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M14-16

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M14-16  [#permalink]

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New post 16 Sep 2014, 00:53
2
7
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

66% (01:41) correct 34% (01:31) wrong based on 82 sessions

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New post 16 Sep 2014, 00:53
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1
Official Solution:


(1) 5 pencils and 3 pens cost 30 cents. Given: \(5x+3y=30\). This equation has only one positive integer solution:\(x=3\) and \(y=5\), so the price of 5 pencils is \(5*3=15\) cents. Sufficient.

(2) 4 pencils and 4 pens cost 32 cents. Given: \(4x+4y=32\), so \(x+y=8\). This equation has more than one positive integer solution, for example: \(x=1\) and \(y=7\) or \(x=3\) and \(y=5\). Not sufficient.


Answer: A
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Re: M14-16  [#permalink]

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New post 30 Oct 2014, 15:40
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Hi Bunuel,

I think this is a great question. I falled for the trap and picked C here. At a very first glance I read this question as:

- Question stem: 5x = ?
- Statement 1: 5x + 3y = 30. I thought that since I had 1 equation and 2 variables I couldn´t solve for this.
- Statement 2: x + y = 8. Again, 1 equation and 2 variables.

So I thought the answer was C.

I understand that what makes Statement 1 sufficient is that the question stem gives us an "Integer Constraint" (“…if the price of each item in cents has a positive integer value…”)

But my question is: how will I realize about this in the next similar problem I have and prevent from falling for this trap again? Should I pay close attention to whether the question stem mentions this "Integer Constraint" (such as in the portion of the question mentioned above)?
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Re: M14-16  [#permalink]

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New post 31 Oct 2014, 05:24
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minwoswoh wrote:
Hi Bunuel,

I think this is a great question. I falled for the trap and picked C here. At a very first glance I read this question as:

- Question stem: 5x = ?
- Statement 1: 5x + 3y = 30. I thought that since I had 1 equation and 2 variables I couldn´t solve for this.
- Statement 2: x + y = 8. Again, 1 equation and 2 variables.

So I thought the answer was C.

I understand that what makes Statement 1 sufficient is that the question stem gives us an "Integer Constraint" (“…if the price of each item in cents has a positive integer value…”)

But my question is: how will I realize about this in the next similar problem I have and prevent from falling for this trap again? Should I pay close attention to whether the question stem mentions this "Integer Constraint" (such as in the portion of the question mentioned above)?


This is a C-Trap question. "C trap" question is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer.

Check for more here: c-trap-questions-177044.html

Hope it helps.
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Re: M14-16  [#permalink]

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New post 13 Jun 2016, 02:16
Very tricky one indeed. One more thing which needs to be highlighted is the fact that there could be another solution to the equation 5x+3y=30 which is x=0 and y=10 but since we are given in the question that the number of pens and pencils are positive integer values, this solution is not possible.
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Re: M14-16  [#permalink]

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New post 10 Aug 2016, 23:03
minwoswoh wrote:
Hi Bunuel,

I think this is a great question. I falled for the trap and picked C here. At a very first glance I read this question as:

- Question stem: 5x = ?
- Statement 1: 5x + 3y = 30. I thought that since I had 1 equation and 2 variables I couldn´t solve for this.
- Statement 2: x + y = 8. Again, 1 equation and 2 variables.

So I thought the answer was C.

I understand that what makes Statement 1 sufficient is that the question stem gives us an "Integer Constraint" (“…if the price of each item in cents has a positive integer value…”)

But my question is: how will I realize about this in the next similar problem I have and prevent from falling for this trap again? Should I pay close attention to whether the question stem mentions this "Integer Constraint" (such as in the portion of the question mentioned above)?


IMO you should be always cautious on similar tasks (number of cakes/pencils/animals). They actually follow a similar pattern, so just solve as much of them as possible. And alway pay attention to INTEGER CONSTRAINT.
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Re: M14-16  [#permalink]

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New post 28 Nov 2016, 17:31
I was not the only one who thought like you..

minwoswoh wrote:
Hi Bunuel,

I think this is a great question. I falled for the trap and picked C here. At a very first glance I read this question as:

- Question stem: 5x = ?
- Statement 1: 5x + 3y = 30. I thought that since I had 1 equation and 2 variables I couldn´t solve for this.
- Statement 2: x + y = 8. Again, 1 equation and 2 variables.

So I thought the answer was C.

I understand that what makes Statement 1 sufficient is that the question stem gives us an "Integer Constraint" (“…if the price of each item in cents has a positive integer value…”)

But my question is: how will I realize about this in the next similar problem I have and prevent from falling for this trap again? Should I pay close attention to whether the question stem mentions this "Integer Constraint" (such as in the portion of the question mentioned above)?
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Re: M14-16  [#permalink]

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New post 01 Dec 2016, 10:28
56 seconds that it.

I found it simple as by looking at statement I, There is only one solution i.e. x=3 and y=5. Only one answer.
Statement II: 4x+4y=32 will have multiple answers for x and y from which fixing one value is not possible.
x=2 then y= 6
x=4 and y=4
x=6 and y=2
Not sufficient.
Hence answer is A.
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Re: M14-16  [#permalink]

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New post 25 May 2017, 07:29
How can I get fast to X=3 and Y=5? should I just plug in numbers? is this the best way to do it?
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New post 25 May 2017, 07:46
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M14-16  [#permalink]

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New post 25 May 2017, 11:35
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cocojiz wrote:
How can I get fast to X=3 and Y=5? should I just plug in numbers? is this the best way to do it?


Hi

Let me see if I can help a bit.

we have 5x + 3y = 30 and we know both x and y have to be positive integers.
5x is obviously a multiple of 5, and on the right side 30 is also a multiple of 5. So, by simple logic, '3y' also must be a multiple of 5.
For '3y' to be a multiple of 5, 'y' must be a multiple of 5.

so we put y=5, and get 5x + 3*5 = 30. This gives us x=3
If we put y=10, we get x =0 which is not possible.

There is another way. 5x + 3y = 30 and we know both x and y have to be positive integers.
3y is obviously a multiple of 3, and on the right side 30 is also a multiple of 3. So, by simple logic, '5x' also must be a multiple of 3.
For '5x' to be a multiple of 3, 'x' must be a multiple of 3.

so we put x=3, and get 5*3 + 3y = 30. This gives us y=5
If we put x=6, we get y =0 which is not possible

This method sometimes helps.
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Re M14-16  [#permalink]

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New post 05 Aug 2017, 06:21
I think this is a high-quality question and I agree with explanation.
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Re: M14-16  [#permalink]

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New post 14 Jun 2018, 13:15
Bunuel wrote:
minwoswoh wrote:
Hi Bunuel,

I think this is a great question. I falled for the trap and picked C here. At a very first glance I read this question as:

- Question stem: 5x = ?
- Statement 1: 5x + 3y = 30. I thought that since I had 1 equation and 2 variables I couldn´t solve for this.
- Statement 2: x + y = 8. Again, 1 equation and 2 variables.

So I thought the answer was C.

I understand that what makes Statement 1 sufficient is that the question stem gives us an "Integer Constraint" (“…if the price of each item in cents has a positive integer value…”)

But my question is: how will I realize about this in the next similar problem I have and prevent from falling for this trap again? Should I pay close attention to whether the question stem mentions this "Integer Constraint" (such as in the portion of the question mentioned above)?


This is a C-Trap question. "C trap" question is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer.

Check for more here: http://gmatclub.com/forum/c-trap-questions-177044.html

Hope it helps.



Hi Bunuel & minwoswoh
I don't remember but i too had a similar question - someone responded/showed this way to solve it --
5x = ?
5x + 3y = 30
5x = 30 - 3y
x = (3/5)*[10 - y]
As it is given that values of x and y are unique integers => (10 - y) has to be multiple of 5, so that denominator can be removed!!!
only value of y which can do it is y=5,
thus y = 5 and x = 3

kudo if this helped :thumbup:
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Re: M14-16  [#permalink]

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New post 27 Mar 2019, 21:15
What if y=0? The problem can still be solved i believe
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Re: M14-16   [#permalink] 27 Mar 2019, 21:15
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