 240

 ⇒ 5x = 10 240

x = 480 km

1. A walks at a uniform rate of 4 km an hour; and 4 hours after his start, B bicycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A ?
1. 16.7 km
2. 18.6 km
3. 21.5 km
4. 26.7 km

1. Distance covered by A in 4 hours = 4 × 4 = 16 km
Relative speed of B with respect to A = 10 – 4 = 6 km/hr
∴  Time taken to catch A

 = 16 = 8 hours 6 3

 ∴ Required distance = 8 × 10 = 80 3 3

= 26.67 km. ≈ 26.7 km

##### Correct Option: D

Distance covered by A in 4 hours = 4 × 4 = 16 km
Relative speed of B with respect to A = 10 – 4 = 6 km/hr
∴  Time taken to catch A

 = 16 = 8 hours 6 3

 ∴ Required distance = 8 × 10 = 80 3 3

= 26.67 km. ≈ 26.7 km