Question 53·200 Super-Hard SAT Math Questions·Advanced Math

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For a certain insulation material, heat flow rate $H$ , in watts; surface area $A$ , in square meters; temperature difference $D$ , in degrees Celsius; and thermal resistance $R$ are related by

\frac{A^2D}{H}=\sqrt[3]{H^2R}

where $A$ , $D$ , $H$ , and $R$ are positive. When $R=48$ , which equation correctly expresses $H$ in terms of $A$ and $D$ ?

A formula with the target variable in both a fraction and a root calls for rearranging a nonlinear formula. With multiple-choice answers, you can assign allowed values to the other letters, use a Desmos regression to find the target, and reject any formula that disagrees. Then check the algebra: clearing the fraction combines powers of the target, so the root in the answer may differ from the root in the question.

Hints

Hint 1
A correct formula must work for every positive $A$ and $D$ . Pick one allowed pair, such as $A=2$ and $D=3$ . Why is a different result at that pair enough to reject a formula?
Hint 2
A regression asks Desmos to find a value that balances an equation. Replace $H$ with $H_1$ and $=$ with $\sim$ , then restrict $H_1$ to positive values, as the question requires.
Hint 3
Use the value from the original equation as a benchmark. Evaluate the proposed formulas at the same $A$ and $D$ . A mismatch rules one out; a match is worth confirming with algebra.

Step-by-step

Approach 1: Test the formulas with Desmos
Step 1
Find a benchmark heat flow
Choose $A=2$ and $D=3$ because both are positive, so a correct formula must work for them. Set $R=48$ in Desmos. In the given equation, change $H$ to $H_1$ and $=$ to $\sim$ so Desmos solves for $H_1$ . Add $\{H_1>0\}$ to keep the value the question allows. Under PARAMETERS, Desmos reports $H_1\approx2.048$ watts.
Step 2
Rule out formulas that miss the benchmark
Type $T=A^6D^3$ as shorthand; Desmos gives $T=1728$ . Enter the first three proposed expressions using $T$ . Desmos prints about $0.488$ , $3.302$ , and $0.435$ , respectively. None equals the $2.048$ watts from the given equation, so none works for every allowed $A$ and $D$ .
Step 3
Check the remaining formula
Type the remaining expression, $\sqrt[5]{T/48}$ . Desmos prints about $2.048$ , matching the original equation at the same input. The other three formulas failed this allowed input, so the equation for heat flow is $H=\sqrt[5]{\frac{A^6D^3}{48}}$ . Choice D.

Approach 2: See why the fifth root appears
Step 1
Substitute the given resistance
The question sets $R=48$ , so replace $R$ in the formula:
$\frac{A^2D}{H}=\sqrt[3]{48H^2}.$
Step 2
Remove the cube root
Cube both whole sides. Cubing undoes a cube root:
$\left(\frac{A^2D}{H}\right)^3=48H^2.$
Step 3
Apply the cube to the fraction
The cube applies to every factor on the left. Raising a power to a power multiplies its exponents:
$\frac{A^6D^3}{H^3}=48H^2.$
Step 4
Clear the denominator
Multiply both sides by $H^3$ ; this is allowed because $H$ is positive:
$A^6D^3=48H^2H^3.$
Step 5
Combine the powers of heat flow
When $H^3$ joins $H^2$ , add their exponents to get $H^5$ . That's the same-base exponent rule: $A^6D^3=48H^{2+3}=48H^5$ . The fifth power is why keeping the original cube root would fail.
Step 6
Isolate the fifth power
Divide both sides by the factor $48$ :
$H^5=\frac{A^6D^3}{48}.$
Step 7
Solve for heat flow
Take the fifth root of both sides to undo the fifth power: $H=\sqrt[5]{\frac{A^6D^3}{48}}$ . This expresses the positive heat flow $H$ in terms of $A$ and $D$ . Choice D.

Question 53·200 Super-Hard SAT Math Questions·Advanced Math

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