I recently have gotten stuck on a problem that appears to be a simultaneous equations DS problem:
At a certain pet shop, 1/3 of the pets are dogs, and 1/5 of the pets are birds. How many of the pets are dogs?
(1) There are 30 birds at the pet shop
(2) There are 20 more birds than dogs at the pet shop
Obviously, statement 1 is sufficient. For statement two, I tried setting up these two simultaneous equations:
Let b = # of birds
Let d = # of dogs
Let x = # of birds and dogs
d = b + 20
(1/3)d + (1/5)b = x
I know the answer is D: Both statements alone, but can't figure out why. Please help if you can!
Let total number perts be N. So we have N/3 dogs and N/5 birds
from (1) N/5 = 30 so N = 150 and N/3 = 50 SUFFF
from (2) N/3 - N/5 = 20 which again gives answer N = 150 and N/3 = 50 SUFF
Btw, I think in (2) the phrase is wrong It has to be 20 birds less. you cant have N/5 - N/3 = 20