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Math  /  Algebra

Question3. A large tank is partially filled with a solution. The tank has a faucet that allows solution to enter the tank rate of 163416 \frac{3}{4} liters per minute. The tank also has a drain that allows solution to leave the tank at a rate 194519 \frac{4}{5} liters per minute. (a) What expression represents the change in volume of solution in the tank in 1 minute? (b) What is the change in volume of the solution after 15 seconds? Show necessary work. (c) What does the change in volume after 15 seconds mean in the real world? Answer in complet sentences. Answer:

Studdy Solution

STEP 1

1. The tank has a faucet and a drain with given rates of flow.
2. The rates are given in liters per minute.
3. We need to find expressions and numerical values for changes in volume over specified time intervals.

STEP 2

1. Determine the net change in volume per minute.
2. Calculate the change in volume over 15 seconds.
3. Interpret the real-world meaning of the change in volume after 15 seconds.

STEP 3

Convert the mixed numbers to improper fractions for easier calculation: - The rate of solution entering the tank is 1634=674 16 \frac{3}{4} = \frac{67}{4} liters per minute. - The rate of solution leaving the tank is 1945=995 19 \frac{4}{5} = \frac{99}{5} liters per minute.

STEP 4

Calculate the net change in volume per minute: - Net change = rate in - rate out = 674−995 \frac{67}{4} - \frac{99}{5} .
To subtract these fractions, find a common denominator: - The least common multiple of 4 and 5 is 20. - Convert 674 \frac{67}{4} to 33520 \frac{335}{20} . - Convert 995 \frac{99}{5} to 39620 \frac{396}{20} .
Net change = 33520−39620=−6120 \frac{335}{20} - \frac{396}{20} = \frac{-61}{20} .

STEP 5

Calculate the change in volume over 15 seconds: - Since 15 seconds is 14 \frac{1}{4} of a minute, multiply the net change per minute by 14 \frac{1}{4} .
Change in volume = −6120×14=−6180 \frac{-61}{20} \times \frac{1}{4} = \frac{-61}{80} .

STEP 6

Interpret the real-world meaning: - The change in volume after 15 seconds is −6180 \frac{-61}{80} liters, which means that the volume of the solution in the tank decreases by 6180 \frac{61}{80} liters in 15 seconds.
The expressions and calculations are as follows: (a) The expression for the change in volume in 1 minute is −6120 \frac{-61}{20} liters. (b) The change in volume after 15 seconds is −6180 \frac{-61}{80} liters. (c) In the real world, this means that the tank is losing solution at a rate that results in a decrease of 6180 \frac{61}{80} liters in 15 seconds.

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