Arthur, Bill, Chris, and David played golf together one Saturday
morning. Eighteen holes were played.
All of them scored 11
pars. The rest of the plays were either eagles, birdies or bogeys.
Create the table below a show how many
eagles, birdies, and bogeys each player scored.
All of them scored 11
pars. So rest of the plays which were either eagles, birdies or bogeys should
add up to seven.
Arthur scored 4 birdies. Bill scored 3 eagles.
Dan did not have any birdie. Chris did not have any eagle.
|
Eagles |
Birdies |
Bogeys |
Arthur |
|
4 |
|
Bill |
3 |
|
|
Chris |
0 |
|
|
Dan |
|
0 |
|
One player eagled four times. One player eagled
once. The only way to have this is
|
Eagles |
Birdies |
Bogeys |
Arthur |
1 |
4 |
|
Bill |
3 |
|
|
Chris |
0 |
|
|
Dan |
4 |
0 |
|
One player birdied twice. One player bogeyed twice.
One player birdied once. The only way to have this is
|
Eagles |
Birdies |
Bogeys |
Arthur |
1 |
4 |
|
Bill |
3 |
2 |
2 |
Chris |
0 |
1 |
|
Dan |
4 |
0 |
|
Filling up the rest of the cells in the table so that
plays added up to seven yields:
|
Eagles |
Birdies |
Bogeys |
Arthur |
1 |
4 |
2 |
Bill |
3 |
2 |
2 |
Chris |
0 |
1 |
6 |
Dan |
4 |
0 |
3 |
|