Solve ((824div9)*0.4)+(0.6*((93+90+100)div3)

$$((824 \div 9) \times 0.4)+(0.6 \times ((93+90+100) \div 3)$$

Evaluate

$$\frac{839}{9}\approx 93.222222222$$

Solution Steps

Convert decimal number 0.4 to fraction \frac{4}{10}. Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$\frac{824}{9}\times \frac{2}{5}+0.6\times \frac{93+90+100}{3}$$

Multiply \frac{824}{9} times \frac{2}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{824\times 2}{9\times 5}+0.6\times \frac{93+90+100}{3}$$

Do the multiplications in the fraction \frac{824\times 2}{9\times 5}.

$$\frac{1648}{45}+0.6\times \frac{93+90+100}{3}$$

Add 93 and 90 to get 183.

$$\frac{1648}{45}+0.6\times \frac{183+100}{3}$$

Add 183 and 100 to get 283.

$$\frac{1648}{45}+0.6\times \frac{283}{3}$$

Convert decimal number 0.6 to fraction \frac{6}{10}. Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$\frac{1648}{45}+\frac{3}{5}\times \frac{283}{3}$$

Multiply \frac{3}{5} times \frac{283}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1648}{45}+\frac{3\times 283}{5\times 3}$$

Cancel out 3 in both numerator and denominator.

$$\frac{1648}{45}+\frac{283}{5}$$

Least common multiple of 45 and 5 is 45. Convert \frac{1648}{45} and \frac{283}{5} to fractions with denominator 45.

$$\frac{1648}{45}+\frac{2547}{45}$$

Since \frac{1648}{45} and \frac{2547}{45} have the same denominator, add them by adding their numerators.

$$\frac{1648+2547}{45}$$

Add 1648 and 2547 to get 4195.

$$\frac{4195}{45}$$

Reduce the fraction \frac{4195}{45} to lowest terms by extracting and canceling out 5.

$$\frac{839}{9}$$

Factor

$$\frac{839}{3 ^ {2}} = 93\frac{2}{9} = 93.22222222222223$$