## Speed, Time and Distance

#### Speed, Time and Distance

1. A person, who can walk down a
 hill at the rate of 4 1 km/hour 2
and up the hill at the rate of 3 km/hour, ascends and comes
down to his starting point in 5 hours. How far did he ascend ?
1. 13.5 km
2. 3 km
3. 15 km
4. 9 km

1. Let the required distance be x km.

 ∴ x + x = 5 9/2 3

 ⇒   x 2 + 1 = 5 9 3

 ⇒  x 2 + 3 = 5 9

 ⇒ x = 5 × 9 = 9 km. 5

##### Correct Option: D

Let the required distance be x km.

 ∴ x + x = 5 9/2 3

 ⇒   x 2 + 1 = 5 9 3

 ⇒  x 2 + 3 = 5 9

 ⇒ x = 5 × 9 = 9 km. 5

1. A man walks a certain distance and rides back in 4 hours 30 minutes. He could ride both
ways in 3 hours. The time required by the man to walk both ways is
1. 4 hours 30 minutes
2. 4 hours 45 minutes
3. 5 hours
4. 6 hours

1. Suppose, time taken while walking be x hours
And, time taken on riding be y hours
∴  According to question

 x + y = 4 1 hours     ...(i) 2

Then, 2y = 3 hours
 y = 1 1 hours 2

From equation (i)
 x = 4 1 − 1 1 = 3 hours 2 2

Time required to walk both ways = 6 hours

##### Correct Option: D

Suppose, time taken while walking be x hours
And, time taken on riding be y hours
∴  According to question

 x + y = 4 1 hours     ...(i) 2

Then, 2y = 3 hours
 y = 1 1 hours 2

From equation (i)
 x = 4 1 − 1 1 = 3 hours 2 2

Time required to walk both ways = 6 hours

1. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they next meet at the starting point again ?
1. 46 minutes 12 seconds
2. 45 minutes
3. 42 minutes 36 seconds
4. 26 minutes 18 seconds

1. Required time = LCM of 252,
308 and 198 seconds.
Now, 252 = 2 × 2 × 3 × 3 × 7
308 = 2 × 2 × 7 × 11
198 = 2 × 3 × 3 × 11
∴  LCM = 2 × 2 × 3 × 3 × 7 × 11
= 36 × 77 seconds

 = 36 × 77 minutes 60

 = 231 = 46 minutes 12 seconds 5

##### Correct Option: A

Required time = LCM of 252,
308 and 198 seconds.
Now, 252 = 2 × 2 × 3 × 3 × 7
308 = 2 × 2 × 7 × 11
198 = 2 × 3 × 3 × 11
∴  LCM = 2 × 2 × 3 × 3 × 7 × 11
= 36 × 77 seconds

 = 36 × 77 minutes 60

 = 231 = 46 minutes 12 seconds 5

1. A man can reach a certain place in 30 hours.
 If he reduces his speed by 1 th, he goes 15
10 km less in that time. Find his speed per hour.
1. 6 km/hr
2.  5 1 km/hr 2
3. 4 km/hr
4. 5 km/hr

1. Let the speed of man be x kmph.

 ∴ 30x − 30 x − x = 10 15

 ⇒  30 x − x + x = 10 15

 ⇒ x = 10 15 30

 ⇒   x = 150 = 5 kmph 30

##### Correct Option: D

Let the speed of man be x kmph.

 ∴ 30x − 30 x − x = 10 15

 ⇒  30 x − x + x = 10 15

 ⇒ x = 10 15 30

 ⇒   x = 150 = 5 kmph 30

1. A and B start at the same time with speed of 40 km/hr and 50 km/hr respectively. If in covering the journey A takes 15 minutes longer than B, the total distance of the journey is :
1. 46 km
2. 48 km
3. 50 km
4. 52 km

1. Let the distance be x km

 Time taken by A = x hrs. 40

 Time taken by B = x hrs. 50

 Now, x − x = 15 40 50 60

 5x − 4x = 15 200 60

 ∴  x = 15 × 200 = 50km 60

Second Method :
 Distance = Product of speed × Diff. in time Diff. of speed

 = 40 × 50 × 15 = 50km 50 − 40 60

##### Correct Option: C

Let the distance be x km

 Time taken by A = x hrs. 40

 Time taken by B = x hrs. 50

 Now, x − x = 15 40 50 60

 5x − 4x = 15 200 60

 ∴  x = 15 × 200 = 50km 60

Second Method :
 Distance = Product of speed × Diff. in time Diff. of speed

 = 40 × 50 × 15 = 50km 50 − 40 60