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

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GMATH Teacher
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A box has only pure material A identical coins, pure material B identi  [#permalink]

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20 Feb 2019, 09:37
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85% (hard)

Question Stats:

39% (02:42) correct 61% (02:35) wrong based on 28 sessions

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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.

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Re: A box has only pure material A identical coins, pure material B identi  [#permalink]

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20 Feb 2019, 10:37
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.

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
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GMATH Teacher
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A box has only pure material A identical coins, pure material B identi  [#permalink]

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20 Feb 2019, 14:15
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.

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

$$? = x + y$$

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

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

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

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

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

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

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

Regards,
Fabio.
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Re: A box has only pure material A identical coins, pure material B identi  [#permalink]

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20 Feb 2019, 17:35
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.
Re: A box has only pure material A identical coins, pure material B identi   [#permalink] 20 Feb 2019, 17:35
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