In order to add these fractions, we must first find a common denominator for all three. Looking at our three different denominators, we can see that their lowest common denominator is 12. Therefore, we need to rewrite each fraction with a denominator of 12:

\(\displaystyle \frac{2}{3}+\frac{3}{4}+\frac{5}{6}=\frac{2(4)}{3(4)}+\frac{3(3)}{4(3)}+\frac{5(2)}{6(2)}=\frac{8}{12}+\frac{9}{12}+\frac{10}{12}\)

Now that the fractions all have a common denominator, we can simply add them together and then simplify the result, recognizing that 27 and 12 have a common factor of 3:

\(\displaystyle \frac{8}{12}+\frac{9}{12}+\frac{10}{12}=\frac{27}{12}=\frac{9}{4}\)

In order to determine which fractions are less than \(\displaystyle \frac{5}{8}\), it's best to convert them to decimals and compare:

\(\displaystyle \frac{5}{8}=0.625\)

\(\displaystyle \frac{1}{2}=0.5\)

\(\displaystyle \frac{3}{4}=0.75\)

\(\displaystyle \frac{2}{3}=0.67\)

\(\displaystyle \frac{5}{6}=0.83\)

\(\displaystyle \frac{7}{6}=1.167\)

Therefore, \(\displaystyle \frac{1}2{}\) is the only choice that is less than \(\displaystyle \frac{5}{8}\).

\(\displaystyle a \ast b = \frac{1}{4}a + b^{2}\)

\(\displaystyle \frac{2}{3} \ast \frac{1}{3}= \frac{1}{4} \cdot \frac{2}{3} + \left (\frac{1}{3} \right ) ^{2}\)

\(\displaystyle = \frac{2}{12} + \frac{1}{9}\)

\(\displaystyle = \frac{1}{6} + \frac{1}{9}\)

\(\displaystyle = \frac{3}{18} + \frac{2}{18}\)

\(\displaystyle = \frac{5}{18}\)

This question tests your ability to read word problems and multiply fractions. Because we know that Olivar has \(\displaystyle \frac{4}{7}\) of what Casius has, we can see that we need to multiply. Perform the following operation to find the answer.

\(\displaystyle \frac{1}{6}*\frac{4}{7}=\frac{4}{42}=\frac{2}{21}\)

\(\displaystyle a \ast b = 2a \div b\)

\(\displaystyle 5\frac{1}{6} \ast 3\frac{1}{3} = 2 \left ( 5\frac{1}{6} \right ) \div 3\frac{1}{3}\)

\(\displaystyle = 2 \left ( \frac{31}{6} \right ) \div \frac{10}{3}\)

\(\displaystyle = \frac{31}{3} \div \frac{10}{3}\)

\(\displaystyle = \frac{31}{3} \cdot \frac{3}{10}\)

\(\displaystyle = \frac{31} {10}\)

\(\displaystyle =3 \frac{1} {10}\)

This rounds to 3.

Seven thirds of eighteen seventieths is 

\(\displaystyle \frac{7}{3} \times \frac{18 }{70} = \frac{7}{3} \times \frac{9}{35} = \frac{1}{1} \times \frac{3}{5}= \frac{3}{5}\)

One half gallon comprises 8 cups. To find the number of oranges needed, divide:

\(\displaystyle 8 \div \frac{1}{3} = 24\) 

\(\displaystyle \frac{15}{75}\) reduces to \(\displaystyle \frac{15 \div 15}{75 \div 15} = \frac{1}{5}\). Raise this to the fourth power:

\(\displaystyle \left ( \frac{1}{5} \right )^{4} = \frac{1}{5^{4}} = \frac{1}{625}\)

\(\displaystyle \left (\frac{3}{7} \right )^{4} = \frac{3^{4} }{7^{4} } = \frac{81 }{2,401}\)

\(\displaystyle \left (-\frac{3}{7} \right )^{4} =\left (\frac{3}{7} \right )^{4} = \frac{81 }{2,401}\)

\(\displaystyle \left (\frac{3}{7} \right )^{4} + \left (-\frac{3}{7} \right )^{4}=\frac{81 }{2,401}+\frac{81 }{2,401}=\frac{162}{2,401}\)

This can actually be solved without any calculation.

A negative number raised to an even power is equal to its absolute value raised to that power, so 

\(\displaystyle \left (-\frac{5}{8} \right )^{4} = \left (\frac{5}{8} \right )^{4}\)

Therefore, we can replace:

\(\displaystyle -\left (\frac{5}{8} \right )^{4} + \left (\frac{5}{8} \right )^{4} = \left (\frac{5}{8} \right )^{4} -\left (\frac{5}{8} \right )^{4} =0\)