## Speed, Time and Distance

#### Speed, Time and Distance

1. The speed of a car is 54 km/hr. What is its speed in m/sec?
1. 15 m/sec
2. 19-44 m/sec
3. 194.4 m/sec
4. 150 m/sec

1.  1 kmph = 5 m/sec 18

 ∴  54 kmph = 5 × 54 18

= 15 m/sec.

##### Correct Option: A

 1 kmph = 5 m/sec 18

 ∴  54 kmph = 5 × 54 18

= 15 m/sec.

1. A car goes 20 metres in a second. Find its speed in km/hr.
1. 18
2. 72
3. 36
4. 20

1.  1 m/sec = 18 kmph 5

 ∴  20 m/sec = 20 × 18 = 72 kmph 5

##### Correct Option: B

 1 m/sec = 18 kmph 5

 ∴  20 m/sec = 20 × 18 = 72 kmph 5

1. A car covers four successive 7 km distances at speeds of 10 km/hour, 20 km/hour, 30 km/
hour and 60 km/hour respectively. Its average speed over this distance is
1. 30 km/hour
2. 20 km/hour
3. 60 km/hour
4. 40 km/hour

1. Total distance = 7 × 4 = 28 km.

 Total time= 7 + 7 + 7 + 7 hours 10 20 30 60

 = 42 + 21 + 14 + 7 hours 60

 = 84 hours = 7 hours 60 5

 ∴  Average speed = Total distance Total time

 = 28 kmph 7/5

 = 28 × 5 = 20 kmph 7

##### Correct Option: B

Total distance = 7 × 4 = 28 km.

 Total time= 7 + 7 + 7 + 7 hours 10 20 30 60

 = 42 + 21 + 14 + 7 hours 60

 = 84 hours = 7 hours 60 5

 ∴  Average speed = Total distance Total time

 = 28 kmph 7/5

 = 28 × 5 = 20 kmph 7

1. A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/h. The distance between the two cities is
1. 540 km
2. 420 km
3. 600 km
4. 620 km

1. Fixed distance = x km and certain speed = y kmph (let).
Case I,

 x = x – 1 y + 10 y

 ⇒ x + 1 = x --- (i) y + 10 y

Case II,
 x = x − 1 − 3 y + 20 y 4

 = x − 4 + 3 y 4

 ⇒ x + 7 = x --- (ii) y + 20 4 y

From equations (i) and (ii),
 x + 1 = x + 7 y + 10 y + 20 4

 ⇒ x − x = 7 − 1 y + 10 y + 20 4

 ⇒ x y + 20 − y − 10 (y + 10)(y + 20)

 = 7 − 4 = 3 4 4

⇒ 3 (y + 10) (y + 20) = 40 x
 ⇒ 3 (y + 10) (y + 20) = x   ---(iii) 40

From equation (i),
 3 (y + 10) (y + 20) + 1 40(y + 10)

 = 3 (y + 10) (y + 20) 40y

⇒  3 (y +20) + 40
 = 3 (y + 10) (y + 20) y

⇒  3y2+ 60y + 40 y = 3(y2 + 30y+ 200)
⇒  3y2 + 100y = 3y2 + 90y + 600
⇒  10y = 600 ⇒ y = 60
Again from equation (i),
 x + 1 = x y + 10 y

 ⇒ x + 1 = x 60 + 10 60

 ⇒ x + 1 = x 70 60

 ⇒ x + 70 = x 70 60

⇒  6x + 420 = 7x
⇒  7x – 6x = 420
⇒  x = 420 km.

##### Correct Option: B

Fixed distance = x km and certain speed = y kmph (let).
Case I,

 x = x – 1 y + 10 y

 ⇒ x + 1 = x --- (i) y + 10 y

Case II,
 x = x − 1 − 3 y + 20 y 4

 = x − 4 + 3 y 4

 ⇒ x + 7 = x --- (ii) y + 20 4 y

From equations (i) and (ii),
 x + 1 = x + 7 y + 10 y + 20 4

 ⇒ x − x = 7 − 1 y + 10 y + 20 4

 ⇒ x y + 20 − y − 10 (y + 10)(y + 20)

 = 7 − 4 = 3 4 4

⇒ 3 (y + 10) (y + 20) = 40 x
 ⇒ 3 (y + 10) (y + 20) = x   ---(iii) 40

From equation (i),
 3 (y + 10) (y + 20) + 1 40(y + 10)

 = 3 (y + 10) (y + 20) 40y

⇒  3 (y +20) + 40
 = 3 (y + 10) (y + 20) y

⇒  3y2+ 60y + 40 y = 3(y2 + 30y+ 200)
⇒  3y2 + 100y = 3y2 + 90y + 600
⇒  10y = 600 ⇒ y = 60
Again from equation (i),
 x + 1 = x y + 10 y

 ⇒ x + 1 = x 60 + 10 60

 ⇒ x + 1 = x 70 60

 ⇒ x + 70 = x 70 60

⇒  6x + 420 = 7x
⇒  7x – 6x = 420
⇒  x = 420 km.

1.  Walking 6 th of his usual speed, a man is 12 minutes 7
late. The usual time taken by him to cover that distance is
1. 1 hour
2. 1 hour 12 minutes
3. 1 hour 15 minutes
4. 1 hour 20 minutes

1. Time and speed are inversely proportional.

 ∴  Usual time × 7 – usual time = 12 minutes 6

 ⇒  Usual time × 1 = 12 minutes 6

∴  Usual time = 72 minutes
= 1 hour 12 minutes
Second Method :
 t = 12 = 1 hrs. 60 5

 Usual time = A × time Diff. of A and B

 = 6 × 1 = 1 1 hrs. (7 − 6) 5 5

= 1 hrs. 12 minutes

##### Correct Option: B

Time and speed are inversely proportional.

 ∴  Usual time × 7 – usual time = 12 minutes 6

 ⇒  Usual time × 1 = 12 minutes 6

∴  Usual time = 72 minutes
= 1 hour 12 minutes
Second Method :
 t = 12 = 1 hrs. 60 5

 Usual time = A × time Diff. of A and B

 = 6 × 1 = 1 1 hrs. (7 − 6) 5 5

= 1 hrs. 12 minutes