Question

choose the sum that is greater than 5. $4\frac{1}{8}+\frac{5}{8}$ $4\frac{3}{8}+\frac{3}{8}$ $4\frac{5}{8}+\frac{7}{8}$ $4\frac{6}{8}+\frac{1}{8}$

Explanation:

Step1: Calculate first sum

Convert mixed number to improper fraction: \(4\frac{1}{8}=\frac{33}{8}\), then add \(\frac{5}{8}\): \(\frac{33}{8}+\frac{5}{8}=\frac{38}{8}=4.75\) (less than 5)

Step2: Calculate second sum

\(4\frac{3}{8}=\frac{35}{8}\), add \(\frac{3}{8}\): \(\frac{35}{8}+\frac{3}{8}=\frac{38}{8}=4.75\) (less than 5) Wait, no, correction: \(4\frac{3}{8}=\frac{35}{8}\), \(\frac{35}{8}+\frac{3}{8}=\frac{38}{8}=4.75\)? Wait, no, \(4\frac{3}{8} + \frac{3}{8}=4 + (\frac{3}{8}+\frac{3}{8})=4+\frac{6}{8}=4.75\)? Wait, no, let's do it as mixed numbers: \(4\frac{1}{8}+\frac{5}{8}=4 + (\frac{1}{8}+\frac{5}{8})=4+\frac{6}{8}=4.75\) (sum 4.75 <5). \(4\frac{3}{8}+\frac{3}{8}=4 + (\frac{3}{8}+\frac{3}{8})=4+\frac{6}{8}=4.75\) (still <5). \(4\frac{5}{8}+\frac{7}{8}=4 + (\frac{5}{8}+\frac{7}{8})=4+\frac{12}{8}=4 + 1.5=5.5\) (5.5>5). \(4\frac{6}{8}+\frac{1}{8}=4 + (\frac{6}{8}+\frac{1}{8})=4+\frac{7}{8}=4.875\) (less than 5). Wait, let's recalculate each:

1. \(4\frac{1}{8}+\frac{5}{8}\): \(4 + \frac{1 + 5}{8}=4+\frac{6}{8}=4.75\) (4.75 <5)
2. \(4\frac{3}{8}+\frac{3}{8}\): \(4 + \frac{3 + 3}{8}=4+\frac{6}{8}=4.75\) (4.75 <5)
3. \(4\frac{5}{8}+\frac{7}{8}\): \(4 + \frac{5 + 7}{8}=4+\frac{12}{8}=4 + 1.5=5.5\) (5.5>5)
4. \(4\frac{6}{8}+\frac{1}{8}\): \(4 + \frac{6 + 1}{8}=4+\frac{7}{8}=4.875\) (4.875 <5)

So the sum greater than 5 is \(4\frac{5}{8}+\frac{7}{8}\)

Answer:

\(4\frac{5}{8}+\frac{7}{8}\)