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A casino pays players with chips that are either turquoise- or violet-

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A casino pays players with chips that are either turquoise- or violet-  [#permalink]

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New post Updated on: 18 May 2018, 06:46
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Question Stats:

33% (01:41) correct 67% (01:26) wrong based on 85 sessions

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A casino pays players with chips that are either turquoise- or violet- colored. If each turquoise- colored chip is worth t dollars, and each violet- colored chip is worth v dollars, where t and v are integers, what is the combined value of four turquoise- colored chips and two violet- colored chips?

(1) The combined value of six turquoise- colored chips and three violet- colored chips is 42 dollars.

(2) The combined value of five turquoise- colored chips and seven violet- colored chips is 53 dollars.

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Originally posted by CAMANISHPARMAR on 18 May 2018, 06:33.
Last edited by Bunuel on 18 May 2018, 06:46, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A casino pays players with chips that are either turquoise- or violet-  [#permalink]

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New post 18 May 2018, 08:31
1
A casino pays players with chips that are either turquoise- or violet- colored. If each turquoise- colored chip is worth t dollars, and each violet- colored chip is worth v dollars, where t and v are integers, what is the combined value of four turquoise- colored chips and two violet- colored chips?

Given

t = Value of one turquoise chip.
v = Value of one violet chip.

we have to find :

4t + 2v = ?


Statement (1) The combined value of six turquoise- colored chips and three violet- colored chips is 42 dollars.

6t + 3v = 42
simplify

3(2t + v ) = 42
2t + v = 14

so if we multiply the above equation by 2 we get

4t + 2v = 28

so Statement 1 is sufficient

Statement(2) The combined value of five turquoise- colored chips and seven violet- colored chips is 53 dollars.

5t + 7v = 53

now if we analyse we see the value v can take are:

v = 1 so 7v = 7
v = 2 so 7v = 14
v = 3 so 7v = 21
v = 4 so 7v = 28
v = 5 so 7v = 35
v = 6 so 7v = 42
v = 7 so 7v = 49


now if we put the value of 7v in the equation 5t + 7v =53 we get

for v = 1 so 7v = 7 ----------> 5t = 46
for v = 2 so 7v = 14 ----------> 5t = 39
for v = 3 so 7v = 21 ----------> 5t = 32
for v = 4 so 7v = 28 ----------> 5t = 25 ---> only possible vaule since t and v are both integer
for v = 5 so 7v = 35 ----------> 5t = 18
for v = 6 so 7v = 42 ----------> 5t = 9
for v = 7 so 7v = 49 ----------> 5t = 4


so from statement 2 we get value of t =5, v = 4 so ve can get the value of 4t + 2v = 28


so OA is D
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Re: A casino pays players with chips that are either turquoise- or violet-  [#permalink]

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New post 28 Nov 2018, 17:33
Hi All,

We're told that a casino pays players with chips that are either turquoise- or violet- colored; each turquoise- colored chip is worth T dollars, and each violet- colored chip is worth V dollars, where T and V are INTEGERS. We're asked for the combined value of four turquoise- colored chips and two violet- colored chips. This question is based on a number of different math skills - as well as your ability to 'play around' with a question and look for Number Properties.

To start, we're asked to find the value of 4T + 2V. This can be rewritten as "what is the value of 2(2T + V)?"

1) The combined value of six turquoise- colored chips and three violet- colored chips is 42 dollars.

The information in Fact 1 gives us the equation 6T + 3V = 42.
The entire equation can be 'divided by 3' to give us 2T + V = 14.
With this equation, we CAN answer the question --> 2(2T+V) = 2(14) = 28
Fact 1 is SUFFICIENT

2) The combined value of five turquoise- colored chips and seven violet- colored chips is 53 dollars.

The information in Fact 2 gives us the equation 5T + 7V = 53.

At first glance, this might appear insufficient, since we have 2 variables, but only one equation. However, the total (53) is relatively small for a sum that is based off of TWELVE coins. Since both T and V are INTEGERS, it's possible that there's only one solution to this equation (and if that's the case, then Fact 2 is also sufficient).

We're essentially adding a multiple of 5 to a multiple of 7 and ending up with 53. You can look for the options rather easily by listing out all of the multiples:
Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Multiples of 7: 7, 14, 21, 28, 35, 42, 49

Looking at these two lists, can you pull a number from each and get a sum of 53...? You would probably find it easiest to start with a multiple of 7, since the 'rest' of the total would have to be from the multiple of 5 (meaning that that 'piece' would end in a 5 or a 0).

With 28 and 25 (re: 4x7 and 5x5), you get a total of 53. There is no other option though. There's just one solution.
Fact 2 is SUFFICIENT

Final Answer:

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A casino pays players with chips that are either turquoise- or violet-  [#permalink]

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New post 28 Nov 2018, 17:55
CAMANISHPARMAR wrote:
A casino pays players with chips that are either turquoise- or violet- colored. If each turquoise- colored chip is worth t dollars, and each violet- colored chip is worth v dollars, where t and v are integers, what is the combined value of four turquoise- colored chips and two violet- colored chips?

(1) The combined value of six turquoise- colored chips and three violet- colored chips is 42 dollars.

(2) The combined value of five turquoise- colored chips and seven violet- colored chips is 53 dollars.

\({\text{chips}}\,\left\{ \begin{gathered}
\,{\text{turqu}}:\$ \,t\,\,{\text{each}} \hfill \\
\,{\text{violet}}:\$ \,v\,\,{\text{each}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\left( {t,v \geqslant 1\,\,\,{\text{ints}}\,\,\left( * \right)} \right)\)

\(? = 4t + 2v\,\,\,\left[ \$ \right]\)


\(\left( 1 \right)\,\,6t + 3v = 42\,\,\,\left[ \$ \right]\,\,\,\,\,\mathop \Rightarrow \limits^{:\,3} \,\,2t + v = 14\,\,\,\left[ \$ \right]\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,2} \,\,\,\,\,? = 28\,\,\,\left[ \$ \right]\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)


\(\left( 2 \right)\,\,5t + 7v = 53\,\,\,\left[ \$ \right]\,\,\,\,\, \Rightarrow \,\,53 - 7v = 5t\,\,,\,\,t\mathop \geqslant \limits^{\left( * \right)} 1\,\,\operatorname{int} \,\,\,\, \Rightarrow \,\,1\mathop \leqslant \limits^{\left( * \right)} v\,\,\left( {\operatorname{int} } \right) \leqslant 7\,\,\,\left( {**} \right)\)

\(\frac{{\left( {3 + 50} \right) - \left( {5v + 2v} \right)}}{5} = \operatorname{int} \,\,\,\, \Leftrightarrow \,\,\,\,\frac{{3 - 2v}}{5} = \operatorname{int} \,\,\,\,\mathop \Leftrightarrow \limits^{\left( {**} \right)} \,\,\,\,v = 4\,\,\,\,\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,t = 5\)

\(\left( {t,v} \right)\,\,{\text{unique}}\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,\,\,{\text{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}{\text{.}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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