Super-Hard SAT Math Question 160 | Free SAT Prep | Aniko

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Question 160·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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An online service increases its monthly fee by $p\%$ . (⁠А‍nіkο.‍aі)

The next month, it decreases the new fee by a percent that is 20 percentage points less than the percent increase.

After both changes, the fee is 12% greater than the original fee.

Which choice is $p$ ?

Translate each percent change into a multiplier and multiply them in the order they occur. Pay special attention to “percentage points,” which means you subtract the numbers first (to get $(p-20)\%$ ) and only then convert to a decimal. After setting the product equal to the final multiplier (here $1.12$ ), solve the resulting equation and choose the meaningful (positive) solution. Prорerty оf An⁠i⁠кo.аі

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Use multipliers, not subtraction

If a fee increases by $p\%$ , multiply by $1+\frac{p}{100}$ (don’t just add $p$ ).

Be careful with “20 percentage points less”

A decrease that is 20 percentage points less than $p\%$ means the decrease is $(p-20)\%$ , not $0.8p\%$ . Рr⁠ореrtу оf⁠ ‍Аn‍і‍кo.aі

Turn the final statement into an equation

“12% greater than the original” means the final fee equals $1.12$ times the original fee.

Desmos Guide

Enter the two expressions to compare

Graph

$y=(1+x/100)(1-(x-20)/100)$
$y=1.12$

Find the intersection

Click the intersection point(s) of the graphs and note the positive $x$ -value (ignore any negative percent increase). Writ‍tеn b‌у Anі⁠ko

Interpret the $x$ -value

The positive intersection’s $x$ -coordinate represents the percent increase $p$ .

Step-by-step Explanation

Write each percent change as a multiplier

Let the original fee be $F$ .

After an increase of $p\%$ , the fee becomes

F\left(1+\frac{p}{100}\right).

The next month it is decreased by $(p-20)\%$ , so multiply by © a‍niko.aі

1-\frac{p-20}{100}.

Set up the equation using the final information

After both changes, the fee is 12% greater than the original, so it equals $1.12F$ :

F\left(1+\frac{p}{100}\right)\left(1-\frac{p-20}{100}\right)=1.12F.

Divide both sides by $F$ (since $F\neq 0$ ).

Rewrite with a single variable

Let $r=\frac{p}{100}$ . Then

\left(1+r\right)\left(1-(r-0.2)\right)=1.12.

Since $1-(r-0.2)=1.2-r$ , the equation becomes

(1+r)(1.2-r)=1.12.

Solve the quadratic

Expand:

\begin{aligned} (1+r)(1.2-r) &= 1.2 - r + 1.2r - r^2 \\ &= 1.2 + 0.2r - r^2. \end{aligned}

Set equal to $1.12$ and rearrange:

\begin{aligned} 1.2 + 0.2r - r^2 &= 1.12 \\ -r^2 + 0.2r + 0.08 &= 0 \\ r^2 - 0.2r - 0.08 &= 0. \end{aligned}

Using the quadratic formula,

\begin{aligned} r &= \frac{0.2\pm\sqrt{(-0.2)^2-4(1)(-0.08)}}{2} \\ &= \frac{0.2\pm\sqrt{0.04+0.32}}{2} \\ &= \frac{0.2\pm 0.6}{2}. \end{aligned}

So $r=0.4$ or $r=-0.2$ .

Convert back to $p$

Since $r=\frac{p}{100}$ and $p$ is a percent increase, use the positive solution $r=0.4$ :

p=100(0.4)=40.

Therefore, $p$ is $40\%$ .

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Step-by-step Explanation

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