Solve (55/6)+(92/3)div(181/3)-40

$$( 5 \frac { 5 } { 6 } ) + ( 9 \frac { 2 } { 3 } ) \div ( 18 \frac { 1 } { 3 } ) - 40$$

Evaluate

$$-\frac{11101}{330}\approx -33.639393939$$

Solution Steps

Multiply 5 and 6 to get 30.

$$\frac{30+5}{6}+\frac{\frac{9\times 3+2}{3}}{\frac{18\times 3+1}{3}}-40$$

Add 30 and 5 to get 35.

$$\frac{35}{6}+\frac{\frac{9\times 3+2}{3}}{\frac{18\times 3+1}{3}}-40$$

Divide \frac{9\times 3+2}{3} by \frac{18\times 3+1}{3} by multiplying \frac{9\times 3+2}{3} by the reciprocal of \frac{18\times 3+1}{3}.

$$\frac{35}{6}+\frac{\left(9\times 3+2\right)\times 3}{3\left(18\times 3+1\right)}-40$$

Cancel out 3 in both numerator and denominator.

$$\frac{35}{6}+\frac{2+3\times 9}{1+3\times 18}-40$$

Multiply 3 and 9 to get 27.

$$\frac{35}{6}+\frac{2+27}{1+3\times 18}-40$$

Add 2 and 27 to get 29.

$$\frac{35}{6}+\frac{29}{1+3\times 18}-40$$

Multiply 3 and 18 to get 54.

$$\frac{35}{6}+\frac{29}{1+54}-40$$

Add 1 and 54 to get 55.

$$\frac{35}{6}+\frac{29}{55}-40$$

Least common multiple of 6 and 55 is 330. Convert \frac{35}{6} and \frac{29}{55} to fractions with denominator 330.

$$\frac{1925}{330}+\frac{174}{330}-40$$

Since \frac{1925}{330} and \frac{174}{330} have the same denominator, add them by adding their numerators.

$$\frac{1925+174}{330}-40$$

Add 1925 and 174 to get 2099.

$$\frac{2099}{330}-40$$

Convert 40 to fraction \frac{13200}{330}.

$$\frac{2099}{330}-\frac{13200}{330}$$

Since \frac{2099}{330} and \frac{13200}{330} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2099-13200}{330}$$

Subtract 13200 from 2099 to get -11101.

$$-\frac{11101}{330}$$

Factor

$$-\frac{11101}{330} = -33\frac{211}{330} = -33.63939393939394$$