Solve 1311/25:13/5/(4frac{3{7}+12/3)*2frac{1}{10}}

$$\frac { 13 \frac { 11 } { 25 } : 1 \frac { 3 } { 5 } } { ( 4 \frac { 3 } { 7 } + 1 \frac { 2 } { 3 } ) \cdot 2 \frac { 1 } { 10 } }$$

Evaluate

$$\frac{21}{32}=0.65625$$

Solution Steps

Divide \frac{13\times 25+11}{25} by \frac{1\times 5+3}{5} by multiplying \frac{13\times 25+11}{25} by the reciprocal of \frac{1\times 5+3}{5}.

$$\frac{\frac{\left(13\times 25+11\right)\times 5}{25\left(1\times 5+3\right)}}{\left(\frac{4\times 7+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Cancel out 5 in both numerator and denominator.

$$\frac{\frac{11+13\times 25}{5\left(3+5\right)}}{\left(\frac{4\times 7+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Multiply 13 and 25 to get 325.

$$\frac{\frac{11+325}{5\left(3+5\right)}}{\left(\frac{4\times 7+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Add 11 and 325 to get 336.

$$\frac{\frac{336}{5\left(3+5\right)}}{\left(\frac{4\times 7+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Add 3 and 5 to get 8.

$$\frac{\frac{336}{5\times 8}}{\left(\frac{4\times 7+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Multiply 5 and 8 to get 40.

$$\frac{\frac{336}{40}}{\left(\frac{4\times 7+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Reduce the fraction \frac{336}{40} to lowest terms by extracting and canceling out 8.

$$\frac{\frac{42}{5}}{\left(\frac{4\times 7+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Multiply 4 and 7 to get 28.

$$\frac{\frac{42}{5}}{\left(\frac{28+3}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Add 28 and 3 to get 31.

$$\frac{\frac{42}{5}}{\left(\frac{31}{7}+\frac{1\times 3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Multiply 1 and 3 to get 3.

$$\frac{\frac{42}{5}}{\left(\frac{31}{7}+\frac{3+2}{3}\right)\times \frac{2\times 10+1}{10}}$$

Add 3 and 2 to get 5.

$$\frac{\frac{42}{5}}{\left(\frac{31}{7}+\frac{5}{3}\right)\times \frac{2\times 10+1}{10}}$$

Least common multiple of 7 and 3 is 21. Convert \frac{31}{7} and \frac{5}{3} to fractions with denominator 21.

$$\frac{\frac{42}{5}}{\left(\frac{93}{21}+\frac{35}{21}\right)\times \frac{2\times 10+1}{10}}$$

Since \frac{93}{21} and \frac{35}{21} have the same denominator, add them by adding their numerators.

$$\frac{\frac{42}{5}}{\frac{93+35}{21}\times \frac{2\times 10+1}{10}}$$

Add 93 and 35 to get 128.

$$\frac{\frac{42}{5}}{\frac{128}{21}\times \frac{2\times 10+1}{10}}$$

Multiply 2 and 10 to get 20.

$$\frac{\frac{42}{5}}{\frac{128}{21}\times \frac{20+1}{10}}$$

Add 20 and 1 to get 21.

$$\frac{\frac{42}{5}}{\frac{128}{21}\times \frac{21}{10}}$$

Multiply \frac{128}{21} times \frac{21}{10} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{42}{5}}{\frac{128\times 21}{21\times 10}}$$

Cancel out 21 in both numerator and denominator.

$$\frac{\frac{42}{5}}{\frac{128}{10}}$$

Reduce the fraction \frac{128}{10} to lowest terms by extracting and canceling out 2.

$$\frac{\frac{42}{5}}{\frac{64}{5}}$$

Divide \frac{42}{5} by \frac{64}{5} by multiplying \frac{42}{5} by the reciprocal of \frac{64}{5}.

$$\frac{42}{5}\times \frac{5}{64}$$

Multiply \frac{42}{5} times \frac{5}{64} by multiplying numerator times numerator and denominator times denominator.

$$\frac{42\times 5}{5\times 64}$$

Cancel out 5 in both numerator and denominator.

$$\frac{42}{64}$$

Reduce the fraction \frac{42}{64} to lowest terms by extracting and canceling out 2.

$$\frac{21}{32}$$

Factor

$$\frac{3 \cdot 7}{2 ^ {5}} = 0.65625$$