recombination frequency problem
I got $26$% as the answer.  

To get a recombination between C and E, there are two possible mechanisms:-   


C and D produce a recombinant, but D and E remain linked, therefore the final genotype will be a recombinant considering C and E(Chiasmata between C and D). Here $P_1=P_{CD}\times P'_{DE}$ where $P$ is the probability of recombination and $P'=1-P$ is the probability of linkage.Hence $P_1=0.08$
C and D remain linked but D and E undergo recombination (Chiasmata between D and E). Here, $P_2=P'_{CD}\times P_{DE}=0.18$ 


If there is a chiasmata between both C and D, and D and E, the resultant will not have a recombination in C and E locus due to double crossing over. If there is no crossing over, there will be no recombination at either loci.

The net probability of recombination $P_{CE}$ will be the sum of these two, as any of these possibilities can occur to give a recombination between C and E. $P_{CE}=P_1+P_2=0.26$  

And Biogirllajja, your calculation (the Edit) was almost correct, but you have to subtract $2$% from both $10$% and $20$%, because, both of them include the probability of crossover at one loci, including the case where the other loci has also crossed over. Hence, you need to subtract it from both, and then add to get $26$%. Tell me if you want me to expand on my explanation or provide some clarifications.  

If we simply add the map distances, we get the actual map distance which disregards the possibility of double recombinations. All map units greater than 10cM have significant chances of double crossovers between them, and hence, the observed recombination frequency would be less than the expected 0.1. Hence, for distances greater than 10cM, the observed frequency of recombination will not be the same as (map distance/100).