Member of the Month

KilaTsunami
October, 2024

Online

Welcome to the Newschoolers forums! You may read the forums as a guest, however you must be a registered member to post. Register to become a member today!

Want a challenge?

profile picture
rvca
Posts: 311 - Karma: 13

This is an 8th grade math problem (for my lil sis' class) I spent like 5 minutes on it and had some trouble. I got an answer but i dont think its right. So I look to NS to help answer my sisters math problem...

The plan is for each individual who can read to spend one year teaching two others to read. After one year that teacher is done but the two new readers must spend one year, each teaching two others to read. If 1000 people can read at the start, how many people will be able to read in ten years?

My answer was 1,277,000 people. But i have a bad feeling about it. A confirmation on that answer or a new one with a good reason would be greatly appreciated... Thanks NS!

Dec 6 2009 6:21PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
1000-----2000-----4000-8000----16000----32000----64000--128000---256000----512000

im scared of sounding like a retard by posting this but doesnt it make sense that it just doubles every year
Dec 6 2009 6:28PM
0 0
profile picture
304NordMan
Posts: 2983 - Karma: 525
ill take his werd for it
Dec 6 2009 6:30PM
0 0
profile picture
rvca
Posts: 311 - Karma: 13
that was my first thought process, but then i remembered that after the first thousand people teach two more, they are done and dont teach anymore people
Dec 6 2009 6:31PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
ya so 1000 people teach 2 people each so now 2000 people can read then 2000 people teach 2 each so now 4000 can read and so on
Dec 6 2009 6:33PM
0 0
profile picture
skithesprings
Posts: 9086 - Karma: 1,349
doubles each year for 10 years, 1024000

the first was a year short i think
Dec 6 2009 6:34PM
0 0
profile picture
t-roy
Posts: 4041 - Karma: 52
i think you are all wrong right from the start.

1000 people start knowing how to read. then each of them teaches 2 people to read, so then 3000 people now know how to read, not just 2000.
Dec 6 2009 6:38PM
0 0
profile picture
rvca
Posts: 311 - Karma: 13
its not that simple though. think about it in smaller numbers like 1 instead of a thousand. if you have one person teach two people, at the end of that year you have three people who can read
Dec 6 2009 6:38PM
0 0
profile picture
t-roy
Posts: 4041 - Karma: 52
my answer is 59,049,000.
Dec 6 2009 6:42PM
0 0
profile picture
sniper_6
Posts: 1919 - Karma: 1,573
                          
Dec 6 2009 6:42PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
ya but that doesnt change much i dont think


1000 people can read they teach 2 each so now 3000 people can read in total but the original 1000 arent teaching anybody anymore so only the 2000 teachers teach 2 people which bring it to 4000 or 6000 but the 2000 who taugh the 4000 dont teach anymore so only 8000 people read i guess u can just add 1000 to the final anwser if you want but im pretty sure it just doubles every year

its grade 8 here guys there not trying to trick anyone
Dec 6 2009 6:43PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
basically the 1000 original people dont teach for 10 years they only teach once then quit
Dec 6 2009 6:44PM
0 0
profile picture
t-roy
Posts: 4041 - Karma: 52
ahh yeah, i missed that aprt in the problem that they stop teaching. so my answer was very wrong. this is a good problem.
Dec 6 2009 6:46PM
0 0
profile picture
charmander
Posts: 10636 - Karma: 370
yep.

i got
2,047,000
Dec 6 2009 6:46PM
0 0
profile picture
rvca
Posts: 311 - Karma: 13
i know what you're saying but the question asks for the total amount of people that can read at the end of the 10 years. not just the new readers
Dec 6 2009 6:47PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
ur wrong
Dec 6 2009 6:47PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
so then just add up all the numbers?
Dec 6 2009 6:49PM
0 0
profile picture
no_steeze
Posts: 23992 - Karma: 4,273
P=A (1+x/n)^nt
Dec 6 2009 6:51PM
0 0
profile picture
RRhighrider
Posts: 2140 - Karma: 1,124
  year0 year1 year2 year3 year4 year5 year6 year7 year8 year9 year10 teachers  1000 2000 4000 8000 16000 32000 64000 128000 256000 512000 1024000 readers 2000 4000 8000 16000 32000 64000 128000 256000 512000 1024000 2048000                             sum 4095000                

i did the sum of the teacher plus the amount of readers in year ten

Dec 6 2009 6:52PM
0 0
profile picture
t-roy
Posts: 4041 - Karma: 52
you need to make sure the teachers who just taught don't get counted in how many new people get taught per year but than you have to add up all the original teachers at the start of each year who didn't taught but still know how to read.
Dec 6 2009 6:52PM
0 0
profile picture
ace_
Posts: 3189 - Karma: 39
i thought it was P derived from A= 3( t-f)^3
Dec 6 2009 6:54PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
i totally solved this first genius genius genius!!
Dec 6 2009 6:54PM
0 0
profile picture
philcote4
Posts: 225 - Karma: 10

this is the way i see it (I really dont know if its right, im just goin with it)

you get this function:

 # of people who know how to read = 1000x + 1000

x being the amount of years after this whole chain starts

i dunno if it works but i tested it for the first 2 years and it seems to work...

Dec 6 2009 6:54PM
0 0
profile picture
philcote4
Posts: 225 - Karma: 10

i meant to say

# of people who know how to read = 2000x +1000

Dec 6 2009 6:56PM
0 0
profile picture
philcote4
Posts: 225 - Karma: 10
nvm, its wrong, im too lazy to fix it though. maybe later
Dec 6 2009 6:57PM
0 0
profile picture
Pow.yuHdig*
Posts: 1903 - Karma: 146
that doesnt work

it goes 1000---2000----4000 so 7000 after 3 years which is right

and your equation goes

1000.1+1000=2000
1000.2+1000=3000
1000.3+1000=4000
1000.4+1000=5000
Dec 6 2009 6:58PM
0 0
profile picture
skithesprings
Posts: 9086 - Karma: 1,349
ok, dont know why but i felt compelled to sit down and figure this out,

in a simpler form,

1+2 = 3
2+4 = 6
4+8 = 12
8+16= 24
ect. adding up the numbers on the right because the people teach two others but are done after that the next year only the people who learned teach. and multiply by 1000 at the end.

for a total of 3069000

check my work though i did most in my head
Dec 6 2009 6:59PM
0 0
profile picture
rvca
Posts: 311 - Karma: 13
so then after 4 years isnt it 17000?
Dec 6 2009 6:59PM
0 0
profile picture
no_steeze
Posts: 23992 - Karma: 4,273
different equations
Dec 6 2009 7:03PM
0 0
profile picture
WorthDaMoney
Posts: 1140 - Karma: 741

3,070,000  people

im pretty sure thats the answer

let us know what the teacher says the answer is

Dec 6 2009 7:04PM
0 0
profile picture
charmander
Posts: 10636 - Karma: 370
maybe

what did you get?

i did it in 1's then multiplied by 1000

people not teaching I people teaching I people that can read
0 1 3
1 2 7
3 4 15
7 8 16+8+7=31
15 16 16(2)=32+16+15=63
21 32 32(2)-64+21+32=117
53 64 128+53=181
117 64 128+117=245
181 64 128+181=309
245 64 128+245=373

373 X 1000 = 373,000

i think i may have messed up somewhere along the line. i feel like there should be a rule to this.
Dec 6 2009 7:04PM
0 0
profile picture
Ben.
Posts: 21463 - Karma: 683
1023000
Dec 6 2009 7:04PM
0 0
profile picture
PAuL_Hellar
Posts: 1355 - Karma: 62
i think you pretty much got it
Dec 6 2009 7:07PM
0 0
profile picture
rvca
Posts: 311 - Karma: 13
there still hasnt been any repeated answers so i feel like the answer is still out there...
Dec 6 2009 7:08PM
0 0
profile picture
WorthDaMoney
Posts: 1140 - Karma: 741
you're missing 1000, i agree completely with your method, thats the way i got my answer : 3,070,000 people
Dec 6 2009 7:08PM
0 0
profile picture
RRhighrider
Posts: 2140 - Karma: 1,124

i though about it some more and there are no readers in year ten.  this this the correct answer

year0 year1 year2 year3 year4 year5 teachers  1000 2000 4000 8000 16000 32000 readers 2000 4000 8000 16000 32000 64000 year6 year7 year8 year9 year10 teachers  64000 128000 256000 512000 1024000 readers 128000 256000 512000 1024000 2047000

Dec 6 2009 7:11PM
0 0
profile picture
philcote4
Posts: 225 - Karma: 10
i think he's got it right. i restarted and thats the same method i was using for the first couple years and i got the same answers. it makes sense.
Dec 6 2009 7:18PM
0 0
profile picture
RRhighrider
Posts: 2140 - Karma: 1,124
0      
Dec 6 2009 7:24PM
0 0
All times are Eastern (-5)