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?

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Once B is rejected, A and C must both be used, then check which remaining pair avoids the C-Y clash.
Updated On: Jul 16, 2026
  • A, C, W and Z
  • A, X, Y and Z
  • A, W, X and Z
  • A, C, Y and Z
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The Correct Option is A

Solution and Explanation

Step 1: Recall the two group sizes.
The chef always picks two ingredients from A, B, C and two from W, X, Y, Z, four in total. So whichever two get dropped from the six, exactly two from each group survive.

Step 2: Use the rejection of B.
Once B is rejected, the only two ingredients left in the taste group are A and C, so both A and C must be in the final mix.

Step 3: Remove options that leave out C.
Option B (A, X, Y and Z) and option C (A, W, X and Z) do not contain C at all, so neither can be right once B is dropped and C is forced in.

Step 4: Apply the C and Y rule to what is left.
That leaves option A (A, C, W and Z) and option D (A, C, Y and Z). The rule says C and Y can never be used together, so option D is out because it pairs C with Y.

Final Answer:
Only A, C, W and Z clears every rule. \[ \boxed{\text{A, C, W and Z}} \]
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