If x is an integer, is x between 27 and 54?

If x is an integer, is x between 27 and 54?

(1) The remainder when x is divided by 7 is 2.
--> x = 7p + 2, for all integers p
--> Many values of x are possible --> Insufficient

(2) The remainder when x is divided by 3 is 2.
--> x = 3q + 2, for all integers q
--> Many values of x are possible --> Insufficient

Combining (1) & (2),
x leaves remainder 2 when divided 7 & 3
LCM (7, 3) = 21
--> x = 21m + 2 for all integers m
--> Possible values = {. . . ,-40, -19, 2, 23, 44, 65 . . . .}
--> Insufficient

IMO Option E

(1) The remainder when x is divided by 7 is 2. insufic
\(x/7=m7+2:2,9,16,23,30,37,44…\)

(2) The remainder when x is divided by 3 is 2. insufic
\(x/3=n3+2:2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47…\)

(1 & 2) insufic
\(x=23,44…\)
or,
\(x=7m+2…x=3n+2…7m+2=3n+2…7m=3n\)
\((m,n)=(3,7;6,14;9,21…):x=7(3)+2=23…x=(23,44,65…)\)

Ans (E)

If x is an integer, is x between 27 and 54?

(1) The remainder when x is divided by 7 is 2.
Taking any random value of x is possible.
if x = 9, R[9/7] = 2; is 27 < x < 54 NO
if x = 30, R[30/7] = 2; is 27 < x < 54 YES

INSUFFICIENT.

(2) The remainder when x is divided by 3 is 2.
Again approaching in a similar manner.
if x = 11, R[11/3] = 2; is 27 < x < 54 NO
if x = 29, R[29/3] = 2; is 27 < x < 54 YES

INSUFFICIENT.

Together 1 and 2
LCM (3,7) = 21
So, values that results in a remainder = 2 are 2, 23, 44, 65 ... so on.
Hence, both possibilities still exists.

INSUFFICIENT.

IMO Answer E.
_________________

Statement 1:
x can be in any form of x = 7i + 2, where i is any nonnegative integer.. We can have x = 9 or x = 37, insufficient.

Statement2:
Similar to above, we can have x in the form of x = 3i + 2, where i is any nonnegative integer. Then x can be 3, 5, or 32. Insufficient.

Combined:
Focus on x - 2, we know x - 2 has a factor of 7 from (1) and a factor of 3 from (2). Then (x-2) must have a factor of 21. We can express x as x = 21*i + 2, where i is a nonnegative integer. So we can have x = 2, x = 23, x = 44, etc. Still insufficient.

Ans: E
_________________

If x is an integer, is x between 27 and 54?

(1) The remainder when x is divided by 7 is 2. —> insufficient: x can be 23 or 44
(2) The remainder when x is divided by 3 is 2.—> insufficient: x can be 23 or 44
Combining 1 & 2 also x can be 23 or 44
So the answer is E

Posted from my mobile device

We are informed that x is an integer between 27 and 54. We are to determine x.

Statement 1: The remainder when x is divided by 7 is 2

Not sufficient. Because the following integers in the given range leave a remainder of 2 when divided by 7:
30, 37, 44, 51.
This implies there is no unique value of x, hence insufficient.

Statement 2: The remainder when x is divided by 3 is 2.
This is also insufficient. We have the following possibilities for x: 29, 32, 35, 38, 41, 44, 47, 51.
Since we have more than one possible value of x, statement 2 is also insufficient.

1+2
combining 1 and 2 lead to 44 and 51. This is still insufficient. We are not able to narrow down to one answer choice. Both 44 and 51 satisfy the conditions in statements 1 and 2.

The answer is, therefore, option E.

X is an integer between 27 and 54

Statement 1- the remainder when x/7=2 eg 9/7=2 no

Eg2- 30/2= 2 yes

Statement 1 is insufficient

Statement 2- X/3=2

Eg 11/3=2 No

32/3=2. Yes


Combining both statements

23/7=2 and 23/3=2. No
44/7=2 and 44/3=2. Yes

Hence E both statements together are insufficient

Posted from my mobile device

(1) The remainder when x is divided by 7 is 2.

Numbers are 9, 16,23, 30,37,44, 51,58, 65, etc

Insufficient

(2) The remainder when x is divided by 3 is 2.

Numbers are 5, 8, 11, 14, 17,20,23,26, etc

Insufficient

Both statements together.

23,44, and the list goes on.

Therefore Answer is E IMO.

1 X= 7k+2 .... x can be 44 & 65 etc
2 x= 3k+2 ..... x can 44 & 65 etc
This implies E is the correct answer

Posted from my mobile device

#1
The remainder when x is divided by 7 is 2.
we get both yes & no
insufficient
#2
again insufficient
from 1&2
LCM= 21
again yes and no
IMO E

i dont get it
the question is asking whether x is between 27 and 54, which means 27<x<54
St-1 - x= 7a+2
so if we use 27<7a+2<54
then, 3.4<a<7.4
so we we get a - 4,5,6,7
no matter what the result the range of x stays within 27 and 54
so why is the option E
cause from what i understand from the question, it is asking if x lies between 27 and 54 and not asking for the exact value of x

What would be the difficulty of this DS problem ?

Posted from my mobile device

You can check difficulty level of a question in the tags just above the first post:

_________________