THE BAG OF NUTS.
Three boys were given a bag of nuts as a Christmas present, and it was
agreed that they should be divided in proportion to their ages, which
together amounted to 171/2 years. Now the bag contained 770 nuts, and
as often as Herbert took four Robert took three, and as often as Herbert
took six Christopher took seven. The puzzle is to find out how many nuts
each had, and what were the boys' respective ages.
It will be found that when Herbert takes twelve, Robert and Christopher
will take nine and fourteen respectively, and that they will have
together taken thirty-five nuts. As 35 is contained in 770 twenty-two
times, we have merely to multiply 12, 9, and 14 by 22 to discover that
Herbert's share was 264, Robert's 198, and Christopher's 308. Then, as
the total of their ages is 171/2 years or half the sum of 12, 9, and 14,
their respective ages must be 6, 41/2, and 7 years.