Question 81·200 Super-Hard SAT Math Questions·Algebra

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A landscaper is spreading 34 pounds of two types of seed on a field. The recommended application rate for grass seed is 1.25 pounds per 16 square yards. The recommended application rate for clover seed is 13.6 ounces per 23 square feet.

(Use the conversions $1\text{ pound} = 16\text{ ounces}$ and $1\text{ square yard} = 9\text{ square feet}$ .)

Let $g$ be the number of square yards planted with grass seed, and let $c$ be the number of square yards planted with clover seed. If all 34 pounds of seed are used, which equation describes all possible values of $g$ and $c$ ?

Different application rates mean you need a unit rate for each seed: pounds per square yard. Convert both rates to those units, multiply each by its planted area, and add the weights to match the total. Desmos can handle the fraction arithmetic, but it can't catch a missing square-foot conversion; check the units before calculating.

Hints

Hint 1
A coefficient, the number multiplying an area, needs to be in pounds per square yard. For grass, divide the given pounds by the given square yards. What happens to the units if you flip that fraction?
Hint 2
Clover's rate is per square foot, but $c$ counts square yards. One square yard contains $9$ square feet, so how many times the per-square-foot amount is needed for one square yard?
Hint 3
After changing clover's ounces to pounds, each rate can be multiplied by its own planted area. The resulting two seed weights must add up to the weight of all the seed used.

Step-by-step

Convert both rates, then add
Step 1
Find the grass rate per square yard
The grass rate is already measured in pounds per square yard. Divide the pounds by the square yards to find the unit rate for one square yard. Type $1.25/16$ in Desmos; it shows $0.078125$ , or $\frac{5}{64}$ using the fraction button. So grass uses $\frac{5}{64}$ pound per square yard.
Step 2
Change the clover area unit
Clover's rate has square feet on the bottom, but $c$ counts square yards. A conversion factor equals $1$ because its top and bottom describe the same area. Put square feet on top so they cancel:
$\frac{13.6\text{ ounces}}{23\text{ square feet}}\left(\frac{9\text{ square feet}}{1\text{ square yard}}\right).$
Using $1/9$ instead would fail to cancel square feet.
Step 3
Change clover's ounces to pounds
The total is in pounds, so clover's ounces must become pounds too. Put ounces on the bottom to cancel them:
$\frac{13.6\text{ ounces}}{23\text{ square feet}}\left(\frac{9\text{ square feet}}{1\text{ square yard}}\right)\left(\frac{1\text{ pound}}{16\text{ ounces}}\right)=\frac{13.6(9)}{23(16)}\frac{\text{pounds}}{\text{square yard}}.$
Leaving off this factor leaves the clover rate in ounces, not pounds.
Step 4
Calculate the clover rate
Type $13.6(9)/(23(16))$ on the next Desmos line. It shows about $0.3326086957$ ; the fraction button shows $\frac{153}{460}$ . This is pounds of clover seed per square yard, not the weight for all $c$ square yards.
Step 5
Write the total-weight equation
Each term is a seed weight: pounds per square yard times square yards gives pounds. Add the grass and clover weights because all $34$ pounds are used:
$\frac{5}{64}g+\frac{153}{460}c=34.$
This equation describes the possible nonnegative planted areas. Choice C.

Question 81·200 Super-Hard SAT Math Questions·Algebra

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