Question 157·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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96 is $p$ % of $\frac{2}{3}p$ , and 80 is $(p-20)$ % of $\frac{2}{3}p$ , where $p>20$ . What is the value of $p$ ?

These are percent part-whole relationships: each amount is a percent of a base. The cue is that both statements use the same base, even though their percents differ. Write both relationships, divide to cancel the base, and use a Desmos regression to find the unknown percent. Keep the subtraction inside $(p-20)\%$ ; it changes the percent, not the base.

Hints

Hint 1
A percent means a fraction out of $100$ . In the first statement, $p$ is the percent, while $\frac{2}{3}p$ is the base, the amount after “of.” What multiplication gives $96$ ?
Hint 2
The second statement uses the same base, but its percent is $p-20$ . That entire expression must be divided by $100$ before you multiply by the base. How does this give $80$ ?
Hint 3
Since $p>20$ , the base isn't zero. Divide the two equations to cancel that base and the $100$ in each percent. What equation remains for $p$ ?

Step-by-step

Cancel the shared percent base
Step 1
Translate the first percent statement
The base is the amount after “of,” so $96$ is $p$ hundredths of $\frac{2}{3}p$ :
$96=\frac{p}{100}\cdot\frac{2p}{3}.$
Step 2
Translate the second percent statement
Use that same base for $80$ . The entire percent is $(p-20)\%$ , so divide $p-20$ by $100$ before multiplying:
$80=\frac{p-20}{100}\cdot\frac{2p}{3}.$
Step 3
Cancel what the equations share
Since $p>20$ , the shared base $\frac{2p}{3}$ isn't zero. Divide amounts that use the same percent base to cancel that base. Divide the first equation by the second:
$\frac{96}{80}=\frac{\frac{p}{100}\cdot\frac{2p}{3}}{\frac{p-20}{100}\cdot\frac{2p}{3}}.$
Cancel $\frac{2p}{3}$ and $\frac{1}{100}$ from the right side:
$\frac{96}{80}=\frac{p}{p-20}.$
Step 4
Solve for the percent number
Type $\frac{96}{80}\sim\frac{p_1}{p_1-20}\left\{p_1>20\right\}$ into Desmos. Use $p_1$ for the unknown and $\sim$ for a regression, which finds a value that makes the two sides match; the braces keep the given restriction. Under PARAMETERS, Desmos shows $p_1=120$ . So the value of $p$ is $120$ . Choice B.

Question 157·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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