# Percent Greater than Problem?

• Jul 27th 2012, 12:16 PM
Sewilliams13
Percent Greater than Problem?
If q is 10% greater than p and r is 10% greater than y, qr is what percent greater than py?
I set up the following equations, Q=10/100(p)+p
R=10/100(y)+y
QR=x/100(py)+py
How do I solve them? I would appreciate any response.
• Jul 27th 2012, 03:22 PM
Soroban
Re: Percent Greater than Problem?
Hello, Sewilliams13!

Quote:

$\text{If }q\text{ is 10\% greater than }p\text{, and }r\text{ is 10\% greater than }y,$
$qr\text{ is what percent greater than }py\,?$

I prefer simpler fractions.

$[1]\;q \:=\:p + \tfrac{1}{10}p \quad\Rightarrow\quad q \:=\:\tfrac{11}{10}p$

$[2]\;r \:=\:y + \tfrac{1}{10}y \quad\Rightarrow\quad r \:=\:\tfrac{11}{10}y$

Multiply [1] and [2]: . $qr \:=\:\left(\tfrac{11}{10}\right)^2py \;=\;\tfrac{121}{100}py \;=\;\left(1 + \tfrac{21}{100}\right)py$

. . . . . . . . . . . . . . . . $qr \;=\;py + \tfrac{21}{100}py$

$\text{Therefore, }qr\text{ is 21\% greater than }py.$

• Jul 27th 2012, 05:14 PM
Sewilliams13
Re: Percent Greater than Problem?
In [1] & [2] why did you take a way the p and the y in the next equation
• Jul 27th 2012, 07:32 PM
Wilmer
Re: Percent Greater than Problem?
They were not "taken away"; they were simply added together; like, 1 + 1/10 = 10/10 + 1/10 = 11/10
• Jul 30th 2012, 12:47 PM
Sewilliams13
Re: Percent Greater than Problem?
-_-Ahh...Thank you so much!