A television show lasted for \(4\frac{2}{3}\) hours. If 1/5th of the total time was spent on advertisements, what was the actual duration of the television show?
Calculating Television Show Duration
This problem asks us to find the actual duration of a television show, given the total time the broadcast lasted and the fraction of that time spent on advertisements. To solve this, we first need to understand the total time in a usable format, then calculate the time spent on advertisements, and finally subtract the advertisement time from the total time.
Understanding the Total Show Time
The total duration of the television show, including advertisements, is given as \(4\frac{2}{3}\) hours. This is a mixed number. It's often easier to perform calculations with fractions when they are in improper fraction form.
To convert a mixed number \(a\frac{b}{c}\) to an improper fraction, we use the formula \(\frac{(a \times c) + b}{c}\).
Applying this to \(4\frac{2}{3}\) hours:
\(4\frac{2}{3} = \frac{(4 \times 3) + 2}{3} = \frac{12 + 2}{3} = \frac{14}{3}\) hours.
So, the total duration of the television show broadcast was \(\frac{14}{3}\) hours.
Calculating Advertisement Time
We are told that \(\frac{1}{5}\)th of the total time was spent on advertisements.
To find the advertisement time, we multiply the fraction of time spent on ads by the total time:
Advertisement Time = \(\text{Fraction of time on ads} \times \text{Total time}\)
Advertisement Time = \(\frac{1}{5} \times \frac{14}{3}\) hours.
Multiplying fractions involves multiplying the numerators together and the denominators together:
Advertisement Time = \(\frac{1 \times 14}{5 \times 3} = \frac{14}{15}\) hours.
Thus, \(\frac{14}{15}\) hours were spent on advertisements during the television show broadcast.
Finding the Actual Duration of the Show
The actual duration of the television show is the total broadcast time minus the time spent on advertisements.
Actual Show Duration = Total time - Advertisement time
Actual Show Duration = \(\frac{14}{3} - \frac{14}{15}\) hours.
To subtract fractions, they must have a common denominator. The denominators are 3 and 15. The least common multiple (LCM) of 3 and 15 is 15.
We need to convert \(\frac{14}{3}\) to an equivalent fraction with a denominator of 15. We multiply both the numerator and the denominator by 5 (since \(3 \times 5 = 15\)):
\(\frac{14}{3} = \frac{14 \times 5}{3 \times 5} = \frac{70}{15}\).
Now we can subtract the fractions:
Actual Show Duration = \(\frac{70}{15} - \frac{14}{15} = \frac{70 - 14}{15} = \frac{56}{15}\) hours.
Converting the Result to a Mixed Number
The actual show duration is \(\frac{56}{15}\) hours. To express this as a mixed number, we divide the numerator (56) by the denominator (15).
\(56 \div 15\).
15 goes into 56 three times (\(15 \times 3 = 45\)).
The remainder is \(56 - 45 = 11\).
So, the improper fraction \(\frac{56}{15}\) can be written as the mixed number \(3\frac{11}{15}\).
Therefore, the actual duration of the television show was \(3\frac{11}{15}\) hours.
Summary of Calculation Steps
- Convert total time to improper fraction: \(4\frac{2}{3} = \frac{14}{3}\) hours.
- Calculate advertisement time: \(\frac{1}{5} \times \frac{14}{3} = \frac{14}{15}\) hours.
- Subtract advertisement time from total time: \(\frac{14}{3} - \frac{14}{15}\).
- Find common denominator and subtract: \(\frac{70}{15} - \frac{14}{15} = \frac{56}{15}\) hours.
- Convert result to mixed number: \(\frac{56}{15} = 3\frac{11}{15}\) hours.
The actual duration of the television show is \(3\frac{11}{15}\) hours.
Revision Table: Television Show Duration Problem
| Concept | Description | How Applied Here |
|---|---|---|
| Mixed Numbers | A number combining a whole number and a fraction. | Total time given as \(4\frac{2}{3}\) hours. |
| Improper Fractions | A fraction where the numerator is greater than or equal to the denominator. | Conversion of mixed number to \(\frac{14}{3}\) for easier calculation. Result \(\frac{56}{15}\) is also an improper fraction. |
| Fraction Multiplication | Multiply numerators and denominators. Used for finding a fraction of a quantity. | Used to calculate advertisement time: \(\frac{1}{5} \times \frac{14}{3}\). |
| Fraction Subtraction | Requires a common denominator. Subtract numerators once denominators are the same. | Used to find actual show duration: \(\frac{14}{3} - \frac{14}{15}\). |
| Common Denominator | A shared multiple of the denominators of two or more fractions. | Used to subtract \(\frac{14}{3}\) and \(\frac{14}{15}\). The common denominator is 15. |
Additional Information on Time and Fractions
Problems involving time often use fractions or mixed numbers to represent durations that are not whole hours. Understanding how to perform arithmetic operations (addition, subtraction, multiplication, division) with fractions is crucial for solving such problems.
- Converting Units: While this problem stayed in hours, sometimes you might need to convert between hours, minutes, and seconds. Remember there are 60 minutes in an hour and 60 seconds in a minute.
- Fractions of a Whole: When you see "a fraction of the total", it implies multiplication. For example, "\(\frac{1}{5}\) of the total time" means \(\frac{1}{5} \times \text{Total Time}\).
- Real-World Application: Calculating net time after deductions (like ads in a broadcast, travel time after stops) is a common application of fraction subtraction.
- Checking your Answer: After finding the actual show duration (\(3\frac{11}{15}\) hours), you could add the advertisement time (\(\frac{14}{15}\) hours) back to see if you get the original total time (\(4\frac{2}{3}\) hours).
- \(3\frac{11}{15} + \frac{14}{15} = \frac{(3 \times 15) + 11}{15} + \frac{14}{15} = \frac{45+11}{15} + \frac{14}{15} = \frac{56}{15} + \frac{14}{15} = \frac{56+14}{15} = \frac{70}{15}\).
- Simplify \(\frac{70}{15}\) by dividing numerator and denominator by their greatest common divisor, 5: \(\frac{70 \div 5}{15 \div 5} = \frac{14}{3}\).
- Convert \(\frac{14}{3}\) back to a mixed number: \(14 \div 3 = 4\) remainder 2, which is \(4\frac{2}{3}\).
- The result matches the original total time, confirming the calculation.
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