# Guess the Boy's Age Puzzle - Solution

### The Puzzle:

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.

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.

### Our Solution:

Answer: 1,276 days. This puzzle can easily be solved by the "trial method". The first girl was just 638 days old, and the boy twice as much, namely 1,276 days. The next day the youngest girl will be 639 days old, and her new recruit 1,915 days, total, 2,554 days, which doubles that of the first boy, who having gained one day, will be 1,277 days old. The next day the boy, being 1,278 days old, brings his big brother, who is 3,834 days old, so their combined ages amount to 5,112 days, which is just twice the ages of the girls, who will now be 640 and 1,916, or 2,556.

The next day, the girls gaining one day each, will represent 2,558 days, which added to 7,670 days of the last recruit, brings up their sum total to 10,228 days, which is just twice that of the two boys, which, with the two points added for the last day, would be increased to 5,114 days.

We arrive at the 7,670 days by saying, the young lady having reached her twenty-first birthday, 21 times 365 equals 7,665 plus 4 days for four leap years, and the extra one day, which, comes with the twenty-first birthday (which is one day towards the twenty-second year).

AN APPROXIMATE SOLUTION USING ALGEBRA

by George Austin

This solution ignores the day joined, so will be a few days wrong.

Let us use x=boy 1's age, y=boy 2's age, p=girl 1's age, q=girl 2's age, and we know that girl 3 is 21

When the 3rd girl joined: 2(x+y)=p+q+21

We also know that p+q=4p, as when the 2nd girl joined the girl ages went from half to double. So: 2(x+y)=4p+21

x+y=2p+10.5 (halve both sides)

x+y=x+10.5 (because 2p=x)

y=10.5 (subtract x from both sides)

y=3x, so: x=3.5 years old (about 1,278 days)

ANOTHER (SIMPLER) SOLUTION USING ALGEBRA

by "gscbiomajor"

Let the first girl be x, the first boy is 2x, the second girl is 3x (since x plus 3x = 4x twice the first boys age) boy three is 6x (6x + 2x = 8x twice girl 1 and 2 ages) and the third girl is 12x (twice boy one and two). Therefore 21 = 12x, 21/12 is 1.75 making the first boys age 3.5 years.

The next day, the girls gaining one day each, will represent 2,558 days, which added to 7,670 days of the last recruit, brings up their sum total to 10,228 days, which is just twice that of the two boys, which, with the two points added for the last day, would be increased to 5,114 days.

We arrive at the 7,670 days by saying, the young lady having reached her twenty-first birthday, 21 times 365 equals 7,665 plus 4 days for four leap years, and the extra one day, which, comes with the twenty-first birthday (which is one day towards the twenty-second year).

AN APPROXIMATE SOLUTION USING ALGEBRA

by George Austin

This solution ignores the day joined, so will be a few days wrong.

Let us use x=boy 1's age, y=boy 2's age, p=girl 1's age, q=girl 2's age, and we know that girl 3 is 21

When the 3rd girl joined: 2(x+y)=p+q+21

We also know that p+q=4p, as when the 2nd girl joined the girl ages went from half to double. So: 2(x+y)=4p+21

x+y=2p+10.5 (halve both sides)

x+y=x+10.5 (because 2p=x)

y=10.5 (subtract x from both sides)

y=3x, so: x=3.5 years old (about 1,278 days)

ANOTHER (SIMPLER) SOLUTION USING ALGEBRA

by "gscbiomajor"

Let the first girl be x, the first boy is 2x, the second girl is 3x (since x plus 3x = 4x twice the first boys age) boy three is 6x (6x + 2x = 8x twice girl 1 and 2 ages) and the third girl is 12x (twice boy one and two). Therefore 21 = 12x, 21/12 is 1.75 making the first boys age 3.5 years.

Puzzle Author: Loyd, Sam

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