Solve [(1/15+1/8)*5-(5/72-frac{1/16)*frac{6}{5}]*frac{2}{19}}{(frac{1}{20}-frac{7}{180}):(frac{1}{20}+frac{1}{40}-frac{23}{360})}

$$\frac { [ ( \frac { 1 } { 15 } + \frac { 1 } { 8 } ) \times 5 - ( \frac { 5 } { 72 } - \frac { 1 } { 16 } ) \times \frac { 6 } { 5 } ] \times \frac { 2 } { 19 } } { ( \frac { 1 } { 20 } - \frac { 7 } { 180 } ) : ( \frac { 1 } { 20 } + \frac { 1 } { 40 } - \frac { 23 } { 360 } ) }$$

Evaluate

$$\frac{1}{10}=0.1$$

Solution Steps

Least common multiple of 15 and 8 is 120. Convert \frac{1}{15} and \frac{1}{8} to fractions with denominator 120.

$$\frac{\left(\left(\frac{8}{120}+\frac{15}{120}\right)\times 5-\left(\frac{5}{72}-\frac{1}{16}\right)\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Since \frac{8}{120} and \frac{15}{120} have the same denominator, add them by adding their numerators.

$$\frac{\left(\frac{8+15}{120}\times 5-\left(\frac{5}{72}-\frac{1}{16}\right)\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Add 8 and 15 to get 23.

$$\frac{\left(\frac{23}{120}\times 5-\left(\frac{5}{72}-\frac{1}{16}\right)\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Express \frac{23}{120}\times 5 as a single fraction.

$$\frac{\left(\frac{23\times 5}{120}-\left(\frac{5}{72}-\frac{1}{16}\right)\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Multiply 23 and 5 to get 115.

$$\frac{\left(\frac{115}{120}-\left(\frac{5}{72}-\frac{1}{16}\right)\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Reduce the fraction \frac{115}{120} to lowest terms by extracting and canceling out 5.

$$\frac{\left(\frac{23}{24}-\left(\frac{5}{72}-\frac{1}{16}\right)\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Least common multiple of 72 and 16 is 144. Convert \frac{5}{72} and \frac{1}{16} to fractions with denominator 144.

$$\frac{\left(\frac{23}{24}-\left(\frac{10}{144}-\frac{9}{144}\right)\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Since \frac{10}{144} and \frac{9}{144} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(\frac{23}{24}-\frac{10-9}{144}\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Subtract 9 from 10 to get 1.

$$\frac{\left(\frac{23}{24}-\frac{1}{144}\times \frac{6}{5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Multiply \frac{1}{144} times \frac{6}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(\frac{23}{24}-\frac{1\times 6}{144\times 5}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Do the multiplications in the fraction \frac{1\times 6}{144\times 5}.

$$\frac{\left(\frac{23}{24}-\frac{6}{720}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Reduce the fraction \frac{6}{720} to lowest terms by extracting and canceling out 6.

$$\frac{\left(\frac{23}{24}-\frac{1}{120}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Least common multiple of 24 and 120 is 120. Convert \frac{23}{24} and \frac{1}{120} to fractions with denominator 120.

$$\frac{\left(\frac{115}{120}-\frac{1}{120}\right)\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Since \frac{115}{120} and \frac{1}{120} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{115-1}{120}\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Subtract 1 from 115 to get 114.

$$\frac{\frac{114}{120}\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Reduce the fraction \frac{114}{120} to lowest terms by extracting and canceling out 6.

$$\frac{\frac{19}{20}\times \frac{2}{19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Multiply \frac{19}{20} times \frac{2}{19} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{19\times 2}{20\times 19}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Cancel out 19 in both numerator and denominator.

$$\frac{\frac{2}{20}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Reduce the fraction \frac{2}{20} to lowest terms by extracting and canceling out 2.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{20}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Least common multiple of 20 and 180 is 180. Convert \frac{1}{20} and \frac{7}{180} to fractions with denominator 180.

$$\frac{\frac{1}{10}}{\frac{\frac{9}{180}-\frac{7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Since \frac{9}{180} and \frac{7}{180} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{1}{10}}{\frac{\frac{9-7}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Subtract 7 from 9 to get 2.

$$\frac{\frac{1}{10}}{\frac{\frac{2}{180}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Reduce the fraction \frac{2}{180} to lowest terms by extracting and canceling out 2.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{1}{20}+\frac{1}{40}-\frac{23}{360}}}$$

Least common multiple of 20 and 40 is 40. Convert \frac{1}{20} and \frac{1}{40} to fractions with denominator 40.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{2}{40}+\frac{1}{40}-\frac{23}{360}}}$$

Since \frac{2}{40} and \frac{1}{40} have the same denominator, add them by adding their numerators.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{2+1}{40}-\frac{23}{360}}}$$

Add 2 and 1 to get 3.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{3}{40}-\frac{23}{360}}}$$

Least common multiple of 40 and 360 is 360. Convert \frac{3}{40} and \frac{23}{360} to fractions with denominator 360.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{27}{360}-\frac{23}{360}}}$$

Since \frac{27}{360} and \frac{23}{360} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{27-23}{360}}}$$

Subtract 23 from 27 to get 4.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{4}{360}}}$$

Reduce the fraction \frac{4}{360} to lowest terms by extracting and canceling out 4.

$$\frac{\frac{1}{10}}{\frac{\frac{1}{90}}{\frac{1}{90}}}$$

Divide \frac{1}{90} by \frac{1}{90} to get 1.

$$\frac{\frac{1}{10}}{1}$$

Anything divided by one gives itself.

$$\frac{1}{10}$$

Factor

$$\frac{1}{2 \cdot 5} = 0.1$$