Let the number of ivy, nightshade and triffid plants be $i$, $n$ and $t$ respectively.

Then:

$2i + 9n + 12t = 120$ and $i + n+ t = 20$, where $i> 0$; $n> 0$; $t> 0$.

Multiplying the second equation by $2$ and subtracting the new equation from the first:

$$7n + 10t=80$$

Thus $$7n = 10(8-t)$$

Therefore $n$ is a multiple of $10$ and since $1 \le n < 20$, $n=10$

Hence $8-t=7$ and therefore $t=1$.

Then:

$2i + 9n + 12t = 120$ and $i + n+ t = 20$, where $i> 0$; $n> 0$; $t> 0$.

Multiplying the second equation by $2$ and subtracting the new equation from the first:

$$7n + 10t=80$$

Thus $$7n = 10(8-t)$$

Therefore $n$ is a multiple of $10$ and since $1 \le n < 20$, $n=10$

Hence $8-t=7$ and therefore $t=1$.

*This problem is taken from the UKMT Mathematical Challenges.**View the archive of all weekly problems grouped by curriculum topic*