Senior class
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. What is the first boy's age (at the start).
3 Feb 2010, 6:10 AM. Express your answer in days.
3 Feb 2010, 6:10 AM. Express your answer in days.
Labels: mathschallenge
posted by Chris at
8:07 PM
23 Comments:
Zaux. You might not have seen my solution to the "How much fabric?" problem.
Question time:
are ages like 10.5 counted :P
i mean like 10 yrs, and 6 months? cuz im gettin an age like tht for the first variable i solved >.<
ok, i am just posting my answers:
I am using the variables like this:
1st boy's age = b1, 2nd boy's age= b2, 1st girl's age = g1.. etc.
my equations:
b1 = 2g1------1
2b1 = g1+g2---------2
b1+b2 = 2g1+2g2----3
2b1+2b2 = g1+g2+g3 (but g3 = 21)
so, 2b1+2b2 = g1+g2+21 ----4
Now, putting 2 in 4
gives b2 = 21/2 (not really necessary)
and putting twice 3rd equation in 4
gives: (g1+g2) = 7
and that in 2 => b1 = 3.5
But, my final doubt:
why is a 21 year old person going to school where 3 yr olds go, and btw, the g1 = 1.25... i dunno, but i think there must b something wrong... but cant find anything wrong with my math
and, now ill go to how i did it on Wagons question
Zaux/Chris...how do you post puzzles so fast?the only way i know how to do it is by e-mailing it to trickofmind@gmail.com and then wait a few days to see if it's posted.but you guys seem to do it at will.HOW!?
The solution I've seen works in days. Unfortunately, it's a terrible solution, it gives the answer and shows how it works out - it doesn't deduce the various ages involved.
I posted this and the Wagons one, in a hurry, to give Zaux a break.
To knightmare, u can join ToM as a member, and then u can post whenever u want:
wht u r doing is this: u give it to Tom, and then posts it whenever he sees it.
what they do:
sign into blogger.com, and post it on the blog, and it comes here directly
I'm not exactly sure how Zaux is doing it. I have a gmail account, and that can be used to make an problem posting account on this site.
Once you've done that, you can edit the original problem and delete (optionally without leaving a tell-tale "This message has been deleted") anyone's post, on your own problem page. You can also delete your own comments on someone elses page, but with the tell-tale left behind.
I think you have to click on the "I Power Blogger" link to make the ToM account, and to login once you've got an account.
ok, i think ill post some questions too.. but question Chris, how do i change my name on blogger? i am a member, but i dont want everyone to see my e-mail id, i doubt ur email id is Chris@gmail..com, or is it?
my real name comes there, i dont want tht. so, ow do i make a nickname?
You should have a control panel for you account profile. If you click on my name (i.e. "Chris") link at the beginning of any of my comments, you can view my profile. You won't find my email address as I've suppressed it. My email is not chris@gmail.com. Even the name (e.g. Chris) is your choice. It just happens that Chris is my real name. I've recently used Arthur Bunyon on another site.
There are a lot of advantages to getting a ToM acount (it's free BTW).
Just a test. I'm Chris. You should see me as Arthur Bunyon (a fake name).
OK I'm Chris again.
ok thanks, btw to knightmare, to become a member, click on Become a member link on the left :)
I got my account bcuz ToM asked if i wanted to join after i sent like 10-12 problems..i guess he got irritated.
Thx Chris, i have posted 5 questions.
btw... is my answer for this question right!!??
Chris...stop it...your scaring the children.
Hello there!
I used an equation approach similar to that of Vago. As the answer is not a round number I've also taken into account the days passed in the problem. Goes as follows:
b1 = 2g1
2(b1 + d) = g1 + d + g2
b1 + 2d + b2 = 2(g1 + 2d + g2 + d)
2(b1 + 3d + b2 + d) = g1 + 3d + g2 + 2d + g3
where every d equals one day.
After simplification and various substitutions it is clear that g1 = 1.75 - 7d/6 and therefore b2 = 3.5 - 7d/3 or
3 years, 5 months, 28 days, and 16 hrs.
Using these values and the original equations it is possible to know the ages of all the kids, wich i found to be...
b1 = 03yr 05mth 28dy 16hrs
b2 = 10yr 06mth 03dy 00hrs
g1 = 01yr 08mth 29dy 08hrs
g2 = 05yr 02mth 29dy 00hrs
g3 = 21yr 00mth 08dy 00hrs
the +8day error in g3 is probably a consequence of using 30-day months and 365-day years only.
see you around
cool, i was going to do somewhat like that, but was too lazy so, i found my method easier. That was the main reason why i posted my first question, and then started on the rest of the variables, but then, i saw that i could get b1 in just 1 step, so posted it :)
Glad my math was correct, i normally make mistakes there :P also, i still dont get why a 1 yr old girl even goes to school!!! at my country, u go to play school after u bcome 2 yrs old. and also if it exists, y does a 21 yr person go to the same school as a 1 yr!!!
Im just thinking.... bored atm :P
Sorry. I forgot to check this problem before going to bed. Totally distracted by Vago's (not so) mini problems.
3.5 years is very close. Work entirely in days, forget months and weeks. Do allow for leap years (which you only need to consider in one case). A century boundary isn't crossed.
Although 3.5 years is very close, it's not close enough for a cigar. Working and thinking in days is very important. This problem is still open. I'm sure you'll see why when the correct solution is revealed. Give the answer in days.
I still havem't got round to solving it myself (working on Vago's problems and fixing computers). The solution from the web site simply states the boys age and then walks you through so that you can check it all hangs together. This site has logos on it saying it is an "Edunet Choice", just shows that they're plonkers - there's little of educational in stating an answer (OK, checking your answer is always a good idea).
This post has been removed by the author.
ok, i got the answer as 1276.5 im taking it as 1276
im not putting the whole solving again, it is the same as b4, this time my equations have changed a bit:
Here, i assume all variables as the ages of people at the first day only, so i add 1 on 2nd, 2 on thirs and 3 on 4th for each of the ppl.
I am just posting the equations.
1- b1 = 2g1
2- 2b1 = g1+g2
3- b1+b2 = 2g1+2g2+4
4- 3 + 2b1+2b2 = g1 +g2+ g3
NOTES: A)g3 = 7670 days.
B) In the equationd, I have canceled the number of days on both sides, eg:
2 will be like this:
2(b1+1) = g1+1 + g2+1
=> 2b1 + 2 = g1 + g2 + 2
2 is canceled and we get the old:
2b1 = g1 + g2
C) To solve, put twice 3 in 4, and put 2 in 4 then. 1st equation isnt actually needed.
I still dont get wht u were saying about " I'm sure you'll see why when the correct solution is revealed." Chris..
1276 days is the answer sought. Those +1s and +2s were missing when you weren't working in days. That's what I was referring to. But looking back, I see that I wasn't paying enough attention (Julio had used days), but you seem to have used years (the first time you posted).
The solution I've seen says that the 21 year old is 21*365 + 4 (leap years) + 1 (as she's on her first birthday => 7670. I'd have gone for 21*365 + 5 for leap years (and 0 or 1 for the day itself), so 7670 or 7671 days is her age.
I've got to go out for the rest of the evening. I'm spent all my timeanswering your mixed bag of questions and fixing PCs, so haven't thoroughly checked the solutions posted here.
I'm going to credit both Vago and Julio as having done it.
Vago, I've just re-read your Feb 3 7:35 AM comment. You've made some mistakes when handling the days. Julio's Feb 2 11:25AM comment has got it right. Both answers could have done with a few more intermediate steps.
I have to retract, and say that only Julio has done it sufficiently well to be the official winner.
hooray!
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