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

Question

(A) –5
(B) 1
(C) 13
(D) 17
(E) 55

Mo2men · Answer

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.

rishi02 · Answer

2x-y=11 & 4x+y=k.

On adding we get 6x=11+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  ?

Abhishek009 · Answer

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

) Check the pattern - 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...

Bunuel · Answer

Edited. Thank you everyone.

stonecold · Answer

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

KanishkM · Answer

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.

KSBGC · Answer

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

D is the correct answer.

fireagablast · Answer

Combine the two equations
(1) 2x-y = 11
(2) 4x+y = z
 
2x-y+4x+y = 11+z
6x = 11+z
x=(11+z)/6

plug in the solutions:
11+17/6 = 28/6 --> doesn't work
17

Papist · Answer

2x-y = 11 can be rewritten as y = 2x-11

Then you can substitute 2x-11 for y in the second equation and get 6x-11

Since we know x and y both have to be integers, we simply have to test y values until we find one that produces a non-integer x value.

A) 6x - 11 = -5...x = 1
B) 6x - 11 = 1...x = 2
C) 6x - 11 = 13...x = 4
D) 6x - 11 = 17...x =  (28/6) which is not an integer 
E) 6x - 11 = 55...x = 11

Therefore Answer Choice  (D)  is correct.

ScottTargetTestPrep · Answer

Solution:

Let 4x + y = z. If we add this with 2x - y = 11, we have: 6x = z + 11. Since x is an integer, we see that z + 11 has to be a multiple of 6.

If we multiply 2x - y = 11 by 2, we have 4x - 2y = 22. If we subtract this from 4x + y = z, we have 3y = z - 22. Since y is an integer, we see that z - 22 has to be a multiple of 3.

Now, let’s look at the given answer choices:

A. -5

If z = -5, we see that -5 + 11 = 6 is a multiple of 6 and -5 - 22 = -27 is a multiple of 3. So 4x + y can be -5.

B. 1

If z = 1, we see that 1 + 11 = 12 is a multiple of 6 and 1 - 22 = -21 is a multiple of 3. So 4x + y can be 1.

C. 13

If z = 13, we see that 13 + 11 = 24 is a multiple of 6 and 13 - 22 = -9 is a multiple of 3. So 4x + y can be 13.

D. 17

If z = 17, we see that 17 + 11 = 28 is NOT a multiple of 6. So 4x + y CAN’T be 17.

Alternate Solution:

Let’s rewrite the first equation as:

2x - y = 11

2x = 11 + y

4x = 22 + 2y

Now substitute 22 + 2y for 4x into the expression 4x + y, obtaining:

22 + 2y + y

22 + 3y

Let’s set 22 + 3y = q and see if we obtain an integer solution for y as we test each of the answer choices.

Choice A. (q = -5): 22 + 3y = -5 becomes 3y = 27. Thus, y = 9, which is an integer.

Choice B. (q = 1): 22 + 3y = 1 becomes 3y = -21. Thus, y = -7, which is an integer.

Choice C. (q = 13): 22 + 3y = 13 becomes 3y = 9. Thus, y = 3, which is an integer.

Choice D. (q = 17): 22 + 3y = 17 becomes 3y = -5. Thus, y = -5/3, which is NOT an integer.

Choice E. (q = 55): 22 + 3y = 55 becomes 3y = 33. Thus, y = 11, which is an integer.

Thus, the correct answer is D.

Answer: D

bumpbot · Answer

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

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

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