1
$\begingroup$

If David runs from home to school at 12 km/hour, he arrives 30 minutes earlier than Ana who walks at 4 km/hour. Find the distance between home and school.

What have I tried? I tried to use the following formula $$\text{time}=\frac{\text{distance}}{\text{velocity}}$$ So I thought that I had to subtract the velocity of Ana from David and use the formula. I am so confused and I need help with solving this problem. Thank you!

$\endgroup$
  • 6
    $\begingroup$ What have you tried? $\endgroup$ – Arthur Aug 6 '19 at 20:43
  • 1
    $\begingroup$ Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. $\endgroup$ – saulspatz Aug 6 '19 at 20:44
2
$\begingroup$

As with most word problems, you need to define variables and write equations that reflect the information you are given. Let the distance to school be $d$. How long does David take to get there? How long does Ana take to get there? What relationship are you given between these two numbers? Now you have an equation to solve.

$\endgroup$
1
$\begingroup$

Note that the distance that David and Ana travel are equal, since the problem doesn't specify otherwise. This means that one variable can be used for the distance, which we'll call d.

Now, let's try to write the formula for David. You seem to know the formula time = distance/velocity, which is good. Just plug in the values you know. We don't know the time, so we'll keep that as a variable, which we'll call t. We do know his velocity, and that's 12 km/hr, and we don't know the distance, so that'll stay as d. Now, plug those into time = distance/velocity, and you get t = d/12.

Let's write the formula for Ana. Her speed is 4 km/hr and her distance is also d. As for her time, we know it's 30 minutes longer than David's time. You might be tempted by this fact to write her time as t + 30, but notice that we keep the other values in terms of hours (e.g. 12 km/hr and 4 km/hr), so we have to keep this expression in hours, too. 30 minutes is half an hour so the expression is actually t + 1/2. We use the same variable t again because we're simply adding half an hour onto t, which is David's time. We have all of the values needed to write the formula for Ana now. Plug the numbers and variables in, and we get t + 1/2 = d/4.

This is important: we now have two equations and two variables, so we can solve the two equations t + 1/2 = d/4 and t = d/12.

Try to solve it yourself. I helped you through the hard part of word problems, which is turning words into numbers.

$\endgroup$
1
$\begingroup$

Yes we do subtract Ana's time from David's time, since Ana took 30 minutes more to get there. So your intuition was almost there. Here is one way to set up this equation.

$$\frac{\text{distance}}{4\frac{\text{km}}{\text{hour}}}-\frac{\text{distance}}{12\frac{\text{km}}{\text{hour}}}=\frac12\text{hour}$$ We aim to find the distance so we must first make the denominators the same $$\frac{3\cdot\text{distance}}{3\cdot4\frac{\text{km}}{\text{hour}}}-\frac{\text{distance}}{12\frac{\text{km}}{\text{hour}}}=\frac12\text{hour}$$ $$\frac{3\cdot\text{distance}}{12\frac{\text{km}}{\text{hour}}}-\frac{\text{distance}}{12\frac{\text{km}}{\text{hour}}}=\frac12\text{hour}$$ Now after subtracting one fraction from another, we have $$\frac{2\cdot\text{distance}}{12\frac{\text{km}}{\text{hour}}}=\frac12\text{hour}$$ I would be doing you a disservice if I just solved the whole problem for you. I trust that you can take it from here. Feel free to ask questions if any.

$\endgroup$
0
$\begingroup$

Call $t$ the time, $v$ the velocity, $d$ the distance, $D$ David and $A$ Anna.

We know that $v_D=12, v_A=4$ and $t_D-t_A = 0.5$ (30 minutes converted in hours).

But $t_D-t_A = \frac{d}{v_D} - \frac{d}{v_A}$ and so we can calculate the distance $d$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.