BTGmoderatorDC wrote:A bowl contains pecans, cashews, and almonds in a ratio of 6 : 10 : 15, respectively. If some of the nuts of one of the three types are removed, which of the following could be the ratio of pecans to cashews to almonds remaining in the bowl?
i. 1 : 2 : 3
ii. 2 : 3 : 4
iii. 4 : 7 : 10
A. I only
B. II only
C. III only
D. I and III only
E. II and III only
Source: Manhattan Prep
$$?\,\,\,:\,\,\,p:c:a\,\,{\text{possible}}\,\,\left( {{\text{when}}\,\,{\text{some}}\,\,{\text{nuts}}\,\,{\text{of}}\,\,{\text{one}}\,\,{\text{type}}\,\,{\text{removed}}} \right)$$
$$p:c:a = 6:10:15\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \matrix{
\,p = 6k \hfill \cr
\,c = 10k \hfill \cr
\,a = 15k \hfill \cr} \right.\,\,\,\,\,\,\left( {k > 0\,\,{\mathop{\rm int}} \left( * \right)} \right)$$
$$\left( * \right)\,\,\left\{ \matrix{
\,{\mathop{\rm int}} - {\mathop{\rm int}} = a - c = 15k - 10k = 5k\,\,{\mathop{\rm int}} \hfill \cr
\,{\mathop{\rm int}} - {\mathop{\rm int}} = p - 5k = 6k - 5k = k\,\,\,{\mathop{\rm int}} \hfill \cr} \right.$$
$$\left( {\text{I}} \right)\,\,\,p:c:a = 1:2:3\,\,\,\, \Rightarrow \,\,\,{\text{possible}}\,\,\left( {k = 1,\,\,{\text{take}}\,\,1\,\,{\text{pecan}}\,\,{\text{nut}}\,\,{\text{out}}\,\,\,\, \Rightarrow \,\,\,\,\left( {p,c,a} \right) = \left( {5,10,15} \right)} \right)$$
$$\,\,\, \Rightarrow \,\,\,\,\,{\text{refute}}\,\,\left( {\text{B}} \right),\left( {\text{C}} \right),\left( {\text{E}} \right)$$
$$\left( {{\text{III}}} \right)\,\,p:c:a = 4:7:10\,\,\,\,\mathop \Rightarrow \limits^{\left( {\text{below}} \right)} \,\,\,{\text{impossible:}}\,\,\,$$
$${\rm{some}}\,\,p\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k - {\rm{some}},10k,15k} \right) \hfill \cr
\,{2 \over 3} = {{10k} \over {15k}} = {c \over a} \ne {7 \over {10}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]$$
$${\rm{some}}\,\,c\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k,10k - {\rm{some}},15k} \right) \hfill \cr
\,{4 \over 7} = {p \over c} = {{6k} \over {10k - {\rm{some}}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,40k - 4 \cdot {\rm{some}} = 42k\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]$$
$${\rm{some}}\,\,a\,\,{\rm{out}}\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left( {p,c,a} \right) = \left( {6k,10k,15k - {\rm{some}}} \right) \hfill \cr
\,{4 \over 7} = {p \over c} = {{6k} \over {10k}} = {3 \over 5}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\left[ {\,k,{\rm{some}}\,\, > 0\,} \right]$$
The correct answer is (A).
(Note that (II) does not need to be evaluated!)
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
