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A trip coordinator bought a certain number of $47 tickets

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A trip coordinator bought a certain number of $47 tickets [#permalink]

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New post 20 Jun 2010, 14:19
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C
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Question Stats:

68% (01:05) correct 32% (00:55) wrong based on 246 sessions

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A trip coordinator bought a certain number of $47 tickets and a certain number of $25 tickets for a concert. How many $47 tickets did she buy?

(1) The trip coordinator spent a total of $595 on tickets for the concert.
(2) She bought half as many $25 tickets as $47 tickets.
[Reveal] Spoiler: OA

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Re: Equations [#permalink]

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New post 20 Jun 2010, 17:59
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Let the number of $47 tickets be x and the number of $25 tickets be y.

Statement 1 says:

\(47*x + 25*y = 595\)

This is clearly insufficient to answer the question, so we can eliminate choices A and D.

So now, look at statement 2.

It says \(y = x/2\)

Now this is also insufficient to answer the question. So we can eliminate answer choice B.

Put them together:

y=x/2

x = 2y

This means:

\(47x + 25x/2 = 595\)

Hence (47+12.5)x = 595

x = \(\frac {595}{59.5}\) = 10

This is solvable. So the answer is C. Hope this answers your question.

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Re: Equations [#permalink]

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New post 21 Jun 2010, 03:08
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anin wrote:
A trip coordinator bought a certain number of $47 tickets and a certain number of $25 tickets for a concert. How many $47 tickets did she buy?

1. The trip coordinator spent a total of $595 on tickets for the concert.
2. She bought half as many $25 tickets as $47 tickets.


This is classic C-trap question: the question which is obviously sufficient if we take statements together. When we see such questions we should become very suspicious.

Let \(x\) be the # of 47$ tickets and \(y\) the # of 25$ tickets (note that \(x\) and \(y\) must be non-negative integers as they represent # of tickets). Question: \(x=?\)

(1) \(47x+25y=595\).

When we have equation of a type \(ax+by=c\) and we know that \(x\) and \(y\) are non-negative integers, there can be multiple solutions possible for \(x\) and \(y\) (eg \(2x+3y=20\)) OR just one combination (eg \(2x+3y=7\)). Hence in some cases \(ax+by=c\) is NOT sufficient and in some cases it is sufficient.

So we should check whether \(47x+25y=595\) has one solution or more than one. How to check this quickly: last digit of \(25y\) will be 0 or 5, so the last digit of \(47x\) must also be 0 or 5 so that their sum to have last digit 5 (595). This reduces the possible values of \(x\) to 3 values: 0, 5, and 10 (if \(x>10\), eg. 15, ... \(47x>595\)).

\(x=0\) -->\(47*0+25y=595\) --> \(25y=595\) --> \(y\neq{integer}\), not a valid solution;
\(x=5\) -->\(47*5+25y=595\) --> \(25y=360\) --> \(y\neq{integer}\), not a valid solution;
At this stage we can already say that statement (1) is sufficient as the last value left for \(x\) -10, must satisfy the equation \(47x+25y=595\) or otherwise the question won't have any solution and question would be flawed. But let's check this anyway:
\(x=10\) -->\(47*10+25y=595\) --> \(25y=125\) --> \(y=5={integer}\).

So equation \(47x+25y=595\) has only one non-negative integer solution: \(x=10\) and \(y=5\). Sufficient.

(2) She bought half as many $25 tickets as $47 tickets --> \(x=2y\). Clearly insufficient.

Answer: A.

Hope it helps.
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Re: Equations [#permalink]

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New post 21 Jun 2010, 03:13
anin wrote:
A trip coordinator bought a certain number of $47 tickets and a certain number of $25 tickets for a concert. How many $47 tickets did she buy?

1. The trip coordinator spent a total of $595 on tickets for the concert.
2. She bought half as many $25 tickets as $47 tickets.

1. 47*x+25*y=595: suff
x, y must be integer.
x < 595/47= 13
x must be divisible by 5 so x can be 0, 5, 10
Only x= 10 makes y an integer

So A is sufficient

2. x=2y: insuf


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Re: Equations [#permalink]

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New post 21 Jun 2010, 04:50
Thanks, Bunuel. I didn't see that trap!! I thought of plugging in numbers initially and decided against it. Maybe I should in the future, just in case :)

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Re: Equations [#permalink]

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New post 21 Jun 2010, 11:22
Thanks, Guys. The OA is A.

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Re: Equations [#permalink]

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New post 21 Jun 2010, 13:26
Nice Question.. Even I was trapped in C...

Thanks Guys

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Re: A trip coordinator bought a certain number of $47 tickets [#permalink]

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New post 19 Nov 2017, 12:52
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Hi All,

While this is an older question, it emphasizes one of the potential patterns that you'll see on Test Day (and the first solution offered actually makes a MISTAKE that many Test Takers will make)...

We're told that a certain number of $47 tickets and a certain number of $25 tickets were purchased for a concert. We're asked for the number of $47 tickets purchased. This question can be solved with a bit of logic and some 'brute-force' arithmetic.

1) The trip coordinator spent a total of $595 on tickets for the concert.

At first glance, the information in Fact 1 implies that there could be multiple solutions, since we have 2 variables and just one equation: 47X + 25Y = 595. However, there is an important Number Property pattern here that limits the possibilities - the number of $25 tickets will add up to a total that is a MULTIPLE of 25. $595 is NOT a multiple of 25 though, so we will have to find a specific multiple of 47 that completes the equation (and that multiple will have to end in either a 5 or a 0)

(47)(5) = 235... which would leave 595 - 235 = 360 for $25 tickets. That is NOT possible though (360 is NOT a multiple of 25), so this is NOT an option.
(47)(10) = 470... which would leave 595 - 470 = 125 for $25 tickets. That IS possible (125 IS a multiple of 25), so this IS an option.
(47)(15) = 705 .... this is NOT an option (the total is TOO HIGH).
Thus, there's only one solution: 10 $47 tickets and 5 $25 tickets
Fact 1 is SUFFICIENT

2) She bought half as many $25 tickets as $47 tickets.

With Fact 2, we could have....
1 $25 ticket and 2 $47 tickets
2 $25 tickets and 4 $47 tickets
Etc.
Fact 2 is INSUFFICIENT

Final Answer:
[Reveal] Spoiler:
A


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Re: A trip coordinator bought a certain number of $47 tickets   [#permalink] 19 Nov 2017, 12:52
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