# Percentage change question

• May 7th 2010, 06:45 AM
clovis
Percentage change question
I'll just set out the numbers first then say what I would like to know.

351 195 65 39 650
54% 30% 10% 6% 100%

would there be a way to figure out the smallest possible whole number values for A-D that would need to be added to make the numbers represent these percentages.

A+351 B+195 C+65 D+39 A+B+C+D+650
35% 32% 22% 11% 100%

please note I'm not a very mathy person so intricate explanations will be wasted on me :P. I'm really just concerned if its possible and how simple/difficult it is to do.

Thanks.
• May 8th 2010, 03:54 AM
sa-ri-ga-ma
Quote:

Originally Posted by clovis
I'll just set out the numbers first then say what I would like to know.

351 195 65 39 650
54% 30% 10% 6% 100%

would there be a way to figure out the smallest possible whole number values for A-D that would need to be added to make the numbers represent these percentages.

A+351 B+195 C+65 D+39 A+B+C+D+650
35% 32% 22% 11% 100%

please note I'm not a very mathy person so intricate explanations will be wasted on me :P. I'm really just concerned if its possible and how simple/difficult it is to do.

Thanks.

$\displaystyle \frac{x}{351} = 0.54$

x= 190.

$\displaystyle \frac{190}{A+352} = 0.35$
Now solve for A.
Repeat the same thing for other numbers.
• May 8th 2010, 07:27 AM
Soroban
Hello, clovis!

If you are "not a very mathy person", you'll hate this approach.
Perhaps others will be amused by the Herculean task I've suggested.

Quote:

I'll just set out the numbers first, then say what I would like to know.

. . $\displaystyle \begin{array}{|c|c|c|c||c|}351 & 195 & 65 & 39 & 650 \\ \hline 54\% & 30\% & 10\% & 6\% & 100\% \end{array}$

Would there be a way to figure out the smallest possible whole number values for $\displaystyle A\text{ - }D$
that would need to be added to make the numbers represent these percentages.

. . $\displaystyle \begin{array}{|c|c|c|c||c|}A+351 & B+195 & C+65 & D+39 & A+B+C+D+650 \\ \hline 35\% & 32\% & 22\% & 11\% & 100\% \end{array}$

From the first column, we have: .$\displaystyle \frac{A+351}{A+B+C+D+650} \:=\:0.35$

. . $\displaystyle A+351 \:=\:0.35(A+B+C+D+650) \quad\Rightarrow$ . $\displaystyle A+351 \:=\:0.35A + 0.35B + 0.35C + 0.35D + 227.5$

. . $\displaystyle -0.65A + 0.35B + 0.35C + 0.35D \:=\:123.5 \quad\Rightarrow\quad -13A + 7B + 7C + 7D \:=\:2470$ .[1]

Second column: .$\displaystyle \frac{B+195}{A+B+C+D+650} \:=\:0.32 \quad\Rightarrow\quad 8A - 17B + 8C + 8D \:=\:-325$ .[2]

Third column: .$\displaystyle \frac{C+65}{A+B+C+D+650} \:=\:0.22 \quad\Rightarrow\quad 11A + 11B - 39C + 11D \:=\:-3900$ .[3]

Fourth column: .$\displaystyle \frac{D+39}{A+B+C+D+650} \:=\:0.11 \quad\Rightarrow\quad 11A + 11B + 11C - 89D \:=\:-3250$ .[4]

Now all you have to do is solve the system of equations.

I'll wait in the car . . .
.