SAT Linear Equations in One Variable Question 9 | Free SAT Prep | Aniko

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Question 9·Hard·Linear Equations in One Variable

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A purification basin initially contained 45,000 liters of water. A pump began draining the basin at a constant rate. After 1.8 hours of pumping, 39,480 liters of water remained in the basin. If the pump continues to drain water at this rate, for how many total hours will the pump have been operating when 28,140 liters of water remain?

For word problems about a quantity changing at a constant rate, first translate the story into a simple rate: compute how much the quantity changes over the given time, then divide to get a per-hour (or per-unit) rate. Next, figure out how much total change is needed to reach the new target amount, and either divide that change by the rate or write a linear equation of the form starting amount ± rate × time = final amount and solve for the time. Work carefully with decimals and fractions, and avoid rounding until the final step to keep your answer accurate.

Hints

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Track how much water is removed

First figure out how many liters of water were pumped out during the first 1.8 hours by comparing the initial amount with the amount remaining.

Use the constant rate

Once you know how many liters were removed in 1.8 hours, divide by 1.8 to find how many liters per hour the pump removes.

Relate the rate to the new target amount

Determine how many total liters must be drained to go from 45,000 liters down to 28,140 liters, then use your draining rate to find how much time that amount of draining takes.

Set up an equation in one variable

You can also write an equation like $45000 - r t = 28140$ , where $r$ is the draining rate you found, and solve for $t$ .

Desmos Guide

Enter the linear model for the water amount

In Desmos, type y = 45000 - (9200/3)x to represent the number of liters of water $y$ remaining in the basin after $x$ hours of pumping.

Graph the target remaining amount as a horizontal line

Type y = 28140 in a new line. This horizontal line represents all points where 28,140 liters of water remain in the basin.

Find the time that matches 28,140 liters

Look at the intersection point of the two graphs (y = 45000 - (9200/3)x and y = 28140). Click the intersection and read off the $x$ -coordinate; that value is the total number of hours the pump has been operating.

Step-by-step Explanation

Find how much water was drained in 1.8 hours

The basin started with 45,000 liters and had 39,480 liters after 1.8 hours.

Amount drained in 1.8 hours:

45000 - 39480 = 5520

So, 5,520 liters were pumped out in 1.8 hours.

Compute the constant draining rate

Use the formula $\text{rate} = \dfrac{\text{amount}}{\text{time}}$ .

\text{rate} = \frac{5520}{1.8}

To simplify:

\frac{5520}{1.8} = \frac{5520}{\tfrac{18}{10}} = \frac{5520 \cdot 10}{18} = \frac{55200}{18} = \frac{9200}{3}

So the pump drains water at a constant rate of $\dfrac{9200}{3}$ liters per hour.

Find how much water must be drained to reach 28,140 liters

Now we want the time when 28,140 liters remain.

Starting amount: 45,000 liters
Desired amount: 28,140 liters

Amount that must be drained:

45000 - 28140 = 16860

So the pump must remove 16,860 liters in total to reach 28,140 liters.

Set up and solve for the total time

Use $\text{time} = \dfrac{\text{amount}}{\text{rate}}$ with amount $16860$ and rate $\dfrac{9200}{3}$ :

\text{time} = \frac{16860}{\tfrac{9200}{3}} = 16860 \cdot \frac{3}{9200}

Simplify the fraction:

\frac{16860}{9200} = \frac{843}{460}

So,

\text{time} = 3 \cdot \frac{843}{460} = \frac{2529}{460} \approx 5.5

Thus, the pump will have been operating for a total of 5.5 hours when 28,140 liters of water remain.

Strategy

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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