Solving Water Tank Capacity Problems Using Algebraic Equations

Understanding and resolving problems related to the capacity of water tanks can help in various practical scenarios, such as ensuring proper water supply in households, industries, and warehouses. In this article, we will discuss how to use algebraic equations to solve these problems and provide detailed steps and solutions.

Problem 1: Tank Half Full and 6 Liters Removed

Let's consider a tank whose capacity is unknown. Initially, the tank is half full. If 6 liters are removed, it becomes two-fifths full. We need to determine the total capacity of the tank.

Given:

The tank is half full initially, i.e., (frac{1}{2}x) liters, where (x) is the total capacity. 6 liters are removed, making the tank two-fifths full, i.e., (frac{2}{5}x) liters.

Formulating the Equation:

From the given information, we can set up the following equation:

(frac{1}{2}x - 6 frac{2}{5}x)

To solve this equation, eliminate the fractions by multiplying every term by 10, the least common multiple of 2 and 5:

(10 left(frac{1}{2}x - 6right) 10 left(frac{2}{5}xright))

This simplifies to:

(5x - 60 4x)

Subtract 4x from both sides:

(5x - 4x 60)

This simplifies to:

(x 60)

Thus, the total capacity of the tank when it is full is (60) liters.

Problem 2: Tank is 2/3 Full After Drawing Off 78 Liters

In this problem, we know that 78 liters are drawn off the tank, and the tank is now two-thirds full. We need to find the total capacity of the tank.

Given:

When 78 liters are drawn off, the tank is two-thirds full. (frac{1}{3}) of the tank's capacity is 78 liters.

Solving the Equation:

Since (frac{1}{3}) of the tank's capacity is 78 liters, the full capacity of the tank can be calculated as follows:

(T 78 times 3 234) liters

Therefore, the total capacity of the tank when it is full is (234) liters.

Problem 3: Tank Initially Two-Thirds Full

In this scenario, we are given that 78 liters are drawn off the tank, and the tank is now two-thirds full. This means that one-third of the tank’s capacity is 78 liters. We need to determine the full capacity of the tank.

Given:

When 78 liters are drawn off, the tank is two-thirds full.

Formulating the Equation:

Let T be the total capacity of the tank. When 78 liters are drawn off, the remaining volume is (T - 78). This remaining volume is two-thirds of the tank's capacity:

[frac{T - 78}{T} frac{2}{3}]

By cross-multiplying, we get:

[3T - 234 2T]

Solving for T:

[3T - 2T 234]

[T 234] liters

Therefore, the total capacity of the tank when it is full is (234) liters.

Conclusion

Understanding how to solve these types of problems using algebraic equations is crucial in various real-life situations. Applying the concepts explained above, one can determine the capacity of tanks and other containers, ensuring efficient resource management and optimization of space.