Question 38·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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In a lab, the concentrations of three solutions $D$ , $E$ , and $F$ are $d$ parts per million, $e$ parts per million, and $f$ parts per million, respectively. If the concentration of solution $D$ is 125% of the concentration of solution $F$ , and the concentration of solution $F$ is 0.006% of the concentration of solution $E$ , which expression represents the value of $d+e$ in terms of $f$ ?

A percent-of relationship takes its percent from the amount after “of.” When every answer is a multiple of one variable, set that variable to $1$ to find the multiplier. Evaluate the first percent in Desmos, then use a regression to work backward through the second. To reverse a percent, divide by it rather than multiply by it. Watch the percent sign on a tiny percent.

Hints

Hint 1
In a percent-of relationship, the amount after “of” is the base, the amount being multiplied. In “D is 125% of F,” which concentration is multiplied by $125\%$ ?
Hint 2
Every answer is a multiple of $f$ , so set $f=1$ to find that multiple. Translate “F is 0.006% of E” into an equation. Which concentration is the unknown amount after “of”?
Hint 3
A percent means “per hundred,” so $0.006\%$ is $0.00006$ , not $0.006$ . Since $f$ is such a small fraction of $e$ , work backward to find $e$ before adding the concentrations.

Step-by-step

Find the multiplier with Desmos
Step 1
Translate D's percent relationship
“D is 125% of F” means $d=125\%\text{ of }f$ .
Step 2
Find D's multiplier
Every answer is a multiple of $f$ , so set $f=1$ to find that multiplier. Type $125\%\operatorname{of}1$ ; Desmos shows $1.25$ . So $d=1.25f$ .
Step 3
Translate F's percent relationship
“F is 0.006% of E” means $f=0.006\%\text{ of }e$ . The unknown $e$ is the amount the percent is taken of, not the result of taking the percent.
Step 4
Work backward to E
With $f=1$ , type $0.006\%\operatorname{of}x_1\sim1$ , where $x_1$ stands for $e$ . The $\sim$ tells Desmos to find the value that makes the relationship true. Under PARAMETERS, it reports $x_1\approx16{,}666.6667$ . That's E's concentration when $f=1$ , not the requested sum.
Step 5
Read E's multiplier as a fraction
Type $x_1$ on its own line and click the fraction button beside its value. Desmos shows $\frac{50{,}000}{3}$ . Since you found this with $f=1$ , the general relationship is $e=\frac{50{,}000}{3}f$ . E being much larger than F fits: F is only $0.006\%$ of E.
Step 6
Add the two concentrations
Type $1.25+x_1$ to add the multipliers you found. Desmos shows about $16{,}667.9167$ ; click its fraction button to see $\frac{200{,}015}{12}$ . Multiply that by $f$ to restore the general case. The requested concentration sum is $d+e=\frac{200{,}015}{12}f$ parts per million. Choice A.

Question 38·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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