Solve 19/20+(6/5-7/40)-7/10

$$\frac{ 19 }{ 20 } +( \frac{ 6 }{ 5 } - \frac{ 7 }{ 40 } )- \frac{ 7 }{ 10 }$$

Evaluate

$$\frac{51}{40}=1.275$$

Solution Steps

Least common multiple of 20 and 5 is 20. Convert \frac{19}{20} and \frac{6}{5} to fractions with denominator 20.

$$\frac{19}{20}+\frac{24}{20}-\frac{7}{40}-\frac{7}{10}$$

Since \frac{19}{20} and \frac{24}{20} have the same denominator, add them by adding their numerators.

$$\frac{19+24}{20}-\frac{7}{40}-\frac{7}{10}$$

Add 19 and 24 to get 43.

$$\frac{43}{20}-\frac{7}{40}-\frac{7}{10}$$

Least common multiple of 20 and 40 is 40. Convert \frac{43}{20} and \frac{7}{40} to fractions with denominator 40.

$$\frac{86}{40}-\frac{7}{40}-\frac{7}{10}$$

Since \frac{86}{40} and \frac{7}{40} have the same denominator, subtract them by subtracting their numerators.

$$\frac{86-7}{40}-\frac{7}{10}$$

Subtract 7 from 86 to get 79.

$$\frac{79}{40}-\frac{7}{10}$$

Least common multiple of 40 and 10 is 40. Convert \frac{79}{40} and \frac{7}{10} to fractions with denominator 40.

$$\frac{79}{40}-\frac{28}{40}$$

Since \frac{79}{40} and \frac{28}{40} have the same denominator, subtract them by subtracting their numerators.

$$\frac{79-28}{40}$$

Subtract 28 from 79 to get 51.

$$\frac{51}{40}$$

Factor

$$\frac{3 \cdot 17}{2 ^ {3} \cdot 5} = 1\frac{11}{40} = 1.275$$