# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/२६४

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68
GAŅIITASĀRASAŃGRAHA.

${\displaystyle {\tfrac {1}{4}}}$ and ${\displaystyle {\tfrac {1}{6}}}$ of itself. The sum of these (quantities so diminished) is ${\displaystyle {\tfrac {1}{2}}}$. What are the unknown fractions here?

134. A certain fraction, diminished (in consecution) by ${\displaystyle {\tfrac {1}{2}},{\tfrac {1}{6}},{\tfrac {1}{5}},{\tfrac {1}{8}}}$ and ${\displaystyle {\tfrac {1}{7}}}$ of itself, become ${\displaystyle {\tfrac {1}{6}}}$. O you, who know the principles of arithmetic, what is that (unknown) fraction?

The rule for finding out any unknown fraction in other required places (than the beginning) :--

135. The optionally split up parts derived from the (given) sum, when divided in order by the simplified known quantities (in the intended Bhāgāpavādha fractions), and (then) subtracted from one (severally), become the unknown (fractional quantities) in the (required) places of our choice.

Thus ends the Bhāgāpavāha variety of fractions.

The rule for finding out the unknown fractions in all the places in relation to a Bhāgānubandha or Bhāgāpavāha variety of fractions (when their ultimate value is known):--

136. Optionally choose your own desired fractions in relation to all unknown places, excepting (any) one. Then by means of the rules mentioned before, arrive at that (one unknown) fraction with the help of these (optionally chosen fractional quantities).

Examples in illustration thereof.

187. A certain fraction combined with five other fractions of itself (in additive consecution) becomes ${\displaystyle {\tfrac {1}{2}}}$; and a certain (other) fraction diminished (by five other fractions of itself in consecution) becomes ${\displaystyle {\tfrac {1}{4}}}$, O friend, give out (all) those fractions.

135. This rule is similar to the rule already given in stanza No. 125.

136. The previous rules here intended are those given in stanzas 122, 125, 132 and 135.

137. In working out the first case in this example, choose the fractions ${\displaystyle {\tfrac {1}{5}},{\tfrac {1}{6}},{\tfrac {1}{7}},{\tfrac {1}{8}}}$ and ${\displaystyle {\tfrac {1}{9}}}$ in places other than the beginning; and then find out, by the rule given in stanza 122, the first fraction which comes to be ${\displaystyle {\tfrac {1}{4}}}$. Or choosing ${\displaystyle {\tfrac {1}{4}},{\tfrac {1}{5}},{\tfrac {1}{6}},{\tfrac {1}{8}}}$ and ${\displaystyle {\tfrac {1}{9}}}$, find out the fraction left out in a place other than the beginning in accordance with the rule given in stanza 125; the result arrived at is ${\displaystyle {\tfrac {1}{7}}}$. Similarly, the second case which involves fractions in dissociation can be worked out with the help of the rules given in stanzas 132 and 135.