raisins, apples and peanuts

Let's call No of bags that contain raisins : R, almonds : A , peanuts : P, a mixture of R&P : RP
From the given information, I set up some equations a below:
R = 10*P ( the no. of bags contains only raisins is ten times the no. of bags contains only peanuts)
A = 20*RP (RP is different from R*P, it's the no. of bags that contain only a mixture of raisins and peanuts)
P = (1/5)*A
A + Axx + Ax = 210 ( Axx is the no. of bags that contain the mixture of almonds, raisins and peanuts - if any. Ax is the no. of bags that contain the mixture of almonds with raisins or with peanuts - if any)
As we already have total no. of bags : 435 = R+P+A + RP + Axx ( or ARP) + Ax ( AR or AP)
So 435 = R+P+RP + 210
Hence R+P+RP = 225 (1)
Now we convert all elements in equation (1) into P
R = 10*P
A = 20*RP, it means RP = (1/20) * A (I)
while P = (1/5) * A ( as above), then A = 5*P (II)
From (I) & (II) RP = (1/20) * A = (1/20) * 5*P = (1/4)*P
Now equation (1) becomes : 10*P + P + (1/4)*P = 225 or 11.25 * P = 225
P = 20
R = 10*P = 200
A = P*5 = 100 (from equation (II))
Total no of bags contain only one kind of them: 200+100+20 = 320, or D