Question in description.

Question

Cody drove to the town hall and back. It took one hour less time to get there then it did to get back. The average speed on the trip there was 30 mph. The average speed on the way back was 24 mph. How many hours did the trip there take?

Answer 1

Hi Aalam, the distance both ways is the same, call it D. Let x be the time it took to get back. Then x - 1 is the time it took to get there.

 

Equation:

 

D(istance) = R(ate) * T(ime)

 

Given: D_t = D_b, D_t is the distance traveled to the town hall. D_b is the same distance returning home.

 

Getting to town hall:

 

D_t = R_t * T_t

 

R_t = x - 1

T_t = 30 mph

 

D_t = 30 * (x - 1)

 

D_t = 30x - 30

 

Returning home:

 

D_b = R_b * T_b

R_b = x
T_b = 24 mph

D_b = 24 * x

D_b = 24x

 

Distance is the same both ways:

 

Since D_t = D_b, substitute for D_t and D_b

 

30x - 30 = 24x

 

Subtract 24x from both sides:

 

6x - 30 = 0

 

Add 30 to both sides:

 

6x = 30 

 

Divide both sides by 6:

 

x = 5 hours

 

It took 5 hours to get back home from the town hall. 

 

It took 4 hours to get from home to the town hall.

 

Questions?

Answer 2

The key to this type of problem is to remember velocity * time = distance, and the distance is the same both ways.

 

v₁t₁ = v₂t₂   and   t₂ = t₁ + 1    ∴   v₁t₁ = v₂(t₁ + 1)  

 

30t₁ = 24(t₁ + 1)

 

6t₁ = 24

 

t₁ = 4 hours