Sunday, July 14, 2013

# How to solve problem with multiple percentage changes on th GRE?

2:31 AM
If a quantity is increased or decreased more than once, you cannot simply add or subtract the percentages.You have to work out each increase or decrease step by step.
Let’s see an example where this important concept is used.

Example:
The population of a town increased by 20% between 1989 and 1995. If the population increased again by 10% from 1995 till 2003, what is the percentage increase in the population between 1989 and 2003?
A) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} 2\%$
B) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} 20\%$
C) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} 30\%$
D) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} 32\%$
E) $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} 40\%$
Solution:

Step 1: This is a two-step percentage change problem. Remember if a measure is increased or decreased more than once, you cannot simply add or subtract the percentages. You have to work out each increase or decrease step by step.
Even though 30% is a very tempting answer, it’s wrong!
Let’s say the original population in 1989 is $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} x$ . Sure we don’t know $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} x$ . However the question wants us to know the percentage increase and therefore it will eventually get canceled.
Since the population increased by 20% between 1989 and 1995, the population at the end of 1995 is$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{20}{100}\times x + x =1.2x$
Since the population increased by 10% between 1995 and 2003, the population at the end of 2003 will be$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{10}{100}\times 1.2x +1.2x =1.32x$ (remember you now need to use the population in 1995 as the original value i.e. $\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \dpi{120} 1.2x$ )
Step 2: % Change in population between 1989 and 2003 =
$\usepackage{color} \definecolor{Myblue}{rgb}{0.27,0.38,0.5} \color{Myblue} \frac{1.32x-x}{x}\times 100=32\%$ .
The correct answer is D

In order to solve percentage change problems quickly, it’s a good idea to remember some common percentage change values. The following tables show Percentage Increases and Percentage decreases for some commonly used numbers.
Original value 100: Percentage change - Final values
 Percentage Change 10 20 30 40 50 60 70 80 90 100 Increase - Final value 110 120 130 140 150 160 170 180 190 200 Decrease - Final value 90 80 70 60 50 40 30 20 10 0
Original value 50: Percentage change - Final values
 Percentage Change 10 20 30 40 50 60 70 80 90 100 Increase - Final value 55 60 65 70 75 80 85 90 95 100 Decrease - Final value 45 40 35 30 25 20 15 10 5 0