Ratio Problems: Practice

The parts total $2 + 3 + 5 = 10$, so $10k = 500$ and $k = 50$. The largest share is the $5$ term.

$10k = 500 \implies k = 50, \quad 5k = 250$

The largest share is $250$ dollars.

The difference is $9k - 5k = 4k$, so $4k = 28$ and $k = 7$. The smaller number is the $5$ term.

$4k = 28 \implies k = 7, \quad 5k = 35$

The shared term $B$ is $3$ in both ratios, so they already agree and chain directly.

$A : B : C = 2 : 3 : 4$

Soda is the $3$ term, so $3k = 12$ and $k = 4$. Juice is the $4$ term.

$3k = 12 \implies k = 4, \quad 4k = 16$

You need $16$ cups of juice.

The parts total $5 + 8 = 13$, so $13k = 91$ and $k = 7$. The larger number is the $8$ term.

$13k = 91 \implies k = 7, \quad 8k = 56$

Solve $3k + 10 = 4k$ for $k$, then green is the $2$ term.

$k = 10, \quad 2k = 20$

There are $20$ green marbles.

The parts total $7 + 3 = 10$, so $10k = 40$ and $k = 4$. Goats is the $3$ term.

$10k = 40 \implies k = 4, \quad 3k = 12$

There are $12$ goats.

The parts total $4 + 5 + 7 = 16$, so $16k = 1600$ and $k = 100$. The middle investor is the $5$ term.

$16k = 1600 \implies k = 100, \quad 5k = 500$

The middle investor puts in $500$ dollars.

The parts total $2 + 3 + 4 = 9$, so $9k = 180$ and $k = 20$. The largest angle is the $4$ term.

$9k = 180 \implies k = 20, \quad 4k = 80$

The largest angle is $80$ degrees.

Trucks is the $3$ term, so $3k = 33$ and $k = 11$. Cars is the $8$ term.

$3k = 33 \implies k = 11, \quad 8k = 88$

There are $88$ cars.

The shared term $B$ is $4$ in both, so $A : B : C = 3 : 4 : 9$. Then $B = 4k = 20$, so $k = 5$ and $C$ is the $9$ term.

$4k = 20 \implies k = 5, \quad 9k = 45$

Students is the $20$ term, so $20k = 560$ and $k = 28$. Teachers is the $1$ term.

$20k = 560 \implies k = 28, \quad 1k = 28$

There are $28$ teachers.