At a local beach, the ratio of little dogs

[guerrero]

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%

OA D

[GMATGuruNY]

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%

OA D

Let L = little dogs, A = average dogs, E = enormous dogs, and T = the total number of dogs.

Original L/A = 2/5.
To double a ratio means to multiply it by 2/1.
Thus:
L/A doubled = (2/1)(2/5) = 4/5.

Original L/E = 2/8 = 4/16.
After L/E increases, the following options are possible:
L/E = 4/15, 4/14, 4/13...

Thus:
Options for L:A:E = 4:5:15, 4:5:14, 4:5:13...

In the ratios above, L=4.
The problem states that T<30.
The percentage of little dogs must be an integer value.
Thus, when L/T is converted to a percentage, the result must be an integer.
Options for L/T:
4/25 = 16/100 = 16%.
4/20 = 20/100 = 20%
4/10 = 40/100 = 40%.
These fractions imply that T=25, T=20, or T=10.

The greatest possible value of T occurs when L:A:E = 4:5:15, implying that L=4, A=5, and E=15.
Here, T=4+5+15=24.
Thus, it is not possible that T=25.

The next greatest option is T=20.
If L:A:E = 4:5:11, then L=4, A=5, and E=11.
Here, T=4+5+11=20.
In this case:
L/T = 4/20 = 20%.
A/T = 5/20 = 1/4 = 25%.
E/T = 11/20 = 55/100 = 55%.

Thus, the new percentage of enormous dogs could be 55%.

The correct answer is D.

[faraz_jeddah]

Thats a super tough question.

What is the source?

[ygdrasil24]

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%

OA D

Let L = little dogs, A = average dogs, E = enormous dogs, and T = the total number of dogs.

Original L/A = 2/5.
To double a ratio means to multiply it by 2/1.
Thus:
L/A doubled = (2/1)(2/5) = 4/5.

Original L/E = 2/8 = 4/16.
After L/E increases, the following options are possible:
L/E = 4/15, 4/14, 4/13...

Thus:
Options for L:A:E = 4:5:15, 4:5:14, 4:5:13...

In the ratios above, L=4.
The problem states that T<30.
The percentage of little dogs must be an integer value.
Thus, when L/T is converted to a percentage, the result must be an integer.
Options for L/T:
4/25 = 16/100 = 16%.
4/20 = 20/100 = 20%
4/10 = 40/100 = 40%.
These fractions imply that T=25, T=20, or T=10.

The greatest possible value of T occurs when L:A:E = 4:5:15, implying that L=4, A=5, and E=15.
Here, T=4+5+15=24.
Thus, it is not possible that T=25.

The next greatest option is T=20.
If L:A:E = 4:5:11, then L=4, A=5, and E=11.
Here, T=4+5+11=20.
In this case:
L/T = 4/20 = 20%.
A/T = 5/20 = 1/4 = 25%.
E/T = 11/20 = 55/100 = 55%.

Thus, the new percentage of enormous dogs could be 55%.

The correct answer is D.

Mitch :

If intially ratio was 2:5:8 , after doubling b/w L and A,
it is 4:5:8 , whats wrong here ?
Why have you created a common 4 to have E equal to 16 ..Pls explain this.
Thanks
Abhinav

[anubhavit]

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%

OA D

Hi,

Let's start with what we have been given

Earlier,
L : A : E = 2 : 5 : 8 - (i)

Later,
L : A = 4 : 5 or - (ii)
it could be 2 / 2.5 - (iii)

L : E = Increases, so in 2/8, either numerator(2) increases or denominator(8) decreases. - (iv)

We need to tell % of E dogs, where both % of L and % of A are integers

Let's start with L : A : E = 4 : 5 : 8 (using i and ii). This also agrees with (iv)

% of L = 4/17 * 100.
% of A = 5/17 * 100.

To make an integer out of above, the denominator should be 20 instead of 17. Most simply, we can change the E from 8 to 11, thereby making the total of L : A : E = 4 : 5 : 11. This still agrees with (iv) because the earlier ratio was 2/8 or 0.25 and now it is 4/11 or ~ 0.36.

Based on L : A : E = 4 : 5 : 11

% of E = 11/20 * 100 => 55 %


Let me know if there is a flaw in my approach.

[BestGMATEliza]

For this problem, I think its easiest to pick numbers. so the ratio starts out 2:5:8 (little:average:enormous)
then the ration of little dogs to average dogs doubles, so it must be 4:5
and all we know is about enormous dogs is that the ration of little to enormous increases

Let's pretend that there are 4 little dogs and 5 average dogs on the beach
the number of enormous dogs must be less than 16 (because the ratio of little to enormous must increase) in addition, it must be such that both 4 and 5 can evenly go into the total number of dogs (i.e. 4+5+#Enormous)

The only number that allows for this is 11, which would make the percentage of enormous dogs 55 or D