When multiplying fractions you simply multiply the numerators to find the numerator of the product, and multiply the denominators to find the denominator of the product.

$\displaystyle \small 3\cdot 4=12$

$\displaystyle \small 5\cdot 15=75$

So, our product is $\displaystyle \small \frac{12}{75}$.

Since both of these numbers are divisible by three, we simplify. 

$\displaystyle \small \frac{12}{3}=4$

$\displaystyle \small \frac{75}{3}= 25$

Our final answer is $\displaystyle \small \frac{4}{25}$.

Rewrite the mixed numbers as improper fractions, then multiply across:

$\displaystyle 3 \frac{1}{3} = \frac{3\times 3 + 1}{3} = \frac{10}{3}$

$\displaystyle 4 \frac{1}{5} = \frac{4 \times 5 + 1}{5} = \frac{21}{5}$

$\displaystyle 3 \frac{1}{3} \times 4 \frac{1}{5} = \frac{10}{3} \times \frac{21}{5} = \frac{2}{1} \times \frac{7}{1} = 14$

$\displaystyle 14 > 12 \frac{1}{5}$

Rewrite the factors as improper fractions, then multiply across:

$\displaystyle 6 \frac{5}{6} = \frac{6\times 6 + 5}{6} = \frac{36 + 5}{6} = \frac{41}{6}$

$\displaystyle 6 = \frac{6}{1}$

$\displaystyle 6 \frac{5}{6} \times 6 =\frac{41}{6}\times \frac{ 6}{1}=\frac{41}{1}\times \frac{ 1}{1} = 41$

(a) $\displaystyle 0.75 t = 0.3$

$\displaystyle 0.75 t \div 0.75 = 0.3 \div 0.75$

$\displaystyle t= 0.3 \div 0.75$

Divide by moving the decimal point right two places in both numbers:

$\displaystyle t= 30 \div 75 = 0.4$

(b) $\displaystyle \frac{7}{4} k = \frac{7}{10}$

$\displaystyle \frac{4} {7} \cdot \frac{7}{4} k =\frac{4} {7} \cdot \frac{7}{10}$

$\displaystyle k =\frac{4} {7} \cdot \frac{7}{10}$

Cross-cancel:

$\displaystyle k =\frac{2} {1} \cdot \frac{1}{5} = \frac{2} {5}$

$\displaystyle 0.4 = \frac{4}{10 } = \frac{4\div 2}{10\div 2 } = \frac{2}{5}$, so $\displaystyle k = t$

$\displaystyle \frac{3}{5} = 3 \div 5 = 0.6$, so 

$\displaystyle \left ( \frac{3}{5} \right )^{3} = 0.6^{3} = 0.6 \times 0.6 \times 0.6 = 0.36 \times 0.6 = 0.216$

$\displaystyle 0.6 ^{2} = 0.6 \times 0.6 = 0.36$

$\displaystyle 0.36 > 0.216$, so  $\displaystyle 0.6^{2} > \left ( \frac{3}{5} \right )^{3}$, making (B) greater.

$\displaystyle \left ( \frac{4}{5} \right )^{3} = 0.8 ^{3} = 0.8 \times 0.8 \times 0,8 = 0.64\times 0,8 = 0.512$

Since $\displaystyle 0.512 > 0.5$, $\displaystyle \left ( \frac{4}{5} \right )^{3} > 0.5$, so (B) is greater.

$\displaystyle \left (\frac{3}{5} \right )^{2} = \frac{3^{2}}{5^{2}} = \frac{9}{25}$

$\displaystyle 1 - \left ( \frac{2}{5} \right )^{2} = 1- \frac{2^{2}}{5^{2}} = 1- \frac{4}{25} = \frac{25}{25}- \frac{4}{25} = \frac{25-4}{25}= \frac{21}{25}$

$\displaystyle \frac{21}{25} > \frac{9}{25}$, so (B) is greater

$\displaystyle \left ( \frac{7}{25} \right )^{2}+ \left ( \frac{24}{25} \right )^{2}$

$\displaystyle =\frac{7^{2}}{25^{2}} + \frac{24^{2}}{25^{2}}$

$\displaystyle =\frac{49}{625 } + \frac{576}{625 } = \frac{49+576}{625 } = \frac{625}{625 } = 1$

$\displaystyle 0.8 ^{2} + 0.6^{2}$

$\displaystyle =0.8 \times 0.8 + 0.6 \times 0.6$

$\displaystyle =0.64 + 0.36 = 1$

The expressions are equal.

Solve for (A):

$\displaystyle ab = 1.5 \cdot 4.5 = 6.75$

Solve for (B):

$\displaystyle a + b = 1.5 + 4.5 = 6$

Since $\displaystyle 6.75>6$, $\displaystyle ab > a+b$ and (A) is greater.

When dealing with problems that involve fractions and whole numbers, first convert your whole number into a fraction. You can easily do this by putting the whole number over 1.

$\displaystyle 12 \rightarrow\tfrac{12}{1}$

$\displaystyle \tfrac{3}{4}\times12=\tfrac{3}{4}\times\tfrac{12}{1}$

When multiplying, multiply the numerators by each other and the denominators by each other. 

$\displaystyle \tfrac{3}{4} \times\tfrac{12}{1} = \tfrac{3\times12}{4\times1}=\tfrac{36}{4}$

Always reduce a fraction when possible. In this case, the numerator and denominator of  $\displaystyle \tfrac{36}{4}$ are both divisible by 4. So, it reduces to 9.

$\displaystyle \tfrac{36}{4}=\tfrac{36\div4}{4\div4}=\tfrac{9}{1}=9$

$\displaystyle \tfrac{36}{4} \rightarrow\tfrac{9}{1}\rightarrow9$