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Exercise 1.4 · Q4

Q.A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer's rule to solve the problem.)

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We model each pump by its rate (tank per minute), turn "together in 10 min" and "B reversed in 30 min" into two linear equations in the rates, and solve by Cramer's rule; the required times are the reciprocals of the rates.

Step 1. Translate the problem into equations. Let aa = rate at which pump A fills the tank (tank/min) and bb = rate at which pump B fills it when running normally.

  • Both pumps together fill the tank in 10 minutes: their rates add, so together they do 110\tfrac1{10} of the tank per minute:

a+b=110a+b=\frac1{10}

  • Pump B reversed (pumping out) together with A fills the tank in 30 minutes: now B's contribution subtracts, giving 130\tfrac1{30} of the tank per minute:

a−b=130a-b=\frac1{30}

Step 2. Write the coefficient determinant.

D=∣111−1∣=1(−1)−1(1)=−2D=\begin{vmatrix}1&1\\1&-1\end{vmatrix}=1(-1)-1(1)=-2

Step 3. Compute DaD_a (replace the aa-column with the constants).

Da=∣1101130−1∣=110(−1)−1(130)=−110−130=−330−130=−430=−215D_a=\begin{vmatrix}\frac1{10}&1\\[2pt]\frac1{30}&-1\end{vmatrix}=\frac1{10}(-1)-1\left(\frac1{30}\right)=-\frac1{10}-\frac1{30}=-\frac{3}{30}-\frac1{30}=-\frac{4}{30}=-\frac2{15}

Step 4. Compute DbD_b (replace the bb-column with the constants).

Db=∣11101130∣=1(130)−110(1)=130−330=−230=−115D_b=\begin{vmatrix}1&\frac1{10}\\[2pt]1&\frac1{30}\end{vmatrix}=1\left(\frac1{30}\right)-\frac1{10}(1)=\frac1{30}-\frac{3}{30}=-\frac2{30}=-\frac1{15}

Step 5. Apply Cramer's rule. …

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