Question:medium
Shared setup (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.
Which of the following must always be true?
I. If C is used, W must be added
II. If Y is used, B must be added
III. If C is not used, W cannot be added
Shared setup (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.
Which of the following must always be true?
I. If C is used, W must be added
II. If Y is used, B must be added
III. If C is not used, W cannot be added
Show Hint
Work out which taste pair each condition forces, then see what that pair locks out.
Updated On: Aug 18, 2026
- I and II only
- II and III only
- I, II and III only
- II only
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Build the forced chains for each taste letter.
There are only three taste pairs possible: A-B, A-C, B-C. Look at what each one locks out from W, X, Y, Z. A-B locks out W. A-C locks out Y. B-C locks out both W and Y.
Step 2: Read off what happens whenever Y appears.
Y can only survive under the A-B pair, since A-C and B-C both remove Y. So using Y forces the taste pair to be A-B, meaning B always comes along. This confirms statement II is always true.
Step 3: Read off what happens whenever C is dropped.
If C is out, the only taste pair left from A, B, C is A-B, which locks out W. So dropping C always locks out W too, confirming statement III is always true.
Step 4: Check statement I with the same chain.
C appears in both A-C, which allows W, and B-C, which locks out W. Since one case allows W and the other blocks it, using C does not always force W in, so statement I fails.
Final Answer:
Only statements II and III are guaranteed every time. \[ \boxed{\text{II and III only}} \]
There are only three taste pairs possible: A-B, A-C, B-C. Look at what each one locks out from W, X, Y, Z. A-B locks out W. A-C locks out Y. B-C locks out both W and Y.
Step 2: Read off what happens whenever Y appears.
Y can only survive under the A-B pair, since A-C and B-C both remove Y. So using Y forces the taste pair to be A-B, meaning B always comes along. This confirms statement II is always true.
Step 3: Read off what happens whenever C is dropped.
If C is out, the only taste pair left from A, B, C is A-B, which locks out W. So dropping C always locks out W too, confirming statement III is always true.
Step 4: Check statement I with the same chain.
C appears in both A-C, which allows W, and B-C, which locks out W. Since one case allows W and the other blocks it, using C does not always force W in, so statement I fails.
Final Answer:
Only statements II and III are guaranteed every time. \[ \boxed{\text{II and III only}} \]
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