A car leaves town heading west at 57 kilometers per hour... (MATH) ?

Question

A second car leaves town 2 hours after the first car, but it is traveling at 72 km/hr. To the nearest hour, when will the second car pass the first?

Can some one explain how to solve this problem?

Answer 1

57x=72(x-2)

57x=72x-144

15x=144

x=9.6

9.6 hours after the first car left.

to the nearest hour, it would be 10 hours after the first car left.

make it a good day

Answer 2

OK, at the time the second car started, the first car was traveling at 57km/h for 2 hours so is 114km in front.

Also you know every hour, the first car travels 57km and the second car 72km, so the second car "catches up" by 15km (72-57) every hour.

So, just do 114 divided by 15 for the number of hours = 7.6 hours = 7 hours 36 mins.

So to the NEAREST hour... 8 hours!

Of course if the second car heads in a different direction like east then it won't pass the first car at all =P

Hope that makes sense.

Answer 3

Let the first Car be Car A, and the second Car B(units in KM/hr)

For the first hour,

Car A = 57   &  Car B = 0  

For the second hour,

Car A = 114    &  Car B = 0

For the third hour,

Car A = 171   &  Car B  =  72

By the seventh hour,

Car  A  =  394   &  Car B = 360

So we can be assured that it will take at least 7 hours.

(As a guideline; you wouldn't want to be wrong)

(114  /  (72 - 57) ) =  7hrs 36mins

The above guideline was in place to prevent errors.

Some people might make a mistake and

add 2 to (114  /  (72 - 57) )

Do take note of it.

Answer 4

second will pass the first after 57*2/(72--57) hours = 7 3/5 hrs = 7h36m