A box has only pure material A identical coins, pure material B identi

Question

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.

Archit3110 · Answer

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

fskilnik · Answer

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

? = x + y

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

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

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

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

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

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

The correct answer is (C).

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

Regards, 
Fabio.

2asc · Answer

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.

Thelegend2631 · Answer

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

fskilnik · Answer

Hi there,  hD13 .

Thank you for your interest in our question, and I am sorry you didn´t mentioned my name (as I did with yours) so that I could come here and help you understand the reasoning.

You did not understand the question stem and you certainly did not take time to understand my solution.

Each statement alone is insufficient to answer the question asked (as shown explicitly), hence there is not the "one unique solution issue" commonly discussed when the right answer is D, which is not the case.

You have all the right to disagree, of course. If you have any doubts related to the problem and the corresponding solution, please feel free to ask.

Success in your studies,
Fabio.

Saupayan · Answer

Is this really meant to be a GMAT representative question? I don't see a way in which I will ever be able to solve it in 2 mins ever after knowing the solution

fskilnik · Answer

Hi,  Saupayan  !

Thank you for your interest in my exercise (and solution).

Please note that the 2min average is what it is: an average. High level candidates are supposed to save time in many questions and use this "residual" wisely.

On the other hand, I am sure this is one of the most demanding problems I have created in more than 20 years teaching for the GMAT. In other words, do not take this problem too "seriously", ok?!

(If you thought about this problem for a while and you understood my solution throughly, the problem's mission was accomplished. And if you liked it, then I am specially happy!)

Best regards and success in your studies,
Fabio.

Saupayan · Answer

Fabio, thanks for the quick response and the brilliant range of questions you have posted on the forum.
With due respect and considerably less experience, I have yet to come across an official problem that couldn't be solved in < 2mins. That is not to say "I am able to solve every one of them in under 2 mins", but when presented with the solution, I can recognize how one could have. If you have examples to the contrary, can you kindly share those? I will be indebted.
In this case, even after knowing the method, if I encounter this problem in the future, I won't be able to. Could that be a me-problem? Quite likely.
That said, The concepts tested are quite nice and I didn't have a lot of difficulty identifying the process. Where I got stuck was the algebraic manipulation and/or number picking. Neither method yielded quick results. Thanks again for the the quick response, and I appreciate your support on the forum

fskilnik · Answer

Dear Saupayan , Thank you for your kind words and your respectful considerations. I guess you are right in terms of official GMAT questions available through GMAT official guides (books or online). In other words, it would be hard to find many questions that would need (say) 4min to be answered (safely) with "best approach". On the other hand, your REAL test will be (I hope) more demanding than that! This is the feedback I get from my top-scorer students (range 95-98 percentile in the quantitative section), just that. The fact that you feel ok about my arguments and techniques presented here it's marvellous. Please take into account what I said (and repeat): this problem is ESPECIALLY time-consuming... In this sense, this is not "really" GMAT-like, although I still believe it is nice for practice and for maturiy development. It's just my humble opinion, I respect who thinks differently. I do not intend to continue this conversation any further. Please accept my apologies if you think they are needed. Thank you again for your interest and understanding, All the best, Fabio. P.S.: I do not work with the GMAT preparation anymore (since 2021). Therefore please do not take it personally if I do not get involved in other questions/solutions topics presented in this forum. I came here because I guessed you deserved my feedback. Your kindness and politeness made sure I was right.

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