How to Calculate the Capacity of a Water Tank Using Algebra
When troubleshooting the capacity of a water tank, the algebraic approach can be a powerful tool. Imagine you have a water tank that is currently 1/3 full and after adding 150 liters of water, the tank becomes full. This scenario poses an interesting conundrum: how can we determine the total capacity of the tank? Let's delve into this problem using algebraic methods and explore the solution step-by-step.
Solving the Problem Using Algebra
Let the total capacity of the tank be denoted as C liters. Initially, the tank is 1/3 full, meaning it contains (frac{1}{3}C) liters of water. After adding 150 liters of water, the tank becomes completely full, leading to the equation:
(frac{1}{3}C 150 C)
Step-by-Step Solution
First, isolate the variable C by subtracting (frac{1}{3}C) from both sides of the equation:(frac{1}{3}C 150 , – , frac{1}{3}C C , - , frac{1}{3}C)
(150 frac{2}{3}C)
Next, solve for C by multiplying both sides of the equation by (frac{3}{2}) to isolate C:(150 times frac{3}{2} C)
(frac{450}{2} C)
(225 C)
Therefore, the capacity of the tank is 225 liters.
Alternative Solutions Explained
Let's also explore some alternative methods to solve this problem, presented in various sources:
Method 1: 1/2 - 1/3 200 liters
[ text{1/2 tank} - text{1/3 tank} 200 ]
[ frac{3}{6} - frac{2}{6} 200 ]
[ frac{1}{6} text{ of the tank} 200 text{ liters} ]
[ text{Full capacity of the tank} 200 div frac{1}{6} 1200 text{ liters} ]
Method 2: 1/2X - 200 1/3X
[ frac{1}{2}X - 200 frac{1}{3}X ]
[ frac{1}{2}X - frac{1}{3}X 200 ]
[ frac{3}{6}X - frac{2}{6}X 200 ]
[ frac{1}{6}X 200 ]
[ X 1200 text{ liters} ]
Method 3: 1/3 4/12
[ frac{7}{12} - frac{4}{12} 3/12 9 text{ liters} ]
[ frac{9}{3} 3 text{ liters} frac{1}{12} text{ of the tank} ]
[ 3 times 12 36 text{ liters} text{capacity of the tank} ]
Method 4: Let the capacity of the tank x
[ frac{x}{3} - 9 frac{7x}{12} ]
[ frac{7x}{12} - frac{x}{3} 9 ]
[ frac{7x - 4x}{12} 9 ]
[ frac{3x}{12} 9 ]
[ x frac{4 times 9}{1} 36 text{ liters} ]
Method 5: Let the capacity of the tank be C litres
[ frac{C}{3} - 9 frac{7C}{12} ]
[ frac{C}{3} - frac{7C}{12} 9 ]
[ frac{4C - 7C}{12} 9 ]
[ frac {-3C}{12} 9 ]
[ -3C 108 ]
[ C 36 text{ liters} ]
Conclusion
Through these various methods, we can see that the capacity of the water tank is consistently determined to be 36 liters. Each method provides a unique perspective on solving the problem, but they all converge on the same solution using fundamental algebraic principles. Understanding these methods can enhance your problem-solving skills and flexibility in dealing with real-world situations involving capacity calculations.