Mahaviracharya/Ganita Sara Samgraha/Chapter VI/236-237

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Mahaviracharya: Ganita Sara Samgraha Chapter $\text {VI}$: Mixed Problems: Problem $236 \text - 237$

$3$ merchants saw in the road a purse containing money.

One said:

If I secure this purse, I shall become twice as rich as both of you together.

Then the second said:

I shall become $3$ times as rich.

Then the third said:

I shall become $5$ times as rich.

What is the value of the money in the purse, as also the money on hand with each of the $3$ merchants?


Solution

The solution is not unique.

The second has $3$ times as much as the first.
The third has $5$ times as much as the first.
The purse holds $15$ times as much as the first owns.


Proof

Let $x$, $y$ and $z$ denote the amount of money owned by the $1$st, $2$nd and $3$rd merchant respectively.

Let $u$ denote the amount of money in the purse.

Let $a$, $b$ and $c$ denote the factor by which $x$, $y$ and $z$ respectively will be richer, were they to get the purse, than the other two combined.


We have:

\(\text {(1)}: \quad\) \(\ds u + x\) \(=\) \(\ds a \paren {y + z}\)
\(\text {(2)}: \quad\) \(\ds u + y\) \(=\) \(\ds b \paren {z + x}\)
\(\text {(3)}: \quad\) \(\ds u + z\) \(=\) \(\ds c \paren {x + y}\)
\(\ds \leadsto \ \ \) \(\ds u + x + y + z\) \(=\) \(\ds \paren {a + 1} \paren {y + z}\) adding $y + z$ to $(1)$
\(\ds \) \(=\) \(\ds \paren {b + 1} \paren {z + x}\) adding $z + x$ to $(2)$
\(\ds \) \(=\) \(\ds \paren {c + 1} \paren {x + y}\) adding $x + y$ to $(3)$


Let $T = u + x + y + z$.

Then:

\(\text {(4)}: \quad\) \(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \paren {y + z}\) \(=\) \(\ds \paren {b + 1} \paren {c + 1}\)
\(\text {(5)}: \quad\) \(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \paren {z + x}\) \(=\) \(\ds \paren {c + 1} \paren {a + 1}\)
\(\text {(6)}: \quad\) \(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \paren {x + y}\) \(=\) \(\ds \paren {a + 1} \paren {b + 1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \times 2 \paren {x + y + z}\) \(=\) \(\ds \paren {b + 1} \paren {c + 1} + \paren {c + 1} \paren {a + 1} + \paren {a + 1} \paren {b + 1}\) adding $(4)$, $(5)$ and $(6)$
\(\text {(7)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \times 2 \paren {x + y + z}\) \(=\) \(\ds S\) where $S$ is defined as it is
\(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \times 2 x\) \(=\) \(\ds S - 2 \paren {b + 1} \paren {c + 1}\) $(7) - 2 \times (4)$
\(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \times 2 y\) \(=\) \(\ds S - 2 \paren {c + 1} \paren {a + 1}\) $(7) - 2 \times (5)$
\(\ds \dfrac {\paren {a + 1} \paren {b + 1} \paren {c + 1} } T \times 2 z\) \(=\) \(\ds S - 2 \paren {a + 1} \paren {b + 1}\) $(7) - 2 \times (6)$
\(\ds \leadsto \ \ \) \(\ds \dfrac x y\) \(=\) \(\ds \dfrac {S - 2 \paren {b + 1} \paren {c + 1} } {S - 2 \paren {c + 1} \paren {a + 1} }\)
\(\ds \dfrac y z\) \(=\) \(\ds \dfrac {S - 2 \paren {c + 1} \paren {a + 1} } {S - 2 \paren {a + 1} \paren {b + 1} }\)

Setting $a = 2$, $b = 3$, $c = 5$ we have:

\(\ds S\) \(=\) \(\ds \paren {3 + 1} \paren {5 + 1} + \paren {5 + 1} \paren {2 + 1} + \paren {2 + 1} \paren {3 + 1}\)
\(\ds \) \(=\) \(\ds 4 \times 6 + 6 \times 3 + 3 \times 4\)
\(\ds \) \(=\) \(\ds 24 + 18 + 12\)
\(\ds \) \(=\) \(\ds 54\)

Hence:

\(\ds \dfrac x y\) \(=\) \(\ds \dfrac {S - 2 \paren {b + 1} \paren {c + 1} } {S - 2 \paren {c + 1} \paren {a + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {54 - 2 \paren {3 + 1} \paren {5 + 1} } {54 - 2 \paren {5 + 1} \paren {2 + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {54 - 48} {54 - 36}\)
\(\ds \) \(=\) \(\ds \dfrac 6 {18}\)
\(\ds \) \(=\) \(\ds \dfrac 1 3\)

and:

\(\ds \dfrac y z\) \(=\) \(\ds \dfrac {S - 2 \paren {c + 1} \paren {a + 1} } {S - 2 \paren {a + 1} \paren {b + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {54 - 2 \paren {5 + 1} \paren {2 + 1} } {54 - 2 \paren {2 + 1} \paren {3 + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {54 - 36} {54 - 24}\)
\(\ds \) \(=\) \(\ds \dfrac {18} {30}\)
\(\ds \) \(=\) \(\ds \dfrac 3 5\)

from which it is seen that:

$x : y : z = 1 : 3 : 5$

and the solution in smallest integers is:

\(\ds a\) \(=\) \(\ds 1\)
\(\ds b\) \(=\) \(\ds 3\)
\(\ds c\) \(=\) \(\ds 5\)
\(\ds p\) \(=\) \(\ds 15\)

$\blacksquare$


Sources

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