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20 Jun 2010, 15:19
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67% (01:35) correct
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wrong
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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.
(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.
21 Jun 2010, 04:08
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anin
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.
General Discussion
whiplash2411
Joined: 09 Jun 2010
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20 Jun 2010, 18:59
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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.
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.
