# help me out to find the distance

• May 23rd 2008, 08:30 PM
neha1
help me out to find the distance
(Doh)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?
• May 23rd 2008, 09:05 PM
sean.1986
Quote:

Originally Posted by hena
(Doh)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.
• May 24th 2008, 04:21 AM
Soroban
Hello, hena!

Nice job, Sean!
. . I have a slightly difference approach . . .

Quote:

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 $d$ = distance to the station.
Let $T$ = exact time to reach the station on time.

We will use: . $\text{(Distance)} \;=\;\text{(Speed)} \times\text{(Time)} \quad\Rightarrow\quad t \:=\:\frac{D}{S}$

At 5 km/hr, it takes him: . $\frac{d}{5}$ hours.
. . But he will be 7 minutes late: . $\frac{d}{5} \:=\:T + \frac{7}{60}\;\;{\color{blue}[1]}$

At 6 km/hr, it takes him: . $\frac{d}{6}$ hours.
. . But he will be 5 minutes early: . $\frac{d}{6} \:=\:T - \frac{5}{60}\;\;{\color{blue}[2]}$

Solve the system of equations.

Subtract [2] from [1]: . $\frac{d}{5} - \frac{d}{6} \;=\;\frac{7}{60} + \frac{5}{60} \quad\Rightarrow\quad \frac{d}{30} \:=\:\frac{1}{5}\quad\Rightarrow\quad \boxed{d \:=\:6}$