Question

Drive to work in 45 min., bus to work takes 1 hr 15 min. If the average rate on the bus is 16 mph slower than his driving rate, how far does he travel to work?

Answer 1

30 miles

Explanation:

Given: Drive to work in 45 min.

Bus would take to travel= 1 hr 15 min

Average rate on the bus is 16 mph slower than his driving rate.

Lets assume the average driving rate be "x" mph.

∴ Average rate of bus=
`(x-16)\ mph`

Now converting unit of time taken to travel to work.

Remember; 1 hours= 60 minutes.

Time taken for drive to work=
`(45\ minutes)/(60\ minutes) = 0.75\ hour`

Time taken in bus to work= 1 hour and 15 minutes=
`60\ minutes+ 15\ minutes= 75\ minutes`

∴ Time taken in bus to work in hours=
`(75\ minutes)/(60\ minutes)= 1.25\ hours`

Considering the distance remain constant to travel for work.

Next, forming an equation to find the value of x.

As we know,
`distance= speed* time`

⇒
`x* 0.75= (x-16)* 1.25`

Using distributive property of multiplication.

⇒
`0.75x= 1.25x-20`

Adding both side by 20

⇒
`0.75x+20= 1.25x`

Subtracting both side by 0.75x

⇒
`20= 0.5x`

Dividing both side by 0.5

⇒
`x= (20)/(0.5)`

∴
`x= 40\ mph`

Hence the average rate for drive to work is 40mph.

Now, subtitiuting the value of x to find the distance travelled to work.

Distance travelled=
`40* 0.75= 30\ miles`

Hence, distance travelled to work is 30 miles.