0

Two cars started driving from two points at the same time. The distance between the points is $750$ kilometres, and it takes $5$ hours for the cars to meet.

One car travels $15$ kilometres in $5$ minutes less than the second car. What is the speed of the two cars?

My attempt

I used ratios and assumed that $15x$ is the speed of the first car and $15y$ is the speed of the second car, but it didn't work as it gave me a decimal answer.

Please show me your methods, making sure to explain clearly. The answer doesn't matter, the explanation matters. (Give me hints before answer)

Community
  • 1
Mork
  • 61
  • what is ish? I cannot understand. – Aniruddha Deshmukh Aug 23 '18 at 11:43
  • 1
    This is not clear...what does "$15$ kilometers $5$ minutes less than" mean? – lulu Aug 23 '18 at 11:43
  • I mean that the first car traveled 15 kilometers in 5 minutes less than the second car? ignore the ish. it's just 750 kilometers – Mork Aug 23 '18 at 11:51
  • Ok....but then the problem makes no sense. Set the speed of the faster car to anything you like, then set the speed of the slower car so that it takes $5$ minutes longer to go $15$ km. – lulu Aug 23 '18 at 11:54
  • Voting to close the question as it is unclear what you are asking. Please edit for clarity. – lulu Aug 23 '18 at 12:22

1 Answers1

1

Say car $B$ takes $t$ minutes to travel $15$ km. Then car $A$, which is the faster car, takes $t - 5$ minutes to travel $15$ km.

By rearranging these, we have that

  1. the speed of car $A$ is $15/(t - 5)$ km/min,
  2. the speed of car $B$ is $15/t$ km/min.

Now $5$ hours is equal to $300$ minutes. If they collectively travel $750$ km after $300$ minutes, then we have (using $\text{distance} = \text{speed} \times \text{time}$) $$ 300\times \left(\frac{15}{t - 5}\right) + 300\times \left(\frac{15}{t}\right) = 750. $$ Using simple albegra, we have $$ \frac{6}{t - 5} + \frac{6}{t} = 1 $$ so that $$ t^{2} - 13t - 30 = 0. $$ I will leave it to you to fill the workings between those three equations.

Hover over the following box for the remainder of the solution, but I would suggest doing so after giving the rest a go yourself.

From here, we use the quadratic formula for $t$ to find $$t = \frac{13 \pm \sqrt{13^{2} - 4(-30)}}{2} = \frac{13 \pm \sqrt{169 + 120}}{2} = \frac{13 \pm 17}{2}.$$ We want $t$ to be positive so we have $t = 15$. Thus $$\begin{array}{ll} 1. & \text{the speed of car $A$ is $15/(15 - 5) = 1.5$ km/min, or $90$ km/hour;}\\ 2. & \text{the speed of car $A$ is $15/15 = 1$ km/min, or $60$ km/hour.}\end{array}$$

Bilbottom
  • 2,658
  • 2
  • 15
  • 33