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

 13.5 km
 3 km
 15 km
 9 km

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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
 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

 4 hours 30 minutes
 4 hours 45 minutes
 5 hours
 6 hours

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Suppose, time taken while walking be x hours
And, time taken on riding be y hours
∴ According to questionx + y = 4 1 hours ...(i) 2
Then, 2y = 3 hoursy = 1 1 hours 2
From equation (i)x = 4 1 − 1 1 = 3 hours 2 2
Time required to walk both ways = 6 hoursCorrect Option: D
Suppose, time taken while walking be x hours
And, time taken on riding be y hours
∴ According to questionx + y = 4 1 hours ...(i) 2
Then, 2y = 3 hoursy = 1 1 hours 2
From equation (i)x = 4 1 − 1 1 = 3 hours 2 2
Time required to walk both ways = 6 hours
 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 ?

 46 minutes 12 seconds
 45 minutes
 42 minutes 36 seconds
 26 minutes 18 seconds

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

 6 km/hr

5 1 km/hr 2  4 km/hr
 5 km/hr

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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
 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 :

 46 km
 48 km
 50 km
 52 km

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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