Emergency! Please help

[gan9606]

I have a no idea how to solve this following question.

Consider an economy in which the average annual growth rate of output per worker rises from 3.0% during the period from 1965 to 1985 to a rate of 5.5% from 1986 to the present. This has led to talk of a new Economy and of sustained higher growth in the future than in the past.

At the old rate (3% growth per year) how much would output per worker have grown-relative to today's level- over 12 years?

 

[Deleted member 4993]

I have a no idea how to solve this following question.

Consider an economy in which the average annual growth rate of output per worker rises from 3.0% during the period from 1965 to 1985 to a rate of 5.5% from 1986 to the present. This has led to talk of a new Economy and of sustained higher growth in the future than in the past.

At the old rate (3% growth per year) how much would output per worker have grown-relative to today's level- over 12 years?

Since you have no idea - let us start with definitions....

What does annual growth rate mean?

You need to use "to today's level" of annual growth rate - what is it?

Tell us what you find......

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (not a statement like "Don't know any of these")

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[gan9606]

Thanks for your reply. I am sorry for just asking. I am a new user and I did not know about the rules. I will keep those rules.

First, Annual growth means that it is increasing every year by growth rate. But how can I know what is the principal number? I do not understand this question at all.

Second, "relative to today's level"... Doesn't it mean 2013? So I need to calculate 1986 to 2013?

Basically, I did not understand question quite well....

I also checked that you said this is similar to compound interest formula. But I think there are not enough info in this question to use the formula.

 

[JeffM]

I have a no idea how to solve this following question.

Consider an economy in which the average annual growth rate of output per worker rises from 3.0% during the period from 1965 to 1985 to a rate of 5.5% from 1986 to the present. This has led to talk of a new Economy and of sustained higher growth in the future than in the past.

At the old rate (3% growth per year) how much would output per worker have grown-relative to today's level- over 12 years?

Thank you for having read the rules and agreeing to abide by them going forward.

The primary idea in algebra is, whenever there is a relevant number that you do not know, label it with a symbol.

What is this question about? (Assuming that you copied it exactly, it is not ideally worded: "At the old rate of 3% growth per year, what would output per worker be after 12 yearsr?")

There are two unknowns in this problem.

w = initial output per worker.

x = output per worker after 12 years if output per worker is growing at 3% annually.

See how defining your terms forces you to translate a word problem into mathematical form.

Do you now see how to solve it?

Edit: By the way, it is EXACTLY similar to a compound interest problem.

 

[gan9606]

I am really sorry but I still do not get it.
Here is what I got so far.
compound interest formula is A=P(1+(r/n))^nt
P=Principal amount
r= annual rate of interest
t=number of years of amount
A=output per worker after 12 years if output per worker is growing at 3% annually.
n=number of times the interest is compounded per year.

Therefore, P is w as you said, r is 3.3%, t is 12, and n is 1. right?? Then how can I get w?? w is unknown.

I am really sorry but would you please explain it little bit more?

Thank you for having read the rules and agreeing to abide by them going forward.

The primary idea in algebra is, whenever there is a relevant number that you do not know, label it with a symbol.

What is this question about? (Assuming that you copied it exactly, it is not ideally worded: "At the old rate of 3% growth per year, what would output per worker be after 12 yearsr?")

There are two unknowns in this problem.

w = initial output per worker.

x = output per worker after 12 years if output per worker is growing at 3% annually.

See how defining your terms forces you to translate a word problem into mathematical form.

Do you now see how to solve it?

Edit: By the way, it is EXACTLY similar to a compound interest problem.

 

[JeffM]

I am really sorry but I still do not get it.
Here is what I got so far.
compound interest formula is A=P(1+(r/n))^nt
P=Principal amount
r= annual rate of interest
t=number of years of amount
A=output per worker after 12 years if output per worker is growing at 3% annually.
n=number of times the interest is compounded per year.

Therefore, P is w as you said, r is 3.3%, t is 12, and n is 1. right?? Then how can I get w?? w is unknown.

I am really sorry but would you please explain it little bit more?

OK Here's the thing: formulas screw you up unless you know why they work.

Suppose I told you that output per worker was 10 last year and increased 3% this year. The number 10 is the base of our computation.Then

output per worker this year would be \(\displaystyle 10 + (0.03 * 10) = 10 + 0.3 = 10.3 = 10 * 1.03.\) Are you with me to here?

Suppose it increases again 3% next year. Well, this year's value is 10.3 so next year's will be

\(\displaystyle 10.3 + (.03 * 10.3) = 10.3 + 0.309 = 10.609.\) Make sense so far?

But there is another way to do this \(\displaystyle 10.3 + (10.3 * .03) = 10.3 * 1.03 = (10 * 1.03) * 1.03 = 10 * 1.03^2.\) Still with me?

How about year 3? That would be \(\displaystyle 10.609 + (10.609 * 0.03) = 10.609 * 1.03 = (10 * 1.03^2) * 1.03 = 10 * 1.03^3.\)

It's exactly the same as the compounding formula with n = 1.

It is in fact the general formula for any system that is increasing at a constant percentage rate each period.

But what if the base is unknown? What if it is w instead of 10? SO WHAT? w is just a number.

At the end of 12 years w will have grown to \(\displaystyle w * 1.03^{12} \approx 1.42576w.\)

And how much bigger than w is 1.42576w? It is 42.576% bigger. The base just drops out if we express the answer as a ratio or a percentage.

Please let me know where I have not been clear.

 

[gan9606]

Thanks a lot!!!

OK Here's the thing: formulas screw you up unless you know why they work.

Suppose I told you that output per worker was 10 last year and increased 3% this year. The number 10 is the base of our computation.Then

output per worker this year would be \(\displaystyle 10 + (0.03 * 10) = 10 + 0.3 = 10.3 = 10 * 1.03.\) Are you with me to here?

Suppose it increases again 3% next year. Well, this year's value is 10.3 so next year's will be

\(\displaystyle 10.3 + (.03 * 10.3) = 10.3 + 0.309 = 10.609.\) Make sense so far?

But there is another way to do this \(\displaystyle 10.3 + (10.3 * .03) = 10.3 * 1.03 = (10 * 1.03) * 1.03 = 10 * 1.03^2.\) Still with me?

How about year 3? That would be \(\displaystyle 10.609 + (10.609 * 0.03) = 10.609 * 1.03 = (10 * 1.03^2) * 1.03 = 10 * 1.03^3.\)

It's exactly the same as the compounding formula with n = 1.

It is in fact the general formula for any system that is increasing at a constant percentage rate each period.

But what if the base is unknown? What if it is w instead of 10? SO WHAT? w is just a number.

At the end of 12 years w will have grown to \(\displaystyle w * 1.03^{12} \approx 1.42576w.\)

And how much bigger than w is 1.42576w? It is 42.576% bigger. The base just drops out if we express the answer as a ratio or a percentage.

Please let me know where I have not been clear.