Mahaviracharya/Ganita Sara Samgraha/Chapter VI/95-96
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Mahaviracharya: Ganita Sara Samgraha Chapter $\text {VI}$: Mixed Problems: Problem $95 \frac 1 2$ and $96 \frac 1 2$
Three puranas formed the pay of one man who is a mounted soldier.
At that rate there were $65$ soldiers in all.
Some among them broke down, and the amount of their pay was given to those who remained in the field.
Of this, each man obtained $10$ puranas.
You tell me, after thinking well, how many remained in the field and how many broke down.
Solution
- $50$ soldiers broke down, while $15$ remained in the field.
Proof
Let $x$ denote the number of soldiers who remained in the field.
Let $y$ denote the number of soldiers who broke down.
We have:
\(\ds x + y\) | \(=\) | \(\ds 65\) | that is, the total number of soldiers | |||||||||||
\(\text {(1)}: \quad\) | \(\ds 3 \paren {x + y}\) | \(=\) | \(\ds 195\) | that is, the total amount of pay to be distributed | ||||||||||
\(\text {(2)}: \quad\) | \(\ds 3 \paren {x + y}\) | \(=\) | \(\ds \paren {3 + 10} x\) | that is, the total pay was divided between the soldiers who remained | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 13 x\) | \(=\) | \(\ds 195\) | substituting for $3 \paren {x + y}$ in $(2)$ from $(1)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 15\) |
which leaves $65 - 15 = 50$ soldiers who broke down.
$\blacksquare$
Sources
- c. 850: Mahaviracharya: Ganita Sara Samgraha: Chapter $\text {VI}$: Mixed Problems: $95 \frac 1 2$ and $96 \frac 1 2$
- 1912: Rao Bahadur M. Rangacharya: The Ganita-Sara-Sangraha of Mahaviracharya: Chapter $\text {VI}$: Mixed Problems: $95 \frac 1 2$ and $96 \frac 1 2$
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Indian Puzzles: $48$