How to obtain variance of a random variable that depends on a hypergeometric variable? I have been given the following problem:

In an assembly line production of industrial robots, gearbox
assemblies can be installed in two minute each if holes have been
properly drilled in the boxes and in nice minutes if the holes must be
be redrilled. Twenty gearboxes are in stock, 2 with improperly drilled
holes. Five gearboxes must be selected from the 20 that are available
for installation in the next five robots.
i) Find the probability that all 5 gearbox will fit properly.
ii) Find the mean, variance, and standard deviation of the time
to install these 5 gearbox

I had no problem with the part (i), however, I was not sure about part (ii), so I checked the solution and I didn't understand where they got the variance of time from. This is the solution:

Any help to understand this will be appreciated.
 A: The time it takes to install each gearbox is 2 minutes if it is properly drilled and 9 minutes if it is not properly drilled. So for any arbitrary gearbox, it takes 2 minutes at least, with an additional 7 minutes if it is improperly drilled. Thus if $Y$ is the number of improperly drilled gearboxes, the time it takes to install 5 gearboxes is
$$T = 10 + 7Y.$$
The variance would then be
$$\mathrm{Var}(T) = \mathrm{Var}(10 + 7Y) = \mathrm{Var}(7Y) = 49\mathrm{Var}(Y).$$
(I'm using the properties of variance: $\mathrm{Var}(X + a) = \mathrm{Var}(X)$ and $\mathrm{Var}(aX) = a^2 \mathrm{Var}(X)$ for any random variable $X$ and constant $a$.)
The variance of the hypergeometric random variable $Y$ is $$\mathrm{Var}(Y) = n\frac{r}{N}\frac{N-r}{N}\frac{N-n}{N-1} = 5\frac{2}{20}\frac{18}{20}\frac{15}{19} = 0.3552632,$$
so the variance of the time $T$ is
$$\mathrm{Var}(T) = 49\mathrm{Var}(Y) = 49(0.3552632) = 17.4079.$$
The standard deviation is the square root of the variance, so $$\mathrm{SD}(T) = \sqrt{\mathrm{Var}(T)} = \sqrt{17.4079} = 4.172277.$$
Note: From the way the solution for part (b)/(ii) is written, it seems like each gearbox would take 1 minute if it is properly drilled and an additional 9 minutes if it is not (so a total of 10 minutes if it is improperly drilled). This is not the information I inferred from the statement of the problem.
Edit: For the proof of the variance of a hypergeometric random variable, go here: http://en.wikibooks.org/wiki/Statistics/Distributions/Hypergeometric
