2019 AMC 10B Problem 11

The two jars are linked by having the same size, so start by finding a jar size that works for both ratios.

1. In Jar $1$, blue to green is $9:1$, so out of every $10$ marbles, $1$ is green. In Jar $2$, out of every $9$ marbles, $1$ is green.
2. The jar size must divide evenly by $10$ and by $9$. Try the smallest such size, $90$ marbles per jar:
- Jar $1$: $9$ green, $81$ blue.
- Jar $2$: $10$ green, $80$ blue.
- Green in total: $9 + 10 = 19$.
3. We need $95$ green marbles in total, and $95 = 5 \cdot 19$. So each jar is really $5$ times as big: $450$ marbles.
4. Scale up:
- Jar $1$: $45$ green, $405$ blue.
- Jar $2$: $50$ green, $400$ blue.
5. Jar $1$ has $405 - 400 = 5$ more blue marbles.

The answer is $\boxed{\textbf{(A)}\ 5}$.

A ratio like $9:1$ tells you what fraction of the whole each color is. When two groups share the same size, choose a trial size that fits both ratios, see what totals it gives, and then scale everything by the same factor to match the real data.

Let each jar hold $N$ marbles. The greens are $\frac{N}{10} + \frac{N}{9} = \frac{19N}{90}$. Setting this equal to $95$ gives $N = 450$. The blues are $\frac{9}{10} \cdot 450 = 405$ and $\frac89 \cdot 450 = 400$, a difference of $5$.

Let each jar hold $N$ marbles. In Jar $1$ the greens are $\frac{1}{10}$ of the marbles, and in Jar $2$ they are $\frac{1}{9}$. Both fractions must be whole numbers, so $N$ is a multiple of $90$; write $N = 90k$.

• Jar $1$: $9k$ green, $81k$ blue.
• Jar $2$: $10k$ green, $80k$ blue.

Total green: $9k + 10k = 19k = 95$, so $k = 5$ (each jar holds $450$ marbles).

Blue difference: $81k - 80k = k = 5$.

The answer is $\boxed{\textbf{(A)}\ 5}$.