Solve 5-(3-17/20)*9/25+1/8

$$5 - ( 3 - 1 \frac { 7 } { 20 } ) \cdot \frac { 9 } { 25 } + \frac { 1 } { 8 }$$

Evaluate

$$\frac{4531}{1000}=4.531$$

Solution Steps

Multiply 1 and 20 to get 20.

$$5-\left(3-\frac{20+7}{20}\right)\times \frac{9}{25}+\frac{1}{8}$$

Add 20 and 7 to get 27.

$$5-\left(3-\frac{27}{20}\right)\times \frac{9}{25}+\frac{1}{8}$$

Convert 3 to fraction \frac{60}{20}.

$$5-\left(\frac{60}{20}-\frac{27}{20}\right)\times \frac{9}{25}+\frac{1}{8}$$

Since \frac{60}{20} and \frac{27}{20} have the same denominator, subtract them by subtracting their numerators.

$$5-\frac{60-27}{20}\times \frac{9}{25}+\frac{1}{8}$$

Subtract 27 from 60 to get 33.

$$5-\frac{33}{20}\times \frac{9}{25}+\frac{1}{8}$$

Multiply \frac{33}{20} times \frac{9}{25} by multiplying numerator times numerator and denominator times denominator.

$$5-\frac{33\times 9}{20\times 25}+\frac{1}{8}$$

Do the multiplications in the fraction \frac{33\times 9}{20\times 25}.

$$5-\frac{297}{500}+\frac{1}{8}$$

Convert 5 to fraction \frac{2500}{500}.

$$\frac{2500}{500}-\frac{297}{500}+\frac{1}{8}$$

Since \frac{2500}{500} and \frac{297}{500} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2500-297}{500}+\frac{1}{8}$$

Subtract 297 from 2500 to get 2203.

$$\frac{2203}{500}+\frac{1}{8}$$

Least common multiple of 500 and 8 is 1000. Convert \frac{2203}{500} and \frac{1}{8} to fractions with denominator 1000.

$$\frac{4406}{1000}+\frac{125}{1000}$$

Since \frac{4406}{1000} and \frac{125}{1000} have the same denominator, add them by adding their numerators.

$$\frac{4406+125}{1000}$$

Add 4406 and 125 to get 4531.

$$\frac{4531}{1000}$$

Factor

$$\frac{23 \cdot 197}{2 ^ {3} \cdot 5 ^ {3}} = 4\frac{531}{1000} = 4.531$$