Question

Alan leaves los angeles at 8:00 am to drive to san francisco, 400 mi away. he travels at a steady 50 mph. beth leaves los angeles at 9:00 am and drives a steady 60 mph. who gets to san francisco first? how long does the first to arrive have to wait for the second?

Answer 1

Given, Alan leaves Los Angeles at 8:00 am to San Fransisco which is 400 miles away from Los Angeles.

He travels at a rate of 50 mph.

So 50 miles distance covered in 1 hour

1 mile distance covered in $\frac{1}{50}$ hour

400 miles distance covered in $\frac{400}{50}$ hours = 8 hours.

Alan takes 8 hours to reach San Fransisco.

The time will be = 8:00 am + 8 hours = 4:00 pm.

Now Beth leaves Los Angeles at 9:00 am.

His speed is 60mph.

So 400 miles distance covered in $\frac{400}{60}$ hours.

We can simplify $\frac{400}{60}$ by dividing 400 and 60 both by 20. We will get $\frac{20}{3}$ there after simplification.

So Beth takes $\frac{20}{3}$ hours to reach San Fransisco.

Let's convert $\frac{20}{3}$ hours to minute. We know that there are 60 minutes in 1 hour.

So $\frac{20}{3}$ hours = $(\frac{20}{3})(60)$ minutes

= $\frac{(20)(60)}{3}$ minutes = $\frac{1200}{3}$ minutes

= $400$ minutes = $(6)(60)+40$ minutes = 6 hours 40 minutes.

So Beth takes 6 hours and 40 minutes to reach San Fransisco.

The time will be = 9:00 am + 6 hours 40 minutes = 3:40 pm.

So we have got Beth gets to San Fransisco first.

He will wait for $(4 pm - 3:40 pm)$ = 20 minutes.

We have got the required answer here.