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Manager  Joined: 09 Feb 2013
Posts: 111
At a local beach, the ratio of little dogs to average dogs  [#permalink]

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Question Stats: 41% (03:18) correct 59% (03:12) wrong based on 296 sessions

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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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Good Questions also deserve few KUDOS.
Intern  Joined: 28 Jul 2013
Posts: 7
GMAT 1: 770 Q51 V42 Re: At a local beach, the ratio of little dogs to average dogs  [#permalink]

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emmak wrote:
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%

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.
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Intern  Joined: 26 May 2010
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Re: At a local beach, the ratio of little dogs to average dogs  [#permalink]

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emmak wrote:
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%

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 Originally posted by bhuwangupta on 14 Jun 2013, 00:17.
Last edited by bhuwangupta on 14 Jun 2013, 00:21, edited 1 time in total.
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Re: At a local beach, the ratio of little dogs to average dogs  [#permalink]

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emmak wrote:
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%

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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Re: At a local beach, the ratio of little dogs to average dogs  [#permalink]

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emmak wrote:
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%

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!!
Senior Manager  Joined: 08 Apr 2012
Posts: 327
Re: At a local beach, the ratio of little dogs to average dogs  [#permalink]

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bhuwangupta wrote:
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

Kudos is the Karma Why is this true?
Why can't x=1?
Am I missing something here?
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Re: At a local beach, the ratio of little dogs to average dogs  [#permalink]

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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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Still not there.

If you're reading this, we've got this. Re: At a local beach, the ratio of little dogs to average dogs   [#permalink] 06 Mar 2019, 07:45
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