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

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.