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In a dog show of poodles and show-dogs,
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20 Feb 2021, 23:42
20 Feb 2021, 23:42
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Question Stats:
80% (01:42) correct
19% (01:39) wrong based on 97 sessions
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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
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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
