Solve 1/5*3/6+3/4:4/10-frac{1}{3}

$$\frac { 1 } { 5 } \cdot \frac { 3 } { 6 } + \frac { 3 } { 4 } : \frac { 4 } { 10 } - \frac { 1 } { 3 }$$

Evaluate

$$\frac{197}{120}\approx 1.641666667$$

Solution Steps

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$\frac{1}{5}\times \frac{1}{2}+\frac{\frac{3}{4}}{\frac{4}{10}}-\frac{1}{3}$$

Multiply \frac{1}{5} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 1}{5\times 2}+\frac{\frac{3}{4}}{\frac{4}{10}}-\frac{1}{3}$$

Do the multiplications in the fraction \frac{1\times 1}{5\times 2}.

$$\frac{1}{10}+\frac{\frac{3}{4}}{\frac{4}{10}}-\frac{1}{3}$$

Divide \frac{3}{4} by \frac{4}{10} by multiplying \frac{3}{4} by the reciprocal of \frac{4}{10}.

$$\frac{1}{10}+\frac{3\times 10}{4\times 4}-\frac{1}{3}$$

Cancel out 2 in both numerator and denominator.

$$\frac{1}{10}+\frac{3\times 5}{2\times 4}-\frac{1}{3}$$

Multiply 3 and 5 to get 15.

$$\frac{1}{10}+\frac{15}{2\times 4}-\frac{1}{3}$$

Multiply 2 and 4 to get 8.

$$\frac{1}{10}+\frac{15}{8}-\frac{1}{3}$$

Least common multiple of 10 and 8 is 40. Convert \frac{1}{10} and \frac{15}{8} to fractions with denominator 40.

$$\frac{4}{40}+\frac{75}{40}-\frac{1}{3}$$

Since \frac{4}{40} and \frac{75}{40} have the same denominator, add them by adding their numerators.

$$\frac{4+75}{40}-\frac{1}{3}$$

Add 4 and 75 to get 79.

$$\frac{79}{40}-\frac{1}{3}$$

Least common multiple of 40 and 3 is 120. Convert \frac{79}{40} and \frac{1}{3} to fractions with denominator 120.

$$\frac{237}{120}-\frac{40}{120}$$

Since \frac{237}{120} and \frac{40}{120} have the same denominator, subtract them by subtracting their numerators.

$$\frac{237-40}{120}$$

Subtract 40 from 237 to get 197.

$$\frac{197}{120}$$

Factor

$$\frac{197}{2 ^ {3} \cdot 3 \cdot 5} = 1\frac{77}{120} = 1.6416666666666666$$