Question 33·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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An irrigation system delivers $r$ liters of water in $m$ minutes through $n$ identical nozzles operating at the same time. Each nozzle receives an equal share of the water flow. A nutrient injector adds 3 cubic centimeters of nutrient solution for every 250 cubic centimeters of water delivered through each nozzle. If these rates remain constant, which expression represents the number of cubic centimeters of nutrient solution added through one nozzle in $k$ hours?

Use $1$ liter $= 1{,}000$ cubic centimeters.

A unit rate gives an amount per one unit of time. When a total flow is split equally, divide by the number of nozzles to find the flow through one nozzle. Then match the time units, convert the water volume to cubic centimeters, and apply the nutrient-to-water ratio. Multiplying by the nozzle count would give more than the total flow, not one nozzle’s share.

Hints

Hint 1
A unit rate tells you how much water flows in one minute. Find the system’s liters per minute first, then divide that flow equally among the nozzles. What fraction goes through one nozzle?
Hint 2
The water rate is per minute, but the requested time is in hours. Each hour has $60$ minutes, so convert $k$ hours to minutes before multiplying by the flow rate.
Hint 3
The injector’s ratio compares nutrient volume with water volume, not with the whole mixture. Once the water through one nozzle is in cubic centimeters, multiply it by $\frac{3}{250}$ .

Step-by-step

Follow the flow through one nozzle
Step 1
Find the system’s flow per minute
A unit rate gives the amount for one minute. Divide the $r$ liters by the $m$ minutes to get the whole system’s rate:
$\frac{r}{m}\text{ liters per minute}.$
Step 2
Find one nozzle’s share
Each nozzle gets an equal share, so divide the system’s flow by $n$ , not multiply by $n$ . The rate through one nozzle is:
$\frac{r/m}{n}=\frac{r}{mn}\text{ liters per minute}.$
Step 3
Change hours to minutes
The flow rate is per minute, so the running time must be in minutes. Use $60$ minutes per hour:
$k\text{ hours}\left(\frac{60\text{ minutes}}{1\text{ hour}}\right)=60k\text{ minutes}.$
Step 4
Find the water through one nozzle
Multiply the one-nozzle rate by the time. The minutes cancel, leaving liters of water:
$\frac{r}{mn}\frac{\text{liters}}{\text{minute}}(60k\text{ minutes})=\frac{60rk}{mn}\text{ liters}.$
Step 5
Convert the water volume
The injector’s ratio uses cubic centimeters, so convert the water, using the given $1{,}000$ cubic centimeters per liter:
$\frac{60rk}{mn}\text{ liters}\left(\frac{1{,}000\text{ cubic centimeters}}{1\text{ liter}}\right)=\frac{60{,}000rk}{mn}\text{ cubic centimeters of water}.$
Step 6
Apply the nutrient-to-water ratio
The injector adds $3$ cubic centimeters of nutrient for every $250$ cubic centimeters of water. Multiply the water volume by that ratio:
$\frac{60{,}000rk}{mn}\cdot\frac{3}{250}\text{ cubic centimeters of nutrient solution}.$
Step 7
Calculate the numerical factor
Type $60{,}000\cdot\frac{3}{250}$ in Desmos. It shows $720$ , the number multiplying $\frac{rk}{mn}$ after the nutrient ratio is applied.
Step 8
State the amount requested
Put that factor with $\frac{rk}{mn}$ . The nutrient solution added through one nozzle in $k$ hours is $\frac{720rk}{mn}$ cubic centimeters. Choice D.

Question 33·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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