3. A straight path is $2 \mathrm{~km}$ in length. Beatrice walks from the beginning of the path to the end of the path at a constant speed of $5 \mathrm{~km} / \mathrm{h}$. Hieu cycles from the beginning of the path to the end of the path at a constant speed of $15 \mathrm{~km} / \mathrm{h}$.
(a) Suppose that Hieu starts 10 minutes later than Beatrice. Determine the number of minutes that it takes Hieu to catch up to Beatrice on the path.
(b) Suppose that Beatrice starts at a time that is exactly $b$ minutes after 9:00 a.m. and that Hieu starts at a time that is exactly $h$ minutes after 9:00 a.m., where $b$ and $h$ are integers from 0 to 59 , inclusive, that are each randomly and independently chosen.
(i) Determine the probability that there is a time at which Beatrice and Hieu are at the same place on the path. (Beatrice and Hieu are at the same place at a given time if they start at the same time or finish at the same time, or are at the same place at some point between the beginning and end of the path at the same time.)
(ii) One day, Beatrice uses a scooter and travels from the beginning of the path to the end of the path at a constant speed of $x \mathrm{~km} / \mathrm{h}$, where $x>5$ and $x<15$. Hieu still cycles from the beginning of the path to the end of the path at a constant speed of $15 \mathrm{~km} / \mathrm{h}$. If the probability that there is a time at which Beatrice and Hieu are at the same place on the path is $\frac{13}{200}$, determine the range of possible values of $x$.
Since 10 minutes is $\frac{1}{6}$ of an hour, when Beatrice walks at $5 \mathrm{~km} / \mathrm{h}$ for 10 minutes, she walk a distance of $5 \mathrm{~km} / \mathrm{h} \times \frac{1}{6} \mathrm{~h}=\frac{5}{6} \mathrm{~km}$. Since Hieu cycles at $15 \mathrm{~km} / \mathrm{h}$ and Beatrice walks at $5 \mathrm{~km} / \mathrm{h}$, then Hieu catches up to In other
In other words, Hieu closes the $\frac{5}{6} \mathrm{~km}$ gap at a rate of $10 \mathrm{~km} / \mathrm{h}$, which takes $\frac{\frac{5}{6} \mathrm{~km}}{10 \mathrm{~km} / \mathrm{h}}$, or $\frac{5}{60} \mathrm{~h}$, or 5 minutes.
Solution 2
When Beatrice walks at $5 \mathrm{~km} / \mathrm{h}$ for 10 minutes, she walks a distance of $5 \mathrm{~km} / \mathrm{h} \times \frac{1}{6} \mathrm{~h}$ or $\stackrel{5}{6} \mathrm{~km}$.
Since Hieu cycles at $15 \mathrm{~km} / \mathrm{h}$, Hieu cycles $15 \mathrm{~km} / \mathrm{h} \times t \mathrm{~h}=15 t \mathrm{~km}$ in total over the course Using similar reasoning, Beatrice walks $5 t \mathrm{~km}$ after her initial $\frac{5}{6} \mathrm{~km}$. When Hieu catches up to Beatrice, they have travelled exactly the same distance, and so $15 t=5 t+\frac{5}{6}$ which gives $10 t=\frac{5}{6}$ and so $t=\frac{5}{60}$. Thus, Hieu catches up to Beatrice after $\frac{5}{60} \mathrm{~h}$, which is equivalent to 5 minutes.
(b) (i) There are 60 integers in the interval from 0 to 59 inclusive.
This means that there are 60 possible values for $b$ and 60 possible values for $h$. Since $b$ and $h$ are chosen randomly and independently, then there are $60 \times 60=3600$ We say that Beatrice and Hieu meet if they are at the same place on the path at the same time, including possibly starting or ending together. pairs of values for $b$ and $h$ for which they meet, and theen divide count the number of Since the path is $2 \mathrm{~km}$ long and Beatrice walks at $5 \mathrm{~km} / \mathrm{h}$, then Beatrice takes $\frac{2 \mathrm{~km}}{5 \mathrm{~km} / \mathrm{h}}$
or $\frac{2}{5} \mathrm{~h}$ or 24 minutes to walk the length of the path.
Since the path is $2 \mathrm{~km}$ long and Hieu cycles at $15 \mathrm{~km} / \mathrm{h}$, then Hieu takes $\frac{2 \mathrm{~km}}{15 \mathrm{~km} / \mathrm{h}}$ or
$\frac{2}{15} \mathrm{~h}$ or 8 minutes to cycle the length of the path.
Since Beatrice walks the path in 24 minutes and Hieu cycles the path in 8 minutes, $24-8=16$. Here is an explanation for why this is true
- If Hieu starts before Beatrice (that is, if $hb$ ) but fewer than 16 minutes
after Beatrice (that is, $h-b16$ ), then Hieu will not catch up to Beatrice. In this case, this is because Hieu is "behind" previous case, Hieu just caught up to Beatrice at the very end of the path.
So we need to count the number of pairs of integer values for $h$ and $b$ with $h$ and $b$
To do this, we work through the values of $b$ from 0 to 59, in each case determining If $b=0$, ther of values of $h$ that satisfy the restrictions.
If $b=1$, then $h$ can be any of 1 to 17 , inclusive. There
$b \leq 43$, we have and $b \leq 43$, then $h$ can be any of $b$ to $b+16$ inclusive, and since required conditions. Note that there are 44 values of $b$ for which there are 17 value
If $b=44$, then $h$ can be any of 44 to 59 , inclusive. We note while $h=60$ is 16 greate than $b=44$, it is not an allowed value for $h$. In this case, there are 16 values for $h$. If $b=45$, then $h$ can be any of 45 to 59 , inclusive. There are 15 values for $h$. As $b$ increases by 1 from 45 to 59 , the number of values of $h$ decreases by 1 in each case, since the maximum value of $h$ does not change while its minimum value increases
When $b=59$, there is 1 value of $h$ (namely, $h=59$ ) that satisfies the conditions.
Therefore, if $N$ is the number of pairs of values of $h$ and $b$ that satisfy the given conditions, then
$$
\begin{aligned}
N & =44 \times 17+16+15+14+\cdots+3+2+1 \\
& =44 \times 17+(16+1)+(15+2)+\cdots+(10+7)+(9+8) \\
& =44 \times 17+8 \times 17 \\
& =52 \times 17 \\
& =884
\end{aligned}
$$

Finally, this means that the probability that Hieu and Beatrice are at the same place on the path at the same time is $\frac{N}{3600}$ or $\frac{884}{3600}$, which simplifies to $\frac{221}{900}$.
(ii) As in (i), there are 3600 possible pairs of values for $b$ and $h$.
If $M$ of these pairs result in Hieu and Beatrice meeting, then the probability that they meet is $\frac{M}{3600}$. meet is $\overline{3600}$
We are told that the probability is $\frac{13}{200}$, and so $\frac{M}{3600}=\frac{13}{200}$ which means that $M$ is equal to $\frac{13 \times 3600}{200}=13 \times 18=234$.
As in (i), Hieu takes 8 minutes to cycle the length of the path.

Since Beatrice rides a scooter at $x \mathrm{~km} / \mathrm{h}$ (with $x>5$ and $x5$, then $\frac{120}{x}8$.
Using a similar analysis to that in (i), Hieu and Beatrice meet whenever $h-b$ is at least 0 and at most $\frac{120}{x}-8$
This means that we need to determine the range of possible values of $x$ for which there are 234 pairs of integer value and 59, inclusive, and $h-b \geq 0$ and $h-b \leq \frac{120}{x}-8$.
Since $h$ and $b$ are integers, then $h-b$ is an integer. Let $t$ be the largest integer less than or equal to 120
Note that, since $t$ is the largest such integer, $t+1$ is greater than $\frac{120}{x}-8$.
Note also that $\frac{120}{x}-8$ is greater than 0 and less than 16 , which means that $t$ is between 0 and 15 , inclusive. This means that subject to the range restrictions on $h$ and $b$, the value of $h-b$ can be fres As in (i), we count the number of pairs of values for $h$.
through the values of $b$. When $b$ of pairs of values for $h$ and $b$ by working systematically .
When $b=1$, there are $t+1$ possible values for $h$, namely the integers from 1 to $t+1$, inclusive.
Similarly, while $b \leq 59-t$, there are $t+1$ possible values for $h$. Since these values of
$b$ run from 0 to $59-t$, inclusive, there are $60-t$ such values of $b$. When $b=(59-t)+1=60-t$, the largest possible value of $h$ is still 59 (which corresponds to $h-b=t-1$ ) and so there are $t$ possible values for $h$. As in (i), the number of possible values of $h$ decreases by 1 each time $b$ increases by 1 until $b=59$. 1 until $b=59$. Since the total number of pairs of values for $b$ and $h$ is 234 , then
$$
\begin{aligned}
234 & =(60-t)(t+1)+t+(t-1)+\cdots+2+1 \\
234 & =(60-t)(t+1)+\frac{1}{2} t(t+1) \\
468 & =2(60-t)(t+1)+t(t+1) \\
468 & =(120-2 t)(t+1)+t(t+1) \\
468 & =(120-t)(t+1)
\end{aligned}
$$

When $t=3$, the right side is equal to $117 \times 4$ which equals 468 .
Expanding, we obtain $468=-t^{2}+119 t+120$ or $t^{2}-119 t+348=0$ Factoring, we
Factoring, we obtain $(t-3)(t-116)=0$. Since $t$ is between 0 and 15, inclusive, then $t=3$.
Since $t$ is 3 , the greatest integer less than or equal to $\frac{120}{x}-8$ is 3 .
This means that $\frac{120}{x}-8 \geq 3$ and $\frac{120}{x}-8<4$.
Thus, $\frac{120}{x} \geq 11$ and $\frac{120}{x}<12$, which tells us that $10<x \leq \frac{120}{11}$, which is the range of possible values for $x$.