TCM Arithmetic 3: Rates and Unit Conversions

By: Tao Steven Zheng (郑涛)

Rates and the Jin you Method

Figure 1. The jin you method or "rule of three"

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.

粟米之法： 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。

The regulated [rates of exchange] for grains: 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 $r_{1}$ and sought rate $r_{2}$, and the given amount $y$, one can determine the sought amount $x$ as follows:

$x={\frac {r_{2}\times y}{r_{1}}}$

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 毫 = 1 厘 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 hao = 1 li 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 圭 = 1 撮 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 gui = 1 cuo 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)

10 絫 = 1 銖 24 銖 = 1 兩 16 兩 = 1 斤 30 斤 = 1 鈞 4 鈞 = 1 石

10 lei = 1 zhu 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).

[2.01] 今有粟一斗，欲為糲米。問：得幾何？ 答曰：為糲米六升。 術曰：以粟求糲米，三之，五而一。

[2.01] Suppose the is 1 dou of millet, and one wishes to exchange it for hulled millet. Question: How much should there be? 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 $x$ 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:

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

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

$x=6\;sheng$

Problem Study 2: From Luoyang to Chang'an

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

[3.33] 今有長安、洛陽相去九百里。車輪一匝一丈八尺。欲自洛陽至長安，問：輪匝幾何？ 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。

[3.33] Suppose the distance between Chang’an and Luoyang is 900 li. One rotation of the cart’s wheel measures 1 zhang 8 chi. Question: If one wishes to travel from Luoyang to Chang’an, how many rotations will the wheel make? 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

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

1 bu = 6 chi

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

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

${\begin{aligned}1\;zhang\;8\;chi&=10\;chi+8\;chi\\&=18\;chi\end{aligned}}$

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

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

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.

[6.14] 今有兔先走一百步，犬追之二百五十步，不及三十步而止。問：犬不止，復行幾何步及之？ 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。

[6.14] Suppose there is a rabbit that runs ahead a distance of 100 bu, and a dog that chases it for 250 bu. The dog stops short when it is 30 bu behind the rabbit. Question: if the dog did not stop, how much farther must the dog run, in bu, to catch the rabbit? Answer: $107{\frac {1}{7}}$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":

$100-30=70$.

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.

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

Therefore, had the dog not stop short it would need to run an additional $107{\frac {1}{7}}$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.

[6.20] 今有鳧起南海，七日至北海；鴈起北海，九日至南海。今鳧鴈俱起。問：何日相逢？ 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。

[6.20] Suppose there is a wild duck that sets out from the southern sea (南海 nan hai), and in 7 days reaches the northern sea (北海 bei hai). A wild goose leaves the northern sea, and in 9 days reaches the southern sea. If the wild duck and the goose sets out at the same time, in how many days will they meet? Answer: $3{\frac {15}{16}}$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 $t$ 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 $a_{1}=7$ for the wild duck, and $a_{2}=9$ for the wild goose. Assuming both birds fly at a constant rate, the duck flies ${\frac {1}{7}}$ of the total distance per day, and the goose flies ${\frac {1}{9}}$ the total distance per day. Together, the two birds travel ${\frac {16}{63}}$ 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:

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

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

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

$t=3{\frac {15}{16}}$.

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

$t={\frac {a_{1}\times a_{2}}{a_{1}+a_{2}}}$

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

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

$t=3{\frac {15}{16}}$.

Therefore, the wild duck and wild goose will meet after $3{\frac {15}{16}}$ 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.

[6.23] 今有一人一日矯矢五十，一人一日羽矢三十，一人一日筈矢十五。今令一人一日自矯、羽、筈，問：成矢幾何？ 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。

[6.23] Suppose a person can straighten 50 arrow shafts in 1 day, or fletch 30 arrows, or install arrowheads for 15 arrows. Assume a person straightens shafts, fletches and installs arrowheads single-handedly. Question: how many arrows can be completed in a day? Answer: $8{\frac {1}{3}}$arrows. Method: If 1 person straightens 50 arrow shafts, then $1{\frac {2}{3}}$persons can fletch 50 arrows and $3{\frac {1}{3}}$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 $x$ 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:

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

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

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

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

$x={\frac {50}{6}}=8{\frac {1}{3}}$.

Therefore, one person can produce $8{\frac {1}{3}}$ 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.

[6.27] 今有人持米出三關，外關三而取一，中關五而取一，內關七而取一，餘米五斗。問：本持米幾何？ 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。

[6.27] Suppose there is a person carrying grain through 3 custom stations. At the outer station 1/3 of the grain is taken as tax, at the middle station 1/5 of the grain is taken as tax, and at the inner station 1/7 of the grain is taken as tax. In the end, there is 5 dou of grain remaining. Question: how much grain was carried originally? Answer: $10\;dou\;9{\frac {3}{8}}\;sheng$. 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 $x$ represent the original amount of grain carried. According to the problem,

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

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

Solving for $x$ gives the solution method in the Jiuzhang Suanshu, which instructs the reader to perform the following calculation:

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

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

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

which gives the result $10\;dou\;9{\frac {3}{8}}\;sheng$.

References

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