Mixed Numbers: Practice

Add the whole parts $2 + 3 = 5$. For the fractions, the least common denominator of $6$ and $4$ is $12$: $\tfrac56 = \tfrac{10}{12}$ and $\tfrac34 = \tfrac{9}{12}$, so $\tfrac{10}{12} + \tfrac{9}{12} = \tfrac{19}{12} = 1\tfrac{7}{12}$.

$5 + 1\tfrac{7}{12} = 6\tfrac{7}{12}$

The fraction parts passed one whole, so carry the $1$ into the whole-number part.

Convert both to improper fractions: $4\tfrac25 = \tfrac{22}{5}$ and $1\tfrac{1}{10} = \tfrac{11}{10}$. Dividing means multiplying by the reciprocal of the divisor.

$\tfrac{22}{5} \times \tfrac{10}{11} = \tfrac{22 \times 10}{5 \times 11} = \tfrac{220}{55} = 4$

Only the divisor is flipped.

Convert both to improper fractions: $3\tfrac13 = \tfrac{10}{3}$ and $1\tfrac45 = \tfrac{9}{5}$. Multiply across and cancel.

$\tfrac{10}{3} \times \tfrac{9}{5} = \tfrac{10 \times 9}{3 \times 5} = \tfrac{90}{15} = 6$

So the product is the whole number $6$.

Write both as fifths. The whole number $3$ is $\tfrac{15}{5}$, and $1\tfrac25 = \tfrac{7}{5}$.

$\tfrac{15}{5} - \tfrac{7}{5} = \tfrac{8}{5} = 1\tfrac{3}{5}$

So $3 - 1\tfrac25 = 1\tfrac35$.

Add the whole parts $5 + 2 = 7$ and the fraction parts $\tfrac16 + \tfrac56 = \tfrac66 = 1$.

$7 + 1 = 8$

The fraction parts make exactly one whole, so carry the $1$ and the answer is the whole number $8$.

Divide $100$ by $7$. Seven goes into $100$ fourteen times, using $98$, leaving $2$.

$100 \div 7 = 14 \text{ remainder } 2$

The quotient $14$ is the whole part and the remainder $2$ over $7$ is the fraction, so $\tfrac{100}{7} = 14\tfrac{2}{7}$.

"How many pieces fit" is a division: $10 \div 1\tfrac14$. Convert the divisor: $1\tfrac14 = \tfrac54$, and $10 = \tfrac{10}{1}$. Multiply by the reciprocal.

$\tfrac{10}{1} \times \tfrac{4}{5} = \tfrac{40}{5} = 8$

So there are $8$ pieces.

Convert both to improper fractions: $1\tfrac23 = \tfrac{5}{3}$ and $2\tfrac14 = \tfrac{9}{4}$. Multiply across and cancel.

$\tfrac{5}{3} \times \tfrac{9}{4} = \tfrac{45}{12} = \tfrac{15}{4} = 3\tfrac{3}{4}$

The last step simplifies by the common factor $3$, then converts to a mixed number.

Add the whole parts $2 + 1 + 1 = 4$. For the fractions, the least common denominator of $6$, $4$, and $12$ is $12$: $\tfrac16 = \tfrac{2}{12}$, $\tfrac34 = \tfrac{9}{12}$, and $\tfrac{1}{12}$ stays.

$\tfrac{2}{12} + \tfrac{9}{12} + \tfrac{1}{12} = \tfrac{12}{12} = 1$

The fractions make one whole, so $4 + 1 = 5$.

The whole parts tie at $3$, so compare the fraction parts $\tfrac25$ and $\tfrac38$ over a common denominator $40$: $\tfrac25 = \tfrac{16}{40}$ and $\tfrac38 = \tfrac{15}{40}$.

$\tfrac{16}{40} > \tfrac{15}{40}$

Since $\tfrac25 > \tfrac38$, the whole comparison gives $3\tfrac25 > 3\tfrac38$.

Convert both to improper fractions: $1\tfrac13 = \tfrac{4}{3}$ and $3\tfrac12 = \tfrac{7}{2}$. Multiply by the reciprocal of the divisor.

$\tfrac{4}{3} \times \tfrac{2}{7} = \tfrac{4 \times 2}{3 \times 7} = \tfrac{8}{21}$

Since $8$ and $21$ share no factor above $1$, this proper fraction is the final answer; the result is less than $1$ because the divisor is larger than what you started with.

Four days at $3\tfrac12$ miles each is $3\tfrac12 \times 4$. Convert: $3\tfrac12 = \tfrac{7}{2}$ and $4 = \tfrac{4}{1}$.

$\tfrac{7}{2} \times \tfrac{4}{1} = \tfrac{28}{2} = 14$

So the runner covers $14$ miles.