# A tank is supplied by two pipes A and B and emptied by a third pipe C

**Situation**

A tank is supplied by two pipes *A* and *B* and emptied by a third pipe *C*. If the tank is initially empty and all pipes are opened, the tank can be filled in 20 hours. If the tank is initially full and *A* and *C* are opened, the tank can be emptied in 4 hours. If the tank is initially full and *B* and *C* are opened, the tank can be emptied in 2 hours. Pipe *A* supplies 50 liters per minute more than *B*.

1. Find the rate of pipe *A* in liters per minute.

A. 120 | C. 110 |

B. 130 | D. 140 |

2. Find the rate of pipe *C* in liters per minute.

A. 170 | C. 150 |

B. 160 | D. 140 |

3. Find the capacity of the tank in liters.

A. 12,000 | C. 11,500 |

B. 12,500 | D. 13,000 |

**Answer Key**

Part 2: [ B ]

Part 3: [ A ]

**Solution**

*V*= capacity of the tank

*A*,

*B*, and

*C*= rate of pipes

*A*,

*B*, and

*C*, respectively

Pipe *A* supplies 50 liters per minute more than *B*

$A = B + 50$

$A - B = 50$ ← Equation (1)

**volume = rate × time**

$1200A + 1200B - 1200C = V$ ← Equation (2)

The tank is initially full and pipes *A* and *C* are opened (4 hours or 240 minutes to empty):

$V + 240A - 240C = 0$

$240A - 240C = -V$ ← Equation (3)

The tank is initially full and pipes *B* and *C* are opened (2 hours or 120 minutes to empty):

$V + 120B - 120C = 0$

$120B - 120C = -V$ ← Equation (4)

Equation (2) + Equation (3)

$1440A + 1200B - 1440C = 0$

$6A + 5B - 6C = 0$ ← Equation (5)

Equation (2) + Equation (4)

$1200A + 1320B - 1320C = 0$

$10A + 11B - 11C = 0$ ← Equation (6)

From Equations (1), (5), and (6)

*answer for part 1*

$B = 60 ~ \text{Lit/min}$

$C = 160 ~ \text{Lit/min}$ ← [ B ] *answer for part 2*

From Equation (2)

$V = 1200(110) + 1200(60) - 1200(160)$

$V = 12,000 ~ \text{Lit}$ ← [ A ] *answer for part 3*