I introduce Japanese Mathematics Tabibito-zan, which is a method to calculate how long does it takes that two objects meets.

## Example

Mr. A and Mr. B are on the same place. A starts walking to northern direction by 60 meters per minutes. After 4 minutes, B starts walking to the same direction by 80 meters per minutes. Now, how long does it takes that two man meets after B’s departure.

### Answer

After 12 minutes.

### Solution

After 4 minutes from A’s departure, A is \( 60 \times 4 = 240 \) meters far from starting point. Then, the distance between A and B is 240 meters, and B starts walking to the same direction after 4 minutes.

Both A and B are moving forward to the same direction, but B is faster than A, so 2 men will meets at somewhere.

Now, think about how 240 ｍeters distance will shrink. A is moving 60 meters per second, B is moving 80 meters per second. Then, it shrinks \( 80 – 60 = 20 \) meters per second.

The distance will be 0 and 2 men meets after

\[ 240 \div 20 = 12 \textrm{[minutes]} . \]## Practice

### Question 1

Mr. X and Mr. Y starts from the same point and go to the place, which is 360 meters far, and come back to the start point. X walks 75 meters per minutes, and Y moves 45 meters per minutes. How long later the 2 men will meet?

#### Answer

6 minutes later.

#### Solution

X is faster than Y, so X meets Y on homeward and Y meets X on outward. The length they walk until they meets is 720 meters.

720 meters shrinks by speeds of X and Y. It is

\[ 75 + 45 = 120 \textrm{[meters per minute]} \]So, they meet after

\[ 720 \div 120 = 6 \textrm{[minutes]} \]#### Another Solution

X arrives the turning point before Y arrives there. Then, proceed time to X’s arrival.

X arrives the turning point after

\[ 360 \div 75 = \frac{24}{5} \textrm{[minutes]} \]Then, Y is far from the staring point by

\[ 45 \times \frac{24}{5} = 216 \textrm{[meters]} \]At that time, the distance between X on the turning point and Y on outward way is

\[ 360 – 216 = 144 \textrm{[meters]} . \]The 2 man walks 144 meters in approaching way. So 2 men meets after

\begin{eqnarray} & & 144 \div \left( 75 + 45 \right) \\ & = & 144 \div 120 \\ & = & \frac{6}{5} \textrm{[minutes]} \end{eqnarray}From the above, 2 men meet after

\[ \frac{24}{5} + \frac{6}{5} = 6 \textrm{[minutes]} \]### Question 2

Johnson runs to the restroom from the office, and James comes back to the office from the restroom. The distance between the office and the restroom is 28 kilometers. If Johnson runs 5 kilometers per hour and James walks 2 kilometers per hour, how long minutes later will the 2 men meets?

#### Answer

240 minutes later.

#### Solution

Johnson and James moves in approaching way. The distance of 2 men, 28 kilometers shrinks ( \( 2 + 5 = \) ) \( 7 \) kilometers per hour.

\[ 28 \div 7 = 4 \]Thus, after 4 hours, 2 men will meet. 4 hours is 240 minutes, so the answer is “240 minutes later”.

## Solve with Equation

Mr. A and Mr. B are on the same place. A starts walking to northern direction by 60 meters per minutes. After 4 minutes, B starts walking to the same direction by 80 meters per minutes. Now, how long does it takes that two man meets after B’s departure.

### Solution

Suppose 2 men will meets \( x \) minutes later after B’s departure. A starts 4 minutes before than B, so A moves for \( x + 4 \) minutes.

\[ 60 \left( x + 4 \right) = 80 x . \]And then,

\[ x = 12 \]