पाटीगणित (श्रीधराचार्य - प्राचीन संस्कृत टीका एवं प्रो. कृपाशंकर शुक्ल सविस्तार व्याख्या)

22 LOGISTICS
Now applying the rule to the uppermost two fractions of this reduced
chain, we get
333
64
1 +
20
1 +
5
Applying the same rule to this chain, it reduces to
6659
1280
1 +
5
and then to
33294 / 6400 or 16647 / 3200
Hence the required result is
16647 / 3200 or 5 647/3200 purâṇas.
Rule for reducing fractions of the bhâgamâtâ class :
42. That (class of fractions) in which fractions of the
'bhâga and other classes occur in combination (of two or
more) is called bhâgamâtâ. In that (class), the result is obtained
by applying the aforesaid rules separately.¹
Ex. 23. What amount is obtained by adding together
½, ¼ of ¼, 1 ÷ ⅓, ½+½ of ½, and ⅓-½ of ⅙ ?
Ex. 24. If you know the method of calculation, say
what amount will be obtained by the addition of ¼, ½ of ⅓,
1 ÷ ½, and 8+½ ?
(23) Rule of three (trairâśika)
Rule of three :
43. In (solving problems on) the rule of three, the
argument (pramâṇa) and the requisition (icchâ), which are
of the same denomination, should be set down in the first and
last places ; the fruit (phala), which is of a different deno-
mination, should be set down in the middle. (This having
¹ Cf. GSS, iii. 138.

RULE OF THREE 23
been done) that (middle quantity) multiplied by the last
quantity should be divided by the first quantity.¹
Ex. 25. If 1 pala and 1 karṣa of sandal wood are ob-
tained for ten and a half paṇas, then for how much will
9 palas and 1 karṣa (of sandal wood of the same quality) be
obtained ?
Here
argument = 1 pala and 1 karṣa
= 1¼ or 5/4 palas ;
fruit = 10½ or 21/2 paṇas ;
and requisition = 9 palas and 1 karṣa
= 9¼ or 37/4 palas
Writing these quantities as directed in the rule, we have
┌───┬────┬────┐
│ 5 │ 21 │ 37 │
│ 4 │ 2 │ 4 │
└───┴────┴────┘
Then applying the rule, the required result
= [(21/2) × (37/4)] / [5/4] = (21×37) / (2×5) paṇas
or 4 purāṇas, 13 paṇas, 2 kākiṇīs and 16 varāṭakas.
Ex. 26. If 1⅓ palas of black pepper are obtained for
1¼ paṇas, then how much of that (black pepper) will be
obtained for (10 - ⅓) paṇas? Say quickly.
Ex. 27. If one and a half droṇas and three kuḍavas of
grain is obtained for 8, say, if you know, for how much will
one khārī and one droṇa (of that grain) be obtained.
Ex. 28. If 60+½ khārīs of grain is obtained for 100+⅓
rūpas, how much of that (grain) will be obtained for a quarter
of a rūpa ? ²
──────────────────────────────────────────────
¹ Cf. Ā, ii. 26; BrSpSi, xii. 10 ; GSS, v. 2(i) ; MSi, xv. 24-25(i) ; GT,
p. 68, vs. 86 ; SiŚe, xiii. 14 ; L (ASS), p. 71, vs. 73 ; GK, I, p. 47, vs. 60.
² Examples similar to Exs. 25-28 occur in GSS, iii. 3-6 ; v. 8, 9, 13,
14 ; GT, p. 68, vs. 87, p. 69, vs. 88 ; p. 70, vs. 90 ; p. 71, vv. 91, 92;
L (ASS), pp. 72-73, vv. 74, 76 ; GK, I, p. 47, lines 13-14, 17-18, p. 48,
lines 8-11 ; and also in Bhāskara I's comm. (629 A.D.) on Ā, ii. 26-27.

24 LOGISTICS
The term rūpa stands for any coin whatever. “By the word rūpaka
is meant a coin of gold, etc., bearing the name paṇa, etc.”¹
Ex. 29. Where one suvarṇa gets 70+⅓ rūpas, say,
friend, what will 1 māṣa as lessened by ⅒ (of a māṣa) get
there.²
Ex. 30. A certain lame person goes to a distance of ⅛
of a yojana in ⅓ of a day, say in how much time will he go
to a distance of 100 yojanas.³
Ex. 31. An insect goes to a distance of ⅙ of an aṅgula in
¼ of a day, in how much time will it go to a distance of 10
and a half yojanas ?⁴
Forward and backward motion:
44(i). On subtracting the backward motion per day
from the forward motion per day, the remainder is the (true)
motion per day.⁵
Ex. 32. The best amongst the elephants goes forward
at the rate of ½(1 + ¼) (1 − ⅓) (1 + ½) of a yojana in
6 × ⅕ × ⅑ × ⅓ (1 + ¼) of a day, and comes back at the rate of
2 (1 − ⅓) yojanas in (1 + ½) days. Say, friend, in how much
time will he go to a distance of 100 yojanas.⁶
Ex. 33. In how much time will a man, earning at the
rate of (8 − ½) rūpas in (1+⅓) days and spending on his food
at the rate of ½ per day, be a lord of 100 (rūpas) ?⁷
¹ रूपकशब्देन पणादिसंज्ञितं स्वर्णादिद्रव्यम् See Parameśvara's commentary
on Ā, ii 30.
² Cf. GT, p. 73, vs. 95.
³ Cf GSS, v. 3, 4 ; GT, p. 72, vs. 93.
⁴ Cf. GSS, v. 5 ; GT, p. 73, vs. 94 ; GK, I, p. 48, lines 16-19 ff.
⁵ Cf. BM. III, L 1, 60 recto ; GSS, v. 23 ; MS, xv. 30.
⁶ Cf. GSS, v. 27. Similar examples occur also in BM. See, for
instance, BM, III, M1, 21 verso ; M4, 36 verso. Also see Pṛthûdaka’s
example quoted by Sudhakara Dvivedi in his comm. on BrSpSi, xii.
10-12, p. 179, lines 14-17; and Bhâskara I's example in his comm. on
Ā, ii. 26-27(i).
⁷ Cf. BM, III, L 1, 60 recto ; M 11, 44 recto and verso ; M 12,
43 recto ; GSS, v. 26.

INVERSE RULE OF THREE 25
(24) Inverse rule of three (vyasta-trairāśika)
Inverse rule of three :
44(ii). When there is change in the unit of measure-
ment, the middle quantity multiplied by the first quantity and
divided by the last quantity gives the result.¹
Ex. 34. When (a given quantity of pearls is) measured
at 8 pearls a necklace, the number of necklaces is twenty ;
say, mathematician, what the number of necklaces would be
(when the same quantity of pearls is measured) at 6 pearls a
necklace.²
Ex. 35. Being measured by the māṣa of 5 raktikās, a
quantity of gold amounts to 300 suvarṇas. Say how much
would that (quantity of gold) amount to, when measured by
the māṣa of 6 raktikās.³
Raktikā is the same as guñjā.
Ex. 36. How much gold of 11 varṇas can be had in
exchange for 168 suvarṇas of 16 varṇas ?⁴
Ex. 37. Quickly say how many blankets of length 6
(cubits) and breadth 2 (cubits) can be made out of (the yarn
which yields) 200 blankets of length 9 (cubits) and breadth
3 (cubits).⁵
Ex. 38. How much gold of 10½ varṇas will be obtained
in exchange for 100 suvarṇas and 8 māṣas (of gold) of
12½ varṇas ?⁶
(25-27) Rules of five, seven and nine (pañca-sapta-nava-rāśika)
Rule of five, seven, and nine :
45. After transposing the fruit from one side to the
other, and then having transposed the denominators (in like
¹ Cf. BrSpSi, xii. 11(i) ; MSi, xv, 25(ii). Also see L (ASS), p. 74,
vs. 77-78 ; GK, I, p. 49, vs. 61.
² The same example occurs in GT, p. 74, lines 8-9.
³ Cf. L (ASS), p. 76, vs. 81 ; GK, I, p. 50, lines 3-4, 7-8. A similar
example occurs in Bhāskara I's comm. on Ā, ii. 26-27(i).
⁴ Cf. GT, p. 74, vs. 96. Also Cf. GSS, v. 18 (i) ; vi. 173½, 174½;
L (ASS), p. 75, vs. 80. For the meaning of varṇa see infra p. 36.
⁵ Cf. GSS, v. 19. ⁶ Cf. GSS, v. 18(ii).

26 LOGISTICS
manner) and multiplied the numbers (so obtained on either
side), divide the side with larger number of quantities (nume-
rators) by the other.¹
The two sides referred to in this rule are known as (i) the argument
side (pramāṇarāśi-pakṣa) and (ii) the requisition side (icchārāśi-pakṣa). In
Ex. 40, for instance, these sides are as follows :
(i) argument side ⎧ 100½ or 201/2
⎨ ⅓
⎩ 1½ or 3/2
(ii) requisition side ⎧ 60¼ or 241/4
⎩ 8 - ½ or 15/2
These are written as
(i) (ii)
┌────────────┬────────────┐
│ 201 │ 241 │
│ 2 │ 4 │
│ │ │
│ 1 │ 15 │
│ 3 │ 2 │
│ │ │
│ 3 │ 0 │
│ 2 │ │
└────────────┴────────────┘
0 being written in place of the desired quantity (unknown).
Then the rule is applied as follows :
Transposing the fruit, we get
(i) (ii)
┌────────────┬────────────┐
│ 201 │ 241 │
│ 2 │ 4 │
│ │ │
│ 1 │ 15 │
│ 3 │ 2 │
│ │ │
│ 0 │ 3 │
│ │ 2 │
└────────────┴────────────┘
─────────────────────────────────────────────────────────────────────────────
¹ Cf. BrSpSi, xii. 11(ii)-12 ; GSS, v. 32 ; MSi, xv. 26, 27 ; GT, p. 74,
vs. 97 ; SiŚe, xiii, 15; L (ASS), pp. 76-77, vs. 82 ; GK, I, p. 50, vs. 62.

RULE OF FIVE, SEVEN, AND NINE 27
Transposing the denominators, we get

¹ Cf. GSS, v. 33, 41 ; GT, p. 75, vs. 98 ; L (ASS), p. 77, vs. 83;
GK, I, p. 50, lines 18-21. Similar examples occur in Bhâskara I's comm.
on Â, ii. 26-27(i).
² Cf. GSS, v. 34 ; GT, p. 76, vs. 99 ; GK, I, p. 51, lines 6-9. Similar
examples occur in Bhâskara I's comm. on Â, ii. 26-27(i).
³ Cf. GSS, v. 35.
⁴ Cf. GSS, v. 36 ; GT, p. 78, vs. 101.

28 LOGISTICS
Ex. 44. If 3 labourers earn 5 rūpas in 2 days, say what
will 8 labourers earn in 9 days ?¹
Ex. 45. If a blanket, whose breadth is 2 (cubits) and
length 8 (cubits), gets 10, what will 2 other (similar) blankets
of breadth 3 (cubits) and length 9 (cubits) get ?²
Ex. 46. If a ( rectangular piece of ) stone with length,
breadth, and thickness equal to 9, 5, and 1 cubits respectively
costs 8, what will two other (rectangular pieces of stone) of
dimensions 10, 7, and 2 cubits cost ?³
Ex. 47. If the diet of an elephant of diameter 2 (cubits),
height 6 (cubits) and length 7 (cubits) is one droṇa, what
should be the diet of an elephant of diameter 3 (cubits), height
9 (cubits) and length 10 (cubits) ?⁴
(28) Barter of commodities (bhāṇḍa-pratibhāṇḍa)
Rule for problems on the barter of commodities :
46(i). In (problems on) the barter of commodities,
having transposed the prices (of the commodities), apply the
previous rule (i.e., rule of five, etc.).⁵
Ex. 48. If 2 palas of dry ginger are obtained for 6
and one pala of long pepper for 9, how much of long
pepper will then be obtained for 6 palas of dry ginger ?⁶

¹ Cf. GT, p. 77, vs. 100.
² This example is the same as that given in GT, p. 78, vs. 102.
³ This example is the same as that given in GT, p. 79, vs. 103.
Also cf. L (ASS), p. 79-80, vs. 85; GK, I, p. 52, lines 2-5.
⁴ Similar examples occur in Bhāskara I's comm. on Ā, ii. 26-27(i)
and in Pṛthûdaka's comm. on BrSpSi, xii. 10-12.
⁵ Cf. BrSpSi, xii. 13(i); MSi, xv. 28(i); GT, p.80, lines 16-17; SiŚe, xiii,
16(i); L (ASS), p. 83 vs. 88; GK, I, p. 53, lines 2-3. Also see GSS, vi. 18.
⁶ Cf. GSS, v. 37, 38 and vi. 19-20; GT, p. 81, lines 6-9.

SALE OF LIVING BEINGS 29
Setting down the argument and requisition sides, we have
┌───┬───┐
│ 6 │ 9 │
│ 2 │ 1 │
│ 6 │ │
└───┴───┘
Transposing the prices of the commodities, we get
┌───┬───┐
│ 6 │ 9 │
│ 1 │ 2 │
│ 6 │ │
└───┴───┘
Now we have to apply the rule of five, etc. But the fruits (i.e. prices)
have been already transposed, and there are no denominators on any
side. Hence the required answer
6 × 1 × 6
= —————————
9 × 2
= 2 palas of long pepper.
Ex. 49. If 16 mangoes are obtained for 2 paṇas and
100 wood-apples for 3 (paṇas), say then how many wood-
apples will be obtained for 6 mangoes.¹
(29) Sale of living beings (jīva-vikraya)
Rule for problems on the sale of living beings :
46(ii). In (problems on) the sale of living beings, the
same rule (of five, etc.) is applied after transposing the ages
(of the living beings).²
Ex. 50. If 5 women of 16 years of age get 200, say
then, O mathematician, how much will 2 women of 20 years
of age get ?³
────────────────────────────────────────────────────────
¹ Cf. GSS, v. 38; GT, p. 80, lines 23-26 ; L (ASS), p. 84, vs. 89;
GK, I, p. 53, lines 5-8. A similar example occurs in Pṛthûdaka’s comm.
on BrSpSi, xii. 13.
² Cf. GT, p. 81, vs. 108; SiŚe, xiii. 16 (ii); GK, I, 53, vs. 63.
³ Cf GSS, v. 40; GT, p. 81, vs. 109; L (ASS), p. 74, vs. 79; GK,
p. 53, lines 15-18.

30 LOGISTICS
Setting down the argument and requisition sides, we have
┌──────────┬──────────┐
│ 5 │ 2 │
│ 16 │ 20 │
│ 200 │ │
└──────────┴──────────┘
Transposing the ages (of the women), we get
┌──────────┬──────────┐
│ 5 │ 2 │
│ 20 │ 16 │
│ 200 │ │
└──────────┴──────────┘
Now we have to apply the rule of five, etc. Therefore, transposing
the fruit, we have
┌──────────┬──────────┐
│ 5 │ 2 │
│ 20 │ 16 │
│ │ 200 │
└──────────┴──────────┘
There being no denominators on any side, the required amount
= (2 × 16 × 200) / (5 × 20) = 64.
Ex. 51. If 3 camels of 10 years of age get 108 purāṇas,
say then what will 5 camels of 9 years of age get.¹
──────────────────────────────────────────────────────────
¹ A closely similar example occurs in GT, p. 82, vs. 110. Also
Cf. GSS, v. 39.

DETERMINATIONS 31
DETERMINATIONS (Vyavahāra)
(1) Determinations pertaining to mixtures of things
(miśraka-vyavahāra)
(i) SIMPLE INTEREST
Rule for finding the capital and the interest, when the
amount and the rate of interest are given :
47. Multiply the argument (pramāṇa-rāśi) by its time,
and the other time by the fruit (phala) ; divide each of those
(products) by their sum, and multiply by the amount (i.e.,
capital plus interest). The results (thus obtained) give the
capital (mūla-dhana) and the interest (vṛddhi-dhana) respectively.¹
That is,
argument × time × amount
Capital = ──────────────────────────────────────────────
argument × time + fruit × other time
fruit × other time × amount
Interest = ────────────────────────────────────────────── ·
argument × time + fruit × other time
Ex. 52. The rate of interest being 5 per cent per
month, a certain sum is seen to amount to 96 in a year. Say,
friend, what is the capital and what the interest ?²
Here argument = 100, time = 1 month, fruit = 5, amount = 96,
and other time = 1 year, i.e., 12 months. Therefore
100 × 1 × 96
Capital = ──────────────────── = 60,
100 × 1 + 5 × 12
5 × 12 × 96
Interest = ──────────────────── = 36.
100 × 1 + 5 × 12
Ex. 53. The interest on 100½ for one month and a
quarter being 1½, a certain sum amounts to 36½ in a period
of 7½ months. (Find the sum and the interest accrued
thereon).
Rule for separating the capital, interest, and the commission
of the surety etc., from the given amount :
48. Divide the product of the argument and its time
──────────────────────────────────────────────────────────────────────────
¹ Cf MSi, xv. 31; GT, p. 82, vs. 111 ; SiŚe, xiii. 17; L (ASS), p. 85,
vs. 90. Also see BrSpSi, xii. 14 (ii); GSS, vi. 21, 23.
² The same example occurs in GT, p. 83, vs. 112. Also see GSS,
vi. 22, 24; L (ASS), p. 86, vs. 91; GK, I, p. 60, lines 10-11. Similar
examples occur also in Pṛthūdaka's comm. on BrSpSi, xii. 14.
Indological Truths

32 DETERMINATIONS
as also the fruit (i.e., interest), etc., as multiplied by the other
time by their sum, and then multiply them by the mixed
amount : then are obtained the capital etc. in their respective
order.¹
This rule is similar to the previous one.
Ex. 54. The rate of interest being 5 per cent per
month, the commission of the surety (bhāvyaka) 1 per cent per
month, the fee of the calculator (vṛtti) ½ per cent per month,
and the charges of the scribe ¼ per cent per month, a certain
sum amounts to 905 in a year. (Find the capital, the interest,
and the shares of the surety, calculator, and the scribe).²
Here argument = 100, time = 1 month, first fruit = 5, second
fruit = 1, third fruit = ½, fourth fruit = ¼ ; amount = 905, and other
time = 1 year, i.e., 12 months.
∴ Capital = (argument × time × amount) / [argument × time + (sum of fruits) × other time]
= (100 × 1 × 905) / [100 × 1 + (5 + 1 + ½ + ¼) × 12]
= 500.
Interest = (first fruit × other time × amount) / [argument × time + (sum of fruits) × other time]
= (5 × 12 × 905) / [100 × 1 + (5 + 1 + ½ + ¼) × 12]
= 300.
Commission of surety = (second fruit × other time × amount) / [argument × time + (sum of fruits) × other time]
= (1 × 12 × 905) / [100 × 1 + (5 + 1 + ½ + ¼) × 12]
= 60.
Similarly, fee of the calculator = 30, and charges of the scribe = 15.
Rule for finding the time in which a sum lent out at simple
interest will be paid back by equal monthly instalments :
49-50. To get the desired time (īpsitakālopanaye) sub-
tract from the capital (mūlāt) the present worths of the first
monthly instalment, the second monthly instalment, etc.,

¹ Cf. GT, p. 83, vs. 113; SiŚe; xiii. 18;
² The same example occurs in GT, p. 83, vs. 114.

MIXTURES. SIMPLE INTEREST 33
(mûla), one after another (pṛthak pṛthak). [This will give
the number of complete months elapsed (gata-mâsa) and a
residue of the capital (mûla-śeṣa), if any]. Now calculate
interest for a month on the residue (śeṣasya-mâsika-phalaṁ),
and subtract that from the amount of monthly instalment
(mâsikopanayât) : by the remainder divide the interest for
a month on the residue (mâsikaphala) as multiplied by the
number of complete months elapsed (mâsa) and increased by
the residue of the capital (mûlaśeṣa).¹ The quotient added
to the number of complete months elapsed (gatamâsa) gives
the time (in months) of recovery of the capital (together
with interest).²
Let C be the capital lent out, i the rate of interest per cent per
month, I the amount of monthly instalment of payment, and T + 1/t,
t > 1, the time in months in which the money is recovered with interest.
Now suppose that the P.W.s of the 1st, 2nd,....... Tth instalments
having been subtracted one after another from C, the residue left is R.
Then evidently, R is the P.W. of I/t for (T + 1/t) months, so that
100 × (I / t)
R = ------------------------ ,
100 + (T + 1/t) i
(Ri / 100) · T + R
whence 1/t = ------------------- ·
I - (Ri / 100)
Hence the required time T + 1/t
(Ri / 100) · T + R
= T + ---------------------- months.
I - (Ri / 100)
Ex. 55-56. A rich man lent to somebody a sum of 100
rûpas at 5 per cent (per month simple interest) and from him
¹ The reading mâsaśeṣayug occurring in the Sanskrit text should be
mûlaśeṣayug.
² Cf. GK, I, p. 72, lines 14-15 ; and p. 73, lines 1-6.

34 DETERMINATONS
took a house bearing a rent of 40 (rûpas) per month. Say,
learned man, after how much time is the debter relieved of
his debt, and what does the rich man get by the gain of bare
accommodation.¹
Rule for converting several bonds into one :
51. The sum (samâsa) of the interests (phala) (accru-
ing on the given bonds) for the elapsed months (gatakâla), being
divided by the sum (aikya) of the interests (on the same
bonds) for one month, gives the time (in months for the equi-
valent single bond); and 100 times the sum of the interests for
one month (on the bonds), being divided by the sum of the
capitals (dhanayoga) (of the bonds), gives the rate of interest
per cent (per month) (for the single bond).²
Let the capitals, rates of interest and the times elapsed for the given
n bonds be as follows :
Capitals : P₁, P₂, P₃,..., Pₙ.
Rates of interest : r₁, r₂, r₃, ... , rₙ per cent per month
Times elapsed : t₁, t₂, t₃, .. , tₙ months.
Then the time elapsed (t) and the rate of interest (r) for the equivalent
single bond are given by
P₁ t₁ r₁ / 100 + P₂ t₂ r₂ / 100 + ... + Pₙ tₙ rₙ / 100
t = —————————————————————————————————————— months,
P₁ r₁ / 100 + P₂ r₂ / 100 + P₃ r₃ / 100 +....+ Pₙ rₙ / 100
( P₁ r₁ / 100 + P₂ r₂ / 100 + P₃ r₃ / 100 +....+ Pₙ rₙ / 100 ) × 100
and r = ——————————————————————————————————————————————————————
P₁ + P₂ + P₃ + .... + Pₙ
per cent per month.
Âryabhaṭa II puts the above results in the following simplified forms : ³
P₁ t₁ r₁ + P₂ t₂ r₂ + ... + Pₙ tₙ rₙ
t = ——————————————————————————— months,
P₁ r₁ + P₂ r₂ + ... + Pₙ rₙ
P₁ r₁ + P₂ r₂ + ... + Pₙ rₙ
and r = ——————————————————————— per cent per month.
P₁ + P₂ + ... + Pₙ

¹ A similar example occurs in GK, I, p. 73, lines 8-10.
² Cf. GT, p. 86, lines 12-15. Also see GSS, vi. 77-77½.
³ See MSi, xv. 33. The reading as emended by S. Dvivedi is
incorrect.
Indological Truths

MIXTURES. SIMPLE INTEREST 35
Ex. 57-58. (There are 4 bonds on which) capitals
amounting to 100, 200, 300, and 400 are given (to some one
on interest) at the rates of 2, 3, 4, and 5 per cent (per month)
in the respective order ; and months amounting to 2, 3, 5, and
4 each multiplied by 2, have passed (since the execution of
the respective bonds). Say, how would a single bond
(eka-patra) be now made out of these.¹
Ex. 59. Also say, O learned (mathematician), how a
single bond be made (out of 4 bonds) with the same capitals
as previously stated and with rates per cent (per month) of
interest augmented by ½ (in each case) and months elapsed
increased by ¼ (in each case).²
Rule for finding the time in which a capital lent out on
interest at a given rate would become a (given) multiple of
itself :
52(i). The product of the time and the argument,
being divided by the fruit and (then) multiplied by the
multiple minus one, gives the (required) time.³
Ex. 60(i). A sum of money is put to interest at 5 per
cent per month. When will it become twice of itself ?⁴
Ex. 60(ii). And when will (another sum of money) put
to interest at 3½ per cent (per month) become 1¼ of itself ?⁵
In Ex. 60(i), time = 1 month, argument = 100, fruit = 5, and
multiple = 2. Therefore the required time
1 × 100
= ------------ × (2 — 1), i.e., 20 months.
5
Similarly in Ex. 60(ii), time = 1 month, argument = 100, fruit
= 3½, i.e. 7/2, and multiple = 1¼. Therefore the required time
1 × 100 50
= ------------ × (1¼ — 1), i.e. -- months
7/2 7
== 7 months, 4 days, 17 ghaṭikās, 8 4/7 caṣakas.

¹ Cf. GSS, vi. 78½ ; GT. p. 86, lines 25-28 (contd. on p. 87 ),
² Cf. GT, p. 87, lines 22-23.
³ Cf. BrSpSi, xii. 14 (i); GT, p. 85, line 19. Both are literally the same.
⁴ A similar example occurs in Pṛthûdaka's comm. on BrSpSi, xii.
14. The same example occurs in GT, p. 85, lines 24-25.
⁵ A similar example occurs in Pṛthûdaka's comm. on BrSpSi, xii.
14. The same example occurs in GT, p. 86, lines 4-5.

36 DETERMINATIONS
(ii) ALLIGATION
Rule for finding the varṇa of the alloy obtained by melting
together a number of pieces of gold of given weights and
varṇas :
52(ii). The sum of the products of weight and varṇa
of the several pieces of gold, being divided by the sum of the
weights of the pieces of gold, gives the varṇa (of the alloy).¹
That is, if n pieces of gold of weights w₁, w₂, w₃, ..., wₙ and
varṇas v₁, v₂, v₃, ..., vₙ respectively are melted together, then the varṇa
v of the alloy is given by
v = (w₁ v₁ + w₂ v₂ + ... + wₙ vₙ) / (w₁ + w₂ + ... + wₙ) .
The term varṇa used above is analogous (though not equivalent) to
the modern term ‘carat’. It is used to denote the fineness of gold. Gold
of 1 varṇa contains 1 part of pure gold and 15 parts of impurities (in the
form of baser metals) ; gold of 2 varṇas contains 2 parts of pure gold and
14 parts of impurities ; and so on. Pure gold, called kalyâṇa-suvarṇa, is of
16 varṇas (or 24 carats). Thus the amount of pure gold in a piece of
gold of weight w and varṇa v is equal to wv/16.
The rationale of the above rule is as follows : The amount of pure
gold in the n pieces of gold mentioned above is equal to
(w₁ v₁)/16 + (w₂ v₂)/16 + ... + (wₙ vₙ)/16 , (1)
and the amount of pure gold in the alloy of those pieces of gold is equal to
((w₁ + w₂ + ... + wₙ) v) / 16 (2)
Since (1) and (2) are equal, therefore
((w₁ + w₂ + ... + wₙ) v) / 16 = (w₁ v₁)/16 + (w₂ v₂)/16 + ... + (wₙ vₙ)/16 .
Hence
v = (w₁ v₁ + w₂ v₂ + ... + wₙ vₙ) / (w₁ + w₂ + ... + wₙ) .
Ex. 61. Of what varṇa do 9, 5, and 17 mâṣas of
gold of 12, 10 and 11 varṇas (respectively) become when
melted together ?

¹ Cf. BM, III, H 1, 16 verso ; GSS, vi. 169 (i), 172½ ; L (ASS),
p. 99, vs. 103 (i); GK, I, p. 76, lines 5-7.

MIXTURES. ALLIGATION 37
Ex. 62. Of what varṇa do (the three pieces of gold of)
5 plus ⅓, 4 plus ⅙, and 7 plus ½ māṣas and 11 plus ½, 10, 8
minus ½ varṇas (respectively) become when mixed in one ?¹
Rule for finding the varṇa of the refined gold (obtained by
mixing up and refining a number of pieces of gold of given
weights and varṇas), when the weight of the refined gold is
known :
53(i). The sum of the products of varṇa and weight
of the several pieces of gold, being divided by (the weight of)
the refined gold, gives the varṇa (of the refined gold).²
That is, if n pieces of gold of weights w₁, w₂, ..., wₙ and varṇas
v₁, v₂, ... , vₙ respectively are mixed up and refined, then the varṇa v
of the refined gold is given by
w₁v₁ + w₂v₂ + ... + wₙvₙ,
v = -------------------------
w
where w is the weight of the refined gold.
The rationale is similar to that of rule 52(ii).
Ex. 63. 5, 8, and 6 suvarṇas of varṇas 12, 9, and 15 minus
½ respectively, being mixed up and refined, are seen to be
reduced to 16 suvarṇas in all. Quickly say the varṇa of that
(refined gold).
Rule for finding the weight of the refined gold, when the
varṇa of the refined gold is known :
53(ii). The same (sum of the products of varṇa and
weight of the several pieces of gold), being divided by the
varṇa (of the refined gold), gives the weight of the refined gold.³
That is, if n pieces of gold of weights w₁, w₂, ... , wₙ and of
varṇas v₁, v₂, .. , vₙ respectively are mixed up and refined, then the weight
w of the refined gold is given by
w₁v₁ + w₂v₂ + ... + wₙvₙ
w = --------------------------- ,
v
where v is the varṇa of the refined gold.
¹ Similar examples occur in BM, III, H 1, 16 verso; H 2, 17 recto
and 17 verso; GSS, vi. 170-171½.
² Cf. MSi, xv. 39 (ii); GK, I, p. 76, vs. 18.
³ Cf. BM, III, H3, 18 recto; GSS, vi. 175½, 176½, MSi, xv. 40;
L (ASS), p. 102, vs. 106; GK, I, pp. 76-77, vs. 19.

38 DETERMINATIONS
Ex. 64. (Three pieces of gold of weights) 10, 7, and 5
mâṣas are seen to be of 9, 8, and 6 varṇas (respectively). On
being mixed up and refined, they become of 11 varṇa. Say the
weight of the refined gold.
Rule for finding the unknown varṇa of one of the pieces of
gold, when the varṇa of the alloy of those pieces of gold is
known :
54. The sum of the weights of (all) the pieces of gold
being multiplied by the varṇa of the alloy (of those pieces
of gold), (then) diminished by the products of weight and
varṇa (of the pieces of gold of known weight and varṇa), and
(then) divided by the weight of the piece of gold of unknown
varṇa, gives the (unknown) varṇa.¹
That is, if n pieces of gold of weights w₁, w₂, ..., wₙ and varṇas
v₁, v₂, ..., vₙ respectively are mixed up with another piece of gold of
weight w and of unknown varṇa, and the varṇa of the alloy is found to
be v', then the unknown varṇa v will be given by
v = ((w₁ + w₂ + .... + wₙ + w) v' - (w₁ v₁ + w₂ v₂ + ... + wₙ vₙ)) / w .
Ex. 65. 1, 2, and 6 suvarṇas of 5, 3, and 4 kṣayas
(respectively) being mixed up with a pala of gold of unknown
varṇa become of 12 varṇa. (Find the unknown varṇa).²
Just as the varṇa of gold indicates the fineness of gold, so the kṣaya
of gold indicates the impurity of gold. Gold of 1 kṣaya contains 1 part of
impurities and 15 parts of pure gold; gold of 2 kṣayas contains 2 parts of
impurities and 14 parts of pure gold; and so on. Thus the varṇa and kṣaya
are related by the formula
varṇa + kṣaya = 16.
Rule for finding the unknown weight of one of the pieces of
gold, when the varṇa of the alloy of those pieces of gold is
known :
55. The sum of the weights of the pieces of gold multi-
plied by the varṇa of the alloy (of those pieces of gold), (then)
¹ Cf. MSi, xv. 39 (i); GK, 1, p. 76, vs. 18.
² Similar examples occur in BM, III, H3, 18 recto; GSS, vi. 178; 179;
L (ASS), p. 102, vs. 107; GK, I, p. 77, lines 4-7.
Indological Truths

MIXTURES. ALLIGATION 39
diminished by the products of weight and varṇa (of the pieces
of gold of known weights and varṇas), and (then) divided by
the varṇa of the piece of gold of unknown weight minus the
varṇa of the alloy, gives the (unknown weight of) gold.¹
That is, if n pieces of gold of weights w₁, w₂, ..., wₙ and varṇas
v₁, v₂, ..., vₙ respectively are mixed up with a piece of gold of varṇa v but
of unknown weight, and the varṇa of the alloy is found to be vʹ, then
the unknown weight w will by given by
w = [(w₁ + w₂ + ... + wₙ) vʹ − (w₁v₁ + w₂v₂ + ... + wₙvₙ)] / (v − vʹ)
Ex. 66. 2, 3, and 4 mâṣas of gold of kṣayas 7, 4, and 5
(respectively) being mixed up with some gold of kṣaya 2, the
alloy is found to be of kṣaya 4. Quickly say with how much
of that gold (of kṣaya 2 have they been mixed up).²
Rule for finding how much of inferior gold (and how much
of pure gold) will have to be taken in constructing a number
of test sticks of progressively decreasing varṇas :
56. The weight in terms of mâṣas (mâṣâtmaka mâna) of
one (test) stick being divided by the difference in terms of
yavas between the varṇas (of pure and inferior gold) and multi-
plied one time, 2 times, etc., of the progressive decrease of
varṇa in terms of yavas, gives the respective amounts of inferior
gold.³
That is, if n test sticks of weight w and varṇas 16 − k, 16 − 2k,...,
16 − nk are to be made out of pure gold and inferior gold of varṇa 16 − K,
then for the construction of those test sticks we will have to take the
following amounts of inferior gold in the respective order :
(w / K) × k, (w / K) × 2k, ......, (w / K) × nk.
¹ Cf. BM, III, H3, 18 verso; GSS, vi. 175½, 180; MSi, xv. 41; L
(ASS), p. 103, vs. 108; GK, I, p. 77, vs. 20.
² Cf. BM, III, H3, 18 verso; GSS, vi. 181; L (ASS), p. 103, vs. 109;
GK, I, p 77, lines 16-19.
³ Cf. GK, I, p. 78, lines 18-21, and p. 79, lines 1-2. Also see GSS,
vi. 192.

40 DETERMINATIONS
Subtracting these amounts severally from w, we will get the respec-
tive amounts of pure gold.
The rationale is as follows :
Let the required amounts (weights) of gold of varṇa 16-K and of
varṇa 16 be respectively
w₁ , w₂ ,......... , wₙ
and (w - w₁) , (w - w₂) , ... , (w - wₙ) .
Then we evidently have
(16 - K) wᵣ + 16 (w - wᵣ) = w(16 - rk),
where r = 1, 2, ... , n ; whence
wᵣ = (w / K) × rk .
The text does not give the relation between a māṣa and a yava,
but the commentator takes one māṣa as equivalent to 16 yavas.
Ex. 67-68. A series of varṇas, from 0 kṣaya to 6 kṣaya,
uniformly increasing by ½ kṣaya, is to be constructed of sticks
of 2 māṣas each. O mathematician, if you know then quickly
say after calculating how much gold of varṇa 16 and how
much gold of varṇa 10 will have to be taken (in each case).¹
Rule for finding the weights of two gold balls of equal value
(i.e., containing equal amounts of pure gold), when the sum
of their weights and also the varṇas of the alloys formed by
mixing each of them with given proportions of the other
are known :
57. The (given) varṇas (severally) divided by the (given)
fractions, each increased by 1, should be mutually diminished
by the given fractions as multiplied by them (i.e., the given
varṇas severally divided by the given fractions, each increased
by 1). The two (results thus obtained) should be divided by
their (own) sum, then their order should be reversed, and
then they should be multiplied by the sum of the weights of
the gold balls. This will give the weights (of the gold balls)
separately.
¹ Cf GK, I, p. 80, lines 2-5.

MIXTURES. ALLIGATION 41
That is, if w be the sum of the weights of the two gold balls,
v₁ the varṇa of the alloy formed by mixing the first gold ball with a/b part
of the second, and v₂ the varṇa of the alloy formed by mixing the second
gold ball with c/d part of the first, then the separate weights w₁ and w₂ of
the first and second gold balls are given by
w₁ = (1/k) { v₂/(1 + c/d) − ((a/b)v₁)/(1 + a/b) } w, (1)
and
w₂ = (1/k) { v₁/(1 + a/b) − ((c/d)v₂)/(1 + c/d) } w, (2)
where k = { v₂/(1 + c/d) − ((a/b)v₁)/(1 + a/b) } + { v₁/(1 + a/b) − ((c/d)v₂)/(1 + c/d) } .
The rationale is as follows :
Let V₁ and V₂ be the varṇas of the first and second gold balls respec-
tively. Then we have
w₁ + w₂ = w , (3)
v₁ = (w₁V₁ + (a/b)w₂V₂) / (w₁ + (a/b)w₂) , (4)
v₂ = ((c/d)w₁V₁ + w₂V₂) / ((c/d)w₁ + w₂) , (5)
and, since the gold balls are of equal value,
w₁V₁ = w₂V₂ . (6)
Dividing (4) by (5), and making use of (6), we have
v₁/v₂ = [(1 + a/b)/(1 + c/d)] × [((c/d)w₁ + w₂) / (w₁ + (a/b)w₂)] ,
whence
w₁ / [v₂/(1 + c/d) − ((a/b)v₁)/(1 + a/b)] = w₂ / [v₁/(1 + a/b) − ((c/d)v₂)/(1 + c/d)]
But by virtue of (3), each of these is equal to w/k. Hence we
have (1) and (2).
Rule for finding the varṇas of two gold balls of equal value,
when the sum of their weights and also the varṇas of the
alloys formed by mixing each of them with given propor-
tion of the other are known :
58. The weights of the two (gold) balls (obtained by
applying the previous rule) should each be increased by the
given fraction of the weight of the other. The two (results