What fraction must be subtracted from the sum of \(2\frac{1}{6}\) and \(2\frac{7}{12}\) to give \(3\frac{1}{4}\)?...
MATHEMATICS
WAEC 2000
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. \(1\frac{1}{2}\)
Correct Answer: D. \(1\frac{1}{2}\)
Explanation
\(2\frac{1}{6} + 2\frac{7}{12}\)
= \(\frac{13}{6} + \frac{31}{12}\)
= \(\frac{26 + 31}{12}\)
= \(\frac{57}{12} = \frac{19}{4}\)
\(\frac{19}{4} - 3\frac{1}{4}\)
= \(\frac{19}{4} - \frac{13}{4}\)
= \(\frac{6}{4}\)
= \(1\frac{1}{2}\)
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