Question 106·200 Super-Hard SAT Math Questions·Advanced Math

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The equation below relates distinct positive real numbers $p$ , $q$ , and $r$ . Which choice correctly expresses $r$ in terms of $p$ and $q$ ?

p = 6q \cdot \sqrt[3]{\left(\frac{r}{25}\right)^{-4}}

A nonlinear formula with the target letter inside a root calls for reversing the operations around that letter. Rewrite the root as a fractional power, then flip the fraction for a negative exponent. A Desmos regression with allowed sample values can rule out choices, but algebra proves the formula works for every allowed value. Keep track of which ratio you flip.

Hints

Hint 1
For a useful numerical check, choose positive, distinct values of $p$ and $q$ with $p\ne6q$ . If $p=6q$ , several choices become the same number, so that sample won't separate them.
Hint 2
A cube root means a power of $\frac13$ . Apply it to the $-4$ exponent, then use the negative exponent to turn $r/25$ into its reciprocal.
Hint 3
To undo a power of $\frac43$ , raise both positive sides to its reciprocal exponent, $\frac34$ . What happens to the fraction containing $r$ when you then take reciprocals?

Step-by-step

Check a sample, then solve the formula
Step 1
Find a numerical value of $r$
For a useful sample, choose $p=12$ and $q=1$ . These are positive and distinct, and $p\ne6q$ . Type those values and the given equation, replacing $r$ with $r_1$ and $=$ with $\sim$ . The $\sim$ asks Desmos to fit $r_1$ ; $\{r_1>0\}$ keeps it positive. Under PARAMETERS, Desmos reports $r_1\approx14.8651$ , which is distinct from $12$ and $1$ .
Step 2
Rule out formulas that fail the sample
Type $25(p/(6q))^{3/4}$ and $25(6q/p)^{4/3}$ on separate lines. Desmos prints about $42.0448$ and $9.9213$ . Neither equals the required $r_1\approx14.8651$ , so this sample rules them out. A sample can't prove that a remaining formula works for every positive $p$ and $q$ , so finish with algebra.
Step 3
Rewrite the cube root as a power
A cube root applies a $\frac13$ power, so the exponent on $r/25$ becomes $-4\cdot\frac13=-\frac43$ . Rewrite the equation:
$p=6q\left(\frac{r}{25}\right)^{-4/3}.$
Step 4
Use the negative exponent
A negative exponent means a reciprocal, not a negative result. Flip $r/25$ :
$p=6q\left(\frac{25}{r}\right)^{4/3}.$
Step 5
Get the power alone
Because $q$ is positive, $6q$ isn't zero. Divide both sides by $6q$ :
$\frac{p}{6q}=\left(\frac{25}{r}\right)^{4/3}.$
Step 6
Undo the $\frac43$ power
Undo a power by raising both sides to its reciprocal exponent. Here $\frac43\cdot\frac34=1$ , and both sides are positive, so raise them to $\frac34$ :
$\left(\frac{p}{6q}\right)^{3/4}=\frac{25}{r}.$
Step 7
Move $r$ to the numerator
Take reciprocals of both sides. The left fraction flips inside its $\frac34$ power, while $25/r$ becomes $r/25$ :
$\left(\frac{6q}{p}\right)^{3/4}=\frac{r}{25}.$
Step 8
Express $r$ in terms of $p$ and $q$
Multiply by $25$ to get the requested formula: $r=25\left(\frac{6q}{p}\right)^{3/4}$ . This expresses $r$ using only $p$ and $q$ . Choice B.

Question 106·200 Super-Hard SAT Math Questions·Advanced Math

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