Pipe A delivers water at a constant rate of cubic feet per second, but only of the water that leaves the pipe actually reaches the field. A farmer wants to cover the entire rectangular field, which is feet long and feet wide, with a 1-inch depth of water. How many hours must the irrigation system run to meet this goal? (Round your answer to the nearest tenth.) .

Solution available with step-by-step explanation

SAT Ratios, Rates & Proportions Question 18 | Free SAT Prep | Aniko

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Question 18·Hard·Ratios, Rates, Proportional Relationships, and Units

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Pipe A delivers water at a constant rate of $2.5$ cubic feet per second, but only $92\%$ of the water that leaves the pipe actually reaches the field. A farmer wants to cover the entire rectangular field, which is $520$ feet long and $330$ feet wide, with a 1-inch depth of water.

How many hours must the irrigation system run to meet this goal? (Round your answer to the nearest tenth.)

$(1\text{ acre}=43,560\text{ square feet})$ .

For rate-and-volume problems like this, first make all units consistent (here, convert inches to feet), then find the needed volume by multiplying area by depth. Next, adjust the given flow rate for any efficiency or percentage loss so you have the actual rate at which the target (the field) receives water. Divide volume by this effective rate to get time in seconds, and finally convert to the units asked for (often hours) using the correct conversion factor (3,600 seconds per hour). Keeping units labeled at each step is the fastest way to avoid common mistakes.

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Think about volume, not just area

Start by finding the area of the rectangular field using its length and width, then remember that you also need to account for the depth of water to get a volume.

Watch your units on the depth

The field dimensions are in feet, but the water depth is given in inches. How many feet is 1 inch, and how does that factor into the volume calculation?

Incorporate the 92% efficiency

Only 92% of the water that leaves the pipe reaches the field. How does this affect the rate at which water actually reaches the field compared with the given 2.5 cubic feet per second?

Convert time correctly

Once you have the time in seconds, remember to convert it into hours using the fact that there are 3,600 seconds in an hour before comparing to the answer choices.

Desmos Guide

Use one expression to compute the time in hours

In Desmos, type the single expression

\frac{520 \times 330 \times (1/12)}{0.92 \times 2.5 \times 3600}

This expression represents (field area) × (depth in feet) ÷ (effective flow rate to the field in cubic feet per second) ÷ (seconds per hour). The numerical result that Desmos displays is the required running time in hours; compare that decimal to the answer choices.

Step-by-step Explanation

Find the volume of water the field must receive

First compute the area of the rectangular field in square feet:

\text{Area} = 520 \times 330 = 171{,}600 \text{ ft}^2

The depth of water is 1 inch. Convert inches to feet:

1 \text{ inch} = \frac{1}{12} \text{ foot}

Now multiply area by depth to get the volume of water that must reach the field:

V_{\text{field}} = 171{,}600 \times \frac{1}{12} = 14{,}300 \text{ ft}^3

Account for the 92% efficiency and find the time in seconds

The pipe delivers $2.5$ cubic feet per second, but only $92\%$ of that reaches the field.

So the effective rate to the field is

\text{Rate}_{\text{field}} = 0.92 \times 2.5 = 2.3 \text{ ft}^3/\text{s}

Time in seconds is volume divided by rate:

t_{\text{seconds}} = \frac{V_{\text{field}}}{\text{Rate}_{\text{field}}} = \frac{14{,}300}{2.3} \approx 6{,}217 \text{ s}

Convert seconds to hours and round

There are $3{,}600$ seconds in 1 hour, so convert the time:

t_{\text{hours}} = \frac{6{,}217}{3{,}600} \approx 1.7 \text{ hours}

Rounded to the nearest tenth, the irrigation system must run for 1.7 hours, which corresponds to choice C).

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation