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### Show Tags

20 Jun 2010, 14:19
5
00:00

Difficulty:

45% (medium)

Question Stats:

66% (01:34) correct 34% (01:36) wrong based on 335 sessions

### HideShow timer Statistics

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.
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Joined: 02 Sep 2009
Posts: 52285

### Show Tags

21 Jun 2010, 03:08
5
4
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.

Hope it helps.
_________________
##### General Discussion
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Joined: 09 Jun 2010
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### Show Tags

20 Jun 2010, 17:59
1
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

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Joined: 05 Jul 2008
Posts: 130
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### Show Tags

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

I need Kudo so so much, guys.
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Joined: 09 Jun 2010
Posts: 1855
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### Show Tags

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: A trip coordinator bought a certain number of $47 tickets [#permalink] ### Show Tags 19 Nov 2017, 12:52 1 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: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: A trip coordinator bought a certain number of $47 tickets [#permalink] ### Show Tags 19 Dec 2017, 05:29 [quote="whiplash2411"]Let the number of$47 tickets be x and the number of $25 tickets be y. Statement 1 says: $$47*x + 25*y = 595$$ If you solve you will get X=10 and Y=1. None of the values other than this will match the equation So now, look at statement 2. It says $$y = x/2$$ This doesnt give any idea about the tickets. It can be (50, 100) (100,200). Answer is 'A'. Re: A trip coordinator bought a certain number of$47 tickets &nbs [#permalink] 19 Dec 2017, 05:29
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