Time and Distance Questions - 1

I start from my home for the school at a certain fixed time. If I walk at the rate of 5 km an hour I am late by 7 minutes. However, if I walk at the rate of 6 km an hour, I reached the college 5 minutes earlier than the scheduled time. Find the distance of the school from my home ?
  • 12 km
  • 8 km
  • 4 km
  • 5 km
  • 6 km
Explanation   

Let the distance be 1 km.

Time taken to cover 1 km at 5 kmph = $\displaystyle \frac{1}{5}$ hours.

Time taken to cover 1 km at 6 kmph = $\displaystyle \frac{1}{6}$ hours.

Difference of times in covering 1 km = $\displaystyle \left(\frac{1}{5} - \frac{1}{6}\right) = \frac{1}{30}$ Hours.

Actual difference of time in covering whole distance = (7 + 5) mts. = 12 minutes = $\displaystyle \frac{1}{5}$ hours.

If the difference of time is ($\displaystyle \frac{1}{30}$) hours. Distance = 1 km.

If the difference of time $\displaystyle \frac{1}{5}$ hours. Distance = 30 $\displaystyle \times \frac{1}{5}$ = 6 km.

Distance between school and my home is 6 km.

Workspace
A person travelled from Town A to Town B at 55 km/hr and returned from Town B to Town A at 45 km/hr. Find average speed of the whole trip ?
  • 50 Km/hr
  • 52.5Km/hr
  • 55 Km/hr
  • 49.5 Km/hr
  • 49 Km/hr
Explanation   

Average speed:

Where speed from point A to point B is at V1 km/hr and point B to point A is at V2 km/hr then

Average speed = $\displaystyle \frac{2 \times \text{V}_1 \times \text{V}_2}{\text{V}_1 + \text{V}_2} = \ \frac{2 \times 55 \times 45}{55 + 45} => \frac{4950}{100}$ = 49.5 km/hr.

Workspace
From his residence walking at 6 Km/hr, a person reaches his office 10 minutes late, or walking at 8 Km/hr he reach his office 5 minutes early. Find distance between residence and office of that person ?
  • 4 Km
  • 6 Km
  • 8 Km
  • 7 Km
  • 5 Km
Explanation   

Let distance be x km.

$\displaystyle \frac{\text{x}}{6} - \frac{\text{x}}{8} = \frac{15}{60} = \frac{\text{x}}{24} = \frac{1}{4}$

$\displaystyle \text{x} = 24 \times \frac{1}{4}$ = 6 km.

OR

$\displaystyle \frac{\text{V}_1 \times \text{V}_2 \times \text{Time}}{\text{V}_1 - \text{V}_2} => \frac{6 \times 8 \times \frac{1}{4}}{6 - 8} => 24 \times \frac{1}{4}$ = 6km.

Workspace
A person travelled half of his journey at 32 Km/hr remaining at 48 Km/hr, total time of the journey is 5 hours. Then total distance covered was ?
  • 192 Km
  • 300 Km
  • 250 Km
  • 240 Km
  • 280 Km
Explanation   

Distance = $\displaystyle \frac{2 \times \text{V}_1 \times \text{V}_2 \times \text{Time}}{\text{V}_1 + \text{V}_2}$

V1, V2 are different speed during the Journey.

Distance = $\displaystyle \frac{2 \times 32 \times 48 \times 5}{80}$ = 192 km.

Workspace
To cover a distance, three cars take time in the ratio 3 : 4 : 5, the speeds of the cars are in the ratio ?
  • 3 : 4 : 5
  • 15 : 20 : 25
  • 25 : 20 : 15
  • 24 : 18 : 15
  • 20 : 15 : 12
Explanation   

Let distance be x => Speed = $\displaystyle \frac{\text{Distance}}{\text{Time}}$

Let speeds of Car1, Car2, Car3 be = $\displaystyle \frac{\text{x}}{3} \ , \frac{\text{x}}{4} \ , \frac{\text{x}}{5} => \frac{20 \ : \ 15 \ : \ 12}{60}$

Therefore ratio of speed of the three cars is 20 : 15 : 12.

Workspace
Subhash travelled $\displaystyle \frac{1}{3}$ rd of his journey by train, $\displaystyle \frac{1}{4}$ th of the remaining by bus, $\displaystyle \frac{1}{3}$ rd of the remaining by car, $\displaystyle \frac{1}{2}$ of the remaining by bike, remaining 40 Km by taxi, total distance covered in the entire journey is ?
  • 360 Km
  • 240 Km
  • 300 Km
  • 280 Km
  • 320 Km
Explanation   

Follow reverse process:

$\displaystyle \frac{1}{2}$ = 40 => 40 $\displaystyle \times \frac{2}{1} = 80 = [\frac{1}{2}]$

80 $\displaystyle \times \frac{2}{1} = 160 = [\frac{3}{4} \text{th}]$

Total distance = 160 $\displaystyle \times \frac{3}{2}$ = 240

Then 240 $\displaystyle \times \frac{1}{3}$ = 80 km [Train].

Remaining = 240 – 80 = 160 $\displaystyle \times \frac{1}{4}$ = 40 km [Bus].

Remaining = 160 – 40 = 120.

120 $\displaystyle \times \frac{1}{3}$ = 40 km [Car].

Remaining = 120 – 40 = 80 km.

80 $\displaystyle \times \frac{1}{2}$ = 40 km [Bike].

Remaining = 80 – 40 = 40 [Taxi].

Workspace
A and B travel same distance at 10 Km and 12 Km respectively, if A takes 15 minutes more time than B takes, then the distance covered is ?
  • 12 Km
  • 13 Km
  • 15 Km
  • 16 Km
  • 17 Km
Explanation   

Let distance be x km.

$\displaystyle \frac{\text{x}}{10} - \frac{\text{x}}{12}$ = 15 min => $\displaystyle \frac{\text{x}}{10} - \frac{\text{x}}{12} = \frac{1}{4}$

=> $\displaystyle \frac{\text{x}}{60} = \frac{1}{4} => 60 \times \frac{1}{4}$ = 15 km.

Workspace
A driver has to reach a town which is 282 Km distant in 7 hours. He covered 120 Km in 4 hours. Then at what speed he must have to travel to reach the town according to the scheduled time ?
  • 60 Km/hr
  • 48 Km/hr
  • 50 Km/hr
  • 63 Km/hr
  • 54 Km/hr
Explanation   

Required speed = $\displaystyle \frac{\text{Remaining distance}}{\text{Remaining time}} = \frac{162 \ \text{km}}{3 \ \text{hours}}$ = 54 km/hr.

Workspace
While travelling a distance of 30 Km a person finds that he has travelled $\displaystyle \frac{1}{5}$th of the remaining distance in 60 minutes, then the total time taken by the person to cover the distance is ?
  • 6 hours
  • 5 hours
  • 7 hours
  • 4 hours
  • 8 hours
Explanation   

By using the method => $\displaystyle \left(\frac{\text{R}}{100 + \text{R}}\right) \times 100$

R = $\displaystyle \frac{1}{5}$ or 20% = $\displaystyle \left(\frac{20}{100 + 20}\right) \times 100 = \frac{1}{6}$ th distance.

$\displaystyle \frac{1}{6}$ th time = 1 hour => total time = 1 $\displaystyle \times \frac{6}{1}$ = 6 hours.

Workspace
Walking at $\displaystyle \frac{3}{4}$th of usual speed a person was 15 minutes late to his office, what is his usual time to reach his office ?
  • 40 minutes
  • 30 minutes
  • 45 minutes
  • 36 minutes
  • 48 minutes
Explanation   

Usual Time = $\displaystyle 1 - \frac{\text{b}}{\text{a}}$ = Late time/Early time.

Here $\displaystyle \frac{\text{a}}{\text{b}} = \frac{3}{4} => 1 – \frac{4}{3}$ = 15 min => $\displaystyle \frac{1}{3}$ = 15 or $\displaystyle \frac{3}{1} \times$ 15 = 45 min.

OR

Usual time = $\displaystyle \frac{\text{a}}{\text{a} - \text{b}} \times$ Late time => $\displaystyle \frac{3}{3 - 4} \times 15$ = 45 min.

Workspace

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Time and Distance Shortcuts and Formulas


Distance = Speed x Time.

Distance = $\displaystyle \frac{2 \times \text{V}_1 \times \text{V}_2 \times \text{Time}}{\text{V}_1 + \text{V}_2}$

V1, V2 are different speed during the Journey.

Time = Distance ÷ Speed

Speed = Distance ÷ Time

Where speed from point A to point B is at V1 km/hr and point B to point A is at V2 km/hr then

Average speed = $\displaystyle \frac{2 \times \text{V}_1 \times \text{V}_2}{\text{V}_1 + \text{V}_2}$

Usual Time = $\displaystyle 1 - \frac{\text{b}}{\text{a}}$ = Late time/Early time.

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