NJSLA Algebra I Quantities. Practice it free.

The number of cast members must divide both totals. The common factors of $54$ and $81$ greater than $1$ are $3$, $9$, and $27$. Only $9$ appears among the choices: $54 \div 9 = 6$ scripts and $81 \div 9 = 9$ kits per member.

Maximize $\dfrac{m}{n}$ with $m = 5$ and $n = 3$, and maximize $\dfrac{1}{p}$ with $p = 8$: $\dfrac{5}{3} \cdot \dfrac{1}{8} = \dfrac{5}{24}$.

Both events coincide every least common multiple of $9$ and $15$ minutes. Since the GCF of $9$ and $15$ is $3\textsf{,}$ divide their product by the GCF to get that interval: $\dfrac{9 \times 15}{3}=45$ minutes. In $90$ minutes after noon, the shared times are at $45$ and $90$ minutes, so there are $2$ additional times.

Multiply by a conversion factor so the old unit cancels: $4\,\text{lb} \times \dfrac{16\,\text{oz}}{1\,\text{lb}} = 64\,\text{oz}$.

The number of team members must divide both totals evenly. The common factors of $50$ and $75$ greater than $1$ are $5$ and $25$. Only $25$ is listed: $50 \div 25 = 2$ packs and $75 \div 25 = 3$ cables per member.

Minimize $\dfrac{m}{n}$ with $m = 1$ and $n = 7$, and minimize $\dfrac{1}{p}$ with $p = 12$: $\dfrac{1}{7} \cdot \dfrac{1}{12} = \dfrac{1}{84}$.

The next shared meeting day is the least common multiple of $6$ and $9\textsf{.}$ Since the GCF of $6$ and $9$ is $3\textsf{,}$ divide their product by the GCF: $\dfrac{6 \times 9}{3}=18\textsf{.}$ So they meet together again in $18$ school days.

Multiply by a conversion factor so the old unit cancels: $2\,\text{km} \times \dfrac{1{,}000\,\text{m}}{1\,\text{km}} = 2{,}000\,\text{m}$.