During their 10 years of marriage, Chuck and Dora had two children, Al and Bob.
Later, by a second wife, Elaine,
Chuck had another son, Frank.
By a
second husband, George, Dora had
a third son, Howard.
At the time of Chuck's marriage to Elaine, Elaine was five times Bob's age and Dora was six times his age. At the time of Dora's marriage to George, George was four times Al's age and Chuck was five times Bob's age. George was 13 years older than Elaine.
On what would have been Chuck and Dora's 20th anniversary, Bob was twice as old as Frank, and Chuck was three times as old as Bob.What was everyone's age at that time?
Solution to the Problem:
Al, 17; Bob, 14; Chuck, 42; Dora, 39; Elaine, 34;
Frank, 7; George, 47; Howard, 4.
(At this time, Chuck and Elaine have been married for nine years,
and Dora and George for seven years.)
B = H + 10
A = F + 10
G = E + 13
The other equations are only true at certain times:
When Elaine and Chuck were married
(more than 10 years after Chuck and Dora were married but less than 20 years):
E = 5B and D = 6B
(more than 10 years after Chuck and Dora were married but less than 20 years):
G = 4A and C = 5B
On the 20th Anniversary,
B = 2F and C = 3B.
Therefore, C = 6F.
F = 1 C = 6
F = 2 C = 12
F = 3 C = 18
F = 4 C = 24
F = 5 C = 30
F = 6 C = 36
F = 7 C = 42
F = 8 C = 48
F = 9 C = 54
We can eliminate the first five and probably six possibilities since this is Chuck's 20th Anniversary. He must have been more than 10 years old! Keep working like this until you find a set of numbers that yields all whole numbers!
Correctly solved by:
| 1. Richard K. Johnson | La Jolla, California |
| 2. David & Judy Dixon | Bennettsville, South Carolina |
| 3. Morgan Vercimak |
Mountain View High School, Mountain View, Wyoming |