Three children P, Q, R and two grown-ups X, Y play a badminton doubles…
2026
Three children P, Q, R and two grown-ups X, Y play a badminton doubles tournament. X and Y are parents to two of the children playing. The child of X is not the same as the child of Y. Exactly one of the children does not have a parent playing in the tournament. The following rules are followed:
(i) A parent and his/her child cannot be on the same team.
(ii) A match can feature at most one parent and his/her child, that is, a maximum of one parent-child pair can play in a match.
The following matches were played:
TEAM 1 | TEAM 2 | |
|---|---|---|
MATCH 1 | P and X | Q and R |
MATCH 2 | P and R | X and Y |
MATCH 3 | R and X | Q and Y |
Which one of the following options is correct?
Answer: C. R does not have any parent playing — In Match 1, X and P are teammates, so X cannot be P's parent. In Match 3, X and R are teammates and Y and Q are teammates, so X cannot be R's parent and Y…
- A.
P does not have any parent playing
- B.
Q does not have any parent playing
- C.
R does not have any parent playing
- D.
X does not have a child playing
Attempted by 9 students.
Show answer & explanation
Correct answer: C
In Match 1, X and P are teammates, so X cannot be P's parent. In Match 3, X and R are teammates and Y and Q are teammates, so X cannot be R's parent and Y cannot be Q's parent. Hence X's child is Q. If Y's child were R, Match 3 would contain both the X-Q and Y-R parent-child pairs, exceeding the allowed maximum of one parent-child pair in a match. Therefore Y's child is P, leaving R as the child without a parent playing.