Price of Pencils

Date: 10/07/2002 at 10:06:13
From: George
Subject: Word problems

The price of pencils has skyrocketed in recent years. Each year for 
the last 7 years the price has increased, and the new price is the sum 
of the prices for the two previous years. Last year a pencil cost 
60 cents. How much does a pencil cost today? How much did a pencil 
cost 7 years ago?

I am trying to help my daughter on this math problem but am having no 
luck. Can you help? My daughter is 11 years old and in 6th grade.

Regards, George

Date: 10/10/2002 at 10:26:15
From: Doctor Peterson
Subject: Re: Word problems

Hi, George.

This is an interesting problem! At first I thought there couldn't be 
enough information, but then I realized that the fact that the price 
increased every year was the key.

The problem involves a Fibonacci sequence, working backward; the most 
orderly way I can see to do it is to use simple algebra. The main idea 
is that, since in each year the price is the sum of the prices of the 
two previous years, the increase in price is always the price the year 
before. For example, if the price started at 5, and the previous year 
it was 2, then the next year it will be 5+2=7, and the following year 
it will be 7+5=12. Working backward, each year's price is the 
difference between the next two years: 12-7=5.

Let's work backward, starting with the unknown price this year, which 
we'll call x. Last year it was 60, so the previous year it must have 
been x-60 (so that the sum of x-60 and 60 gives the next price, x). 
Continuing this process backwards gives us a sequence of expressions:

    x
    60
    x-60
    120-x
    2x-180
    300-3x
    5x-480
    780-8x
    13x-1260

I've continued this to the eighth year back, since that price is the 
increase from the seventh year.

Now we use the key: all of these increases must be positive. That 
gives us a sequence of inequalities, each of which can be solved to 
find a range for x:

    x > 0
    60 > 0
    x-60 > 0        x > 60
    120-x > 0                   x < 120
    2x-180 > 0      x > 90
    300-3x > 0                  x < 100
    5x-480 > 0      x > 96
    780-8x > 0                  x < 97.5
    13x-1260 > 0    x > 96.92

Notice that alternate inequalities give lower and upper bounds on what 
x can be, getting tighter and tighter; by the end, we see that

    96.92 < x < 97.5

and the only whole number value x can have is 97. Now, plugging this 
into the expressions for the prices (or just working backward 
numerically by subtracting, which is easier), we get

    x        = 97
    60       = 60
    x-60     = 37
    120-x    = 23
    2x-180   = 14
    300-3x   = 9
    5x-480   = 5
    780-8x   = 4
    13x-1260 = 1

Seven years ago, the price was 4 cents, and it has increased every 
year since, following the indicated pattern.

Of course, if you just have an answer, you can check it by working 
forward and seeing if the results are correct. And I suspect that the 
method an 11-year-old is expected to use is more like trial and error: 
try setting the original price at 1, 2, 3, 4 and trying various small 
increments for the next year until you find one that works. Basically, 
this is just a long addition problem.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/