An animal shelter began the day Tuesday with a ratio of 5 cats for eve

Any statement provide enough information about both Dog and Cats is qualified.
1/ Know the number of Cats. (Not Dogs)
2/ The Q did not say: all adopted leave. It just said: if leave, must be adopted => Know nothing about the number of Dogs.
=> E

No cats were adopted on Tuesday.

Since we don’t know how many dogs were adopted on Tuesday, we can’t determine the ratio of cats to dogs at the end of Tuesday.

Statement one alone is not sufficient.

Statement Two Alone:

4 dogs were adopted on Tuesday.

Since we don’t know how many cats were adopted on Tuesday, we can’t determine the ratio of cats to dogs at the end of Tuesday.

Statement two alone is not sufficient.

Statements One and Two Together:

Even though we know no cats and 4 dogs were adopted on Tuesday, we still can’t determine the ratio of cats to dogs at the end of Tuesday. That is because we don’t know the number of cats and dogs at the beginning of the day (recall that we know only that the ratio of cats to dogs at the beginning of the day was 5:11). For example, if there were 5 cats and 11 dogs at the beginning of the day, then there would be 5 cats and 7 dogs at the end of the day, making the cats-to-dogs ratio of 5:7. However, if there were 10 cats and 22 dogs at the beginning of the day (still satisfying the ratio of 5 : 11), then there would be 10 cats and 18 dogs at the end of the day, making the cats-to-dogs ratio of 10:18 or 5 to 9.

Answer: E

\(\begin{array}{*{20}{c}}\\
{{\text{cats}} = \,\,5\,k} \\ \\
{{\text{dogs}} = 11\,k} \\
\end{array}\,\,\,\,\,\,\left( {k > 0\,\,\,\operatorname{int} } \right)\,\,\,\,\,\,\,\left( * \right)\)

\(\left( * \right)\,\,{\mkern 1mu} k\,{\mkern 1mu} {\mkern 1mu} = \,\,{\mkern 1mu} {\mkern 1mu} 11k - 2\left( {5k} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\text{ = }}{\mkern 1mu} {\mkern 1mu} \operatorname{int} - \operatorname{int} \,\,\, = \,\,\,\operatorname{int}\)

That´s the k technique, a "killer" tool of our method (when dealing with ratios)!

\(\left( {1 + 2} \right)\,\,\,\,\,{\text{?}}\,\,\, = \,\,\,\frac{{5k}}{{\,11k - 4\,}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,k = 1\,\,\,\, \Rightarrow \,\,\,? = \frac{5}{7} \hfill \\\\
{\text{Take}}\,\,k = 2\,\,\,\, \Rightarrow \,\,\,? = \frac{{10}}{{18}} \ne \frac{5}{7} \hfill \\ \\
\end{gathered} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

Bunuel : Two quick questions

Q1) If the question had given the actual initial numbers for cats and dogs ...and the new ratio after giving away the dogs, turned out to be instead 5 : 9 (completely made up) ...

I think i would have still picked (E) because

5: 9 can be rewritten as
-- 10:18
-- 15:27
-- 20: 36
-- 30:36


Aren't these examples of several different ratio's, thus calling for option (E) as well in this scenario even with the initial number of cats and dogs given

------------------------

Q2)

If the actual number of cats to dogs is 10 : 22

If i remove 4 dogs, i get a ratio of 10 : 18 (22-4) or 5: 9

But why can't i do the following step

-- take the 10 : 22
-- divide by 2 first to bring it to 5: 11
-- then remove 4 dogs from the 11
-- new ratio will be 5: 7 (11-4)

Isn't this a way to also remove 4 as we are removing it from 22 in 10:22 to give me 10 : 18

Thank you !

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