Perform the indicated operation.

[tex]\[1 \frac{2}{3} + 5 \frac{3}{8}\][/tex]

A. [tex]\[6 \frac{1}{24}\][/tex]
B. [tex]\[6 \frac{5}{24}\][/tex]
C. [tex]\[7 \frac{1}{24}\][/tex]



Answer :

To add the mixed numbers [tex]\( 1 \frac{2}{3} \)[/tex] and [tex]\( 5 \frac{3}{8} \)[/tex], we can follow these steps:

1. Convert the mixed numbers to improper fractions:

[tex]\[ 1 \frac{2}{3} = \frac{5}{3} \quad \text{(1 whole and 2/3 to improper fraction: \( 1 \times 3 + 2 = 5 \) over 3)} \][/tex]

[tex]\[ 5 \frac{3}{8} = \frac{43}{8} \quad \text{(5 whole and 3/8 to improper fraction: \( 5 \times 8 + 3 = 43 \) over 8)} \][/tex]

2. Find a common denominator for the two fractions. The least common multiple (LCM) of 3 and 8 is 24.

3. Convert the fractions to have the common denominator:

[tex]\[ \frac{5}{3} = \frac{5 \times 8}{3 \times 8} = \frac{40}{24} \][/tex]

[tex]\[ \frac{43}{8} = \frac{43 \times 3}{8 \times 3} = \frac{129}{24} \][/tex]

4. Add the fractions:

[tex]\[ \frac{40}{24} + \frac{129}{24} = \frac{169}{24} \][/tex]

5. Convert the improper fraction back to a mixed number. Divide 169 by 24:

[tex]\[ 169 \div 24 = 7 \quad \text{(whole number)} \][/tex]

The remainder is:

[tex]\[ 169 - 7 \times 24 = 169 - 168 = 1 \][/tex]

So the remainder is 1, making the fraction part [tex]\(\frac{1}{24}\)[/tex].

Therefore, the final result in mixed number format is:

[tex]\[ 7 \frac{1}{24} \][/tex]

So, the correct answer is [tex]\( 7 \frac{1}{24} \)[/tex].

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