Question

There were $x$ pigeons and $y$ mynahs in a cage. One fine morning $p$ of them escaped to freedom. If the bird keeper, knowing only that $p = 7$, was able to figure out without looking into the cage that at least one pigeon had escaped, then which of the following does not represent a possible $(x, y)$ pair?

• A. (10, 8)
• B. (7, 2)
• C. (25, 6)
• D. (12, 4)

Hint:

For such certainty problems, compare the other group’s count to the escape count; if the other group’s count is less than the escape number, you guarantee some from the target group escaped.

Answer 1

The Correct Option is A

If the keeper can conclude that at least one pigeon escaped without seeing, then it must be impossible for all $p=7$ escaped birds to be mynahs. This means there are fewer than 7 mynahs in the cage initially.
Mathematically: if $y<p$, then at least $(p-y)$ pigeons must have escaped. If $y \ge p$, then it is possible that all escaped birds were mynahs, so the keeper couldn’t be sure any pigeon escaped.
Now check each option:
- (10, 8): $y=8 \ge 7$, so all 7 could have been mynahs. Keeper could not be sure — this violates the condition.
- (7, 2): $y=2<7$, so at least $7-2=5$ pigeons escaped — keeper can be sure. Valid.
- (25, 6): $y=6<7$, so at least $1$ pigeon escaped — valid.
- (12, 4): $y=4<7$, so at least $3$ pigeons escaped — valid.
Thus, the only impossible pair is (10, 8).