lets take $ value in relation C>B>A
we need to find value for ABC
#1
A+B+2C= 28
in sufficient
#2
2A+2B+c= 21

in sufficient
from 1 & 2
A+B=28-2C
so
2*(28-2C)+C=21
c = 11.6$ ~12 $
and we can determine values for A& B
IMO C

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


\(? = x + y\)


\(\left( 1 \right)\,\,\,\left\{ \matrix{
\,28 = 2 \cdot 10 + 1 \cdot 6 + 1 \cdot 2\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,10,1,6,1,2} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 16 \hfill \cr
\,28 = 2 \cdot 10 + 1 \cdot 5 + 1 \cdot 3\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,10,1,5,1,3} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 15 \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,\left\{ \matrix{
\,21 = 2 \cdot 8 + 2 \cdot 2 + 1 \cdot 1\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,8,2,2,1,1} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 10 \hfill \cr
\,21 = 2 \cdot 8 + 1 \cdot 3 + 2 \cdot 1\,\,\,\, \Rightarrow \,\,\,\left( {A,x,B,y,C,z} \right) = \left( {2,8,1,3,2,1} \right)\,\,{\rm{viable}}\,\,\,\,\, \Rightarrow \,\,\,? = 11 \hfill \cr} \right.\)


\(\left( {1 + 2} \right)\,\,\,{\rm{I}}.\,\,\left\{ \matrix{
\,2x + y + z = 28 \hfill \cr
\,x + 2y + 2z = 21 \hfill \cr} \right.\,\,\,\,\,{\rm{or}}\,\,\,\,\,{\rm{II}}.\,\,\left\{ \matrix{
\,2x + y + z = 28 \hfill \cr
\,x + y + 3z = 21 \hfill \cr} \right.\,\,\,\,\,{\rm{or}}\,\,\,\,\,{\rm{III}}.\,\,\left\{ \matrix{
\,x + 2y + z = 28 \hfill \cr
\,x + y + 3z = 21 \hfill \cr} \right.\)

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

\({\rm{II}}{\rm{.}}\,\,\left\{ \matrix{
\,2x + y + z = 28 \hfill \cr
\,x + y + 3z = 21\,\,\left( * \right) \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,x - 2z = 7\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,z = 1\,\,\, \Rightarrow \,\,\,x = 9\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y = 9\,\,\,\,\,{\rm{impossible}} \hfill \cr
\,z = 2\,\,\, \Rightarrow \,\,\,x = 11\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y = 4\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 15 \hfill \cr
\,z \ge 3\,\,\, \Rightarrow \,\,\,x \ge 13\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y < 0\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\)

\({\rm{III}}.\,\,\,\left\{ \matrix{
\,x + 2y + z = 28 \hfill \cr
\,x + y + 3z = 21\,\,\left( * \right) \hfill \cr} \right.\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,y - 2z = 7\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,x + 5z = 14\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,z = 1\,\,\, \Rightarrow \,\,\,x = 9\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,y = 9\,\,\,{\rm{impossible}} \hfill \cr
\,z \ge 2\,\,\, \Rightarrow \,\,\,x \le 4\,\,\, \Rightarrow \,\,\,\left( {x,y,z} \right) = \left( {4,3,2} \right)\,\,\,{\rm{impossible}} \hfill \cr} \right.\)


The correct answer is (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________

Figured this out through another method, greatly appreciate it if anyone can test my logic.

I think its fairly evident that none of the options alone can be sufficient to get the solution, so let's use them together.
1) suppose we take an extra of coin: A: 2A+B+C=28

2) since we have taken one extra coin, and the total amount is less than that of the earlier selection of only 4 coins, we can deduce that the new selection did not include the same additional coin from the first time.
two cases comes out of this:
1 extra of each of B & C: A+2B+2C=21
2 extra of B or C (lets suppose its C): A+B+3C=21

combine the equations we get from 1) with each we get from 2)
Scenario #1
2A+B+C=28
A+2B+2C=21

simplify two equations:
(2A+B+C) - (A+2B+2C) = 28-21
A=B+C+7

Sub into 2A+B+C=28
2(B+C+7)+B+C=28
3(B+C)=14

Since the question specifies that the dollar amount is an integral number, this case is invalid as 14 cannot be divided by 3

Scenario #2
2A+B+C=28
A+B+3C=21

Do the same with equations above through subtracting equations & Sub into equation 1
(2A+B+C)-(A+B+3C)=28-21
A=2C+7

2(2C+7)+B+C=28
5C+B=14

This could potentially work, so let's test out numbers.
C=1, B=9, so A=2C+7=9. Dollar value has to be different on each coin, so A cannot equal to B, so invalid.
C=2, B=4, so A=2C+7=11.

Plug in the three values in to the two equations to verify
2(11)+4+2=28
11+4+3(2)=21

Max value from two coins: 11+4=15
C is correct.

Th1is "seems" like a good Question but it isn't.


Using 1 we can find out that maximum value for coins would be 26 , since the smallest possible integral value would be 1
Hence, it is sufficient.

However, Statement 2 gives a condition which makes Statement 1 is insufficient only when ST.2 is considered.

Even Statement 2 should be independently sufficient giving a A,B as 15,2 and in no other condition do we get a higher value.


An ideal GMAT question would answer in one way and with one solution unlike this Question.


IN SHORT : What is the maximum dollar value Jacob could take out from the box if he takes out exactly two different coins from it?

The above question is answered using either of statements.

thoughts ? VeritasKarishma chetan2u