fskilnik wrote:

GMATH practice exercise (Quant Class 14)

A box has only pure material A identical coins, pure material B identical coins, and pure material C identical coins. Each coin is worth a certain integral number of dollars and coins of different materials don´t have the same dollar value. What is the maximum dollar value Jacob could take out from the box if he takes out exactly two different coins from it?

(1) Abraham took four coins out from the box, at least one coin of each material, and he realized they were worth a total of 28 dollars.

(2) Isaac took five coins out from the box, at least one coin of each material, and he realized they were worth a total of 21 dollars.

\(\left. \matrix{

{\rm{A}}\,\,{\rm{coins}}\,\,{\rm{material}}\,\,{\rm{A}}\,\,\, \to \,\,\,\$ x\,\,{\rm{worth}}\,\,{\rm{each}} \hfill \cr

{\rm{B}}\,\,{\rm{coins}}\,\,{\rm{material}}\,\,{\rm{B}}\,\,\, \to \,\,\,\$ y\,\,{\rm{worth}}\,\,{\rm{each}} \hfill \cr

{\rm{C}}\,\,{\rm{coins}}\,\,{\rm{material}}\,\,{\rm{C}}\,\,\, \to \,\,\,\$ z\,\,{\rm{worth}}\,\,{\rm{each}} \hfill \cr} \right\}\,\,\,\,1 \le z < y < x\,\,\,{\rm{ints}}\,\,\,\left( {{\rm{without}}\,\,{\rm{loss}}\,\,{\rm{of}}\,\,{\rm{generality}}!} \right)\)

\(? = x + y\)

\(\left( 1 \right)\,\,\,\left\{ \matrix{

\,28 = 2 \cdot 10 + 1 \cdot 6 + 1 \cdot 2\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,10,1,6,1,2} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 16 \hfill \cr

\,28 = 2 \cdot 10 + 1 \cdot 5 + 1 \cdot 3\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,10,1,5,1,3} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 15 \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,\left\{ \matrix{

\,21 = 2 \cdot 8 + 2 \cdot 2 + 1 \cdot 1\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,8,2,2,1,1} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 10 \hfill \cr

\,21 = 2 \cdot 8 + 1 \cdot 3 + 2 \cdot 1\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,8,1,3,2,1} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 11 \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,{\rm{I}}.\,\,\left\{ \matrix{

\,2x + y + z = 28 \hfill \cr

\,x + 2y + 2z = 21 \hfill \cr} \right.\,\,\,\,\,{\rm{or}}\,\,\,\,\,{\rm{II}}.\,\,\left\{ \matrix{

\,2x + y + z = 28 \hfill \cr

\,x + y + 3z = 21 \hfill \cr} \right.\,\,\,\,\,{\rm{or}}\,\,\,\,\,{\rm{III}}.\,\,\left\{ \matrix{

\,x + 2y + z = 28 \hfill \cr

\,x + y + 3z = 21 \hfill \cr} \right.\)

\({\rm{I}}{\rm{.}}\,\,\mathop {\, \Rightarrow }\limits^{\left( + \right)} \,\,\,x + y + z = {{28 + 21} \over 3} = {{49} \over 3} \ne {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}!\)

\({\rm{II}}{\rm{.}}\,\,\left\{ \matrix{

\,2x + y + z = 28 \hfill \cr

\,x + y + 3z = 21\,\,\left( * \right) \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,x - 2z = 7\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{

\,z = 1\,\,\, \Rightarrow \,\,\,x = 9\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y = 9\,\,\,\,\,{\rm{impossible}} \hfill \cr

\,z = 2\,\,\, \Rightarrow \,\,\,x = 11\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y = 4\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 15 \hfill \cr

\,z \ge 3\,\,\, \Rightarrow \,\,\,x \ge 13\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y < 0\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\)

\({\rm{III}}.\,\,\,\left\{ \matrix{

\,x + 2y + z = 28 \hfill \cr

\,x + y + 3z = 21\,\,\left( * \right) \hfill \cr} \right.\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,y - 2z = 7\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,x + 5z = 14\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{

\,z = 1\,\,\, \Rightarrow \,\,\,x = 9\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,y = 9\,\,\,{\rm{impossible}} \hfill \cr

\,z \ge 2\,\,\, \Rightarrow \,\,\,x \le 4\,\,\, \Rightarrow \,\,\,\left( {x,y,z} \right) = \left( {4,3,2} \right)\,\,\,{\rm{impossible}} \hfill \cr} \right.\)

The correct answer is (C).

We follow the notations and rationale taught in the

GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik ::

GMATH method creator (Math for the GMAT)

Our high-level "quant" preparation starts here:

https://gmath.net