The fuel efficiency E, in miles per gallon, of a certain vehicle is modeled by the equation below, where s...
GMAT Advanced Math : (Adv_Math) Questions
The fuel efficiency \(\mathrm{E}\), in miles per gallon, of a certain vehicle is modeled by the equation below, where \(\mathrm{s}\) is the speed of the vehicle in miles per hour, for \(\mathrm{s \geq 50}\).
\(\mathrm{E = 48 - 0.4(s - 50)}\)
Which of the following equations expresses the speed of the vehicle in terms of its fuel efficiency?
1. TRANSLATE the problem information
- Given equation: \(\mathrm{E = 48 - 0.4(s - 50)}\)
- Need to find: \(\mathrm{s}\) in terms of \(\mathrm{E}\) (meaning \(\mathrm{s}\) = some expression involving \(\mathrm{E}\))
- What this tells us: We need to solve for \(\mathrm{s}\) by rearranging the equation
2. INFER the approach
- This is an inverse function problem requiring algebraic manipulation
- Strategy: Isolate \(\mathrm{s}\) by undoing each operation step-by-step
- Start by getting the \(\mathrm{s}\)-containing term by itself
3. SIMPLIFY to isolate the s-containing term
- Start with: \(\mathrm{E = 48 - 0.4(s - 50)}\)
- Subtract 48 from both sides: \(\mathrm{E - 48 = -0.4(s - 50)}\)
4. SIMPLIFY to isolate the parenthetical expression
- Divide both sides by -0.4: \(\mathrm{\frac{E - 48}{-0.4} = s - 50}\)
- This is equivalent to multiplying by -2.5: \(\mathrm{-2.5(E - 48) = s - 50}\)
- Distribute: \(\mathrm{-2.5E + 120 = s - 50}\)
5. SIMPLIFY to isolate s completely
- Add 50 to both sides: \(\mathrm{s = -2.5E + 120 + 50}\)
- Combine constants: \(\mathrm{s = -2.5E + 170}\)
- Reorder: \(\mathrm{s = 170 - 2.5E}\)
Answer: C. \(\mathrm{s = 170 - 2.5E}\)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Making sign errors when dividing by the negative coefficient -0.4
Students often struggle with the step \(\mathrm{\frac{E - 48}{-0.4} = s - 50}\) and may incorrectly get positive values instead of the correct -2.5 multiplier. They might calculate \(\mathrm{\frac{E - 48}{0.4} = 2.5(E - 48)}\) instead of \(\mathrm{-2.5(E - 48)}\), leading to \(\mathrm{s = 2.5E - 120 + 50 = 2.5E - 70}\).
This may lead them to select Choice A (\(\mathrm{s = 2.5E - 70}\)).
Second Most Common Error:
Poor INFER reasoning about the solution strategy: Students may attempt to substitute answer choices back into the original equation rather than systematically solving algebraically
This approach can work but is time-consuming and error-prone. Students may make calculation mistakes during substitution or become confused about which direction to test, leading to random guessing among the choices.
The Bottom Line:
Success requires careful attention to signs during algebraic manipulation, particularly when dividing by negative coefficients. The key insight is recognizing this as a systematic "undo the operations" problem rather than a guess-and-check scenario.