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9 a factor of 2x

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by venmic » Sat Jun 04, 2011 8:08 pm
9 factor of 2x


if 9 is a factor of 2x then which of the following may not be an integer?

A) 6x/54 + 2x/3
B) (4x-18)/9
C) (2x+27)/9
D) (81-4x²)/81
E) (2x-3)/3

[spoiler]Answer A
[/spoiler]
what if X is 54 then A IS an integer

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by Frankenstein » Sat Jun 04, 2011 8:17 pm
venmic wrote:9 factor of 2x


if 9 is a factor of 2x then which of the following may not be an integer?

A) 6x/54 + 2x/3
B) (4x-18)/9
C) (2x+27)/9
D) (81-4x²)/81
E) (2x-3)/3

[spoiler]Answer A
[/spoiler]
what if X is 54 then A IS an integer

Source Kaplan test
Hi,
'may not' is not the same as 'will not'. we are not concerned about when it is an integer. We only need to check if the value can ever not be an integer. So, if we take 2x as odd multiples of 9, op A won't be an integer. But all other options will be integers irrespective of what you take the value of x.
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by cans » Sat Jun 04, 2011 8:20 pm
9 is factor of 2x. x=is multiple of 9/2
a)6x/54 + 2x/3 = 7x/9 as x is multiple of 9/2, not sure(7/2 is not integer)
b)(4x-18)/9 2x is multiple of 9 and 18 is also multiple of 9. thus integer
c) (2x+27)/9 = 2x and 27 both are multiples of 9. thus integer
d) 2x is multiple of 9, thus 4*x^2 is multiple of 81. Thus integer
e) 2x is multiple of 9. thus integer/
IMO A
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by Anurag@Gurome » Sat Jun 04, 2011 10:12 pm
venmic wrote:If 9 is a factor of 2x then which of the following may not be an integer?

A) 6x/54 + 2x/3
B) (4x-18)/9
C) (2x+27)/9
D) (81-4x²)/81
E) (2x-3)/3
The problem has already been answered but I want to add something.

The critical point for solving this problem is we don't know whether x is integer or not. Usually in this kind of problems the unknown is declared as integer and that may subconsciously lead someone to a false assumption.

If x is integer, then 9 is a factor of 2x means 9 is a factor of x also. But if x is not an integer, then no such conclusion can be made.
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by sunilrawat » Sat Jun 04, 2011 10:28 pm
2x being a miltiple of 9, is also a multiple of 3
a) (3*2x)/(9*6) + 2x/3
Here, the second term is completely divisible by 3 leaving remainder as 0. But in the first term on simplification, we are left with (x/9) which may or may not result in 0 remainder. Thus option A.

Rest all options give 0 remainder and give an integer as the answer on simplification.
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by sandeep800 » Sun Jun 05, 2011 10:40 am
IMO A.... Thanx Anurag for additional inputs..
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by aftableo2006 » Sun Jun 05, 2011 11:06 pm
this is a good question a is the answer
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by Gurpinder » Sat Jun 11, 2011 8:46 am
2x/9 = int which also means that 2x is a multiple of 9. Therefore the minimum value of x = 9. So just plug that value in and estimate which answer doesn't workout.

Answer: A

(E) 2(9)-3/3 - since all are multiples of 3, it will workout
(D) since all are multiples of 9, it will workout
(C) since all are multiples of 3 and 9, all will workout
(B) again, all multiples of 3,9--> will workout
(A) This is all you are left with.
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by smackmartine » Sat Jun 11, 2011 9:16 am
IMO A
As soon as we see there is no restriction on x,we can think of x to be a fraction or integer.
After closely watching each option, we find that all options, but A have a even integer multiplied to it after we solve each expression to its basic form i.e 6x/54 = x/9 .

At x=4.5 , 9 is a factor of 2x.

a)x/9 + 2x/3

2x/3 ---> Integer as 2x is a factor of 9 ,but x/9 =4.5/9 is not.
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