Calculate the percentage abundance of Lithium-6 and Lithium-7 given their relative atomic mass is 6.94.

Let $x$ be the percentage abundance of Lithium-6.
Then the percentage abundance of Lithium-7 will be $(100 - x)$.
The formula for relative atomic mass (Ar) is:

$$Ar = \frac{(\text{mass of isotope 1} \times \text{abundance 1}) + (\text{mass of isotope 2} \times \text{abundance 2})}{100}$$

Given:
Ar = 6.94
Mass of Lithium-6 = 6
Mass of Lithium-7 = 7

Multiply both sides by 100:

$$6.94 \times 100 = 6x + 7(100 - x)$$ $$694 = 6x + 700 - 7x$$ $$694 = 700 - x$$

Subtract 700 from both sides:

$$694 - 700 = -x$$ $$-6 = -x$$ $$x = 6$$

Percentage abundance of Lithium-6 = $x = 6\%$
Percentage abundance of Lithium-7 = $100 - x = 100 - 6 = 94\%$

The percentage abundance of Lithium-6 is $\boxed{\text{}6\%\text{}}$ and Lithium-7 is $\boxed{\text{}94\%\text{}}$.

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The formula for relative atomic mass (Ar) is:

$$Ar = \frac{(\text{mass of isotope 1} \times \text{abundance 1}) + (\text{mass of isotope 2} \times \text{abundance 2})}{100}$$

Given:
Ar = 30
Abundance of isotope 1 = 65%
Mass number of isotope 1 = $x$
Abundance of isotope 2 = 35%
Mass number of isotope 2 = 31

Multiply both sides by 100:

$$30 \times 100 = 65x + (31 \times 35)$$ $$3000 = 65x + 1085$$

Subtract 1085 from both sides:

$$3000 - 1085 = 65x$$ $$1915 = 65x$$

Divide by 65:

$$x = \frac{1915}{65}$$ $$x = 29.46$$

Since mass numbers are typically whole numbers, we can round this to the nearest whole number.

$$x \approx 29$$

The mass number represented by $x$ is approximately $\boxed{\text{}29\text{}}$ (or 29.46 if not rounded).