Solve 1/frac{1{40}+1/20+1/1.7}

$$\frac { 1 } { \frac { 1 } { 40 } + \frac { 1 } { 20 } + \frac { 1 } { 1.7 } }$$

Evaluate

$$\frac{680}{451}\approx 1.507760532$$

Solution Steps

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

$$\frac{1}{\frac{1}{40}+\frac{2}{40}+\frac{1}{1.7}}$$

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

$$\frac{1}{\frac{1+2}{40}+\frac{1}{1.7}}$$

Add 1 and 2 to get 3.

$$\frac{1}{\frac{3}{40}+\frac{1}{1.7}}$$

Expand \frac{1}{1.7} by multiplying both numerator and the denominator by 10.

$$\frac{1}{\frac{3}{40}+\frac{10}{17}}$$

Least common multiple of 40 and 17 is 680. Convert \frac{3}{40} and \frac{10}{17} to fractions with denominator 680.

$$\frac{1}{\frac{51}{680}+\frac{400}{680}}$$

Since \frac{51}{680} and \frac{400}{680} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{51+400}{680}}$$

Add 51 and 400 to get 451.

$$\frac{1}{\frac{451}{680}}$$

Divide 1 by \frac{451}{680} by multiplying 1 by the reciprocal of \frac{451}{680}.

$$1\times \frac{680}{451}$$

Multiply 1 and \frac{680}{451} to get \frac{680}{451}.

$$\frac{680}{451}$$

Factor

$$\frac{5 \cdot 17 \cdot 2 ^ {3}}{11 \cdot 41} = 1\frac{229}{451} = 1.507760532150776$$