If x is an integer, is x2 > 25? (1) |x – 3| > 5 (2) (x + y)2 > 49, whe

Given : x is an integer.

Question : Is x < - 5 or x > 5 ?

St 1: |x - 3| > 5

x - 3 > 5 or x > 8

Or x - 3 < -5 or x < -2

Therefore, 8 > 5 but -2 is not less the -5 .

(x + y) ^ 2 > 49

x + y > +/-7

x + y < - 7 --- (I) or x + y > 7 (II)

|y| < 2

y = -1, 0, 1

Put y = -1 and 1 in equation (I)

x - 1 < -7, so x < -6

x +1 < -7, so x < -8

Put y = -1 and 1in equation (II)

x - 1 > 7, so x > 8

x+1 > 7, so x > 6


So from statement 2, we get all the values of x in the range x < - 5 or x > 5

Hence Sufficient (B)

Given:
x^2 > 25?
x<-5 or x >5?

statement 1:
x-3> 5 or x-3<-5
x>8 or x<-2
x can be -3 falling outside the range
x can be -6 falling within the range
not sufficient

statement 2:
-2 < y < 2
(x + y) > 7 or x+y<-7
x > 7-y or x < -7-y
x > 7 - y, in order to get the minimum value take maximum value of y: 1
x > 7 -1 or x> 6

x < -7-y, in order to get the maximum value take minimum value of y: -1
x < -7+1 or x < -6

x > 6 or x < -6
this is sufficient

you assume y to be an integer in statement 2 when it's not mentioned explicitly. y could be 1.9 or -1.9. Doesn't change the answer in this case though

It is not given in the question stem but it is given in statement 2.

hey arunsharma

can we write statement 2 as x+y>7?

hi Kritisood,

statement 2 is of the form \(a^2>b^2\)
if we know that a & b are not 0 and have a positive value then yes, I can write a > b.
otherwise I'll have to solve this inequality by finding the possible range for \((a+b)(a-b)>0\).
in this question we have to follow the latter approach as we do not know signs of x & y.

got it thanks
Also from st 2, we get -6<x<6 how is this sufficient to answer the question? we have to answer if -5<x<5
as per -6<x<6 the answer may (x=4) or may not (x=5/x=-5) be in the range of -5<x<5
am i missing something? chetan2u

\(x^2 > 25\) MEANS we are looking for whether x<-5 or x>5

(1) |x – 3| > 5
x-3>5.....x>8....So surely x>5
Or 3-x>5.....x<-2....x could be between -2 and -5, then NO, and ix x<-5 , yes
Insuff

(2) (x + y)^2 > 49, where y is an integer such that |y| < 2
\(\sqrt{(x + y)^2} > \sqrt{49}.\)
Two possibilities
a) x+y<-7, but -2<y<2....so y could be -1, 0, 1
Max value of x is when y is the least, so let us take it y=-1, we get x+(-1)<-7....x<-6.
b) x+y>7, but -2<y<2...so y could be -1, 0, 1
Min value of x is when y is the greatest, so let us take it y=1, we get x+y+1>7....x>6.
Thus if the min value of x>5, we can again say yes.
Suff

B

Kritisood, now on your query.
If statement II gives you the range - x<-6 and x>6, and you have to answer whether x<-5 and x>5, your answer will be yes.
So you are looking for x<-5, and statement II tells you that x can be -7, -8 or less .... x will always be less than -5.
Or you are looking for x>5, and statement II tells you that x can be 7, 8 or more.... x will always be greater than 5.

got it, chetan2u thanks for explaining.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.