GRE Percent Change Problems
Definitions and Symbols
Let $x_i$ and $x_f$ denote the initial and final values of $x,$ respectively, with $x_i\neq0.$
The
It follows that \begin{equation} \label{eq3} x_f=x_i+\Delta x. \end{equation}
If we divide both sides of equation (\ref{eq3}) by $x_i$ we obtain \begin{equation} \label{eq16} \frac{x_f}{x_i}=1+\frac{\Delta x}{x_i} \end{equation}
We define the
Substituting this symbol into equation (\ref{eq16}) yields the following equivalent forms:
\begin{equation} \label{eq15} \frac{x_f}{x_i}=1+{}_x\Delta\% \end{equation}
\begin{equation} \label{eq14} 1+{}_x\Delta\%=\frac{x_f}{x_i} \end{equation}
\begin{equation} \label{eq4} {}_x\Delta\%=\frac{x_f}{x_i}-1 \end{equation}
If ${}_x\Delta\%\gt0,$ then ${}_x\Delta\%$
is said to be a
When the change in only one variable $x$ is considered, it is conventional to write
\begin{equation} \label{eq5} \Delta\%={}_x\Delta\%. \end{equation}
Percent Change and the GRE
Problems on the GRE can involve the percent change of up to three variables $x, y$ and $z$ in the same problem. To study these, it will be useful to let $\alpha=a\%=\frac{a}{100}={}_x\Delta\%$ and $\beta=b\%=\frac{b}{100}={}_y\Delta\%$ and $\gamma=c\%=\frac{c}{100}={}_z\Delta\%,$ so that, $a=\alpha\times100$ and $b=\beta\times100$ and $c=\gamma\times100.$ With this done, we can simplify our language and discuss the problems only in terms of $a,b,c$ and $\alpha,\beta,\gamma$ without reference to the underlying variables $x,$ $x_i,$ $x_f,$ $\Delta x,$ ${}_x\Delta\%,$ $y,$ $y_i,$ $y_f,$ $\Delta y,$ ${}_y\Delta\%,$ $z,$ $z_i,$ $z_f,$ $\Delta z,$ ${}_z\Delta\%.$ From now on, this shall be done. Also, and unless stated otherwise, we assume that $a\gt0$ and $b\gt0.$
The Problems
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To increase a number by $a\%,$ multiply it by $1+\alpha.$
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To decrease a number by $a\%,$ multiply it by $1−\alpha.$
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If an $a\%$ increase is followed by a $b\%$ increase, then the overall $c\%$ increase is given by
\begin{equation} \label{eq6} \gamma=(1+\alpha)(1+\beta)−1. \end{equation}
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If a final amount is the result of increasing an initial amount by $a\%$, divide the final amount by $1+\alpha$ to find the initial amount. That is, if $A_2=(1+\alpha)A_1,$ then
\begin{equation} \label{eq7} A_1=\frac{A_2}{(1+α)} \end{equation}
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If a final amount results from decreasing an initial amount by $a\%,$ divide the final amount by $1−\alpha$ to find the initial amount. That is, if $A_2=(1−\alpha)A_1,$ then
\begin{equation} \label{eq8} A_1=\frac{A_2}{(1−α)} \end{equation}
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If $A_1\lt A_2,$ then the $a\%$ increase from $A_1$ to $A_2$ is greater than the $b\%$ decrease from $A_2$ to $A_1.$ That is, if $A_2=(1+\alpha)A_1$ and $A_1=(1−\beta)A_2,$ then $\alpha\gt\beta$ and $a\gt b.$
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An $a\%$ decrease followed by a $b\%$ decrease is smaller than a single decrease of $(a+b)\%.$ In particular, the difference is $\alpha\beta.$ That is,
\begin{equation} \label{eq9} (1−\alpha)(1−\beta)\lt(1−\alpha−\beta). \end{equation}
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An $a\%$ increase followed by a $b\%$ increase is larger than a single increase of $(a+b)\%.$ In particular, the difference is $\alpha\beta.$ That is,
\begin{equation} \label{eq10} (1+\alpha)(1+\beta)\gt(1+\alpha+\beta). \end{equation}
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Corollary. An increase (or decrease) of $a\%$ followed by another increase (decrease) of $a\%$ is larger (smaller) than a single increase (decrease) of $2a\%.$ In particular, the difference is $\alpha^2.$ That is,
\begin{equation} \label{eq11} (1+\alpha)^2\gt(1+2\alpha). \end{equation}
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Corollary. An $a\%$ increase followed by an $a\%$ decrease is less than the initial value. In particular, the difference is $\alpha^2.$ That is,
\begin{equation} \label{eq12} (1+\alpha)(1−\alpha)=1−\alpha^2\lt1 \end{equation}
whereas
\begin{equation} \label{eq13} (1+\alpha−\alpha)=1. \end{equation}
Sources
See p. 149 of Princeton Review.