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3 age problems (Posted on 2006-03-24) Difficulty: 3 of 5
I A domestic partnership of two consenting adults has three children, Alice, Bob, and Charlie, and the difference between the adults' ages was the same as between Alice and Bob and between Bob and Charlie. The product of Alice and Bob’s ages equaled one parent's age. The product of Bob and Charlie’s ages equaled the other parent's age. The sum of the five ages amounted to ninety years. What was the age of each person?

II A woman has nine children(!), all born at regular intervals, and the sum of the squares of their ages (in years) was equal to the square of her own. What was the age of each child?

III Boadicea died one hundred and twenty-nine years after Cleopatra was born. Their combined life-span was one hundred years. Cleopatra died in 30 B.C.E. When was Boadicea born?

See The Solution Submitted by Jer    
Rating: 3.6667 (3 votes)

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Solution Other solutions | Comment 5 of 8 |

I The adults are ages 36 and 36 with Alice, Bob & Charlie 6 year-old triplets.

II There are a myriad of possible answers.  The equation for the mother's age can be given by SQRT(9a2 + 72ad + 204d2), where a is the age of the youngest child and d is the interval in years between each child.

Mother's age:   Children's ages:
48                   2, 5, 8, 11, 14, 17, 20, 23, 26
96                   4, 10, 16, 22, 28, 34, 40, 46, 52

I know 92 is a little old for being a mother, but it is a possibility. If a cure for old age and menopause is ever discovered, the solution can be extended.

III Cleopatra was born in 69 BCE, and given that she died in 30 BCE, she was 39 when she died.  Boadicea therefore was 61 years old when she died.  Keeping in mind that there is no 0 BCE or 0 CE, it can be calculated that she died in 61 CE and was born in 1 BCE.

Edited on March 25, 2006, 1:38 am
  Posted by Dej Mar on 2006-03-24 23:12:55

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