Solve 3:{1+1/3*[(1/4):1/48-1/10]*frac{1}{3}-frac{1}{6}}

$$3 : \{ 1 + \frac { 1 } { 3 } \times [ ( \frac { 1 } { 4 } ) : \frac { 1 } { 48 } - \frac { 1 } { 10 } ] \times \frac { 1 } { 3 } - \frac { 1 } { 6 } \}$$

Evaluate

$$\frac{135}{97}\approx 1.391752577$$

Solution Steps

Divide \frac{1}{4} by \frac{1}{48} by multiplying \frac{1}{4} by the reciprocal of \frac{1}{48}.

$$\frac{3}{1+\frac{1}{3}\left(\frac{1}{4}\times 48-\frac{1}{10}\right)\times \frac{1}{3}-\frac{1}{6}}$$

Multiply \frac{1}{4} and 48 to get \frac{48}{4}.

$$\frac{3}{1+\frac{1}{3}\left(\frac{48}{4}-\frac{1}{10}\right)\times \frac{1}{3}-\frac{1}{6}}$$

Divide 48 by 4 to get 12.

$$\frac{3}{1+\frac{1}{3}\left(12-\frac{1}{10}\right)\times \frac{1}{3}-\frac{1}{6}}$$

Convert 12 to fraction \frac{120}{10}.

$$\frac{3}{1+\frac{1}{3}\left(\frac{120}{10}-\frac{1}{10}\right)\times \frac{1}{3}-\frac{1}{6}}$$

Since \frac{120}{10} and \frac{1}{10} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{1+\frac{1}{3}\times \frac{120-1}{10}\times \frac{1}{3}-\frac{1}{6}}$$

Subtract 1 from 120 to get 119.

$$\frac{3}{1+\frac{1}{3}\times \frac{119}{10}\times \frac{1}{3}-\frac{1}{6}}$$

Multiply \frac{1}{3} times \frac{119}{10} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{1+\frac{1\times 119}{3\times 10}\times \frac{1}{3}-\frac{1}{6}}$$

Do the multiplications in the fraction \frac{1\times 119}{3\times 10}.

$$\frac{3}{1+\frac{119}{30}\times \frac{1}{3}-\frac{1}{6}}$$

Multiply \frac{119}{30} times \frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{1+\frac{119\times 1}{30\times 3}-\frac{1}{6}}$$

Do the multiplications in the fraction \frac{119\times 1}{30\times 3}.

$$\frac{3}{1+\frac{119}{90}-\frac{1}{6}}$$

Convert 1 to fraction \frac{90}{90}.

$$\frac{3}{\frac{90}{90}+\frac{119}{90}-\frac{1}{6}}$$

Since \frac{90}{90} and \frac{119}{90} have the same denominator, add them by adding their numerators.

$$\frac{3}{\frac{90+119}{90}-\frac{1}{6}}$$

Add 90 and 119 to get 209.

$$\frac{3}{\frac{209}{90}-\frac{1}{6}}$$

Least common multiple of 90 and 6 is 90. Convert \frac{209}{90} and \frac{1}{6} to fractions with denominator 90.

$$\frac{3}{\frac{209}{90}-\frac{15}{90}}$$

Since \frac{209}{90} and \frac{15}{90} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{\frac{209-15}{90}}$$

Subtract 15 from 209 to get 194.

$$\frac{3}{\frac{194}{90}}$$

Reduce the fraction \frac{194}{90} to lowest terms by extracting and canceling out 2.

$$\frac{3}{\frac{97}{45}}$$

Divide 3 by \frac{97}{45} by multiplying 3 by the reciprocal of \frac{97}{45}.

$$3\times \frac{45}{97}$$

Express 3\times \frac{45}{97} as a single fraction.

$$\frac{3\times 45}{97}$$

Multiply 3 and 45 to get 135.

$$\frac{135}{97}$$

Factor

$$\frac{3 ^ {3} \cdot 5}{97} = 1\frac{38}{97} = 1.3917525773195876$$