Japanese Mathematics TABIBITO-ZAN


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(zero) 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,

\[ x = 12 \]


Japanese Mathematics TSURUKAME-ZAN


Tsuru-kame-zan is a method of calculation based on figuring the number of cranes and tortoises from the totals of their legs. Let’s try it.

It is introduced in the rakugo Sanshi Katsura created. In ancient time, it was appeared in China as a method to calculate rabbits and sheep, I heard.

Example

There are cranes and tortoises. The number of them is 34, and the number of legs is 78. Now, how many cranes are there? And how about tortoises?

Answer

There are 29 cranes and 5 tortoises.

Solution

It’s important to think about difference.

Both crane and tortoise have 2 legs at least. So, there are \( 34 \times 2 \) legs, at least.

\[ 34 \times 2 = 68 ( \textrm{legs} ) \]

Now, there are 78 legs and difference from 68 is 10.

Why does the difference 10 appear? It’s because tortoise have 4 legs. All of the difference 10 is tortoises’ legs.

One tortoise generate 2 legs difference, so

\[ 10 \div 2 = 5 ( \textrm{tortoises}) . \]

It’s the number of tortoises. Then cranes are \( 34 – 5 = 29 \) .

Practice

Question 1

There are two kind of books. One has 230 pages and the other has 180 pages. Now, there are 9 books and 1920 pages. Calculate the number of each kind of books.

Question 2

There are two kind of bags. One contains 3 kg of wheat, the other contains 2 kg of sugar. Now, there are 45 bags and weighs 113 kg. Calculate the weight of all wheat.

Answer

69 kg

Solve with Equation

Now, I’ll introduce solution by equation. There are cranes and tortoises. The number of them is 34, and the number of legs is 78.

Solution

It’s okey to use \( x \) , \( y \) and create multi equation, but solve with simple equation.

Denote the number of tortoises as \( x \). The number of tortoises and cranes is 34, then the number of cranes is \( \left( 34 – x \right) \) .

The number of legs are 78, then

\[ 4 x + 2 \left( 34 – x \right) = 78 . \]

So,

\[ x = 5 . \]

Thus, there are 5 tortoises and \( 34 – 5 = 29 \) cranes.


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