Fractions (Dividing... Maybe?) The question... (With mixed fractions)

$$3 \frac {3}{4} > 1 \frac{19}{21}$$
  How many times $3 \frac {3}{4}$ is bigger than $1 \frac{19}{21}$?

I assume you divide the first fraction by the second but I cant seem to do it, could someone maybe answer the question in steps so I can see what you did therefore understand this a bit better? That would be greatly appreciated, thanks!
I know this is stupid, I've been revising for 6 hours straight. It's difficult to get my head around this when I'm already terrible at multiplying and dividing fractions.
 A: You deduced (correctly, I might add) that the quantity you are looking for is $$\frac{3\frac{3}{4}}{1\frac{19}{21}}.$$ The first step is to combine the whole numbers with their fraction: $$3\frac{3}{4}=3+\frac{3}{4}=\frac{12}{4}+\frac{3}{4}=\frac{15}{4},$$ $$1\frac{19}{21}=1+\frac{19}{21}=\frac{21}{21}+\frac{19}{21}=\frac{40}{21}.$$ The problem becomes finding $$\frac{\frac{15}{4}}{\frac{40}{21}}.$$ Next, we realize that dividing by a number is the same as multiplying by it reciprocal: $$\frac{\frac{15}{4}}{\frac{40}{21}}=\frac{15}{4}\times\frac{21}{40}.$$ Now do normal fraction multiplication to finish the calculation.
A: A rule to memorize and know by rote is that the fraction $\frac a b$ divided by the fraction $\frac c d$ is the same answer as the $\frac a b$ times the reciprical of the fraction $\frac d c $.
So $\frac a b \div \frac cd = \frac {\frac a b }{\frac cd} = \frac a b \times \frac d c = \frac {ad}{bc} .$
The hard part is understanding or explaining why.
It can make sense it you think of it as dividing into c/d number of pieces but it can get confusing.  The best way to convince yourself is by this "trick".
$\frac a b \div \frac c d = \frac {\frac ab}{\frac cd} = \frac {\frac ab}{\frac cd}\times \frac {\frac dc}{\frac dc}=\frac {\frac ab \times \frac dc}{\frac cd \times \frac dc} = \frac {\frac ab \times \frac dc} 1 =\frac ab \times \frac dc $.
