this is a word problem about rates

Question

Jennifer drove from Town A to Town B at a certain speed. On her way back, she travelled 10km/h faster and took 12.5% less time to complete the journey. Find the speed that Jennifer travelled at while driving from Town A to Town B.

Answer 1

Lets write out the equations given from the problem.

 

1.) She traveled from Town A to Town B.

Let s₁ be the speed she traveled. Speed = distance / time.

So let s₁ = d/t₁ where d is distance from town A to town B and t₁ is time it took.

 

2.) On way back she travelled 10 km/h faster.

So s₂ = s₁ + 10

 

3.) It took 12.5 percent less time to complete the journey.

So t₂ = (1-0.125)*t₁ = 0.875t₁

 

Solve for d in # 1.

We have d = s₁t₁ (*)

 

Similarly d = s₂t₂

 

Set equal and we have

s₁t₁ = s₂t₂ = (s₁+10)(0.875)t₁

                 = 0.875s₁t₁ + 8.75t₁

 

Solving you get

0.125s₁t₁ = 8.75t₁

 

Divide by t₁ and 0.125 so you get s₁ = 70 km/hr.

 

Using #2 the speed back is 10 km/hr more so it is

s₂ = 80 km/hr.

Hope that is helpful.

Best - David