Which expression is equivalent to h^(16) q^(7)/h^(2) q^(21), where h gt 0 and q gt 0?
GMAT Advanced Math : (Adv_Math) Questions
Which expression is equivalent to \(\frac{\mathrm{h}^{16} \mathrm{q}^{7}}{\mathrm{h}^{2} \mathrm{q}^{21}}\), where \(\mathrm{h} \gt 0\) and \(\mathrm{q} \gt 0\)?
1. TRANSLATE the problem information
- Given: \(\frac{\mathrm{h}^{16} \mathrm{q}^{7}}{\mathrm{h}^{2} \mathrm{q}^{21}}\) where \(\mathrm{h} \gt 0\) and \(\mathrm{q} \gt 0\)
- Find: Equivalent simplified expression
2. INFER the approach
- Since we have the same bases in numerator and denominator, use quotient rule for exponents
- Apply the rule separately to each variable: h and q
- Handle any negative exponents that result
3. SIMPLIFY using quotient rule for exponents
- For h terms: \(\frac{\mathrm{h}^{16}}{\mathrm{h}^{2}} = \mathrm{h}^{16-2} = \mathrm{h}^{10}\)
- For q terms: \(\frac{\mathrm{q}^{7}}{\mathrm{q}^{21}} = \mathrm{q}^{7-21} = \mathrm{q}^{-14}\)
- Combined result: \(\mathrm{h}^{10} \times \mathrm{q}^{-14}\)
4. SIMPLIFY the negative exponent
- Since \(\mathrm{q}^{-14} = \frac{1}{\mathrm{q}^{14}}\)
- The expression becomes: \(\mathrm{h}^{10} \times \frac{1}{\mathrm{q}^{14}} = \frac{\mathrm{h}^{10}}{\mathrm{q}^{14}}\)
Answer: A. \(\frac{\mathrm{h}^{10}}{\mathrm{q}^{14}}\)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY skill: Students make arithmetic errors when subtracting exponents, particularly with the q terms where \(7-21 = -14\). Some students might calculate this as \(21-7 = 14\) instead, leading to \(\mathrm{q}^{14}\) in the numerator rather than \(\mathrm{q}^{-14} = \frac{1}{\mathrm{q}^{14}}\) in the denominator.
This may lead them to select Choice C (\(\mathrm{h}^{10} \mathrm{q}^{14}\)).
Second Most Common Error:
Conceptual confusion about negative exponents: Students correctly subtract exponents but don't remember that negative exponents create reciprocals. They might leave \(\mathrm{q}^{-14}\) as is or mishandle the conversion.
This leads to confusion and guessing between the remaining choices.
The Bottom Line:
This problem tests two fundamental exponent rules in sequence - students must both apply the quotient rule correctly AND properly handle the negative exponent that results. Missing either step leads to an incorrect answer.