algebra word prob: A man can walk from A to B and back in a certain time at 7km/hr.

[fizzy19]

A man can walk from A to B and back in a certain time at 7km/hr. If he walks from A to B at 8km/h and returns from B to A at 6km/hr, he takes 2 minutes more. Find the distance between A and B.

Have been working on this problem for hours now! i just can't seem to form the right equation. Can anyone please form an equation and explain to me how they did it?
thanks soo much

 

[Steven G]

A man can walk from A to B and back in a certain time at 7km/hr. If he walks from A to B at 8km/h and returns from B to A at 6km/hr, he takes 2 minutes more. Find the distance between A and B.

Have been working on this problem for hours now! i just can't seem to form the right equation. Can anyone please form an equation and explain to me how they did it?
thanks soo much

R*T =D

Let D= distance from A to B and from B to A

A to B at 8km/h: (8km/hr)T = D
B to A at 6km/hr, he takes 2 minutes more: (6km/hr)(T+2) = D

Since the two D's are equal to one another we can say that (8km/hr)T= (6km/hr)(T+1/30) or simply 8T=6(T+1/30). Solve for T and then find D


EDIT:T must be in hours (or minutes)
EDIT#2: I solved a different problem! I thought that the return trip at 6km/h took two more minutes than the trip from A to B at 8km/hr.

 

[soroban]

Hello, fizzy19!

A man can walk from A to B and back in a certain time at 7 km/h.

If he walks from A to B at 8 km/h and returns from B to A at 6 km/h,

he takes 2 minutes more. \(\displaystyle \;\)Find the distance between A and B.

Careful!

The speed is given in kilometers per hour.

The difference in time is given in minutes.

\(\displaystyle \quad 2\text{ min.} = \frac{1}{30}\text{ hr.}\)


Let \(\displaystyle d\) = distance from \(\displaystyle A\) to \(\displaystyle B.\)


He walks \(\displaystyle 2d\) km at 7 km/h.

\(\displaystyle \quad\) This takes \(\displaystyle T_1 \:=\:\frac{2d}{7}\) hours.


He walks from A to B at 6 km/hr.

\(\displaystyle \quad\) This takes\(\displaystyle \frac{d}{6}\) hours.

He walks from B to A at 8 km/hr.

\(\displaystyle \quad\) This takes \(\displaystyle \frac{d}{8}\) hours.

His total time is: \(\displaystyle \:T_2 \:=\:\frac{d}{6} + \frac{d}{8}\) hours.

\(\displaystyle \quad\)And this time is \(\displaystyle \frac{1}{30}\) more than \(\displaystyle T_1.\)



There is our equation! \(\displaystyle \;\;\;\dfrac{d}{6} + \dfrac{d}{8} \:=\:\dfrac{2d}{7} + \dfrac{1}{30}\)