Find all positive integers $a$ and $b$ such that $a / 5$ and $b / 7$ are less than 1 and satisfy the condition

$$
1<\frac{a}{5}+\frac{b}{7}<2
$$
If a fraction is positive and less than 1, its numerator must be less than its denominator. Thus, we must have

$$
0<a<5 \quad \text { and } \quad 0<b<7
$$

Since $\frac{1}{5} a+\frac{1}{7} b=\frac{1}{35}(7 a+5 b)$, the given condition is equivalent to

$$
1<\frac{7 a+5 b}{35}<2, \quad \text { or } \quad 35<7 a+5 b<70
$$

Thus, we see that the problem consists of finding all integers $a$ and $b$ such that

$$
0<a<5, \quad 0<b<7 \quad \text { and } \quad 35<7 a+5 b<70
$$

Let's examine each of the four options for $a$, from 1 to 4, with corresponding possibilities for $b$, among the integers $1,2,3,4,5$ and 6.

- $a=1$. Then $7 a+5 b=7+5 b$ and, from $35<7+5 b<70$, it follows that $28<5 b<63$, or $6 \leq b \leq 12$. Since $b \in\{1,2,3,4,5,6\}$, we conclude that the only option is $b=6$.
- $a=2$. Then $7 a+5 b=14+5 b$ and, from $35<14+5 b<70$, it follows that $21<5 b<56$, or $5 \leq b \leq 11$. Since $b \in\{1,2,3,4,5,6\}$, we conclude that the only options are $b=5$ and $b=6$.
- $a=3$. Then $7 a+5 b=21+5 b$ and, from $35<21+5 b<70$, it follows that $14<5 b<49$, or $3 \leq b \leq 9$. Since $b \in\{1,2,3,4,5,6\}$, we conclude that the only options are $b=3,4,5$ or 6.
- $a=4$. Then $7 a+5 b=28+5 b$ and, from $35<28+5 b<70$, it follows that $7<5 b<42$, or $2 \leq b \leq 8$. Since $b \in\{1,2,3,4,5,6\}$, we conclude that the only options are $b=2,3,4,5$ or 6.

We conclude by displaying the twelve solutions $a, b$ in the following table.

| $a$ | $b$ | $1<\frac{a}{5}+\frac{b}{7}<2$ |  | $b$ | $1<\frac{a}{5}+\frac{b}{7}<2$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 6 | $\frac{1}{5}+\frac{6}{7}=\frac{37}{35}$ | 2 | 5 | $\frac{2}{5}+\frac{5}{7}=\frac{39}{35}$ |
|  | 2 | $\frac{4}{5}+\frac{2}{7}=\frac{38}{35}$ |  | 6 | $\frac{2}{5}+\frac{6}{7}=\frac{44}{35}$ |
|  | 3 | $\frac{4}{5}+\frac{3}{7}=\frac{43}{35}$ | 3 | 3 | $\frac{3}{5}+\frac{3}{7}=\frac{36}{35}$ |
|  | 4 | $\frac{4}{5}+\frac{4}{7}=\frac{48}{35}$ |  | 4 | $\frac{3}{5}+\frac{4}{7}=\frac{41}{35}$ |
|  | 5 | $\frac{4}{5}+\frac{5}{7}=\frac{53}{35}$ |  | 5 | $\frac{3}{5}+\frac{5}{7}=\frac{46}{35}$ |
|  | 6 | $\frac{4}{5}+\frac{6}{7}=\frac{58}{35}$ |  | 6 | $\frac{3}{5}+\frac{6}{7}=\frac{51}{35}$ |