What fraction must be subtracted from the sum of \(2\frac{1}{6}\) and \(2\frac{7}{12}\) to give \(3\frac{1}{4}\)?

  • A \(\frac{1}{3}\)
  • B \(\frac{1}{2}\)
  • C \(1\frac{1}{6}\)
  • D \(1\frac{1}{2}\)
waec 2000mathematics

The correct answer is D. \(1\frac{1}{2}\)

\(2\frac{1}{6} + 2\frac{7}{12}\)

= \(\frac{13}{6} + \frac{31}{12}\)

= \(\frac{26 + 31}{12}\)

= \(\frac{57}{12} = \frac{19}{4}\)

\(\frac{19}{4} - 3\frac{1}{4}\)

= \(\frac{19}{4} - \frac{13}{4}\)

= \(\frac{6}{4}\)

= \(1\frac{1}{2}\)

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