A Percent Larger?

diaste · Postby **diaste** » Sat Jan 17, 2009 2:31 pm

Hi all! New guy here just so you know.
I'm stuck and I need help.

I'm working a percent larger problem.

I have an original cell. I have to find how much larger the consecutive cells are. The percentage is 6.3. Every succeeding cell is 6.3% larger than the previous.
When I solve for the second cell I use x + .063x = x If I substitute 10 for x the second cell = 10.63. The third cell then is 10.63 + .063(10.63)= 11.29 Yes? No?
But when I consider any two consecutive cells and attempt to find the percent smaller of a previous cell the percent is approx. 5.9%. The book says this is correct but how is it possible?

Now I have 18 cells to consider. The book is telling me that the 18th cell is going to about 2.85 times larger than the original cell. Fine. They ask if (1.06) to the 18th power or (1.06) to the 19th power describes the relative size of the 18th cell to the original. I know the answer is (1.06) to the 18th but why? .063 to the 18th power is no where close.

DAiv · Postby **DAiv** » Sat Jan 17, 2009 6:21 pm

I have an original cell. I have to find how much larger the consecutive cells are. The percentage is 6.3. Every succeeding cell is 6.3% larger than the previous.
When I solve for the second cell I use x + .063x = x If I substitute 10 for x the second cell = 10.63. The third cell then is 10.63 + .063(10.63)= 11.29 Yes? No?

Hi Diaste

That is virtually correct, but if you factor out the $x$ in your equation, you'll get:

$\begin{eqnarray}&x \,+\, 0.063x \\ =& x(1 \,+\, 0.063) \\ =& x(1.063) \\ =& 1.063x\end{eqnarray}$

... which ends up being simpler, and this is where their less precise 1.06 comes from.

To put this into a general formula, we can use the subscripts $_n$ and $_{n+1}$ to differentiate between any given $x$ and the $x$ that follows it:

$x_{\small{n+1}} \,=\, 1.063x_{\small{n}$

So, this is basically saying, 'whatever size cell number $n$ is at any given time, the next cell (cell number ${n+1}$ ) will be 1.063 times larger'.

Now I have 18 cells to consider. The book is telling me that the 18th cell is going to about 2.85 times larger than the original cell. Fine. They ask if (1.06) to the 18th power or (1.06) to the 19th power describes the relative size of the 18th cell to the original. I know the answer is (1.06) to the 18th but why? .063 to the 18th power is no where close.

Well, the original cell, let's call its size $x_{\small{0}}$ , is the size every other cell is relative to.

Since the question is using the less precise value of 1.06, I'll use that value here.

$\begin{array}{lllllllr} \text{The original, } & x_{\small 0} &=& x_{\small 0} &=& x_{\small 0} &=& x_{\small 0} \\ \text{The first copy, } & x_{\small 1} &=& x_{\small 0} \,\times\, 1.06 &=& x_{\small 0} \,\times\, 1.06^1 &=& 1.06\,x_{\small 0} \\ \text{The second copy, } & x_{\small 2} &=& x_{\small 0} \,\times\, 1.06 \,\times\, 1.06 &=& x_{\small 0} \,\times\, 1.06^2 &=& 1.12\,x_{\small 0} \\ \text{The third copy, } & x_{\small 3} &=& x_{\small 0} \,\times\, 1.06 \,\times\, 1.06 \,\times\, 1.06 &=& x_{\small 0} \,\times\, 1.06^3 &=& 1.19\,x_{\small 0} \\ . &&& . && . && . \\ . &&& . && . && . \\ . &&& . && . && . \\ \text{The eighteenth copy, }&x_{\small 18} &=& x_{\small 0} \,\times\, 1.06 \,\times\, 1.06 \,\times\, 1.06 \,\times\, ... \,\times\, 1.06 \,\times\, 1.06 &=& x_{\small 0} \,\times\, 1.06^{18} &=& 2.85\,x_{\small 0} \\ \end{array}$

Therefore, the eighteenth copy will be $1.06^{18}$ = 2.85 times larger than the original cell.

But when I consider any two consecutive cells and attempt to find the percent smaller of a previous cell the percent is approx. 5.9%. The book says this is correct but how is it possible?

Each cell is 5.9% (and not 6.3%) smaller than the subsequent cell because the difference in size between the two cells represents a smaller proportion of the larger cell compared to the smaller one.

It might be easier to see if we use a different example.

Let's say each subsequent cell is 100% larger than the previous one, that is, each subsequent cell is twice as large. Now look at it from the other direction. If each previous cell is half the size of the subsequent one, each previous cell will be 50% smaller. If it were 100% smaller, it would have a size of zero, which clearly can't be right.

Back to our original values, and then rearranging to make $x_n$ the subject:

$\begin{eqnarray} x_{\small n+1} &=& 1.063 x_n \\ \frac{x_{\small n+1}}{1.063} &=& x_n \\ x_{\small n+1} \,\times\, \frac{1}{1.063} &=& x_n \\ 0.941 \, x_{\small n+1} &=& x_n \\ \end{eqnarray}$

So, the smaller cell is 94.1% the size of the larger one, or (100% - 94.1% = 5.9%) smaller.

DAiv

diaste · Postby **diaste** » Sat Jan 17, 2009 10:31 pm

Thanks for the reply. I see the solution now. It was messing me up they arrived at 1.06 because I couldn't make the solutions equate.

Thanks again!

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