Mixed Numbers: Additional Practice (Set 1)

Add the whole parts $4 + 2 = 6$. For the fractions, the common denominator of $8$ and $4$ is $8$: $\tfrac34 = \tfrac68$, so $\tfrac58 + \tfrac68 = \tfrac{11}{8} = 1\tfrac38$.

$6 + 1\tfrac{3}{8} = 7\tfrac{3}{8}$

The fractions passed one whole, so carry the $1$.

The fraction being subtracted is larger, so convert to improper fractions: $8\tfrac16 = \tfrac{49}{6}$ and $3\tfrac56 = \tfrac{23}{6}$.

$\tfrac{49}{6} - \tfrac{23}{6} = \tfrac{26}{6} = \tfrac{13}{3} = 4\tfrac{1}{3}$

The last step simplifies by the common factor $2$, then converts to a mixed number.

Divide $47$ by $6$. Six goes into $47$ seven times, using $42$, leaving $5$.

$47 \div 6 = 7 \text{ remainder } 5$

The quotient $7$ is the whole part and the remainder $5$ over $6$ is the fraction, so $\tfrac{47}{6} = 7\tfrac{5}{6}$.

Write both over a denominator of $8$. The whole number $5$ is $\tfrac{40}{8}$, and $2\tfrac38 = \tfrac{19}{8}$.

$\tfrac{40}{8} - \tfrac{19}{8} = \tfrac{21}{8} = 2\tfrac{5}{8}$

So $5 - 2\tfrac38 = 2\tfrac58$.

Add the whole parts $1 + 2 = 3$. For the fractions, the common denominator of $6$ and $3$ is $6$: $\tfrac23 = \tfrac46$, so $\tfrac56 + \tfrac46 = \tfrac96 = 1\tfrac12$.

$3 + 1\tfrac{1}{2} = 4\tfrac{1}{2}$

The fractions passed one whole, so carry the $1$.

Convert both to improper fractions: $7\tfrac12 = \tfrac{15}{2}$ and $2\tfrac12 = \tfrac{5}{2}$. Multiply by the reciprocal of the divisor and cancel.

$\tfrac{15}{2} \times \tfrac{2}{5} = \tfrac{30}{10} = 3$

So $2\tfrac12$ fits into $7\tfrac12$ exactly $3$ times.

Add the whole parts $4 + 3 = 7$. For the fractions, the least common denominator of $6$ and $4$ is $12$: $\tfrac16 = \tfrac{2}{12}$ and $\tfrac34 = \tfrac{9}{12}$.

$\tfrac{2}{12} + \tfrac{9}{12} = \tfrac{11}{12}$

The fraction total is proper, so nothing carries: $4\tfrac16 + 3\tfrac34 = 7\tfrac{11}{12}$.

The fraction being subtracted is larger, so convert to improper fractions: $9\tfrac14 = \tfrac{37}{4}$ and $5\tfrac23 = \tfrac{17}{3}$. The least common denominator is $12$.

$\tfrac{111}{12} - \tfrac{68}{12} = \tfrac{43}{12} = 3\tfrac{7}{12}$

So $9\tfrac14 - 5\tfrac23 = 3\tfrac{7}{12}$.

Convert both to improper fractions: $1\tfrac18 = \tfrac{9}{8}$ and $2\tfrac23 = \tfrac{8}{3}$. Multiply across and cancel.

$\tfrac{9}{8} \times \tfrac{8}{3} = \tfrac{72}{24} = 3$

So the product is the whole number $3$.

Add the whole parts $2 + 1 = 3$. For the fractions, the least common denominator of $4$ and $6$ is $12$: $\tfrac34 = \tfrac{9}{12}$ and $\tfrac56 = \tfrac{10}{12}$, so $\tfrac{9}{12} + \tfrac{10}{12} = \tfrac{19}{12} = 1\tfrac{7}{12}$.

$3 + 1\tfrac{7}{12} = 4\tfrac{7}{12}$

The fractions passed one whole, so carry the $1$.

Convert both to improper fractions: $10\tfrac12 = \tfrac{21}{2}$ and $3\tfrac12 = \tfrac{7}{2}$. Multiply by the reciprocal of the divisor and cancel.

$\tfrac{21}{2} \times \tfrac{2}{7} = \tfrac{42}{14} = 3$

So $3\tfrac12$ fits into $10\tfrac12$ exactly $3$ times.

The fraction being subtracted is larger, so convert to improper fractions: $6\tfrac25 = \tfrac{32}{5}$ and $2\tfrac45 = \tfrac{14}{5}$.

$\tfrac{32}{5} - \tfrac{14}{5} = \tfrac{18}{5} = 3\tfrac{3}{5}$

So $6\tfrac25 - 2\tfrac45 = 3\tfrac35$.