JAMB - Mathematics (1978 - No. 2)
The sum of \(3\frac{7}{8}\) and \(1\frac{1}{3}\) is less than the difference between \(\frac{1}{8}\) and \(1\frac{2}{3}\) by:
3\(\frac{2}{3}\)
5\(\frac{1}{4}\)
6\(\frac{1}{2}\)
8
8\(\frac{1}{8}\)
Explanation
\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\)
= \(4\frac{29}{24}\)
\(\equiv 5\frac{5}{24}\)
\(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\)
= \(\frac{40 - 3}{24}\)
= \(\frac{37}{24}\)
\(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\)
= \(\frac{11}{3}\)
= 3\(\frac{2}{3}\)
Comments (0)
