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Guess The Boy's Age by Sam Loyd


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It appears that an ingenious or eccentric teacher being desirous of bringing together a number of older pupils into a class he was forming, offered to give a prize each day to the side of boys or girls whose combined ages would prove to be the greatest.

Well, on the first day there was only one boy and one girl in attendance, and, as the boy's age was just twice that of the girl's, the first day's prize went to the boy.

The next day the girl brought her sister to school, and it was found that their combined ages were just twice that of the boy, so the two girls divided the prize.

When school opened the next day, however, the boy had recruited one of his brothers, and it was found that the combined ages of the two boys were exactly twice as much as the ages of the two girls, so the boys carried off the honors of that day and divided the prizes between them.

The battle waxed warm and on the fourth day the two girls appeared accompanied by their elder sister; so it was then the combined ages of the three girls against the two boys, and the girls won off course, once more bringing their ages up to just twice that of the boys'. The struggle went on until the class was filled up, but as our problem does not need to go further than this point, to tell the age of that first boy, provided that the last young lady joined the class on her twenty-first birthday. Now, guess the first boy's age.

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If the total age of the girls plus 21 is twice the age of the boys, and the age of the boys is twice the age of the girls, then the age of the boys is 14 and the age of the girls was 7 (before Miss 21 arrived).

2b = g + 21

b = 2g

4g = g + 21

3g = 21

g = 7

b = 14

That's groovy and all, but if the total age of the girls was 7 before Miss 21 arrived, and that total is twice what the boys used to be, then the total age of the boys was three and a half (before a ten and a half year old boy joined the class). I realize some kids run around saying that they are so many and a half years old, but I would assume that the answer should be an integer (otherwise, we could continue dividing the youngest boys age by 4 indefinitely to find a younger boy).

If that is correct, then there were only 3 people in that class: two girls, aged 21 and 7; and one boy aged 14.

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Stating off, boy;s age --=x, therefore giri=x/2.. 2nd girl brings age total to 2x, 2nd boy brings it to 4x, the 3rd girl, who is 21, brings it to 8x, So 21 + 2x=8x, 6x=21, x=3.5, Which also makes the first girl only


21 months old. Must be a child prodigy
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If the last lady joined to join did so on Day 4 then the age of 1st girl is 1.75 yrs and 1st Boy 3.5. Actually works out like this:



Last girl to join was on Day 2 - Age of 1st girl - 7 , age of 1st Boy 14 yrs
Last girl to join was on Day 4 - Age of 1st girl -1.75, age of 1st Boy 3.5 yrs
Last girl to join was on Day 6 - Age of 1st girl -.4375, age of 1st Boy .875 yrs
Last girl to join was on Day 8 - Age of 1st girl - .109375, age of 1st boy .21875 yrs
and so on.....
If you imagine that the poor school master did not run a post natal child care unit and given that on day 4 an elder sister was brought in, most logical answer looks like 1.75 as age of 1st girl and age of 1st boy 3.5 and the class filled up on day 4 itself or the next day.




Edited by nakulendu
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