Question

Matthew jogged to his friend’s house 12 miles away and then got a ride back home. It took him 2 hours longer to jog there than ride back. His jogging rate was 25 mph slower than the rate when he was riding. What was his jogging rate?

Answer 1

the answer is D

Explanation:

Answer 2

The jogging rate of Matthew is 5 mph.

Explanation:

Let the jogging rate of Matthew be x mph.

It is given that his jogging rate was 25 mph slower than the rate when he was riding. So, the riding rate is (x+25) mph.

The distance between Matthew and his friend's house is 12 miles.

`Speed=(Distance)/(Time)`

`Time=(Distance)/(Speed)`

The time taken by Matthew in jogging is
`(12)/(x)` and the time taken by Matthew in riding is
`(12)/(x+25)`.

It took him 2 hours longer to jog there than ride back.

`(12)/(x)=(12)/(x+25)+2`

`(12)/(x)-(12)/(x+25)=2`

`(12(x+25)-12x)/(x(x+25))=2`

`12x+300-12x=2x(x+25)`

`300=2x^2+50x`

`0=2x^2+50x-300`

`0=x^2+25x-150`

`0=x^2+30x-5x-150`

`0=(x+30)(x-5)`

Equate each factor equal to 0.

`x=5,-30`

The speed cannot be negative, therefore the jogging rate of Matthew is 5 mph.