The process merely consists in giving 100a + 10b + c + 250, where a, b, and c represent the three throws.  The result is obvious.

387.—­THE VILLAGE CRICKET MATCH.

[Illustration: 

| Mr. Dumkins >>-->
|------------------------>    |
|    <-------------------     |
|     ------------------->    |
1 |<-----------------------     |
|                             |
|    <------------------------|
|     ------------------->    |
|    <-------------------     |
|     ----------------------->|
|            <--<< Mr. Podder |

| Mr. Luffey >>-->
|------------------------>    |
|    <-------------------     |
|     ----------------------->|
2 |                             |
|<-----------------------     |
|     ------------------->    |
|    <------------------------|
<--<< Mr. Struggles |

]

The diagram No. 1 will show that as neither Mr. Podder nor Mr. Dumkins can ever have been within the crease opposite to that from which he started, Mr. Dumkins would score nothing by his performance.  Diagram No. 2 will, however, make it clear that since Mr. Luffey and Mr. Struggles have, notwithstanding their energetic but careless movements, contrived to change places, the manoeuvre must increase Mr. Struggles’s total by one run.

388.—­SLOW CRICKET.

The captain must have been “not out” and scored 21.  Thus:—­

2 men (each lbw) 19 4 men (each caught) 17 1 man (run out) 0 3 men (each bowled) 9 1 man (captain—­not out) 21 —­ —­ 11 66

The captain thus scored exactly 15 more than the average of the team.  The “others” who were bowled could only refer to three men, as the eleventh man would be “not out.”  The reader can discover for himself why the captain must have been that eleventh man.  It would not necessarily follow with any figures.

389.—­THE FOOTBALL PLAYERS.

The smallest possible number of men is seven.  They could be accounted for in three different ways:  1.  Two with both arms sound, one with broken right arm, and four with both arms broken. 2.  One with both arms sound, one with broken left arm, two with broken right arm, and three with both arms broken. 3.  Two with left arm broken, three with right arm broken, and two with both arms broken.  But if every man was injured, the last case is the only one that would apply.

390.—­THE HORSE-RACE PUZZLE.

The answer is:  L12 on Acorn, L15 on Bluebottle, L20 on Capsule.

391.—­THE MOTOR-CAR RACE.

The first point is to appreciate the fact that, in a race round a circular track, there are the same number of cars behind one as there are before.  All the others are both behind and before.  There were thirteen cars in the race, including Gogglesmith’s car.  Then one-third of twelve added to three-quarters of twelve will give us thirteen—­the correct answer.