Two cars travel the same distance. The first car travels at a rate of 41 mph and reaches its destination in t hours. The second car travels at a rate of 54 mph and reaches its destination in 3.7 hours earlier than the first car. How long does it take for the first car to reach its destination? Answer in units of hours.

Guest Jun 25, 2015

#2**+5 **

Let's call the time it takes the first car to travel the dstance, t.

Then, the time the second car travels is (t - 3.7) hrs. ..... and since the distances are the same, we have

54(t - 3.7) = 41t simplify

54t - 199.8 = 41t rearrange

54t - 41t = 199.8 simplify

13t = 199.8 divde both sides by 13

t = about 15.369 hrs

CPhill Jun 25, 2015

#1**0 **

41_{mph} * t_{hours} = Total distance

54_{mph} * 3.7_{hours} = Total distance

Therefore: 41_{mph} * t_{hours} = 54_{mph} * 3.7_{hours}

*Divide both sides by 41 to make "t" the subject of the equation.*

t = (54 * 3.7) / 41

= 199.8 / 41

= 4.8731707317073171

= **4.9 _{hours} (to 1 d.p.)**

Sir-Emo-Chappington Jun 25, 2015

#2**+5 **

Best Answer

Let's call the time it takes the first car to travel the dstance, t.

Then, the time the second car travels is (t - 3.7) hrs. ..... and since the distances are the same, we have

54(t - 3.7) = 41t simplify

54t - 199.8 = 41t rearrange

54t - 41t = 199.8 simplify

13t = 199.8 divde both sides by 13

t = about 15.369 hrs

CPhill Jun 25, 2015

#3**0 **

Once again I give incorrect answers because I skim over a detail >_<'

I shall make it my task to read things in their entirety from now on!

Sir-Emo-Chappington Jun 25, 2015