536 Puzzles and Curious Problems

(Elliott) #1
232 Answers


  1. THEIR AGES


Thirty years and twelve years, respectively.


  1. BROTHER AND SISTER


The boy's age was ten and his sister's four.


  1. A SQUARE FAMILY


The ages of the nine children were respectively 2, 5, 8, II, 14, 17,20,23,26,
and the age of the father was 48.

42. IN THE YEAR 1900

The man was born in 1856 and died in 1920, aged 64 years. Let x = age at
death. Then 29x = date of birth. The date of birth + age = date of death,
so that 29x + x = 30x, or date of death. Now, from the question he was clearly
alive in 1900, and dead by 1930. So death occurred during or between those
dates, and as the date is 30x, it is divisible by 30. The date can only be 1920,
which, divided by 30, gives 64. So in 1900 he was 44 years of age.


  1. FINDING A BIRTHDAY


The man must have been born at midday on February 19, 1873, and at
midday on November 11, 1928, he had lived 10,17616 days in each century.
Of course, the century ended at midnight on December 31, 1900, which was
not a leap year, and his age on November 11, 1928, was 55 years and nearly
9 months.


  1. THE BIRTH OF BOADICEA


There were 129 years between the birth of Oeopatra and the death of
Boadicea; but, as their united ages amounted to 100 years only, there must
have been 29 years when neither existed-that is, between the death of
Oeopatra and the birth of Boadicea. Therefore Boadicea must have been
born 29 years after the death of Cleopatra in 30 B.C., which would be in the
year 1 B.C.

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