GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
In a dog show of poodles and show-dogs,
[#permalink]
20 Feb 2021, 23:42
20 Feb 2021, 23:42
1
17
Bookmarks
00:00
Question Stats:
79% (01:40) correct
20% (01:40) wrong based on 86 sessions
Hide Show timer Statistics
In a dog show of poodles and show-dogs, \(\frac{1}{6}^{th}\) of show-dogs are poodles and \(\frac{1}{7}^{th}\) of poodles are show-dogs
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Minimum number of dogs in the dog show |
10 |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
In a dog show of poodles and show-dogs,
[#permalink]
21 Feb 2021, 04:56
21 Feb 2021, 04:56
1
Solution:
The most important this to pay attention to is that \(\frac{1}{6}\) show-dogs are poodles and \(\frac{1}{7}\) show-dogs are poodles are equal.
So let the total poodles be P & Show dogs be S
Thus, \(\frac{1}{6}\)P= \(\frac{1}{7}\)S
i.e. \(\frac{P}{S}\)= \(\frac{6}{7}\)
Hence, the minimum number of poodles and show dogs has to be a multiple of 7+6=13 and 13 being the lowest total number.
13 or any other multiple of 13>10
QTY A> QTY B
IMO A
Hope this helps!
The most important this to pay attention to is that \(\frac{1}{6}\) show-dogs are poodles and \(\frac{1}{7}\) show-dogs are poodles are equal.
So let the total poodles be P & Show dogs be S
Thus, \(\frac{1}{6}\)P= \(\frac{1}{7}\)S
i.e. \(\frac{P}{S}\)= \(\frac{6}{7}\)
Hence, the minimum number of poodles and show dogs has to be a multiple of 7+6=13 and 13 being the lowest total number.
13 or any other multiple of 13>10
QTY A> QTY B
IMO A
Hope this helps!
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: In a dog show of poodles and show-dogs,
[#permalink]
21 Feb 2021, 12:26
21 Feb 2021, 12:26
1
2
Bookmarks
OA Explanation:
Let Show-dogs be \(S\) and Poodles be \(P\)
As per the question,
\(P = \frac{1}{6}S\)
\(S = 6P\)
\(S = \frac{1}{7}P\)
\(P = 7S\)
Remember, we have been asked about the minimum number of dogs, which is only possible when the intersection is least
i.e. dogs which are Poodle as well as Show-dog should be \(1\)
S = (6)(1) = \(6\)
P = (7)(1) = \(7\)
Don't forget, 1 dog is both Show-dog and Poodle
So,
Only Show-dog = \(5\)
Only Poodle = \(6\)
Both Poodle and Show-dog = \(1\)
Total number of dogs =\( 6 + 5 + 1 = 12\)
Col. A: \(12\)
Col. B: \(10\)
Col. A > Col. B
Hence, option A
Let Show-dogs be \(S\) and Poodles be \(P\)
As per the question,
\(P = \frac{1}{6}S\)
\(S = 6P\)
\(S = \frac{1}{7}P\)
\(P = 7S\)
Remember, we have been asked about the minimum number of dogs, which is only possible when the intersection is least
i.e. dogs which are Poodle as well as Show-dog should be \(1\)
S = (6)(1) = \(6\)
P = (7)(1) = \(7\)
Don't forget, 1 dog is both Show-dog and Poodle
So,
Only Show-dog = \(5\)
Only Poodle = \(6\)
Both Poodle and Show-dog = \(1\)
Total number of dogs =\( 6 + 5 + 1 = 12\)
Col. A: \(12\)
Col. B: \(10\)
Col. A > Col. B
Hence, option A
