The relationship between the temperature in degrees Fahrenheit, F, and the temperature in degrees Celsius, C, is given by the...
GMAT Advanced Math : (Adv_Math) Questions
The relationship between the temperature in degrees Fahrenheit, \(\mathrm{F}\), and the temperature in degrees Celsius, \(\mathrm{C}\), is given by the formula \(\mathrm{F} = \frac{9}{5}\mathrm{C} + 32\). Which of the following formulas correctly expresses the temperature in degrees Celsius, \(\mathrm{C}\), in terms of the temperature in degrees Fahrenheit, \(\mathrm{F}\)?
1. TRANSLATE the problem information
- Given formula: \(\mathrm{F = \frac{9}{5}C + 32}\)
- Need to find: A formula for C in terms of F
- This means we need to solve for C (isolate C on one side)
2. INFER the solution strategy
- To isolate C, we need to "undo" the operations affecting it
- C is being multiplied by 9/5, then 32 is being added
- We'll undo these operations in reverse order: first subtract 32, then divide by 9/5 (or multiply by 5/9)
3. SIMPLIFY by subtracting 32 from both sides
\(\mathrm{F = \frac{9}{5}C + 32}\)
\(\mathrm{F - 32 = \frac{9}{5}C + 32 - 32}\)
\(\mathrm{F - 32 = \frac{9}{5}C}\)
4. SIMPLIFY by multiplying both sides by 5/9
- Since C is multiplied by 9/5, we multiply by its reciprocal 5/9
- \(\mathrm{\frac{5}{9}(F - 32) = \frac{5}{9} \times \frac{9}{5}C}\)
- \(\mathrm{\frac{5}{9}(F - 32) = 1 \times C}\)
- \(\mathrm{C = \frac{5}{9}(F - 32)}\)
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students make sign errors when moving the constant term, adding 32 instead of subtracting it.
When students see \(\mathrm{F = \frac{9}{5}C + 32}\), they might incorrectly think "to get rid of the +32, I need to add 32 to the other side" rather than recognizing they need to subtract 32 from both sides. This leads them to \(\mathrm{F + 32 = \frac{9}{5}C}\), and ultimately to \(\mathrm{C = \frac{5}{9}(F + 32)}\).
This may lead them to select Choice B (\(\mathrm{C = \frac{5}{9}(F + 32)}\)).
Second Most Common Error:
Conceptual confusion about reciprocals: Students use 9/5 instead of 5/9 when trying to isolate C.
Students correctly subtract 32 but then think "C is being multiplied by 9/5, so I multiply both sides by 9/5" instead of recognizing they need the reciprocal 5/9 to cancel out the 9/5. This gives them \(\mathrm{C = \frac{9}{5}(F - 32)}\).
This may lead them to select Choice C (\(\mathrm{C = \frac{9}{5}(F - 32)}\)).
The Bottom Line:
This problem requires careful attention to inverse operations - students must remember that to undo multiplication by a fraction, you multiply by its reciprocal, and to undo addition, you subtract.