Let's synthesize the given ratios to find a single, unified ratio for all four roles: Trainees (\(T\)), Assistants (\(A\)), Nurses (\(N\)), and Coordinators (\(C\)).
We are given the base ratio:
\(\frac{T}{A} = \frac{3}{4}\)From this, we can derive the other two relationships:
(1) \(\frac{T}{N} = \frac{3}{4} \times (\frac{3}{4}) = \frac{9}{16}\)
(2) \(\frac{A}{C} = 2 \times (\frac{3}{4}) = \frac{3}{2}\)
To merge these into one continuous ratio (\(T : A : N : C\)), we need common terms. Let's scale \(T : A\) from \(3 : 4\) to
\(9 : 12\) so it perfectly matches the \(9\) in the \(T : N\) ratio.
Scaling \(A : C\) by 4 gives us
\(12 : 8\), neatly matching our newly scaled \(A\) value.
This reveals our master unified ratio:
\(T : A : N : C = 9 : 12 : 16 : 8\)This means the actual number of people in each role can be represented as \(9x, 12x, 16x,\) and \(8x\) (where \(x\) is a positive integer).
Now for the constraint: The combined number of trainees and coordinators (\(9x + 8x = 17x\)) must be strictly less than the smallest three-digit positive integer with distinct digits.
The smallest 3-digit number is 100 (digits are not distinct), followed by 101 (digits are not distinct). The smallest one with fully distinct digits is
102 (using 1, 0, and 2).
Setting up the inequality:
\(17x < 102\)
\(x < 6\)
Since \(x\) must be a positive integer, the maximum possible value for \(x\) is
5. *(Note: If you fall for the trap of treating the inequality as "less than or equal to", you will incorrectly get x = 6 and pick option D!)*
Finally, we maximize the combined number of assistants and nurses (\(A + N\)):
\(12x + 16x = 28x\)
Maximum value = \(28 \times 5 =\)
140.
Correct Answer: C