Solve lfloor(18+32)div8+3rfloordiv6

$$\lfloor (18+32) \div 8+3 \rfloor \div 6$$

Evaluate

$$\frac{3}{2}=1.5$$

Solution Steps

Add 18 and 32 to get 50.

$$\frac{\lfloor \frac{50}{8}+3\rfloor }{6}$$

Reduce the fraction \frac{50}{8} to lowest terms by extracting and canceling out 2.

$$\frac{\lfloor \frac{25}{4}+3\rfloor }{6}$$

Convert 3 to fraction \frac{12}{4}.

$$\frac{\lfloor \frac{25}{4}+\frac{12}{4}\rfloor }{6}$$

Since \frac{25}{4} and \frac{12}{4} have the same denominator, add them by adding their numerators.

$$\frac{\lfloor \frac{25+12}{4}\rfloor }{6}$$

Add 25 and 12 to get 37.

$$\frac{\lfloor \frac{37}{4}\rfloor }{6}$$

Dividing 37 by 4 gives 9 and remainder 1. Rewrite \frac{37}{4} as 9+\frac{1}{4}.

$$\frac{\lfloor 9+\frac{1}{4}\rfloor }{6}$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 9+\frac{1}{4} is 9.

$$\frac{9}{6}$$

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$\frac{3}{2}$$

Factor

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$