The sum of \(3\frac{7}{8}\) and \(1\frac{1}{3}\) is less than the difference between \(\frac{1}{8}\) and \(1\frac{2}{3}\)...

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:

A. 3\(\frac{2}{3}\)
B. 5\(\frac{1}{4}\)
C. 6\(\frac{1}{2}\)
D. 8
E. 8\(\frac{1}{8}\)

\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\)

= \(4\frac{29}{24}\)

\(\equiv 5\frac{5}{24}\)

\(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\)

= \(\frac{40 - 3}{24}\)

= \(\frac{37}{24}\)

\(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\)

= \(\frac{11}{3}\)

= 3\(\frac{2}{3}\)

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