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

42 DETERMINATIONS
thus obtained) being divided by the (same) fractions, each
increased by 1, then multiplied by the (respective) varṇas,
and then divided by the (respective) weights of the gold
balls, give the varṇas (of the two gold balls).¹
That is, if _w_₁ and _w_₂ be the weights of two gold balls whose
varṇas are _V_₁ and _V_₂ respectively, _v_₁ the varṇa of the alloy formed by mixing
the first gold ball with a/b part of the second, and _v_₂ the varṇa of the
alloy formed by mixing the second gold ball with c/d part of the first, then
_V_₁ = [_w_₁ + (a/b) _w_₂] / [(1 + a/b) _w_₁] · _v_₁,
_V_₂ = [_w_₂ + (c/d) _w_₁] / [(1 + c/d) _w_₂] · _v_₂.
Of these results, the first follows from equations (4) and (6) of the
previous rule, and the second from equations (5) and (6).
Ex. 69. There are two small balls of gold of equal
worth (i.e., having equal quantities of pure gold), whose com-
bined weight is 5 māṣas. When they are respectively combined
with ⅔ part and ½ part of the other, they become of 10 and 9
varṇas respectively. (Find their weights and varṇas separa-
tely).
Ex. 70. O learned one, there are two small balls of
gold of equal value. When taken together, their weight is 12
māṣas ; and when they are respectively combined with ⁵/₇ part
and ⅕ part of the other, they become of 12 and 10½ varṇas
respectively. (Find their weights and varṇas separately).
(iii) PARTNERSHIP
Rule for finding the shares of the partners in the produce :
59(i). To obtain the individual shares (of the partners)
in the produce (phalâvâptyai), the seeds (contributed by the
partners) (prakṣepa), as divided by their sum, should be severally
multiplied by the produce (phala).²

¹ See also GSS, vi 209-212; and GK, I, p. 81, lines 9-18.
² Cf. BrSpSi, xii. 16 (i); SiŚe, xiii. 19 (i); L (ASS), p. 90, vs. 94; GK,
I, p. 54, lines 11-12. Also see GSS, vi, 79½ ; MSi, xv. 36.

MIXTURES. PURCHASE AND SALE 43
The prakṣepa, according to the commentator, is "that which is thrown,
scattered, or sown as seed" and therefore "seed", and the phala is "what is
produced out of that".
Ex. 71. Two, three, five, and four prasthas of seeds (are
the contributions of the partners) and 210 (prasthas of grain)
is the produce; what are the shares (of the partners)
separately ?
Ex. 72. ½ prastha is (the contribution) of one, ⅓ (prastha)
of another, ⅑ (prastha) of still another, and 1700 (prasthas) is
the produce. Say what are their shares (in the produce)
separately.
(iv) PURCHASE AND SALE
Rule for obtaining the quantities and prices of each of a
number of commodities separately, when they are purchased
in a specified proportion for a given amount of money :
59(ii). Having divided the (given) rate-prices (of the
commodities) by the (corresponding) rate-quantities (of those
commodities) and (then) having multiplied them by the
corresponding proportions (in which those commodities are
purchased), apply the previous rule.¹
Âryabhaṭa II states the rule more explicitly. He writes :²
"Set down the rate-prices as multiplied by the respective propor-
tional parts and divided by the rate-quantities of the respective commodi-
ties in two places. Multiply the results set down in one place as well as the
proportional parts by the mixed price and divide by the sum of the results
set down in the other place. These give the prices (paid for the respective
commodities) and the quantities (of the respective commodities that are
obtained) respectively."
Ex. 73-74. 7 kuḍavas of mudga ("seeds of Phaseolus
mungo") are obtained for 9 paṇas, and ½ kuḍava of rice is
obtained for one paṇa.

¹ Cf. SiŚe, xiii. 19 (ii); GK, I, p. 57, vs. 2. Also see GSS, vi. 87½-89½.
² MSi, xi. 37-38 (i). Also see L (ASS), p. 92, vs. 98.

44 DETERMINATIONS
Then, O merchant, take 3 paṇas and a half and quickly
give me one part of rice and two parts of mudga.
Solution. Dividing the rate-prices of mudga and rice by the corres-
ponding rate-quantities of those commodities, we get
9/7 and 2/1.
Multiplying these by the corresponding proportions in which they
are purchased (i.e., by 2 and 1), we get
18/7 and 2.
Now we apply the previous rule. Thus dividing 18/7 and 2 by their
sum, and multiplying them by the total price (i.e., 3½ paṇas), we get
63/32 and 49/32 paṇas
These are the prices of mudga and rice respectively.
The quantities of mudga and rice obtained for 63/32 paṇas and 49/32 paṇas
respectively can now be obtained by the rule of three. These come out to
be 49/32 kuḍavas and 49/64 kuḍavas respectively.
Ex. 75. ½ pala of asafoetida (hiṅgu), 2 palas of long
pepper (pippalī), and 7 palas of dry ginger (śuṇṭhī), are each
obtained for one rūpa. Give me equal quantities (of each
of them) for one rūpa.¹
Rule for finding the rates of purchase and sale when mer-
chants investing unequal capitals become equally rich by
purchasing and selling articles at the same rates :
60. Reduce the capitals (vitta) to a common denomina-
tor, and then remove the denominators. This being done,
the greatest amount increased by an optional number (1, 2, 3,
etc.) gives the rate of sale (i.e., the rate at which the
articles are sold for one rūpa ; vikraya).
That (i.e., the rate of sale) multiplied by the price at
which each remnant article is sold (antyārgha), being dimini-
────────────────────────────────────────────────────────────
¹ Cf. GK, I, p. 57, lines 16-19.

MIXTURES, PURCHASE AND SALE 45
shed by 1 and then multiplied by the common denominator
(of the capitals), gives the rate of purchase (i.e., the rate at
which the articles are purchased for one rūpa ; kraya).¹
That is, if n persons, whose capitals reduced to a common deno-
minator are
C₁/D, C₂/D, C₃/D, ..., Cₙ/D rūpas respectively,
invest their capitals in purchasing certain articles at the rate of x articles
per rūpa, and if by selling those articles in multiples of y at the rate of y
articles per rūpa, and the remnant articles, at the rate of R rūpas per
article, they become equally rich, then
x = D(Ry – 1),
and y = C_λ + k,
where C_λ is the greatest of all C's, and k an arbitrary integer.
The rationale of this rule is as follows :
Suppose that by selling the articles in multiples of y, the n persons
earn P₁, P₂, P₃ ..., Pₙ rūpas respectively. Then the total amounts
earned by the n persons by selling all the articles (including the remnant
articles) are respectively
P₁ + R((C₁/D)x – P₁y), P₂ + R((C₂/D)x – P₂y), ... ,
and Pₙ + R((Cₙ/D)x – Pₙy) rūpas,
or (RC₁x/D) – P₁(Ry – 1), (RC₂x/D) – P₂(Ry – 1), ...,
and (RCₙx/D) – Pₙ(Ry – 1) rūpas.
Since these amounts are equal to one another, therefore we have
(RC₁x/D) – P₁(Ry – 1) = (RC₂x/D) – P₂(Ry – 1) = ...
= (RCₙx/D) – Pₙ(Ry – 1).
or RC₁x – P₁D(Ry – 1) = RC₂x – P₂D(Ry – 1) = ...
= RCₙx – PₙD(Ry – 1) = M say. (1)
To solve (1), let x = M = D(Ry – 1);
─────────────────────────────────────────────────────────────
¹ Also see GSS, vi. 102½ ; GK, I, pp. 94-97, vv. 36 (ii) – 38(i).

46 DETERMINATIONS
then
RCᵣ - Pᵣ = 1 , (2)
for r = 1, 2, ..., n .
But (Cᵣ/D)x / y = Pᵣ + Aᵣ / y ,
where Aᵣ is the remainder when (Cᵣ/D)x is divided by y , so that,
when x = D(Ry - 1),
Pᵣ = Cᵣ x / Dy - Aᵣ / y
= Cᵣ (Ry - 1) / y - Aᵣ / y
= RCᵣ - (Cᵣ + Aᵣ) / y (3)
Therefore, we must have
y > Cᵣ , for all r ,
or y > C_λ ,
where C_λ is the greatest of all C's.
Hence a solution of (1) is
x = D(Ry - 1),
y = C_λ + k ,
k being an arbitrary integer.
Sub-rule for finding another rate of purchase, when the capitals
are integral and the articles which are left after selling at
the general rate as also the amounts of money acquired by
selling the articles at the general rate have a common factor :
61. The articles which are severally left after selling
(at the general rate) (i.e., the remnant articles) as also the
amounts of money which are severally acquired by selling the
articles (at the general rate), being abraded by their common
factor (lit. divisor), the resulting equal amounts of money
(i.e., the equal sale proceeds) give another rate of purchase.
This rule relates to the case when D = 1 , and may be expressed
analytically as follows :
If d be the common factor of
P₁, P₂, ..., Pₙ
and ( C₁x - P₁y ), ( C₂x - P₂y ), ..., ( Cₙx - Pₙy ),
Indological Truths

MIXTURES, PURCHASE AND SALE 47
then another rate of purchase is x' articles per rûpa, where
x' = P₁/d + R( C₁x - P₁y )/d
= P₂/d + R( C₂x - P₂y )/d
...
= Pₙ/d + R( Cₙx - Pₙy )/d ,
where x = Ry - 1, y = C_λ + k, C_λ being the greatest of all C's and
k an arbitrary integer; the rate of sale being the same as before
( i.e., C_λ+k articles per rûpa ).
The rationale of this rule is as follows :
When D = 1 , the equations (1) of the previous rule may be
written as
RC₁x - P₁(Ry - 1) = RC₂x - P₂(Ry - 1) = ...
= RCₙx - Pₙ(Ry - 1) = M. (1)
Dividing throughout by d , and writing x' for x/d , P_r' for P_r/d,
and M' for M/d, we have
RC₁x' - P₁'(Ry - 1) = RC₂x' - P₂'(Ry - 1) = ....
= RCₙx' - Pₙ'(Ry - 1) = M'. (2)
The values of x , y satisfying (1) give one set of rates of purchase
and sale, and the values of x' , y satisfying (2) give another set of rates
of purchase and sale.
When M = Ry - 1 , a solution of (1) is, as before,
x = Ry - 1 ⎫
⎬ (3)
y = C_λ+k ⎭
and the corresponding solution of (2) is
x' = (Ry - 1)/d , i.e., M' ⎫
⎬ (4)
y = C_λ+k . ⎭
The rates of purchase and sale given by (3) are those already given
in Rule 60 ; those given by (4) are those mentioned in the present rule.
Since, from (2),
M' = P₁/d + R(C₁x - P₁y)/d = P₂/d + R(C₂x - P₂y)/d
= ... = Pₙ/d + R(Cₙx - Pₙy) ,
it is clear that the value of x' given by (4) is the same as that stated
in the above rule. It can be easily seen that M' is an integer.
Note. When D ≠ 1, then another rate of purchase will be D(Ry - 1)/d
articles per rûpa, where y = C_λ + k.

48 DETERMINATIONS
Ex. 76. The capitals of (three) men are 1, 3, and 5
(rûpas) or ⅓, ¼ and ½ (rûpas) (respectively). By purchasing and
selling (certain articles) at the same rates and by selling the
remnant articles at the rate of 1 for 3 (rûpas), they become
possessed of equal riches. (Find the rates of purchase and
sale).¹
Sub-rule for finding the rates of purchase and sale when each
of the remnant articles is sold for a fractional price :
62. When the remnant-price (i.e., the price at which
each remnant article is sold) is fractional (lit. has a denomi-
nator), reduce it along with the capitals to a common
denominator (and remove the denominators as before), and
multiply the rates of purchase and sale resulting from them
(by applying Rule 60) by the common denominator.²
That is, if the capitals of the n persons and the remnant-price
reduced to a common denominator, be as follows
Capitals : C₁/D , C₂/D , C₃/D , ... , Cₙ/D rûpas
Remnant-price : (R/D) rûpas,
then x = D² ( (R/D)y - 1 ) ,
and y = D (C_λ + k) ,
C_λ being the greatest of all C's and k an arbitrary integer.
The rationale of this rule is as follows :
As before, equating the wealths of the n persons after all the articles
have been sold away, we get [see equations (1) of Rule 60 (notes)]
(R/D) · C₁x - P₁D ( (R/D)y - 1 ) = (R/D) · C₂x - P₂D ( (R/D)y - 1 )
= ... = (R/D) · Cₙx - PₙD ( (R/D)y - 1 ) ,
¹ Cf. GSS, vi. 103½, 104½; GK, I, p. 96, lines 2-6, and p. 97, lines 7-9.
² Cf. GSS, vi. 107½. For another rule, see GSS, vi. 109½; GK, I, p. 98,
lines 2-7.

MIXTURES. PURCHASE AND SALE 49
or RC₁x - P₁D² ( R/D y - 1 ) = RC₂x - P₂D² ( R/D y - 1 )
= ... = RCₙx - PₙD² ( R/D y - 1 ) = N, say. (1)
Let x = N = D²( R/D y - 1 ), then
RCᵣ - Pᵣ = 1, (2)
for r = 1, 2, ..., n.
But ((Cᵣ/D)x) / y = Pᵣ + Aᵣ/y,
where Aᵣ is the remainder when Cᵣx/D is divided by y, so that, when
x = D²( R/D y - 1 ),
Pᵣ = Cᵣx / Dy - Aᵣ / y
= [ Cᵣ D² ( R/D y - 1 ) ] / Dy - Aᵣ / y
= RCᵣ - (Cᵣ D + Aᵣ) / y. (3)
Therefore we must have
y > CᵣD, for all r,
or y > C_λD,
where C_λ is the greatest of all C's.
Hence a solution of (1) is
x = D²( R/D y - 1 ),
y = D(C_λ + k),
k being an arbitrary integer.
Ex. 77. The capitals of four men are 1½, 2, 3, and 5
rûpas. By purchasing and selling (certain articles) at the same
rates, and by selling the remnant articles at the rate of 1 for ½
(of a rûpa), they become possessed of equal riches. (Find the
rates of purchase and sale).¹

¹ Cf. GSS, vi. 108½, 110½; GK, I, p. 99, lines 2-5 and 15-18.

50 DETERMINATIONS
Rule for finding out how a specified number of creatures of
given rates can be bought for a specified price :
63-64. By the price of one creature of any variety
multiply the rate-creatures (of other varieties) in the order in
which they have been stated (in the problem) (and also the
number of creatures to be bought). From the products (corres-
ponding to the rate-creatures) severally subtract the respec-
tive rate-prices of the creatures (and from the product
corresponding to the number of creatures to be bought sub-
tract the specified price). Now multiply the various remain-
ders, excepting that obtained by subtracting the specified
price, by optional numbers (multipliers) which are to be
chosen in such a way that (i) the resulting products when
added together may yield (the remainder obtained by sub-
tracting) the specified price as sum, and (ii) on taking the
products of those multipliers and the respective rate-prices,
a negative number or zero may not be obtained for the multi-
plier of the creature which is without a multiplier. (The multi-
pliers for the various creatures, obtained in this way, when
multiplied by the respective rate-creatures, will give the
number of creatures of the different varieties that will be
bought for the specified price ; and the same multipliers
when multiplied by the rate-prices of the respective creatures
will give the prices that will be paid for the creatures of the
respective varieties).¹
For the explanation of this rule see under Ex. 78-79.
Ex. 78-79. Pigeons are sold at the rate of 5 for 3 (rûpas),
cranes at the rate of 7 for 5 (rûpas), swans at the rate of 9 for
7 (rûpas), and peacocks at the rate of 3 for 9 (rûpas). Knowing
the rates as stated above, bring 100 birds for 100 rûpas
for the amusement of the prince.²
¹ Cf. GK, I, p. 92, vv. 34(ii)-35. For other rules see GSS, vi. 146½, 151.
² The same example occurs in GSS, vi. 152-153; BBi (ASS), p. 163,
vs. 138; GK, I, p. 93, lines 2-5. A similar problem occurs in BM, III, E 3,
58 verso, which may be stated as : ‘A man earns 3 maṇḍas in a day, a
woman 1½ maṇḍas in a day, and a sûḍha ½ maṇḍa in a day. If 20 of
them earn 20 maṇḍas in a day, how many of each category are there?’

MIXTURES. PURCHASE AND SALE 51
Let the number of pigeons, cranes, swans, and peacocks bought for
100 rûpas, and the prices paid for the respective birds be as follows :
Pigeons Cranes Swans Peacocks
Number 5x 7y 9z 3u } (1)
Price 3x 5y 7z 9u
Then we have
5x+7y+9z+3u = 100 , (2)
3x+5y+7z+9u = 100 . (3)
Multiplying (2) by 3 (which is the price of one peacock) and
subtracting (3) therefrom, we have
12x+16y+20z = 200 . (4)
This is an indeterminate equation having an indefinitely large
number of solutions. But we are in the present case concerned only with
those positive and non-zero solutions of (4), which make
5x , 7y , 9z , and 3u
positive integers.
Proceeding by trial, we obtain the following sixteen such solutions
of (4) :
(1) x=3, y=4, z=5 . (9) x=4, y=7, z=2 .
(2) x=11, y=3, z=1 . (10) x=8, y=4, z=2 .
(3) x=1, y=8, z=3 . (11) x=12, y=1, z=2 .
(4) x=6, y=3, z=4 . (12) x=5, y=5, z=3 .
(5) x=2, y=6, z=4 . (13) x=9, y=2, z=3 .
(6) x=4, y=2, z=6 . (14) x=7, y=1, z=5 .
(7) x=3, y=9, z=1 . (15) x=1, y=3, z=7 .
(8) x=7, y=6, z=1 . (16) x=2, y=1, z=8 .
The corresponding values of u are obtained from (2).
Substituting these values of x, y, z, and u in (1), we obtain sixteen
valid solutions of the problem. (For actual solutions, see the answer).
The above analysis explains the basis of the foregoing rule (Rule
63-64). It may be observed that the multipliers mentioned in the rule
refer to the variables x, y, z, and u.
Indological Truths

52 DETERMINATIONS
Ex. 80. (The rates of sale) of pomegranates, mangoes,
and wood-apples are respectively 1 for 2 (rûpas), 5 for 3 (rûpas),
and 2 fruits for 1 (rûpa). Bring 100 (fruits) for 80 (rûpas).¹
(v) MEETING OF TWO TRAVELLERS
Rule for finding the time in which the fast traveller, who starts
travelling on the same track after the slow traveller has
already covered a specified distance, would overtake the slow
traveller :
65. By the difference between the speeds per day of the
fast and slow travellers divide the distance already travelled
by the slow traveller. This gives the time (in days in which
the fast traveller would overtake the slow one).²
Ex. 81-82. When a person, travelling ( at the speed of )
8 yojanas per 5 minus ½ days, has already travelled for 6 minus
¼ days, another person, who travels ( at the speed of) 3 yojanas
a day, starts travelling (from the same place) along the same
track. Say, after calculating, when the latter traveller would
overtake the former.³
Rule for finding the time when two travellers, one fast and
the other slow, who, starting simultaneously from the
same place, are destined to go to a specified distance
and then to come back by the same track, will meet each other
on the way, one going ahead and the other coming back :
66(i). The (length of the) track divided by half the
sum of the speeds per day (of the two travellers) gives the
time (in days) at meeting.⁴
The time at meeting is evidently measured since the start of travel.
Ex. 83. One man travels ( at the speed of) 8 yojanas
(a day) and another (at the speed of) 2 yojanas (a day).
(They start simultaneously from the same place, and after
reaching the destination come back by the same track).
¹ Cf. GSS, vi. 147½-149, 150.
² Cf. BM, III, A 13, 3 recto; GSS, vi. 326½.
³ Cf. BM, III, B4, 4 recto; E 2, 53 recto; E 2, 53 verso; GSS, vi. 327½.
⁴ Cf. GSS, vi. 319 (ii).

MIXTURES. WAGES AND PAYMENTS 53
The (length of the) track is 100 yojanas. Say where is the
meeting (of the two), one going ahead and the other coming
back.¹
Sub-rule for finding the length of the track :
66(ii). The time of meeting multiplied by half the sum
of the speeds per day (of the two travellers) gives the length of
the track.
Sub-rule for finding the speed of one of the two travellers :
67(i). The (length of the) track as divided by the time
of meeting, being doubled and then diminished by the speed
(of one of the two travellers) gives the other speed (i.e., the
speed of the other traveller).
(vi) WAGES AND PAYMENTS
Rule for finding the wages for carrying a bottle of oil when
some oil falls down on the way due to a hole in the bottle :
67(ii). Give half the wages for the quantity fallen down
(on the way) and full for the rest.²
It can be easily seen that the wages prescribed in this rule corres-
pond to the average load carried by the porter.
Ex. 84-85. While a leathern oil-bottle (kutapa), filled
with 200 palas of oil, was being carried (by a porter) to a
distance of 8 yojanas for 5 paṇas (as wages), a hole hap-
pened to occur in the bottom of it through which the oil
leaked out (on the way) continuously. If 20 palas of oil be left
(in the bottle), what wages should be paid (to the porter)?³
Rule for finding the amounts to be paid to a dancing party
by the spectators, who see the dance for different parts of the
day, the amount to be paid to the dancing party for the whole
day being given:
68. The succeeding parts of the day, each diminished
by the preceding one, should be multiplied by the ‘fruit’
¹ Cf. BM, III, 9 verso ; GSS, vi. 321-321½
² Cf. GK, I, p. 102, lines 6-7.
³ Read bhṛte for dhṛte in the text. For a similar example, see GK, I,
p. 103, lines 2-5.

54 DETERMINATIONS
(phala); then having divided them severally by the respec-
tive numbers of spectators, add the preceding results to
the succeeding ones. The results thus obtained being multi-
plied by the numbers of spectators who go away (at the end
of the respective parts of the day), give the amounts to be paid
(to the dancing party by the spectators who go away at the
end of the respective parts of the day).¹
By the word ‘fruit’ (phala) is meant here the amount to be paid to
the dancing party for the whole day.
The above rule is based on the assumption that the spectators
have to pay the dancing party in proportion to the time of seeing the
dance. That is, if the spectators who see the dance for the whole day
pay 96 rūpas in all, then the spectators who see the dance for ¼ of a day
will have to pay 96/4, i.e., 24 rūpas in all.
For the illustration of the rule see under the next example.
Ex. 86-87. One man saw a dance for one quarter of the
day, another for two quarters of the day, another for
three quarters of the day, and (yet) another till the end of the
day. The dancing party has to be paid by them a sum of
ninety six rūpas in all. If payment is to be made in propor-
tion to the time of seeing the dance, how much of that (sum)
should be paid by each of them separately?
Solution. Writing down the parts of the day during which the
various spectators saw the dance, we have
1/4 2/4 3/4 4/4
Diminishing the succeeding part by the preceding one, we get
1/4 1/4 1/4 1/4
Multiplying each of these by the fruit (i.e., by 96), we get
24 24 24 24
Dividing these numbers by the number of spectators who saw the
dance during the respective parts of the day, we get
6 8 12 24
¹ See GSS, vi. 230.

MIXTURES. THE CISTERN PROBLEM 55
Adding the preceding numbers to the succeeding ones, we get
6 14 26 50
Multiplying by the number of spectators who go away at the end of
the respective parts of the day, we get
6 14 26 50
Hence the spectator who saw the dance for ¼ of the day will have
to pay 6 rûpas ; the spectator who saw the dance for ²⁄₄ of the day will have
to pay 14 rûpas ; the spectator who saw the dance for ¾ of the day will
have to pay 26 rûpas ; and the spectator who saw the dance for the whole
day who will have to pay 50 rûpas.
Ex. 88. A palanquin is to be carried to a distance of
3 krośas by 10 men for 100 (rûpas as wages). Of those (men),
2, 3, and 5 stop away after going over 1, 2, and 3 krośas respec-
tively. (Calculate the wages of each of them separately).¹
Ex. 89-90. Five Brâhmaṇas, enchanters of stotras
(“hymns in praise of gods”), were invited by a certain person
to take part in the worship of the five faces of the five-faced
god Śiva on (a remuneration of) 300 rûpas. And they, on the
completion of worship of one, two, three, four, and five faces
respectively, went away (from the place of worship) one by
one.² Say what are their remunerations (separately).
(vii) THE CISTERN PROBLEM
Rule for finding the time in which a cistern would be filled
up by a number of drains flowing into it simultaneously, when
the times in which the cistern is filled up by each drain
separately are given :
69. Having divided unity severally by the (given)
fractions (of time), take the sum of the quotients and by that
sum divide unity (again) : this will give the time in which
the cistern is filled up (when all the drains are simultaneously
opened to flow into it).³
¹ Cf. GSS, vi. 231-232.
² What is meant is this : when the worship of the first face was com-
pleted by all the five Brâhmaṇas, one Brâhmaṇa left the place of worship ;
when the worship of the second face was completed by the remaining four
Brâhmaṇas, one more Brâhmaṇa left the place of worship ; when the
worship of the third face was completed by the remaining three Brâhmaṇas,
one more Brâhmaṇa left the place of worship ; and so on.
³ Cf. MSi, xv. 43; L (ASS), p. 91, vs. 96 ; GK, I, p. 94, line 2-3.

56 DETERMINATIONS
That is, if n drains severally fill up a cistern in
a₁/b₁, a₂/b₂, ... , aₙ/bₙ of a day,
then all the drains working together will fill up the cistern in
1 / (b₁/a₁ + b₂/a₂ + ... + bₙ/aₙ) of a day.
Ex. 91. In what time will the (four) drains, which
severally fill up a cistern in ½, ¼, ⅕, and ⅙ of a day, fill up
that cistern if they are opened simultaneously (to flow into
it) ? 1
(viii) WAGES PAID FROM THE COMMODITY
Rule for finding the wages of a porter who carries a commo-
dity over a part of the stipulated distance, when the wages
are to be paid out of the commodity itself :
70. (To obtain the wages for a part of the stipulated
distance) divide half the product of the commodity and the
stipulated distance as diminished by the square root extracted
from the square of half the product of the commodity and the
stipulated distance as diminished by the (continued) product
of the distance (already) gone over, the distance to be gone
over, the commodity, and the stipulated wages, by the
distance which is still to be gone over.²
Assuming that the (stipulated) wages for carrying a articles over the
(stipulated) distance d are w articles (out of those a articles), let the
wages for carrying those a articles over a distance d₁, where d₁ < d, be
x articles (out of those a articles). Then calculating at the same rate, the
wages for carrying the remaining a - x articles over the remaining distance
d - d₁ are equal to
(a - x) (d - d₁)x / (ad₁) articles.
But the remaining wages being w - x articles, we have
(a - x) (d - d₁)x / (ad₁) = w - x ,
1 Similar examples occur in Pṛthûdaka Svâmî's (860 A.D.) comm.
on BrSpSi, xii. 9; L (ASS), p. 91, vs. 97 ; GK, I, p. 94, lines 5-6.

• Cf. GSS, vi. 226. For another rule, see GK, I, p. 103, lines 10-15.

• MIXTURES. WAGES PAID FROM COMMODITY 57
  or, (d - d₁) x² - adx + awd₁ = 0 ,
  whence x = [ad/2 ± √((ad/2)² - awd₁ (d - d₁))] / (d - d₁) .
  Taking the negative sign before the radical, we have the rule stated
  above. The positive sign is rejected because it makes x > w.
  Ex. 92. If for carrying 24 jack-fruits over a distance of
  5 krośas (a porter) is to get 9 of those jack-fruits, what will
  he get if he carries them over a distance of 2 krośas (only) ?¹
  Rule for finding the distances over which two porters carry
  a commodity in turn, when the wages paid to the porters
  out of the commodity carried by them are given :

71. The first and the second loads (i.e., the loads carried
    by the first and the second porters), each multiplied by
    the wages for the other, being added together, the resul-
    ting sum is (to be taken as) the divisor (of the following
    results) : the results obtained by multiplying the products
    of the wages received (for each of the loads) and the
    (total) distance (gone over by the porters), each by the other
    load. This gives the distances (gone over by the first and the
    second porters separately).²
    Suppose that the first porter carries a articles over a distance x
    (unknown) and gets b articles out of those a articles as wages, and that
    the second porter carries the remaining a - b articles over the remaining
    distance d - x and gets c articles as wages. Then, we have
    c = b(a - b)(d - x) / ax ,
    whence x / ((a - b)b) = (d - x) / ac = d / (ac + (a - b)b)
    Hence the above rule.
    Ex. 93-94. Twenty-four jack-fruits were carried (over
    a certain distance) by one (person) for 4 out of those (24) jack-
    fruits as wages ; the remaining jack-fruits were carried (over
    the remaining distance) by another person for 5 of them (as
    wages). The load was thus carried by the two (person) over

¹ Cf. GSS, vi. 227. Also see GK, I, p. 103, lines 17-20.
² See GSS, vi. 228.

58 DETERMINATIONS
a distance of 5 krośas (in all). Say, O learned, how much of
that distance was gone over by each of them ?¹
(ix) COMBINATIONS OF SAVOURS
Rule for finding the number of combinations that can be
formed out of a given number of savours (or things) by
taking 1, 2, 3, ...., all at a time :
72. Writing down the numbers beginning with 1 and
increasing by 1 up to the (given) number of savours in the
inverse order, divide them by the numbers beginning with 1
and increasing by 1 in the regular order, and then multiply
successively by the preceding (quotient) the succeeding one.
(This will give the number of combinations of the savours
taken 1, 2, 3, ... , all at a time respectively).²
Let the given number of savours be six. Then the numbers
1, 2, 3,..., 6 are written down in the inverse order thus :
6 5 4 3 2 1
These are divided by the numbers 1, 2, 3, ... 6 respectively, so that
we get
6/1, 5/2, 4/3, 3/4, 2/5, 1/6.
Now multiplication is made successively by the preceding quotient of the
succeeding one. Thus we get
6/1, 6/1 × 5/2, 6/1 × 5/2 × 4/3, 6/1 × 5/2 × 4/3 × 3/4, 6/1 × 5/2 × 4/3 × 3/4 × 2/5,
6/1 × 5/2 × 4/3 × 3/4 × 2/5 × 1/6.
These, according to the rule, are respectively the numbers of combinations
of the six savours taken 1, 2, 3,..., 6 at a time.
Ex. 95. Friend, a cook prepares varieties of food with
the six savours, pungent, bitter, astringent, acid, saline, and
sweet. Say what is the (possible) number of varieties ?³

¹ A similar problem occurs in GSS, vi. 229.
² Cf. GSS, vi. 218 ; MSi, xv. 45(ii)-46 ; L (ASS), p. 106, vv. 112-113
(i); GK, II, pp. 318-319, vs. 58.
³ The same example occurs in GSS, vi. 219 ; GK, II, p. 319, Ex. 22.

MIXTURES. SPECIAL PROBLEMS 59
Rule for writing down the combinations of 2 or more savours
in a serial order :
73. To get the 2-savour combinations add the pre-
ceding savour to all the succeeding one’s in order ; and to get
the combinations of 3 or more savours add the preceding
savour to the (succeeding) combinations of 2 or more savours
which do not contain the preceding savour.
Let the six savours be denoted by
a b c d e f .
Then the 2-savour combinations are :
ab, ac, ad, ae, af ; bc, bd, be, bf ;
cd, ce, cf ; de, df ; ef.
The 3-savour combinations are :
abc, abd, abe, abf, acd, ace, acf, ade, adf, aef ;
bcd, bce, bcf, bde, bdf, bef ; cde, cdf, cef ; def.
The 4-savour combinations are :
abcd, abce, abcf, abde, abdf, abef, acde, acdf ;
acef, adef ; bcde, bcdf, bcef, bdef, cdef.
The 5-savour combinations are :
abcde, abcdf, abcef, abdef, acdef, bcdef.
The 6-savour combination is :
abcdef.
(x) CERTAIN SPECIAL TYPES OF PROBLEMS
Rule for solving problems on pillars (stambhoddeśaka¹) or
problems involving remainders (śeṣoddeśaka) :
74(i). To solve problems on pillars or problems involv-
ing remainders, divide the visible quantity (dṛśya) by one minus
(the sum of) the fractional parts (of the whole).²
¹ Mahāvīra calls stambha-jāti by the name bhāga-jāti, and Śrīpati
by the name dṛśya-jāti. See GSS, iv. 4(i) and GT, p. 41, vs. 55.
² Cf. GSS, iv. 4(i); GT, p. 41, vs. 55. Also see GK,I, p. 17, lines 14-19.
Mahāvīra gives a separate rule for the śeṣa-jāti. See GSS, iv. 4(ii). The
same rule is given by Āryabhaṭa II and Śrīpati. See MSi, xv. 20; GT,
p. 44, line 11.

50 DETERMINATIONS
Ex. 96. One-fourth, one-third, and one-sixth of a pillar
are respectively buried under the water, mud, and sand of a
river, and three cubits (of the pillar) are visible. Give out
the measure (of the length) of that (pillar).¹
Here the given fractions (of the pillar) are ¼, ⅓, and ⅙ ; and the
visible quantity (i.e., the visible part of the pillar) is equal to ‘ 3 cubits.’
Hence the length of the pillar
3
= ————————————————— , i.e., 12 cubits.
1 - (¼ + ⅓ + ⅙)
Ex. 97. After giving away one-half of a quantity, then
⅔ of what remains, then ¾ of what remains thereafter, and
then ⅘ of what remains thereafter, the residue left is 3.
(What is that quantity ?)²
Here the fractions (of the whole quantity) given away are
½, ⅔ of (1 - ½), ¾ of (1 - ½) (1 - ⅔) , and
⅘ of (1 - ½) (1 - ⅔) (1 - ¾) ,
i.e., ½, ⅓, ⅛, and ¹/₃₀ ;
and the visible quantity (i.e., the residue left) is 3 .
Hence the required quantity
3
= ——————————————————————— , i.e., 360 .
1 - (½ + ⅓ + ⅛ + ¹/₃₀)
Rule for solving problems involving differences (viśeṣoddeśaka):
74(ii). On subtracting the smaller quantity from the
greater quantity what remains is called the difference
(viśeṣa).³ (This having been done) the procedure (to be adop-
ted is the same as the other one (i.e., the previous one).⁴
Ex. 98. Of a herd of cows, one-half went away towards
the east and one-fourth towards the west, the difference of the

¹ Cf. GSS, iv. 5; GT, p. 41, vs. 56. For other examples see GSS, iv.
6-22 ;GT, p. 42, vs. 57 ; L (ASS), pp 47-48, vs. 53; GK, I, p. 20, lines 4-7,
10-13, 16-19.
² For other examples, see GSS, iv. 29-30, 31, 32; GT, p. 44, lines
20-23 ; p. 45, lines 16-19; L (ASS), pp. 49-51, vs. 54.
³ Mahāvīra does not differentiate between bhāga-jāti and viśeṣa-jāti
(or viśleṣa-jāti)). So he sets problems on the latter under the bhāga-jāti.
See GSS, iv. Ex. 23-27.
⁴ Cf. GT, p. 46, lines 10-13.

MIXTURES. SPECIAL PROBLEMS 61
two as multiplied by 2 and divided by 5 went away
towards the north, and three (cows) are left. (What is the
numerical strength of the herd ?)¹
Here the fractions (of the herd of cows) that went away in the
various directions are
½, ¼, and ⅖(½ - ¼)
i.e., ½, ¼, and ⅒ ;
and the visible quantity (i e., the number of cows left) is 3. Hence the
numerical strength of the herd
= 3 / [1 - (½ + ¼ + ⅒)] , i.e., 20 cows.
Rule for solving problems involving remainders due to sub-
traction of square root, etc. (mūlādiśeṣoddeśaka) : ²
75. When the visible quantity stands near a square
root, multiply the visible quantity by 4, then increase that by
the square of the pada (i.e., the co-efficient of the square root
of the unknown), then extract the square-root of that and
increase that by the pada, and then take the square of half of
that ; (and when the visible quantity stands near a fraction)
divide the visible quantity by one minus the fraction.³
The first part of this rule relates to the solution of the quadratic
equation of the type
x - p√x = d,
where p is the pada, d the visible quantity (dṛśya), and √x the positive
square root of x. The solution is correctly stated as
x = [ {√(4d + p²) + p} / 2 ]²

¹ For similar examples see GSS, iv. 23-27; GT, p. 46, lines 20-25;
L (ASS), p. 53, vs. 55.
² Mahāvīra and Śrīpati have called mūlādiśeṣa-jāti by the name
śeṣamūla-jāti.
³ Cf. GSS, iv. 40; GT, p. 48, lines 24-25.