Every 12 minutes, Bus A completes a trip from $P$ to $X$ to $S$ to $X$ to $P$. Every 20 minutes, Bus B completes a trip from $Q$ to $X$ to $T$ to $X$ to $Q$. Every 28 minutes, Bus $\mathrm{C}$ completes a trip from $R$ to $X$ to $U$ to $X$ to $R$. At 1:00 p.m., Buses $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ depart from $P, Q$ and $R$, respectively, each driving at a constant speed, and each turning around instantly at the endpoint of its route. Each bus runs until 11:00 p.m. At how many times

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-017.jpg?height=309&width=396&top_left_y=1615&top_left_x=1298)
between 5:00 p.m. and 10:00 p.m. will two or more buses arrive at $X$ at the same time?
(A) 18
(B) 19
(C) 20
(D) 21
(E) 22
Bus A takes 12 minutes to complete one round trip that begins and ends at $P$.

Since $P X=X S$, it takes Bus A $12 \div 4=3$ minutes to travel from $P$ to $X, 6$ minutes to travel from $X$ to $S$ to $X$ ( 3 minutes from $X$ to $S$ and 3 minutes from $S$ to $X$ ), and 6 minutes to travel from $X$ to $P$ to $X$.

That is, Bus A first arrives at $X$ at 1:03 and then continues to return to $X$ every 6 minutes. We write times that Bus A arrives at $X$ in the table below.

| Bus A | $1: 03$ | $1: 09$ | $1: 15$ | $1: 21$ | $1: 27$ | $1: 33$ | $1: 39$ | $1: 45$ | $1: 51$ | $1: 57$ | $2: 03$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |

Notice that Bus A arrives at $X$ at 1:03 and exactly one hour later at 2:03.

This makes sense since Bus A returns to $X$ every 6 minutes and 60 minutes (one hour) is divisible by 6 .

This tells us that Bus A will continue to arrive at the same number of minutes past each hour, or 2:03, 2:09, 2:15, .., 3:03, 3:09, .., 5:03, 5:09, . . , 9:03, 9:09, . ., 9:51, 9:57.

Bus B takes 20 minutes to complete one round trip that begins and ends at $Q$.

Since $Q X=X T$, it takes Bus B $\frac{20}{4}=5$ minutes to travel from $Q$ to $X, 10$ minutes to travel from $X$ to $T$ to $X$ ( 5 minutes from $X$ to $T$ and 5 minutes from $T$ to $X$ ), and 10 minutes to travel from $X$ to $Q$ to $X$.

That is, Bus B first arrives at $X$ at 1:05 and then continues to return to $X$ every 10 minutes. We write times that Bus B arrives at $X$ in the table below.

| Bus B | $1: 05$ | $1: 15$ | $1: 25$ | $1: 35$ | $1: 45$ | $1: 55$ | $2: 05$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |

Notice that Bus B arrives at $X$ at 1:05 and exactly one hour later at 2:05.

This makes sense since Bus B returns to $X$ every 10 minutes and 60 minutes (one hour) is divisible by 10 .

This tells us that Bus B will continue to arrive at the same number of minutes past each hour, or $2: 05,2: 15,2: 25, \ldots, 3: 05,3: 15, \ldots, 5: 05,5: 15, \ldots, 9: 05,9: 15, \ldots, 9: 45,9: 55$.

From the two tables above, we see that Bus A and Bus B both arrive at $X$ at 15 minutes and 45 minutes past each hour.

Thus between 5:00 p.m. and 10:00 p.m., these two buses will meet $2 \times 5=10$ times at $X$.

These times are: 5:15, 5:45, 6:15, 6:45, 7:15, 7:45, 8:15, 8:45, 9:15, and 9:45.

Bus $\mathrm{C}$ takes 28 minutes to complete one round trip that begins and ends at $R$.

Since $R X=X U$, it takes Bus $\mathrm{C} \frac{28}{4}=7$ minutes to travel from $R$ to $X, 2 \times 7=14$ minutes to travel from $X$ to $U$ to $X$, and 14 minutes to travel from $X$ to $R$ to $X$.

That is, Bus $\mathrm{C}$ first arrives at $X$ at 1:07 and then continues to return to $X$ every 14 minutes. Unlike Bus A and Bus B, Bus C will not arrive at $X$ at consistent times past each hour since 60 is not divisible by 14 .

What is the first time after 5:00 p.m. that Bus $\mathrm{C}$ arrives at $X$ ?

Since 238 is a multiple of $14(14 \times 17=238)$, Bus C will arrive at $X 238$ minutes after first arriving at $X$ at 1:07 p.m.

Since 238 minutes is 2 minutes less than 4 hours $(4 \times 60=240)$, Bus $\mathrm{C}$ will arrive at $X$ at 5:05 p.m. (This is the first time after 5:00 p.m. that Bus C arrives at $X$.)

Bus B also arrives at $X$ at 5:05 p.m.

Are there other times after 5:05 p.m. (and before 10:00 p.m.) that Bus B and Bus C arrive at $X$ at the same time?

Bus B arrives at $X$ every 10 minutes and Bus $\mathrm{C}$ arrives at $X$ every 14 minutes.

Since the lowest common multiple of 10 and 14 is 70, then Bus B and Bus C will each arrive at $X$ every 70 minutes after 5:05 p.m., or at 6:15 p.m., 7:25 p.m., 8:35 p.m., and at 9:45 p.m.

Next we determine if there are times when Bus A and Bus C arrive at $X$ at the same time.

Bus $\mathrm{C}$ arrives at $X$ every 14 minutes after 5:05 p.m., or 5:19 p.m., 5:33 p.m., and so on.

Bus A also arrives at $X$ at 5:33 p.m.

Are there other times after 5:33 p.m. (and before 10:00 p.m.) that Bus A and Bus C arrive at $X$ at the same time?

Bus A arrives at $X$ every 6 minutes and Bus $\mathrm{C}$ arrives at $X$ every 14 minutes.

Since the lowest common multiple of 6 and 14 is 42 , then Bus A and Bus C will each arrive at $X$ every 42 minutes after 5:33 p.m., or at 6:15 p.m., 6:57 p.m., 7:39 p.m., 8:21 p.m., 9:03 p.m., and at 9:45 p.m.

The times when each pair of buses meet at $X$ at the same time between 5:00 p.m. and 10:00 p.m. are listed below.

Bus A and Bus B: 15 and 45 minutes past each hour

Bus B and Bus C: 5:05 p.m., 6:15 p.m., 7:25 p.m., 8:35 p.m., 9:45 p.m.

Bus A and Bus C: 5:33 p.m., 6:15 p.m., 6:57 p.m., 7:39 p.m., 8:21 p.m., 9:03 p.m., 9:45 p.m.

Finally, we determine the number of different times that two or more buses arrive at $X$ at the same time.

Bus $\mathrm{A}$ and Bus B arrive at $X$ at 10 different times.

Bus $\mathrm{B}$ and Bus $\mathrm{C}$ arrive at $X$ at 5 different times; however 2 of these times (6:15 p.m. and 9:45 p.m.) have already been counted, so there are 3 new times.

Bus A and Bus C arrive at $X$ at 7 different times; however 2 of these times (6:15 p.m. and 9:45 p.m.) have already been counted, so there are 5 new times.

The number of times that two or more buses arrive at $X$ between 5:00 p.m. and 10:00 p.m. is $10+3+5=18$.

ANSWER: (A)