Alan drives an average of 100 miles each week. His car can travel an average of 25 miles per gallon...
GMAT Algebra : (Alg) Questions
Alan drives an average of \(\mathrm{100}\) miles each week. His car can travel an average of \(\mathrm{25}\) miles per gallon of gasoline. Alan would like to reduce his weekly expenditure on gasoline by \(\$5\). Assuming gasoline costs \(\$4\) per gallon, which equation can Alan use to determine how many fewer average miles, \(\mathrm{m}\), he should drive each week?
1. TRANSLATE the problem information
- Given information:
- Current driving: 100 miles per week
- Car efficiency: 25 miles per gallon
- Gas price: \(\$4\) per gallon
- Desired reduction: \(\$5\) in weekly spending
- Find: equation for miles reduction (m)
2. INFER the approach needed
- To find how many miles to reduce, I need to know the cost per mile
- Then I can set up: (cost per mile) × (miles reduced) = money saved
3. TRANSLATE to find cost per mile
- Cost per mile = (cost per gallon) ÷ (miles per gallon)
- Cost per mile = \(\frac{\$4}{25 \text{ miles}} = \frac{\$4}{25} \text{ per mile}\)
4. INFER the final equation setup
- If Alan drives m fewer miles, he saves: \(\frac{4}{25} \times m\) dollars
- He wants to save \(\$5\), so: \(\frac{4}{25} \times m = 5\)
Answer: D. \(\frac{4}{25}m = 5\)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students flip the fraction, calculating miles per dollar instead of dollars per mile.
They might think: "I have \(\$4\) per gallon and 25 miles per gallon, so that's 25/4." This gives them miles per dollar rather than the needed dollars per mile. This leads them to select Choice A (\(\frac{25}{4}m = 95\)) or Choice B (\(\frac{25}{4}m = 5\)).
Second Most Common Error:
Poor TRANSLATE reasoning: Students misinterpret the "\(\$5\) reduction" as meaning Alan wants to drive 95 miles instead of reducing his spending by \(\$5\).
They set up the equation thinking Alan wants to spend money on 95 miles of driving rather than understanding he wants to reduce his spending by \(\$5\). This leads them to select Choice A (\(\frac{25}{4}m = 95\)) or Choice C (\(\frac{4}{25}m = 95\)).
The Bottom Line:
This problem requires careful attention to units and what quantity is being reduced. Students must distinguish between the cost savings (\(\$5\)) and potential miles driven, while also ensuring their rate calculation gives dollars per mile, not miles per dollar.