To solve this, let's use the principle of inclusion-exclusion and denote the following variables:

• Let:
  • $S$: Students who passed Science.
  • $H$: Students who passed History.
  • $G$: Students who passed Geography.
• We know:
  • $|S| = 90$
  • $|H| = 120$
  • $|G| = 100$

Given that none of the students failed in all subjects, we have:

1. The total number of students, $|S \cup H \cup G|$, is 180.
2. The number of students who passed at least one subject is at most 50% more than those who passed in at least two subjects.

Let’s set up our problem with variables and calculate:

• Let $x$ be the number of students who passed all three subjects, $|S \cap H \cap G|$.
• Let $|S \cap H| = a$, $|S \cap G| = b$, and $|H \cap G| = c$.

Step 1: Inclusion-Exclusion Principle

Using the inclusion-exclusion principle:

$|S \cup H \cup G| = |S| + |H| + |G| - (|S \cap H| + |S \cap G| + |H \cap G|) + |S \cap H \cap G|$

Substitute known values:

$180 = 90 + 120 + 100 - (a + b + c) + x$

Simplifying:

$180 = 310 - (a + b + c) + x$

$a + b + c - x = 130$

Step 2: Condition on Students Passing at Least Two Subjects

The number of students who passed at least two subjects is given by $(a + b + c - 3x)$.

We are told that the number of students who passed at least one subject is at most 50% more than the number who passed at least two subjects. Therefore:

$|S \cup H \cup G| \leq 1.5 \times (a + b + c - 3x)$

Substituting $|S \cup H \cup G| = 180$:

$180 \leq 1.5 \times (a + b + c - 3x)$

Now, we will solve this system of equations to determine the maximum value of $x$.

Continuing from above, let's refine our approach for maximizing $x$, the number of students who passed all three subjects.

We have:

1. From the inclusion-exclusion principle: $a + b + c - x = 130$

2. Condition on students passing at least two subjects: $180 \leq 1.5 \times (a + b + c - 3x)$

Step 3: Substitute $a + b + c$

From the first equation, we can express $a + b + c$ in terms of $x$: $a + b + c = 130 + x$

Substitute this into the inequality from the condition on students passing at least two subjects: $180 \leq 1.5 \times (130 + x - 3x)$

Simplify the inequality: $180 \leq 1.5 \times (130 - 2x)$

$180 \leq 195 - 3x$

Rearrange to isolate $x$: $3x \leq 15$

$x \leq 5$

Conclusion

The maximum number of students who passed all three subjects is therefore $\boxed{5}$.

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Would you like more details or have any questions? Here are some related questions to consider:

1. How does the inclusion-exclusion principle work in set theory?
2. How can we apply this method to similar problems with different numbers of sets?
3. What if some students could have failed all three subjects? How would that change the approach?
4. How would changing the "50% more" condition impact the answer?
5. Can we derive a formula for problems involving more than three subjects?

Tip: For problems with conditions involving multiple overlapping groups, the inclusion-exclusion principle is a powerful tool for managing over-counting.