You are not logged in.
Pages: 1
Suppose 1,000 people taught 2 others how to read in a year and then stopped teaching. Then the 2,000 new readers taught 2 more each the next year and then stopped. If the pattern continues, how many total readers would there be at the conclusion of 10 years?
Hi Raven17;
Welcome to the forum!
2000 at the end of 1st year
4000 at the end of 2nd year
.
.
.
512 000 at the end of 9 years
1 024 000 at the end of 10 years
or by formula:
What you have is 1000 * 2^y where y is the year.
1000 * 2^10 = 1 024 000
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Thank you for your response.
Would the idea of "total readers" mean that we should add the old to the new at the end of each year? Isn't the 1,024,000 just the number of readers "created" in the last year?
Would the sum not be 10000+2000+4000+8000+ . . . 1024000 or 2,047,000?
Yes! You're Right!!
It should be 1000.(2[sup]11[/sup]-1)
Last edited by ZHero (2010-05-21 17:14:20)
If two or more thoughts intersect, there has to be a point!
Offline
Hi Raven17
Yes, I have left that for you.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Hi Raven17;
The general solution is:
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Suppose 1,000 people taught 2 others how to read in a year and then stopped teaching. Then the 2,000 new readers taught 2 more each the next year and then stopped. If the pattern continues, how many total readers would there be at the conclusion of 10 years?
Number of Readers Initially = 1000 = 1000×2[sup]0[/sup]
Number of New Readers..
at the end of 1[sup]st[/sup] year = (1000)×2 = 1000×2[sup]1[/sup]
at the end of 2[sup]nd[/sup] year = (1000×2)×2 = 1000×2[sup]2[/sup]
at the end of 3[sup]rd[/sup] year = (1000×2[sup]2[/sup])×2 = 1000×2[sup]3[/sup]
.
.
.
at the end of 10[sup]th[/sup] year = (1000×2[sup]9[/sup])×2 = 1000×2[sup]10[/sup]
Total Number of Readers at the End of 10 years...
⇒ 1000×2[sup]0[/sup]+1000×2[sup]1[/sup]+1000×2[sup]2[/sup]+1000×2[sup]3[/sup]+...+1000×2[sup]10[/sup]
⇒ 1000×(2[sup]0[/sup]+2[sup]1[/sup]+2[sup]2[/sup]+2[sup]3[/sup]+...+2[sup]10[/sup])
⇒ 1000×(2[sup]11[/sup]-1)
NOTE: This an example of Exponential Growth!
If two or more thoughts intersect, there has to be a point!
Offline
Pages: 1