In an exam consisting of three subjects — English, History, and Mathematics — with 120, 60, and 100 questions respectively, student C attempted a total of 192 questions. If C attempted 54 questions in History, and the number of questions attempted in English is more than those in History but less than those in Mathematics, what could be the possible number of questions attempted by C in Mathematics?

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**Solution:** 1. Total questions attempted by C = 192. 2. Questions attempted in History = 54. 3. Let the number of questions attempted in Mathematics be *m* and in English be *e*. 4. Since total questions attempted by C are 192, we have: $$m + e + 54 = 192 \ m + e = 138$$ 5. Given that English questions attempted are more than History but less than Mathematics: $$e > 54 \ e < m$$ 6. Substitute $e = 138 - m$ into the inequalities: - From $e > 54$: $$138 - m > 54 \ 138 - 54 > m \ 84 > m$$ - From $e < m$: $$138 - m < m \ 138 < 2m \ 69 < m$$ 7. Combining both inequalities, the range for *m* is: $$69 < m < 84$$ 8. Therefore, the possible integer values for *m* (Mathematics questions attempted) lie between 70 and 83. 9. Among the options given, $70$, $75$, and $81$ fall within this range. **Answer:** Options $70$, $75$, and $81$ are all possible values for the number of questions attempted in Mathematics by C.

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