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$A$ and $B$ walk back and forth along the path between their houses at uniform rates, with $B$ walking faster.

One evening, they leave home at the same time and pass each other 55 meters from $A$'s house. After $B$ turns at $A$'s house, he catches up to her before she reaches his house ($B$'s), at a point 85 meters from her ($A$'s) house.

There is an ice cream truck 25 meters from $B$'s house and smoothie truck near $A$'s house, and at the beginning of their walk they passed the trucks near their respective homes at the same time.

How far apart are the smoothie truck and and ice cream truck?

I tried to set up some equations, letting $x$ be the distance between their homes.

In time $t$, $B = x-55$, and $A = 55$.

In time $t+y$ (second meeting), $B = 2x-140$, $A = x-85$.

Then using $d = vt$, and $t= d/v$ I wrote out, with $a$ as $A$'s speed and $b$ as $B$'s speed

$$\frac{55}{a} \label{a}\tag{1}$$

$$\frac{x-55}{b} \label{b}\tag{2}$$

$$\frac{x-85}{a} \label{c}\tag{3}$$

$$\frac{2x-140}{b} \label{d}\tag{4} $$

When I tried to use a system of equations for these, making \ref{a} and \ref{b} and \ref{c} and \ref{d} equal since they had the same time variable, it did not work.

Any ideas? Any and all help is appreciated.

Hans Hüttel
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user978757
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1 Answers1

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You need to use all the information given. You already have two equations:

$$ \frac{55}{a} = \frac{x-55}{b} \label{e}\tag{5} $$

$$ \frac{x-85}{a} = \frac{2x -140}{b} \label{f}\tag{6}$$

From these, you get two equations for the ratio $\frac{a}{b}$. You can use these to find $x$ and thereby the ratio $\frac{a}{b}$.

But you also need to use the information about the trucks. Let $s$ be the distance from $B$'s house to the smoothie truck. We know that

$$ \frac{25}{b} = \frac{s}{a} \label{g} \tag{7}$$

and it is now easy to see that $s = 25 \frac{a}{b}$.

Hans Hüttel
  • 4,271