The idea is to construct a probability distribution on the 10 K-Prim type questions. The probability distribution for a single question is $$\begin{align*} p_0 = \Pr[S = 0] &= \sum_{k=0}^2 \binom{4}{k} (0.5)^k (0.5)^{4-k} = \frac{11}{16}, \\ p_1 = \Pr[S = 0.5] &= \binom{4}{3} (0.5)^3 (0.5)^1 = \frac{1}{4}, \\ p_2 = \Pr[S = 1] &= \binom{4}{4} (0.5)^4 (0.5)^0 = \frac{1}{16}. \end{align*} $$ This assumes that for each such question, the choice of True/False is equally likely for each of the four answers, and the each answer is independent, thus the number of correct answers follows a ${\rm Binomial}(4,0.5)$ distribution.
Next, the distribution of the sum of the scores of 10 K-Prim questions can be derived from the multinomial distribution, though it is somewhat tedious to compute: let $X_0$, $X_1$, $X_2$ be random variables that count the number of $0$-point, $0.5$-point, and $1$-point scores out of the 10 questions. Then $$\Pr[(X_0, X_1, X_2) = (a,b,c)] = \frac{10!}{a! b! c!} p_0^{a} p_1^b p_2^c.$$ Then we can tabulate the sum; we do this in Mathematica:
Flatten[Table[{b/2 + c, PDF[MultinomialDistribution[10, {11/16, 1/4, 1/16}],
{10 - b - c, b, c}]}, {b, 0, 10}, {c, 0, 10 - b}], 1]
Table[{k, Total[Select[%, #[[1]] == k &]][[2]]}, {k, 0, 10, 1/2}]
which gives us the desired probability distribution for these 10 questions. Call this random variable $K$. Now, for the remaining 20 questions, the total point count is simple; it is simply $A \sim {\rm Binomial}(20, 0.2)$. So the probability that the total score is at least $18$ out of $30$ is $$\sum_{k=0}^{20} \Pr[K = k/2]\Pr[A \ge 18 - k/2].$$ Again, we use Mathematica:
Sum[%[[k, 2]] (1 - CDF[BinomialDistribution[20, 1/5], 18 - k/2]),
{k, 1, Length[%]}]
This gives us $$\frac{8327843221553613}{2^9 \cdot 10^{20}} \approx 1.62653 \times 10^{-7}.$$ This is so small that it is unlikely that a naive simulation approach will be able to approximate it.
Best Answer
Consider flipping a fair coin 10 times vs 1000 times. Is it more likely to get 7 or more heads in the first case, or 700 or more heads in the second case? I think you can intuit that the former is more likely than the latter, and you'd be exactly right. This is the Law of Big Numbers: the bigger your sample size, the closer you can expect the observed frequency to get to the probability. As you yourself started to suspect, this is indeed all about the distribution: the distribution curve gets more narrow the greater the sample size is. So, with a probability of 0.5, you can expect to get the number of heads with 1000 flips to be closer to 0.5 than if you just flip 10 times.
So: if the probability of you guessing the answer an individual question correctly is greater than the percentage of question you need to get correctly, then you should go for the test with more questions. If the probability of guessing correctly is smaller than the needed percentage, then you should go for the smaller test.