This problem can be solved with a shortcut which the GMAT typically expects. If you consider the percentage of enormous dogs as x, then the percentage of the rest of the dogs becomes 100-x. Now this percentage is split between the 2 types of dogs in the ratio of 4:5. We also know that the percentages have to be integers. Meaning 100-x will have to be divisible by 9

A quick look at the solutions tells us that 100-x is divisible by 9 only in the case if D.

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

Given ratio of Little Dogs:Average Dogs = 2:5. When this ratio doubles, it becomes 4:5. Ratio between Little dogs: Enormous dogs = 2:8.

Also, this increases from 2:8-->1:4 to 4:(8+x) , where x is an non-negative integer(Maximum value which x can assume is 7, under the given conditions)

New % of little dogs =\(\frac{4}{(4+5+8+x)}*100\)= \(\frac{4}{(17+x)}*100\) The minimum value of x, for which this is an integer ; x = 3.

Similarly, % of average dogs =\(\frac{5}{(17+x)}*100\). Again, x = 3 for an integral value.

Thus, the new percentage of enormous dogs =\(\frac{(8+3)}{20}*100\)= 55 %

D.
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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!!

Why is this true?
Why can't x=1?
Am I missing something here?

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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Can someone explain why C is wrong?

18 total dogs

4 Little, 5 Average, 9 Enormous = 18

- 18 is less than 30
- 4:5 L/A ratio remains in tact
- 2:8 L/E ratio is increased (to 4/9)

Initial ratio - 2:5:8

Then the ratio of L/A doubles = 4:5

Now the percentages need to be integers

Total dogs = 20 works

E = 11 or 55% D