Solve 8/15div[5/32*(9/10+1/8)]

$$\frac { 8 } { 15 } \div [ \frac { 5 } { 32 } \times ( \frac { 9 } { 10 } + \frac { 1 } { 8 } ) ]$$

Evaluate

$$\frac{2048}{615}\approx 3.330081301$$

Solution Steps

Least common multiple of 10 and 8 is 40. Convert \frac{9}{10} and \frac{1}{8} to fractions with denominator 40.

$$\frac{\frac{8}{15}}{\frac{5}{32}\left(\frac{36}{40}+\frac{5}{40}\right)}$$

Since \frac{36}{40} and \frac{5}{40} have the same denominator, add them by adding their numerators.

$$\frac{\frac{8}{15}}{\frac{5}{32}\times \frac{36+5}{40}}$$

Add 36 and 5 to get 41.

$$\frac{\frac{8}{15}}{\frac{5}{32}\times \frac{41}{40}}$$

Multiply \frac{5}{32} times \frac{41}{40} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{8}{15}}{\frac{5\times 41}{32\times 40}}$$

Do the multiplications in the fraction \frac{5\times 41}{32\times 40}.

$$\frac{\frac{8}{15}}{\frac{205}{1280}}$$

Reduce the fraction \frac{205}{1280} to lowest terms by extracting and canceling out 5.

$$\frac{\frac{8}{15}}{\frac{41}{256}}$$

Divide \frac{8}{15} by \frac{41}{256} by multiplying \frac{8}{15} by the reciprocal of \frac{41}{256}.

$$\frac{8}{15}\times \frac{256}{41}$$

Multiply \frac{8}{15} times \frac{256}{41} by multiplying numerator times numerator and denominator times denominator.

$$\frac{8\times 256}{15\times 41}$$

Do the multiplications in the fraction \frac{8\times 256}{15\times 41}.

$$\frac{2048}{615}$$

Factor

$$\frac{2 ^ {11}}{3 \cdot 5 \cdot 41} = 3\frac{203}{615} = 3.330081300813008$$