Bunuel
At the beginning of the day, a pet store had dogs and cats in the ratio of 5 to 3. If no new animals were added to the store and the only animals that left were those that were sold, what was the ratio of dogs to cats at the end of the day?
(1) The store sold 2 dogs that day.
(2) The store sold 1/3 of its cats that day.
If no new animals were added to the store and the only animals that left were those that were sold, what was the ratio of dogs to cats at the end of the day?This sentence is a little confusing, but shouldn't it be the following. "The only animals that were left were those that were NOT sold?"
With regards to the question.
Old Ratio
D:C = 5:3 => D = 5x and C = 3x
New Ratio
D:C =?
1) 2 dogs sold that day
If x = 1, then D = 5 and C = 3
2 dogs sold, so the new ratio D : C = 3 : 3
If x = 2, then D = 10 and C = 6
2 dogs sold, so the new ratio D : C = 8 : 6
Inconsistent answers. Insufficient.
2) (1/3) of total cats were sold.
If x = 1, then D = 5 and C = 3
(1/3) of 3 = 1, so the new ratio D : C = 5 : 2
If x = 2, then D = 10 and C = 6
(1/3) of 6 = 2, so the new ratio D : C = 10 : 4
If x = 3, then D = 15 and C = 9
(1/3) of 9 = 3, so the new ratio D : C = 15 : 6
This is keeping the number of dogs constant. If even 1 dog is sold, the ratio changes.
Insufficient.
(1 + 2)
Same above.
x= 1 ; D = 5 , C = 3
New Ratio = 3:2
x = 2 : D = 10 , C = 6
New Ratio = 8 : 4
Insufficient.
Answer is E.
Edited Post.