Variance of the sum of two poisson random variables

Hi, I have this problem, the solution of which is unclear to me. It's as follows.

Let the Poisson random variable U be the number of calls for technicals assistance received by a computer firm during the firm's normal 9 hour day. Suppose the average number of calls per hour is 7.0 and that each call costs the company $50. Let V be a Poisson random variable representing the number of calls for assistance received during the remaining 15 hours of the day. Suppose the average number of calls per hour for this time period is 4.0 and that each call costs the firm $60. Find the expected cost and the variance of the cost associated with calls received during a 24-hour day.

Now, the expectation and the variance for a Poisson random variable are the same, right? And as these are independent random variable the expected cost of the sum should be the sum of the individual expectations. i.e.



This squares with the solution manual. However, it gives the variance as $373500.

But if these are independent random variables, why isn't the variance the sum of the individual variances, and in this case the same as the expectation? i.e. $6750

This has got to be a typo right? Or am I missing something. My experience with presumed typos in other maths books is it's usually the latter ;-) so I thought I'd check with you cats, MD.

p.s. shouldn't the units of the variance be too?

The expectation and the variance for a Poisson random variable are the same. But the poisson-distributed random variables here are the number of calls, not the cost. Both the expectation and variance for the number of calls are dimensionless.
If you multiply a random variable with a constant factor, the variance gets scaled with the square of this factor. Therefore, if u is the number of calls during daytime and U the costs, Var(U)=(50$)^2 * var(u) = (50$)^2 * E(u) = (50$) * E(U) = 157500$^2
Similarly, Var(V)=60$*E(V)=216000$^2 and Var(U+V)=373500$^2.