Question 5: Find the sum of [tex]\(8 \frac{5}{6} + 2 \frac{4}{9}\)[/tex].

A. [tex]\(10 \frac{23}{18}\)[/tex]
B. [tex]\(11 \frac{3}{5}\)[/tex]
C. [tex]\(11 \frac{5}{18}\)[/tex]
D. [tex]\(12 \frac{7}{8}\)[/tex]



Answer :

Sure, let's find the sum of [tex]\(8 \frac{5}{6}\)[/tex] and [tex]\(2 \frac{4}{9}\)[/tex] step by step.

1. Convert the mixed numbers to improper fractions:

For [tex]\(8 \frac{5}{6}\)[/tex]:
[tex]\[ 8 \frac{5}{6} = \frac{8 \times 6 + 5}{6} = \frac{48 + 5}{6} = \frac{53}{6} \][/tex]

For [tex]\(2 \frac{4}{9}\)[/tex]:
[tex]\[ 2 \frac{4}{9} = \frac{2 \times 9 + 4}{9} = \frac{18 + 4}{9} = \frac{22}{9} \][/tex]

2. Find a common denominator for the fractions:

The denominators are 6 and 9. The least common multiple (LCM) of 6 and 9 is 18.

3. Convert the improper fractions to have the common denominator:

For [tex]\(\frac{53}{6}\)[/tex]:
[tex]\[ \frac{53}{6} = \frac{53 \times 3}{6 \times 3} = \frac{159}{18} \][/tex]

For [tex]\(\frac{22}{9}\)[/tex]:
[tex]\[ \frac{22}{9} = \frac{22 \times 2}{9 \times 2} = \frac{44}{18} \][/tex]

4. Add the fractions:

[tex]\[ \frac{159}{18} + \frac{44}{18} = \frac{159 + 44}{18} = \frac{203}{18} \][/tex]

5. Convert the resulting improper fraction back to a mixed number:

We know
[tex]\[ \frac{203}{18} \approx 11 \text{ R } 5 \][/tex]
where 11 is the quotient and 5 is the remainder, so:
[tex]\[ \frac{203}{18} = 11 \frac{5}{18} \][/tex]

6. Check if the fractional part can be simplified:

The greatest common divisor (gcd) of 5 and 18 is 1, so the fraction [tex]\(\frac{5}{18}\)[/tex] is already in its simplest form.

Therefore, the sum of [tex]\(8 \frac{5}{6}\)[/tex] and [tex]\(2 \frac{4}{9}\)[/tex] is:
[tex]\[ 11 \frac{5}{18} \][/tex]

Thus, the correct answer is:
[tex]\[ \boxed{11 \frac{5}{18}} \][/tex]

The answer is [tex]\( \text{(c)} \)[/tex].