A 5-item menu for lunch is to be decided from amongst five dishes P, Q, R, S…
2023
A 5-item menu for lunch is to be decided from amongst five dishes P, Q, R, S and T and four beverages W, X, Y and Z on the following conditions: (i) P and Z have to be together. (ii) W cannot be put with Y. (iii) S and X cannot be selected together. (iv) R and T have to be together. (v) Y cannot be selected with Q. If two of the items have to be dishes (and hence three beverages), what will be the lunch menu?
- A.
P Q W X Z
- B.
P S X Y Z
- C.
Q S X Z Y
- D.
R T W X Z
Attempted by 40 students.
Show answer & explanation
Correct answer: A
Answer: P Q W X Z
Reasoning:
Two dishes must be chosen. If R or T were chosen, the rule that R and T have to be together would force both R and T to occupy the two dish slots.
If R and T are the two dishes, P is not selected, so Z cannot be selected (P and Z must be together). That would leave only W, X and Y as possible beverages, but W and Y cannot be together, so no valid 3-beverage set exists with R and T. Therefore R and T cannot be the chosen dishes.
Thus the two dishes must be selected from P, Q and S. Because P and Z must be together, any selection that includes P forces Z to be a beverage.
Consider P and Q as the two dishes: Q forbids Y, so Y cannot be a beverage. With P requiring Z, the three beverages must be Z plus W and X. W and X are compatible, so the set P, Q, W, X, Z satisfies all constraints.
Consider P and S: S forbids X, so the beverages would need to be Z, W and Y; but W and Y cannot be together, so this is impossible.
Consider Q and S: without P, Z cannot be chosen; Q forbids Y and S forbids X, leaving only W available, so it is impossible to get three beverages.
Since all other dish-pair choices fail, the unique valid menu is P, Q as the two dishes and W, X, Z as the three beverages.