Question

Shared setup for this question and the next (chef and ingredients): A chef is preparing a recipe using four ingredients out of six liquids. Two of these ingredients must come from the taste group A, B and C, and the other two must come from the group W, X, Y and Z. Three combinations are never allowed: B with W, C with Y, and Y with Z.

If the chef rejected B because of its possible side effects but decided to use Z, which is a possible combination of the four ingredients in the recipe?

• A. A, C, W and Z
• B. A, X, Y and Z
• C. A, W, X and Z
• D. A, C, Y and Z

Hint:

Once B is rejected, A and C must both be used, then check which remaining pair avoids the C-Y clash.

Answer 1

The Correct Option is A

Step 1: Write the three ban rules where you can see them.
No B with W. No C with Y. No Y with Z. Also remember, two ingredients always come from A, B, C and two from W, X, Y, Z.

Step 2: Test each option against "C must be present".
Since B is rejected, A and C are the only ones left in the taste group, so any valid answer must contain both. Option B (A, X, Y, Z) fails right away because C is missing. Option C (A, W, X, Z) fails too, C is missing there as well.

Step 3: Test what remains against the ban rules.
Option A (A, C, W, Z) has A and C, plus W and Z. None of the three banned pairs (B-W, C-Y, Y-Z) shows up here, so it passes clean.

Step 4: Check option D the same way.
Option D (A, C, Y, Z) has C sitting right next to Y, a banned pair, so this option breaks a rule and gets rejected.

Final Answer:
Option A is the only combination that keeps every rule intact. \[ \boxed{\text{A, C, W and Z}} \]