11. Homework for 25 Sep — Fractions

Homework for Friday 25 September: parts 31 a and c and 32 c and e. Try them before opening the solution.

31. Combined operations

Find the irreducible fraction that results from the following operations with fractions.

a) $\displaystyle \dfrac{3}{8} - \dfrac{15}{16} : \dfrac{3}{4} + \dfrac{7}{20}$

The division comes before the addition and subtraction:

\dfrac{15}{16} : \dfrac{3}{4} = \dfrac{15 \cdot 4}{16 \cdot 3} = \dfrac{(3 \cdot 5) \cdot 2^2}{2^4 \cdot 3} = \dfrac{5}{4}

What is left is additions and subtractions. $\text{LCM}(8, 4, 20) = 40$:

\dfrac{3}{8} - \dfrac{5}{4} + \dfrac{7}{20} = \dfrac{15}{40} - \dfrac{50}{40} + \dfrac{14}{40} = -\dfrac{21}{40}

Common mistake: working out $\dfrac{3}{8} - \dfrac{15}{16}$ first. Until the division is done, the $\dfrac{15}{16}$ cannot be subtracted.

-\dfrac{21}{40}

c) $\displaystyle \left(\dfrac{90}{35} + \dfrac{54}{28}\right) : \left(\dfrac{16}{20} - 3\right)$

First bracket. Simplify $\dfrac{90}{35} = \dfrac{18}{7}$ and $\dfrac{54}{28} = \dfrac{27}{14}$:

\dfrac{18}{7} + \dfrac{27}{14} = \dfrac{36}{14} + \dfrac{27}{14} = \dfrac{63}{14} = \dfrac{9}{2}

Second bracket. Simplify $\dfrac{16}{20} = \dfrac{4}{5}$:

\dfrac{4}{5} - 3 = \dfrac{4}{5} - \dfrac{15}{5} = -\dfrac{11}{5}

Now the division, crosswise:

\dfrac{9}{2} : \left(-\dfrac{11}{5}\right) = -\dfrac{9 \cdot 5}{2 \cdot 11} = -\dfrac{45}{22}

-\dfrac{45}{22}

Find the irreducible fraction after carrying out the following combined operations. Remember the order of operations to reach the right result.

c) $\displaystyle 5 \cdot \left(\dfrac{6}{5} - \dfrac{9}{8} : \dfrac{1}{12}\right)$

Inside the bracket, the division comes before the subtraction:

\dfrac{9}{8} : \dfrac{1}{12} = \dfrac{9 \cdot 12}{8 \cdot 1} = \dfrac{108}{8} = \dfrac{27}{2}

\dfrac{6}{5} - \dfrac{27}{2} = \dfrac{12}{10} - \dfrac{135}{10} = -\dfrac{123}{10}

Finally, the multiplication:

5 \cdot \left(-\dfrac{123}{10}\right) = -\dfrac{5 \cdot 123}{2 \cdot 5} = -\dfrac{123}{2}

-\dfrac{123}{2}

e) $\displaystyle \dfrac{6}{32} : \dfrac{5}{8} - \dfrac{5}{4} : \left(\dfrac{8}{7} - \dfrac{10}{8}\right)$

The bracket first. Simplify $\dfrac{10}{8} = \dfrac{5}{4}$; $\text{LCM}(7, 4) = 28$:

\dfrac{8}{7} - \dfrac{5}{4} = \dfrac{32}{28} - \dfrac{35}{28} = -\dfrac{3}{28}

Then the two divisions, simplifying $\dfrac{6}{32} = \dfrac{3}{16}$:

\dfrac{3}{16} : \dfrac{5}{8} = \dfrac{3 \cdot 8}{16 \cdot 5} = \dfrac{3}{10} \qquad \dfrac{5}{4} : \left(-\dfrac{3}{28}\right) = -\dfrac{5 \cdot 28}{4 \cdot 3} = -\dfrac{35}{3}

And the subtraction. Minus times minus is plus; $\text{LCM}(10, 3) = 30$:

\dfrac{3}{10} - \left(-\dfrac{35}{3}\right) = \dfrac{3}{10} + \dfrac{35}{3} = \dfrac{9}{30} + \dfrac{350}{30} = \dfrac{359}{30}

\dfrac{359}{30}