Solve (181/4+21/6):11/4

$$( 18 \frac { 1 } { 4 } + 2 \frac { 1 } { 6 } ) : 1 \frac { 1 } { 4 }$$

Evaluate

$$\frac{49}{3}\approx 16.333333333$$

Solution Steps

Multiply 18 and 4 to get 72.

$$\frac{\frac{72+1}{4}+\frac{2\times 6+1}{6}}{\frac{1\times 4+1}{4}}$$

Add 72 and 1 to get 73.

$$\frac{\frac{73}{4}+\frac{2\times 6+1}{6}}{\frac{1\times 4+1}{4}}$$

Multiply 2 and 6 to get 12.

$$\frac{\frac{73}{4}+\frac{12+1}{6}}{\frac{1\times 4+1}{4}}$$

Add 12 and 1 to get 13.

$$\frac{\frac{73}{4}+\frac{13}{6}}{\frac{1\times 4+1}{4}}$$

Least common multiple of 4 and 6 is 12. Convert \frac{73}{4} and \frac{13}{6} to fractions with denominator 12.

$$\frac{\frac{219}{12}+\frac{26}{12}}{\frac{1\times 4+1}{4}}$$

Since \frac{219}{12} and \frac{26}{12} have the same denominator, add them by adding their numerators.

$$\frac{\frac{219+26}{12}}{\frac{1\times 4+1}{4}}$$

Add 219 and 26 to get 245.

$$\frac{\frac{245}{12}}{\frac{1\times 4+1}{4}}$$

Multiply 1 and 4 to get 4.

$$\frac{\frac{245}{12}}{\frac{4+1}{4}}$$

Add 4 and 1 to get 5.

$$\frac{\frac{245}{12}}{\frac{5}{4}}$$

Divide \frac{245}{12} by \frac{5}{4} by multiplying \frac{245}{12} by the reciprocal of \frac{5}{4}.

$$\frac{245}{12}\times \frac{4}{5}$$

Multiply \frac{245}{12} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{245\times 4}{12\times 5}$$

Do the multiplications in the fraction \frac{245\times 4}{12\times 5}.

$$\frac{980}{60}$$

Reduce the fraction \frac{980}{60} to lowest terms by extracting and canceling out 20.

$$\frac{49}{3}$$

Factor

$$\frac{7 ^ {2}}{3} = 16\frac{1}{3} = 16.333333333333332$$