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.
Average speed:
Where speed from point A to point B is at V_{1} km/hr and point B to point A is at V_{2} 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.
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.
Distance = $\displaystyle \frac{2 \times \text{V}_1 \times \text{V}_2 \times \text{Time}}{\text{V}_1 + \text{V}_2}$
V_{1}, V_{2} are different speed during the Journey.
Distance = $\displaystyle \frac{2 \times 32 \times 48 \times 5}{80}$ = 192 km.
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.
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].
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.
Required speed = $\displaystyle \frac{\text{Remaining distance}}{\text{Remaining time}} = \frac{162 \ \text{km}}{3 \ \text{hours}}$ = 54 km/hr.
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.
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.
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Distance = Speed x Time.
Distance = $\displaystyle \frac{2 \times \text{V}_1 \times \text{V}_2 \times \text{Time}}{\text{V}_1 + \text{V}_2}$
V_{1}, V_{2} are different speed during the Journey.
Time = Distance ÷ Speed
Speed = Distance ÷ Time
Where speed from point A to point B is at V_{1} km/hr and point B to point A is at V_{2} 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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