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At a speed of \(\mathrm{50}\text{ miles per hour}\), a certain car uses \(\mathrm{1}\text{ gallon}\) of gasoline every \(\mathrm{30}\text{ miles}\). If the car starts with a full \(\mathrm{12}\text{ gallon}\) tank of gasoline and travels for \(\mathrm{5}\text{ hours}\) at \(\mathrm{50}\text{ miles per hour}\), the amount of gasoline used would be what fraction of a full tank?
Let's break down what we know in everyday terms:
- The car drives at a constant speed of \(50\) miles per hour
- It will drive for \(5\) hours total
- At this speed, the car burns through \(1\) gallon of gas for every \(30\) miles it travels
- The tank holds \(12\) gallons when full
- We need to find what fraction of this full tank gets used during the trip
Think of it like planning a road trip - we know how fast we're going, how long we'll drive, and how thirsty our car is for gas. We want to know what portion of our full tank we'll use up.
Process Skill: TRANSLATE - Converting the problem's real-world scenario into clear mathematical relationships
When something travels at a steady speed for a certain time, the total distance is simply:
\(\mathrm{Distance} = \mathrm{Speed} \times \mathrm{Time}\)
In our case:
- Speed = \(50\) miles per hour
- Time = \(5\) hours
- Distance = \(50 \times 5 = 250\) miles
So the car will travel \(250\) miles during this \(5\)-hour trip.
Now we need to figure out how much gas the car will drink during these \(250\) miles.
We know the car's "appetite": it uses \(1\) gallon for every \(30\) miles.
To find total gas consumption:
If \(30\) miles requires \(1\) gallon, then \(250\) miles requires:
\(250 \div 30 = \frac{25}{3}\) gallons
Let's verify this makes sense: \(\frac{25}{3} = 8\frac{1}{3}\) gallons, which seems reasonable for a \(250\)-mile trip.
Finally, we compare the gas used to the tank's full capacity.
- Gas used = \(\frac{25}{3}\) gallons
- Full tank capacity = \(12\) gallons
- Fraction used = (Gas used) ÷ (Full capacity)
Fraction = \(\left(\frac{25}{3}\right) \div 12 = \left(\frac{25}{3}\right) \times \left(\frac{1}{12}\right) = \frac{25}{36}\)
This means the car uses \(\frac{25}{36}\) of its full tank during the trip.
The amount of gasoline used would be \(\frac{25}{36}\) of a full tank.
This matches answer choice E. \(\frac{25}{36}\)
To verify: \(\frac{25}{36} \approx 0.69\) or about \(69\%\) of the tank, which makes sense for an \(8\frac{1}{3}\) gallon consumption from a \(12\)-gallon tank.
1. Misinterpreting the fuel efficiency relationship
Students might confuse the given information "\(1\) gallon of gasoline every \(30\) miles" and think it means the car travels \(1\) mile per gallon instead of \(30\) miles per gallon. This fundamental misunderstanding of the fuel efficiency would lead to completely incorrect calculations.
2. Confusing what fraction to find
The question asks for "what fraction of a full tank" is used, but students might mistakenly try to find what fraction remains unused, or what fraction of the total distance corresponds to fuel consumption. This misinterpretation of the target quantity leads to solving the wrong problem entirely.
1. Arithmetic errors in division
When calculating \(250 \div 30\), students often make mistakes converting this to the fraction \(\frac{25}{3}\). They might incorrectly calculate it as \(8.33\) and then struggle to work with decimals, or make errors in the long division process.
2. Incorrect fraction division
When computing \(\left(\frac{25}{3}\right) \div 12\), students frequently forget to multiply by the reciprocal. They might incorrectly calculate it as \(\frac{25}{3 \times 12} = \frac{25}{36}\), getting the right answer by accident, or make other fraction manipulation errors like \(\frac{25 \times 12}{3}\).
3. Unit conversion confusion
Students might get confused about whether they're working with gallons, miles, or hours, especially when setting up the fuel consumption calculation. They might incorrectly multiply or divide by the time factor when it's not needed.
1. Selecting the decimal equivalent instead of the fraction
After correctly calculating \(\frac{25}{36} \approx 0.694\), students might look for a decimal answer choice or try to match their decimal to an incorrect fraction that's close in value, rather than recognizing that \(\frac{25}{36}\) is exactly what they calculated and appears as choice E.