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In a dog show of poodles and show-dogs,

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In a dog show of poodles and show-dogs, [#permalink] New post   20 Feb 2021, 23:42
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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

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.
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In a dog show of poodles and show-dogs, [#permalink] New post   21 Feb 2021, 04:56
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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!
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Re: In a dog show of poodles and show-dogs, [#permalink] New post   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
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Re: In a dog show of poodles and show-dogs, [#permalink] New post   21 Feb 2021, 12:28
Ks1859 wrote:
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!


Ks1859
Your Answer is correct but the total would be 12 instead of 13.
You have probably counted a dog which is Poodle as well as Show-dog twice!
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Re: In a dog show of poodles and show-dogs, [#permalink] New post   21 Feb 2021, 19:05
KarunMendiratta wrote:
Ks1859 wrote:
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


Your Answer is correct but the total would be 12 instead of 13.
You have probably counted a dog which is Poodle as well as Show-dog twice!


Ah! Got you

Thanks!!

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Re: In a dog show of poodles and show-dogs, [#permalink] New post   04 Mar 2024, 23:45
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Re: In a dog show of poodles and show-dogs, [#permalink]
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