Problem 48 - Basic College Mathematics

Question

A tank had \(20 \mathrm{~L}\) of gasoline in it when it was \(\frac{4}{5}\) full. How much could it hold when full?

Step-by-Step Solution

Answer

The tank's total capacity is \(25 \mathrm{~L}\).

Step 1: Understand the given information

The tank is \(\frac{4}{5}\) full and contains \(20 \mathrm{~L}\) of gasoline. This means \(20 \mathrm{~L}\) represents \(\frac{4}{5}\) of the tank's total capacity.

Step 2: Set up the equation

Let the total capacity of the tank be \(C\, in \mathrm{~L}\). Since \(20 \mathrm{~L}\) is \(\frac{4}{5}\) of the total capacity: \(\frac{4}{5} C = 20\).

Step 3: Solve the equation for C

To find \C\, multiply both sides of the equation by \(\frac{5}{4}\): \(C = 20 \times \frac{5}{4} \).

Step 4: Simplify the solution

Calculate the right-hand side: \(C = 20 \times \frac{5}{4} = 25 \).

Step 5: State the final answer

The tank's total capacity when full is \(25 \mathrm{~L}\).

Key Concepts

solving equations with fractions

Many practical problems involve fractions, especially when dealing with parts of a whole. In our exercise, we're given a tank that's \(\frac{4}{5}\) full, containing \(20 \, \mathrm{L}\) of gasoline. This fraction helps us find the tank's full capacity.
To set up our equation, we let the total capacity be C (in liters). The fraction equation is: \(\frac{4}{5} C = 20\).
This equation states that \(20 \, \mathrm{L}\) is \(\frac{4}{5}\) of the full capacity. To isolate C (the total capacity), multiply both sides by the reciprocal of \(\frac{4}{5}\), which is \(\frac{5}{4}\).
The equation transforms into: \(C = 20 \times \frac{5}{4}\).
Simplifying the right-hand side gives us \(25 \, \mathrm{L}\), so the tank's total capacity is \(25 \, \mathrm{L}\). Solving equations with fractions is crucial for determining quantities and proportions in everyday problems.

ratio and proportion

Ratios and proportions simplify comparisons between quantities. In our exercise, \(20 \, \mathrm{L}\) is to \(\frac{4}{5}\) of the tank’s capacity. This is a practical example of using a ratio in a proportional relationship.
The ratio \(\frac{4}{5}\) tells us that the \(20 \, \mathrm{L}\) represents \(\frac{4}{5}\) of the tank's total capacity. By expressing this as an equation, \(\frac{4}{5} C = 20\), we're setting up a proportion to identify the missing quantity (full capacity).
Understanding proportions is valuable in solving many real-life problems, like cooking, mixing solutions, or determining travel distances using scaled maps. Recognizing and using ratios helps to balance and adjust quantities accurately.

unit conversion in word problems

Unit conversions in word problems often come up in exercises related to measurements, currency, or other units of comparison. In our tank capacity problem, we don't change units, but understanding capacity and volume units (liters, milliliters, gallons) is useful.
Suppose the problem involved different units, like converting gallons to liters. This adds a layer where you must remember conversion factors: \(1 \, \text{gallon} = 3.785 \, \mathrm{L}\).
For example, if the tank's capacity was given in gallons, we'd convert that to liters using the multiplication of the capacity in gallons by \(3.785\) to get the volume in liters. Mastering unit conversions enhances solving multi-step problems accurately. It’s helpful to keep a list of common conversion factors as a reference.