Solve 31/10*3/10+7/5div1/20=?

$$\frac { 31 } { 10 } \times \frac { 3 } { 10 } + \frac { 7 } { 5 } \div \frac { 1 } { 20 } = ?$$

Evaluate

$$\frac{2893}{100}=28.93$$

Solution Steps

Multiply \frac{31}{10} times \frac{3}{10} by multiplying numerator times numerator and denominator times denominator.

$$\frac{31\times 3}{10\times 10}+\frac{\frac{7}{5}}{\frac{1}{20}}$$

Do the multiplications in the fraction \frac{31\times 3}{10\times 10}.

$$\frac{93}{100}+\frac{\frac{7}{5}}{\frac{1}{20}}$$

Divide \frac{7}{5} by \frac{1}{20} by multiplying \frac{7}{5} by the reciprocal of \frac{1}{20}.

$$\frac{93}{100}+\frac{7}{5}\times 20$$

Express \frac{7}{5}\times 20 as a single fraction.

$$\frac{93}{100}+\frac{7\times 20}{5}$$

Multiply 7 and 20 to get 140.

$$\frac{93}{100}+\frac{140}{5}$$

Divide 140 by 5 to get 28.

$$\frac{93}{100}+28$$

Convert 28 to fraction \frac{2800}{100}.

$$\frac{93}{100}+\frac{2800}{100}$$

Since \frac{93}{100} and \frac{2800}{100} have the same denominator, add them by adding their numerators.

$$\frac{93+2800}{100}$$

Add 93 and 2800 to get 2893.

$$\frac{2893}{100}$$

Factor

$$\frac{11 \cdot 263}{2 ^ {2} \cdot 5 ^ {2}} = 28\frac{93}{100} = 28.93$$