Is x an integer?
1. x+2 = x3
2. x>0
answer is [spoiler]A
Thanks![/spoiler]
DS  integers + absolute value
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Solving this question requires us to first recognize the 4 situations in which two absolute values can equal each other. I'll show this with 4 examples:
4=4 both values inside are already positive
4=4 both values inside are negative
4=4 one value is negative and the other is positive
4=4 one value is positive and the other is negative
So, (1) gives us 4 possible situations:
a. x+2 = x3 (both sides already yield positive outcomes)
b. (x+2) = (x3) (both sides yield negative outcomes, which we "undo" or make positive, by negating both sides, which is the same as finding the abs value of each side)
c. (x+2) = x3 (the left side yields a negative outcome, which we "undo" by negating)
d. x+2 = (x3) (the right side yields a negative outcome, which we "undo" by negating)
Note: c is actually impossible since it could never be the case that x+2 is negative while x3 is positive, but it doesn't really matter here.
If we solve each of the 4 equations we find that a and b cannot be solved. c and d yield x=1/2
Since x=1/2, we can conclude that (1) is sufficient.
(2) is not sufficient.
Answer: A
4=4 both values inside are already positive
4=4 both values inside are negative
4=4 one value is negative and the other is positive
4=4 one value is positive and the other is negative
So, (1) gives us 4 possible situations:
a. x+2 = x3 (both sides already yield positive outcomes)
b. (x+2) = (x3) (both sides yield negative outcomes, which we "undo" or make positive, by negating both sides, which is the same as finding the abs value of each side)
c. (x+2) = x3 (the left side yields a negative outcome, which we "undo" by negating)
d. x+2 = (x3) (the right side yields a negative outcome, which we "undo" by negating)
Note: c is actually impossible since it could never be the case that x+2 is negative while x3 is positive, but it doesn't really matter here.
If we solve each of the 4 equations we find that a and b cannot be solved. c and d yield x=1/2
Since x=1/2, we can conclude that (1) is sufficient.
(2) is not sufficient.
Answer: A

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From what I learnt on BTG, I think we dont need to solve all the 4 conditions.Brent Hanneson wrote:Solving this question requires us to first recognize the 4 situations in which two absolute values can equal each other. I'll show this with 4 examples:
4=4 both values inside are already positive
4=4 both values inside are negative
4=4 one value is negative and the other is positive
4=4 one value is positive and the other is negative
Answer: A
for example: if we are given
x y = a b
we need to solve only the foll cases
(x y) = + (ab) AND
(xy) = (a b)
Let me know if my understanding is wrong.
thanks
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You're absolutely right, vittalgmat.
I did more work than was necessary.
Good catch!
I did more work than was necessary.
Good catch!
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Or, using a distance approach, from Statement 1 we have x  (2) = x  3, which says, in words, "the distance between x and 2 is equal to the distance between x and 3". So x must be halfway between 2 and 3, and x = 1/2.joanjgonzalez wrote:Is x an integer?
1. x+2 = x3
2. x>0
answer is [spoiler]A
Thanks![/spoiler]
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
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Thanks Ian
I was also thinking something like that, but was not able to strike the problem in the same way
If we use the nonconventional methods like this, we save lot of time.
Karan
I was also thinking something like that, but was not able to strike the problem in the same way
If we use the nonconventional methods like this, we save lot of time.
Karan

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Thats superb Ian.Ian Stewart wrote:Or, using a distance approach, from Statement 1 we have x  (2) = x  3, which says, in words, "the distance between x and 2 is equal to the distance between x and 3". So x must be halfway between 2 and 3, and x = 1/2.joanjgonzalez wrote:Is x an integer?
1. x+2 = x3
2. x>0
answer is [spoiler]A
Thanks![/spoiler]