The Admiral and Lady Hyena scored 9 points between them, as did Lord Giraffe and Mr. Gnu.

Since Lady Hyena scored the least, this means that she can have scored
no more than three points; she therefore shot *no more than*
three beasts.

She must have shot at least two beasts, since she shot more than everyone else. But if she only shot two beasts, that would mean that everyone else shot no more than one animal. Mr. Gnu shot a lion, so the three gnu in this case must have been taken one apiece by each of the others. This would mean that Lady Hyena shot a gnu and some other beast; her score would then have been greater than that of the Admiral and Lord Giraffe.

Lady Hyena, therefore, shot three beasts and scored three points. She must have shot three hyenas, which were worth a single point each.

The Admiral scored six points from no more than two beasts.

Could he have gained this score from a single beast? No, for that would mean that a giraffe or gnu (he didn't shoot a lion) was worth six points, pushing the score for a lion to above six. The only person to shoot a lion was Mr. Gnu, who scored no more than 5 points.

The admiral must therefore have shot two beasts. Possibilities: two giraffe (at three points each), two gnu (at three points each), a giraffe and a gnu (at four and two points, respectively). His score could not have come from anything plus a hyena, since again that would push the value for a lion too high.

Mr. Gnu shot a lion, for at least four points. Since he scored at most five points, this means that he could not have also bagged a gnu, which must be worth at least two points.

The Admiral and Lord Giraffe must therefore have taken the three gnu between them.

If Lord Giraffe took a single gnu then the Admiral must have bagged two of them for six points total. Lord Giraffe must have shot a gnu and a hyena for four points, and Mr. Gnu's lion would then score him five points. This, however, cannot be a valid solution since the bag would not have included a giraffe. This is the condition which Simon overlooked in his

If, on the other hand, Lord Giraffe took two gnu then they must have been worth two points each. In this case the Admiral must have shot a giraffe (for four points) and a gnu (for a total of six) and Mr. Gnu must have only bagged the lion, for five points.

Lady Hyena shot three hyenas at one point each.

Mr. Gnu Hyena shot a lion for five points.

Lord Giraffe shot two gnu (two points each).

Admiral Lion shot a gnu (two points) and a giraffe (four points).

Three beasts were shot by their namesakes; Lady Hyena bagged them all.

Sean, for the first correct answer. James, for the correct answer (with an accurate analysis), and for pointing out that there are lots of solutions if you permit fractional scores for the animals (such a possibility can be ruled out by inference from the question; the people on the shoot are all upper-class twits and probably don't even know about the number zero). Simon, for getting the wrong answer but for trying to get Excel to produce it for him; finally, a big well-done to Spike who produced the most comprehensive analysis.