if a man walks at the rate of 5kmph,he misses a train by only 7 min.however if he walks at the rate of 6kmph he reaches the station 5 minutes before the arrival of train.find the distance covered by him to reach the station?
if a man walks at the rate of 5kmph,he misses a train by only 7 min.however if he walks at the rate of 6kmph he reaches the station 5 minutes before the arrival of train.find the distance covered by him to reach the station?
Let d be the distance travelled to the train in kilometres.
Let s be the speed travelled to reach the train on time in km/h.
Let t be the time taken to reach the train on time, in hours.
d = st
d = 5(t+7/60)
d = 6(t-5/60)
5(t+7/60) = 6(t-5/60)
5t + 35/60 = 6t - 30/60
t = 65/60
d = 5(72/60) = 360/60 = 6 kilometres.
Hello, hena!
Nice job, Sean!
. . I have a slightly difference approach . . .
If a man walks at the rate of 5 km/hr, he misses a train by only 7 minutes.
However, if he walks at the rate of 6 km/hr, he reaches the station 5 minutes early.
Find the distance covered by him to reach the station?
Let $\displaystyle d$ = distance to the station.
Let $\displaystyle T$ = exact time to reach the station on time.
We will use: . $\displaystyle \text{(Distance)} \;=\;\text{(Speed)} \times\text{(Time)} \quad\Rightarrow\quad t \:=\:\frac{D}{S}$
At 5 km/hr, it takes him: .$\displaystyle \frac{d}{5} $ hours.
. . But he will be 7 minutes late: .$\displaystyle \frac{d}{5} \:=\:T + \frac{7}{60}\;\;{\color{blue}[1]} $
At 6 km/hr, it takes him: .$\displaystyle \frac{d}{6}$ hours.
. . But he will be 5 minutes early: .$\displaystyle \frac{d}{6} \:=\:T - \frac{5}{60}\;\;{\color{blue}[2]} $
Solve the system of equations.
Subtract [2] from [1]: .$\displaystyle \frac{d}{5} - \frac{d}{6} \;=\;\frac{7}{60} + \frac{5}{60} \quad\Rightarrow\quad \frac{d}{30} \:=\:\frac{1}{5}\quad\Rightarrow\quad \boxed{d \:=\:6}$