TCM Arithmetic 3: Rates and Unit Conversions

**By: Tao Steven Zheng (郑涛)**

## **Rates and the Jin you Method**

The concept of rate (率 *lü*)[1] was highly important in ancient Chinese mathematics. One important method discussed in Chapter 2 of the *Jiuzhang Suanshu* is the *jin you* method (今有術 *jin you shu*) or *jin you* rule. The *jin you* method most likely arose from the arithmetic used for commercial transactions of antiquity, for every problem in this chapter dealt with the exchange of different grains that followed a defined market rate. This method involves solving for an unknown quantity given three known quantities. In the *Jiuzhang Suanshu*, the three known quantities are give specific titles: the sought rate (所求率 *suo qiu lü*), the given amount (所有數 suo you shu), and the given rate (所有率 *suo you lü*). At the beginning of Chapter 2 of the *Jiuzhang Suanshu*, there is a table of market rates of various grains and beans provided for the reader, as well as a detailed description of the *jin you* method. The reader then uses the market rates to solve for the unknown quantity of grain that is required by a given problem.

[1] In ancient China, the mathematical concepts of ratio, rate, and proportion share the same character 率. This differs from the modern usage: 比 for ratio, 比率 for rate, and 比例 for proportion.

粟率五十；糲米三十
粺米二十七；糳米二十四
御米二十一；小䵂十三半
大䵂五十四；糲飯七十五
粺飯五十四；糳飯四十八
御飯四十二；菽、荅、麻、麥各四十五
稻六十；豉六十三
飧九十；熟菽一百三半
櫱一百七十五

今有術曰：以所有數乘所求率為實，以所有率為法，實如法而一。

Unhusked millet 50; Hulled millet 30
Milled millet 27; Refined millet 24
Imperial millet 21; Refined wheat 13 ½
Coarse wheat 54; Cooked coarse wheat 75
Cooked milled millet 54; Cooked refined millet 48
Cooked imperial millet 42; Soy beans, Small beans, Sesame seed, Wheat 45
Paddy rice 60; Fermented soy beans 63
Porridge 90; Cooked beans 103 ½
Fermented grain 175

*Jin you* method: Take the given amount multiplied by the sought rate as the dividend. The given rate is the divisor. Divide the dividend by the divisor.

According to the above description, the *jin you* method states that given two rates, the given rate

r
1

{\displaystyle r\_1 }

 and sought rate

r
2

{\displaystyle r\_2 }

, and the given amount

y

{\displaystyle y}

, one can determine the sought amount

x

{\displaystyle x}

 as follows:

# x

r
2

×
y

r
1

{\displaystyle x = \frac{r\_2 \times y}{r\_1} }

We will later discover that the *jin you* method , as well as the notion of rates is vital in explaining the reasoning behind traditional Chinese arithmetic, linear equations, and geometry. Over the centuries, knowledge of the *jin you* method travelled along the Eurasian trade routes (often called the Silk Road), making its way to India, the Middle East, and Europe. In ancient India, the astronomer-mathematician Aryabhata (476 - 550 AD) formulated a method for calculating rates equivalent to the *jin you* method. Indian mathematicians called it *Trairāśika*, which literally translates to the "rule of three quantities". For centuries, the mathematicians and merchants of Muslim empires and Renaissance Europe still refer this arithmetic method as the "rule of three".

## **Unit Conversions**

The *jin you* method is an abstract arithmetic principle related to rates and proportionality. Knowing how the *jin you* method can be used for calculating rates, one can apply this principle to the conversion of standardized units. In ancient China, there were three fundamental measures: linear measure (度 *du*), capacity measure (量 *liang*), and weight measure (衡 *heng*). The three tables below lists the conversion rates of progressively larger units, which will later be used in the calculations for the problem studies.

**Table 1. Linear Measure (度 *du*)**

10 厘 = 1 分
10 分 = 1 寸
10 寸 = 1 尺
10尺 = 1 丈
6 尺 = 1 步
40 尺 = 1 匹（疋）
50 尺 = 1 端
300 步 = 1 里
10 *li* = 1 *fen*
10 *fen* = 1 *cun*
10 *cun* = 1 *chi*
10 *chi* = 1 *zhang*
6 *chi* = 1 *bu*
40 *chi* = 1 *pi*
50 *chi* = 1 *duan*
300 *bu* = 1 *li*

**Table 2. Capacity Measure (量 *liang*)**

10 撮 = 1 抄
10 抄 = 1 勺
10 勺 = 1 合
10 合 = 1 升
10 升 = 1 斗
10 斗 = 1 斛
10 *cuo* = 1 *chao*
10 *chao* = 1 *shao*
10 *shao* = 1 *ge*
10 *ge* = 1 *sheng*
10 *sheng* = 1 *dou*
10 *dou* = 1 *hu*

**Table 3. Weight Measure (衡 *heng)***

24 銖 = 1 兩
16 兩 = 1 斤
30 斤 = 1 鈞
4 鈞 = 1 石
24 *zhu* = 1 *liang*
16 *liang* = 1 *jin*
30 *jin* = 1 *jun*
4 *jun* = 1 *dan*[2]

[2] The character 石 is pronounced *shi* to mean stone, but is pronounced *dan* to mean the unit of measurement.

### **Problem Study 1: Exchanging Grains**

The following problem is from the *Jiuzhang Suanshu* (Chapter 2, Problem 1).

答曰：為糲米六升。

術曰：以粟求糲米，三之，五而一。

Answer: 6 *sheng* of husked millet are exchanged.

Method: Using [the given amount of] millet to determine [the sought amount of] hulled millet, multiply by 3, and divide by 5.

**Solution for *Jiuzhang Suanshu* [2.01]**

Let

x

{\displaystyle x}

 denote the sought amount of hulled millet. From the table of capacity measures (Table 2), we find that 10 *sheng* (升) makes 1 *dou* (斗). Thus, the given amount is 10 *sheng* of millet. Referencing the market rates from Chapter 2 of the *Jiuzhang Suanshu*, we find that the exchange rate for millet is 50, and the exchange rate for hulled millet is 30. Substituting the given rates and the given amount of millet into the *jin you* rule yields:

# x

30
×
10

s
h
e
n
g

50

{\displaystyle x = \frac{30 \times 10 \; sheng}{50} }

# x

3
×
10

s
h
e
n
g

5

{\displaystyle x = \frac{3 \times 10 \; sheng}{5} }

# x
6

s
h
e
n
g

{\displaystyle x = 6 \; sheng }

### **Problem Study 2: From Luoyang to Chang'an**

The following problem is from the *Sunzi Suanjing* (Chapter 3, Problem 33).

答曰：九萬匝。

術曰：置九百里，以三百步乘之，得二十七萬步。又以六尺乘之，得一百六十二萬尺。以車輪一丈八尺為法。除之，即得。

Answer: 90,000 rotations.

Method: Put down 900 *li*, multiply by 300 *bu* to obtain 270,000 *bu*. Next multiply by 6 *chi* to obtain 1,620,000 *chi*. Let [the circumference of ] the wheel, 1 *zhang* 8 *chi*, be the divisor (法 fa). Divide to get the answer.

**Solution for *Sunzi Suanjing* [3.33]**

(1) Convert the distance between Luoyang (洛阳) and Chang’an (长安) from the unit *li* (里) to *chi* (尺).

1 *li* = 300 *bu*

900

l
i
×

300

b
u

1

l
i

=
270000

b
u

{\displaystyle 900 \; li \times \frac{300 \; bu}{1 \; li} = 270000 \; bu }

1 *bu* = 6 *chi*

270000

b
u
×

6

c
h
i

1

b
u

=
1620000

c
h
i

{\displaystyle 270000 \; bu \times \frac{6 \; chi}{1 \; bu} = 1620000 \; chi }

(2) Convert one rotation of the wheel into *chi*.

1

z
h
a
n
g

8

c
h
i

=
10

c
h
i
8

c
h
i

=
18

c
h
i

{\displaystyle
\begin{align}
1 \; zhang \; 8 \; chi &= 10 \;chi + 8 \; chi \\
&= 18 \; chi
\end{align}
}

(3) Divide the distance between Luoyang and Chang’an by the circumference of one rotation.

1620000

c
h
i

18

c
h
i

/

r
o
t
a
t
i
o
n
s

=
90000

r
o
t
a
t
i
o
n
s

{\displaystyle \frac{1620000 \; chi}{18 \;chi/rotations} = 90000 \; rotations }

## **Complex Rate Problems**

Chapter 6 of the *Jiuzhang Suanshu*, entitled *jun shu* (均輸), introduces several classes of rate problems that involve multiple rates. Unlike the *jin you* problems of Chapter 2, *jun shu* problems do not follow a set formula. The problems are more complex in nature due to the fact that the rates are often combined in different manners depending on the given scenario. Each problem is concerned with a different practical application, including the transport of grains, the conscription of soldiers, the problem of pursuit, shared work, and taxation. We will investigate the different applications in the following problem studies.

### **Problem Study 3: Rabbit and Dog**

*Jiuzhang Suanshu* Chapter 6 Problem 14 is a classic pursuit problem.

答曰：一百七步、七分步之一。

術曰：置兔先走一百步，以犬走不及三十步減之，餘為法。以不及三十步乘犬追步數為實，實如法得一步。

Answer:

107

1
7

{\displaystyle 107\frac{1}{7} }

*bu*.

Method: Set the 100 *bu* by which the rabbit is ahead, then subtract the 30 *bu* by which the dog stops short (不及 *bu ji*), and let the difference remaining be the divisor (法 *fa*). Multiply the distance stopped short, 30 *bu*, by the distance the dog chased, and let this be the dividend (實 *shi*). Divide the the dividend by the divisor to obtain the result in *bu*.

**Solution for *Jiuzhang Suanshu* [6.14]**

The difference between the 100 *bu* by which the rabbit has run ahead at the start and the 30 *bu* the dog stopped short is treated as the "run ahead rate":

100
−
# 30
70

{\displaystyle 100 - 30 = 70 }

.

Treating the 250 *bu* travelled by the dog while chasing the rabbit as "catching up rate", one can use the *jin you* rule to calculate the remaining distance the dog needed to run in order to catch the rabbit.

30
×
250

70

=
107

1
7

{\displaystyle \frac{30\times 250}{70} = 107\frac{1}{7} }

Therefore, had the dog not stop short it would need to run an additional

107

1
7

{\displaystyle 107\frac{1}{7} }

*bu* to catch the rabbit.

### **Problem Study 4: Duck and Goose**

*Jiuzhang Suanshu* Chapter 6 Problem 20 is a prototype of the class of problems that involve shared work.

答曰：三日、十六分日之十五。

術曰：并日數為法，日數相乘為實，實如法得一日。

Answer:

3

15
16

{\displaystyle 3\frac{15}{16} }

days.

Method: Add the number of days together and let this be the divisor. Multiply the number of days together and let this be the dividend. Divide the dividend by the divisor to obtain the number of days.

**Solution for *Jiuzhang Suanshu* [6.20]**

Let

t

{\displaystyle t}

 denote the number of days it takes the wild duck and wild goose to meet. For this problem, the rates of flying for each bird would be

a
1

=
7

{\displaystyle a\_1 = 7 }

 for the wild duck, and

a
2

=
9

{\displaystyle a\_2 = 9 }

 for the wild goose. Assuming both birds fly at a constant rate, the duck flies

1
7

{\displaystyle \frac{1}{7}}

 of the total distance per day, and the goose flies

1
9

{\displaystyle \frac{1}{9}}

 the total distance per day. Together, the two birds travel

16
63

{\displaystyle \frac{16}{63} }

 of the total distance per day. Thus, invert the result to calculate the number of days it takes both birds to meet. In modern mathematical notation, one can express the problem as follows:

1
t

=

1
7

1
9

{\displaystyle \frac{1}{t} = \frac{1}{7} + \frac{1}{9} }

1
t

=

16
63

{\displaystyle \frac{1}{t} = \frac{16}{63} }

# t

63
16

{\displaystyle t = \frac{63}{16} }

# t
3

15
16

{\displaystyle t = 3\frac{15}{16} }

.
This calculation gives exactly the same answer as the method prescribed in the *Jiuzhang Suanshu*:

# t

a
1

×

a
2

a
1

a
2

{\displaystyle t = \frac{a\_1 \times a\_2}{a\_1 + a\_2} }

# t

7
×
9

7
9

{\displaystyle t = \frac{7 \times 9}{7 + 9} }

# t

63
17

{\displaystyle t = \frac{63}{17} }

# t
3

15
16

{\displaystyle t = 3\frac{15}{16} }

.
Therefore, the wild duck and wild goose will meet after

3

15
16

{\displaystyle 3\frac{15}{16} }

 days of continuous flying.

### **Problem Study 5: Making Arrows**

*Jiuzhang Suanshu* Chapter 6 Problem 23 is a another shared work; this time with more than two rates.

答曰：八矢、少半矢。

術曰：矯矢五十，用徒一人。羽矢五十，用徒一人、太半人。筈矢五十，用徒三人、少半人。并之，得六人，以為法。以五十矢為實。實如法得一矢。

Answer:

8

1
3

{\displaystyle 8\frac{1}{3} }

arrows.

Method: If 1 person straightens 50 arrow shafts, then

1

2
3

{\displaystyle 1\frac{2}{3} }

persons can fletch 50 arrows and

3

1
3

{\displaystyle 3\frac{1}{3} }

persons can install arrow heads for 50 arrows. Adding gives 6 persons as the divisor. Take 50 arrows as the dividend. Divide the dividend by the divisor to obtain the number of arrows.

**Solution for *Jiuzhang Suanshu* [6.23]**

Let

x

{\displaystyle x}

 represent the number of arrows that one person can produce. This problem can be solved in a similar fashion to the Duck and Goose problem:

1
x

=

1
50

1
30

1
15

{\displaystyle \frac{1}{x} = \frac{1}{50} + \frac{1}{30} + \frac{1}{15} }

50
x

=

50
50

50
30

50
15

{\displaystyle \frac{50}{x} = \frac{50}{50} + \frac{50}{30} + \frac{50}{15} }

50
x

=
1
1

2
3

3

1
3

{\displaystyle \frac{50}{x} = 1 + 1\frac{2}{3} + 3\frac{1}{3} }

50
x

=
6

{\displaystyle \frac{50}{x} = 6 }

# x

50
6

=
8

1
3

{\displaystyle x = \frac{50}{6} = 8\frac{1}{3} }

.
Therefore, one person can produce

8

1
3

{\displaystyle 8\frac{1}{3} }

 arrows if he straightens shafts, fletches and installs arrowheads single-handedly.

### **Problem Study 6: Taxation on Rice**

*Jiuzhang Suanshu* Chapter 6 Problem 27 is a problem of compounded taxation, with three different taxation rates at each step.

答曰：十斗九升、八分升之三。

術曰：置米五斗。以所稅者三之，五之，七之，為實。以餘不稅者二、四、六相乘為法。實如法得一斗。

Answer:

10

d
o
u

9

3
8

s
h
e
n
g

{\displaystyle 10 \; dou \; 9\frac{3}{8} \;sheng }

.

Method: Set 5 *dou* of grain, and the amounts that are taxed. Multiply this by 3, 5, and 7, and let this be the dividend (實 *shi*). Take the remainder that was not taxed, multiply 2, 4, 6 together and let this be the divisor (法 *fa*). Divide the dividend by the divisor to obtain the answer in *dou*.

**Solution for *Jiuzhang Suanshu* [6.27]**

Let

x

{\displaystyle x}

 represent the original amount of grain carried. According to the problem,

x

(

1
−

1
3

)

(

1
−

1
5

)

(

1
−

1
7

)

=
5

{\displaystyle x\left(1-\frac{1}{3} \right) \left(1-\frac{1}{5}\right) \left(1-\frac{1}{7}\right) = 5 }

x

(

2
3

)

(

4
5

)

(

6
7

)

=
5

{\displaystyle x\left(\frac{2}{3} \right) \left(\frac{4}{5}\right) \left(\frac{6}{7}\right) = 5}

Solving for

x

{\displaystyle x}

 gives the solution method in the *Jiuzhang Suanshu*, which instructs the reader to perform the following calculation:

5

d
o
u

(

3
×
5
×
7

2
×
4
×
6

)

=

525
48

d
o
u

=
10

15
16

d
o
u

{\displaystyle \begin{align} 5 \; dou \; \left(\frac{3 \times 5 \times 7}{2 \times 4 \times 6} \right) &= \frac{525}{48} \; dou \\
&= 10\frac{15}{16} \;dou
\end{align} }

Since 1 *dou* equals 10 *sheng*, the answer makes the conversion

15
16

d
o
u

(

10

s
h
e
n
g

1

d
o
u

)

=
9

3
8

s
h
e
n
g

{\displaystyle \frac{15}{16} \; dou \left(\frac{10\; sheng}{1 \; dou}\right) = 9\frac{3}{8} \; sheng }

which gives the result

10

d
o
u

9

3
8

s
h
e
n
g

{\displaystyle 10 \; dou \; 9\frac{3}{8} \;sheng }

.

## **References**

[1] 钱宝琮主编.《中国数学史》，北京：科学出版社，1964年。

[2] 纪志刚《孙子算经、张邱建算经、夏侯阳算经导读》， 武汉：湖北教育出版社，1999年。

[3] 郭书春《九章算术译注》，上海：上海古籍出版社，2009年。

[4] 李兆华《中国数学史基础》，天津：天津教育出版社，2010年。

[5] Martzloff, Jean-Claude. A History of Chinese Mathematics (English Translation). Spinger-Verlag. 1997, 2006.

[6] Yong, Lam Lay, & Ang, Tian Se. *Fleeting Footsteps: Tracing the Conception of Arithmetic and Algebra in Ancient China, Revised Edition*. World Scientific Publishing Company. 2004

[7] Wilkinson, Endymion. *Chinese History: A New Manual, 4th Edition*. Harvard University Press. 2015.