At a local beach, the ratio of little dogs to average dogs to enormous dogs is 2:5:8. Late in the afternoon, the ratio of little dogs to average dogs doubles and the ratio of little dogs to enormous dogs increases. If the new percentage of little dogs and the new percentage of average dogs are both integers and there are fewer than 30 total dogs at the beach, which of the following represents a possible new percentage of enormous dogs?

A. 25%
B. 40%
C. 50%
D. 55%
E. 70%
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Little Dogs(L), Average Dogs(A) and Enormous Dogs (E)
The initial ratio for L:A:E :: 2:5:8
Initial Total dogs = 15X ( x assumed; 2+5+8= 15), Since the total dogs are less than 30 therefore initial total value has to be 15

L = 2, A = 5 E = 8

L:A= 2:5

This ratio doubles

Hence New Dog count is

L= 4 , A = 5 E= X: Also 4+5+x<30

We need to Find X*100/(4+5+X)

Now it says that new percentage of little dogs and Average dogs is an integer

%L = 4*100/(9+x) %A = 5*100/(9+x); Only Value for X is 11 ; 9+x<30 and % integer


Therefore, Enormus Dogs % is = 11*100/(20) = 55%

D is the Ans

Kudos is the Karma

I like to go straight forward and with the question stem. This is how I solved analyzing the question sentence by sentence.

1) Initially the ratio of Local : Average : Enormous is 2:5:8.
2) After afternoon ratios become :- 4:5:x (since we do not know in what ratio are enormous dogs in)
3) Initially Local : Enormous ratio is 1:4. Now, Ques stem says that this ratio increases. Therefore new ratio i.e. 4:x > 1:4 ; this gives us x < 16.

4) Now, percentage of Local dogs in afternoon is \(\frac{(4*100)}{(9+x)}\) and average dogs is \(\frac{(5*100)}{(9+x)}\)

5) These ratios will be integers only when the denominators of these terms are common factors of 400 and 500 and the multiplication of those factors such as 2,5,10,20,50 etc. Since x is less than 16, x can only be 1 and 11 so that the denominators are 10 and 20 respectively.

6) Therefore the possible percentages of Enormous dogs can be 10% (i.e. (1/(9+1))*100 considering x = 1) and 55% (i.e. (11/(9+11))*100 considering x = 11)

Only 55% is in the options hence D

Consider Kudos if the post helped!!

A sufficient algebraic approach has already been provided by previous users, I used a rather unorthodox way of solving this.

So by afternoon we have a new ratio of l:a of 4/5 and a greater ratio of l:e compared to the morning one. In the morning, it was 2:8, so i fathomed it to be e.g. 3:8.

Next, i combined the ratios to get a total of 12:15:32 ratio with a total of 59 dogs.

Now, the question states that there are fewer than 30 dogs, so in our scenario we are good to continue (59<60).

32/59 ~ 32/60 ~= 0,53 so it needs to be a bit higher due to all of our rounding up/downs, so 0,55 or 55%.

Not the best way to go about it, neither the most foul-proof, but if it works - it works.
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