Question

water is being pumped into a 20-foot-tall cylindrical tank at a constant rate.

• the depth of the water is increasing linearly.
• at 10:30 a.m., the water depth was 6.8 feet.
• it is now 3:00 p.m., and the depth of the water is 10.4 feet.

what will the depth (in feet) of the water be at 5:00 p.m.?

12 ft
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Explanation:

Step1: Calculate time elapsed

From 10:30 a.m. to 3:00 p.m., the time elapsed is \( 4.5 \) hours (since 3:00 p.m. - 10:30 a.m. = 4 hours 30 minutes = \( 4.5 \) hours). The depth increased from \( 6.8 \) feet to \( 10.4 \) feet, so the change in depth is \( 10.4 - 6.8 = 3.6 \) feet.

Step2: Find the rate of depth increase

The rate of depth increase (slope) is \( \frac{3.6}{4.5} = 0.8 \) feet per hour.

Step3: Calculate time from 3:00 p.m. to 5:00 p.m.

From 3:00 p.m. to 5:00 p.m., the time elapsed is \( 2 \) hours.

Step4: Find the depth increase in 2 hours

The depth will increase by \( 0.8 \times 2 = 1.6 \) feet.

Step5: Calculate the depth at 5:00 p.m.

At 3:00 p.m., the depth is \( 10.4 \) feet. So at 5:00 p.m., the depth is \( 10.4 + 1.6 = 12 \) feet.

Answer:

12