Solve lfloor(18+3.2)div8+.3rfloordiv6

$$\lfloor (18+3.2) \div 8+.3 \rfloor \div 6$$

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Solution Steps

Add 18 and 3.2 to get 21.2.

$$\frac{\lfloor \frac{21.2}{8}+0.3\rfloor }{6}$$

Expand \frac{21.2}{8} by multiplying both numerator and the denominator by 10.

$$\frac{\lfloor \frac{212}{80}+0.3\rfloor }{6}$$

Reduce the fraction \frac{212}{80} to lowest terms by extracting and canceling out 4.

$$\frac{\lfloor \frac{53}{20}+0.3\rfloor }{6}$$

Convert decimal number 0.3 to fraction \frac{3}{10}.

$$\frac{\lfloor \frac{53}{20}+\frac{3}{10}\rfloor }{6}$$

Least common multiple of 20 and 10 is 20. Convert \frac{53}{20} and \frac{3}{10} to fractions with denominator 20.

$$\frac{\lfloor \frac{53}{20}+\frac{6}{20}\rfloor }{6}$$

Since \frac{53}{20} and \frac{6}{20} have the same denominator, add them by adding their numerators.

$$\frac{\lfloor \frac{53+6}{20}\rfloor }{6}$$

Add 53 and 6 to get 59.

$$\frac{\lfloor \frac{59}{20}\rfloor }{6}$$

Dividing 59 by 20 gives 2 and remainder 19. Rewrite \frac{59}{20} as 2+\frac{19}{20}.

$$\frac{\lfloor 2+\frac{19}{20}\rfloor }{6}$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 2+\frac{19}{20} is 2.

$$\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$\frac{1}{3}$$

Factor

$$\frac{1}{3} = 0.3333333333333333$$