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# If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be

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Math Expert
Joined: 02 Sep 2009
Posts: 60678
If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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22 May 2016, 14:42
00:00

Difficulty:

55% (hard)

Question Stats:

61% (02:03) correct 39% (02:19) wrong based on 174 sessions

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If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be

(A) –5
(B) 1
(C) 13
(D) 17
(E) 55

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Posts: 2344
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Re: If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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22 May 2016, 17:18
1
Bunuel wrote:
If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be

(A) –5
(B) 1
(C) 13
(D) 17
(E) 551

Hi Bunuel,

2x-y=11....y=2x-11

4x+y=4x+2x-11=6x-11

6x-11=-5...x=1
6x-11=1... x=2
6x-11=13...x=4
6x-11=17..X is not integer
6x-11=551..X is not integer

I think the choice E is 55 not 551. Otherwise both D & E CANNOT be solution.
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Re: If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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23 May 2016, 09:38
1
2x-y=11 & 4x+y=k.

x = (11+k)/6

Insert each option and verify.

Both D and E do not satisfy the equation though. One should be removed ? Bunuel ?
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Re: If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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23 May 2016, 11:11
1
Bunuel wrote:
If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be

(A) –5
(B) 1
(C) 13
(D) 17
(E) 551

Will use a rather unusual way ( As I can not think algebraically , using X and Ys )

Check the pattern -

Attachment:

Capture.PNG [ 2.83 KiB | Viewed 3601 times ]

Can you find something useful ? Yes you are correct the numbers are divisible by 6....

Given 2x–y= 11 and 4x+ y ( Results in the options ) must be a multiple of 6 , check out..

(A) –5

11-5 =6 (Multiple of 6)

(B) 1

11 + 1 = 12 (Multiple of 6)

(C) 13

11 + 13 = 24 (Multiple of 6)

(D) 17

11 + 17 = 28 (Not a Multiple of 6)

(E) 551

11 + 551 = 562 (Not a Multiple of 6)

I request Bunuel to kindly edit the post, either (D) or (E) must be changed...
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Math Expert
Joined: 02 Sep 2009
Posts: 60678
Re: If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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28 Oct 2016, 01:42
Abhishek009 wrote:
Bunuel wrote:
If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be

(A) –5
(B) 1
(C) 13
(D) 17
(E) 551

Will use a rather unusual way ( As I can not think algebraically , using X and Ys )

Check the pattern -

Attachment:
Capture.PNG

Can you find something useful ? Yes you are correct the numbers are divisible by 6....

Given 2x–y= 11 and 4x+ y ( Results in the options ) must be a multiple of 6 , check out..

(A) –5

11-5 =6 (Multiple of 6)

(B) 1

11 + 1 = 12 (Multiple of 6)

(C) 13

11 + 13 = 24 (Multiple of 6)

(D) 17

11 + 17 = 28 (Not a Multiple of 6)

(E) 551

11 + 551 = 562 (Not a Multiple of 6)

I request Bunuel to kindly edit the post, either (D) or (E) must be changed...

Edited. Thank you everyone.
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Re: If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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28 Oct 2016, 01:49
Here is how i would solve it =>
2x-y=11
2x=11+y
4x=22+2y

now we need to check the value of 4x+4 => 22+2y+y=22+3y
hence 22+3y=value(in the options)
so y=value-22/3
so checking the options only D is unfit
Hence D
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Re: If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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27 Jan 2019, 10:12
Bunuel wrote:
If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be

(A) –5
(B) 1
(C) 13
(D) 17
(E) 55

If x and y are integers and 2x–y= 11

2x - 11 = y

now just get the values for x and y, they are integers you can plug in

x-----y
1-----(-9)
2-----(-7)
3-----(-5)
4-----(-3)
5-----(-1)

Now just substitute them back and get the value which is not possible

16 - 3 = 13
20 - 1 = 19

D doesn't exist.
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If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be  [#permalink]

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29 Jan 2019, 13:53
Bunuel wrote:
If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be

(A) –5
(B) 1
(C) 13
(D) 17
(E) 55

Given,

2x - y = 11

2x -11 =y

we are looking for the value of 4x + y.

4x + y

4x + 2x -11

6x -11 = ?

*** Trail and error method is suitable now.

6*1 - 11 = -5
6*2 -11 =1
6*3 -11 =7
6*4 -11 =13
6*5 -11 = 19
6*6 - 11 = 25

look at the answer pattern........Value is increasing .

17 is not in the list and no chance to get this integer.

6*11 -11 =55

If x and y are integers and 2x–y= 11, then 4x+ y CANNOT be   [#permalink] 29 Jan 2019, 13:53
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