7 of 30 = 1/ [ ] of 20

Here is the solution for :

: $\frac{\square}{\square} \text{ of } 30 = \frac{1}{\square} \text{ of } 20$

Step 1: Represent the missing digits with variables. Let the first fraction be $\frac{a}{b}$ and the missing denominator on the right side be $c$. The equation becomes: $\frac{a}{b} \times 30 = \frac{1}{c} \times 20$

Step 2: Simplify the equation. $\frac{30a}{b} = \frac{20}{c}$ Divide both sides by 10: $\frac{3a}{b} = \frac{2}{c}$ Cross-multiply to eliminate denominators: $3ac = 2b$

Step 3: Find single-digit integer values for $a, b, c$ that satisfy $3ac = 2b$. Since $2b$ is an even number, $3ac$ must also be even. This implies that either $a$ or $c$ (or both) must be even. Let's try $a=1$. Then $3c = 2b$. For $b$ and $c$ to be single digits, we can test values for $c$: If $c=1$, $3(1) = 2b \Rightarrow 3 = 2b \Rightarrow b = 1.5$ (not an integer). If $c=2$, $3(2) = 2b \Rightarrow 6 = 2b \Rightarrow b = 3$. This gives us $a=1, b=3, c=2$. These are all single digits.

Step 4: Substitute these values back into the original problem to verify. $\frac{1}{3} \text{ of } 30 = \frac{1}{2} \text{ of } 20$ Calculate both sides: Left side: $\frac{1}{3} \times 30 = 10$ Right side: $\frac{1}{2} \times 20 = 10$ Both sides are equal, so the values are correct.

The missing digits are $\boxed{\frac{1}{3}}$ and $\boxed{2}$.