3.An 1145 L tank contains 295 L of water and is being filled with brine from a full 1280 L tank at a rate of 1.55 U.S. gallons every 6 seconds. Will the smaller tank ever contain three times the amount of liquid as the larger tank? If so, when? If not, explain why.
After t seconds, we want
295+1.55*t/6 = 3(1280 - 1.55*t/6)
t = 3430.6
check:
after 3430.6 seconds, 886.2 gallons have drained
295+886.2 = 1181.2
1280-886.2 = 393.8
If the smaller tank were big enough to hold enough, after 3430.6 seconds it would hold 3 times the fluid of the larger tank.
However, it's too small, so we need to check to see when the larger tank has drained to 1145/3 gallons.
That occurs after 3477.4 seconds. It takes longer, because the small tank overflows, and the larger tank has to drain longer.
If the smaller tank is sealed, then it will never hold 3 times as much as the larger tank, because flow will stop too soon.
To determine if the smaller tank will ever contain three times the amount of liquid as the larger tank, we can calculate the amount of liquid in each tank over time.
Let's break it down step by step:
1. Convert the volume units:
- The smaller tank contains 295 L of water.
- The larger tank contains 1145 L.
2. Calculate the rate at which the larger tank is being filled:
- The rate is given as 1.55 U.S. gallons every 6 seconds.
- To convert gallons to liters, we can use the conversion factor 1 U.S. gallon = 3.78541 liters.
- So, the rate becomes: 1.55 U.S. gallons * 3.78541 liters/U.S. gallon / 6 seconds = 0.99903 liters/second.
3. Determine the time it takes for the smaller tank to contain three times the amount of liquid as the larger tank:
- Let's assume the time is represented by 't' in seconds.
- The amount of liquid in the smaller tank at time 't' is 295 L + (0.99903 L/second * t seconds).
- The amount of liquid in the larger tank at time 't' is 1145 L + (0.99903 L/second * t seconds).
4. Solve the equation for when the smaller tank contains three times the amount of liquid as the larger tank:
- We're looking for a situation where the amount of liquid in the smaller tank is three times the amount in the larger tank, so we can set up the equation: 295 L + (0.99903 L/second * t seconds) = 3 * (1145 L + (0.99903 L/second * t seconds)).
- Simplifying the equation, we get: 295 L + 0.99903 L/second * t seconds = 3 * 1145 L + 3 * (0.99903 L/second * t seconds).
- Cancelling out the units of seconds, we have: 295 L + 0.99903 L * t = 3 * 1145 L + 3 * 0.99903 L * t.
- Solving for 't', we can bring like terms to one side of the equation: 0.99903 L * t - 3 * 0.99903 L * t = 3 * 1145 L - 295 L.
- Simplifying further, we get: -2.00006 L * t = 3435 L - 295 L.
- Now, dividing both sides by -2.00006 L, we obtain: t = (3435 L - 295 L) / (-2.00006 L).
- Evaluating the right-hand side, we have: t = 3139 L / (-2.00006 L).
- Finally, dividing 3139 L by -2.00006 L gives us the time in seconds when the smaller tank contains three times the amount of liquid as the larger tank.
5. Calculate the final result:
- Plugging in the values, we find: t = -1569.3945 seconds.
Based on the calculation, we obtained a negative value for 't', which implies that the time required for the smaller tank to contain three times the amount of liquid as the larger tank is not achievable. Therefore, the smaller tank will never contain three times the amount of liquid as the larger tank.