Solve (1+1/7)*7/32

$$( 1 + \frac { 1 } { 7 } ) \times \frac { 7 } { 32 }$$

Evaluate

$$\frac{1}{4}=0.25$$

Solution Steps

Convert 1 to fraction \frac{7}{7}.

$$\left(\frac{7}{7}+\frac{1}{7}\right)\times \frac{7}{32}$$

Since \frac{7}{7} and \frac{1}{7} have the same denominator, add them by adding their numerators.

$$\frac{7+1}{7}\times \frac{7}{32}$$

Add 7 and 1 to get 8.

$$\frac{8}{7}\times \frac{7}{32}$$

Multiply \frac{8}{7} times \frac{7}{32} by multiplying numerator times numerator and denominator times denominator.

$$\frac{8\times 7}{7\times 32}$$

Cancel out 7 in both numerator and denominator.

$$\frac{8}{32}$$

Reduce the fraction \frac{8}{32} to lowest terms by extracting and canceling out 8.

$$\frac{1}{4}$$

Factor

$$\frac{1}{2 ^ {2}} = 0.25$$