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

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