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gsingh0711
Joined: 02 Mar 2018
Last visit: 15 Aug 2018
Posts: 42
Given Kudos: 29
Location: India
Schools: IIMA PGPX"20 IIMA PGPX"20
GMAT 1: 640 Q51 V26
GPA: 3.1
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Updated on: 30 Jul 2018, 20:44
Originally posted by gsingh0711 on 30 Jul 2018, 14:51.
Last edited by Bunuel on 30 Jul 2018, 20:44, edited 1 time in total.
Last edited by Bunuel on 30 Jul 2018, 20:44, edited 1 time in total.
Renamed the topic and edited the question.
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Given that x is a real number, is −1 < x < 3?
(1) ||x − 1| − 1| < 1
(2) (x + 1)(x − 3) < 0
(1) ||x − 1| − 1| < 1
(2) (x + 1)(x − 3) < 0
gsingh0711
Joined: 02 Mar 2018
Last visit: 15 Aug 2018
Posts: 42
Given Kudos: 29
Location: India
Schools: IIMA PGPX"20 IIMA PGPX"20
GMAT 1: 640 Q51 V26
GPA: 3.1
Products:
30 Jul 2018, 14:53
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Explanation:
As we know | x | = x if x > 0, and | x | = −x if x < 0.
Considering statement 1:
| | x − 1 | − 1 | < 1
⇒ −1 < | x − 1 | − 1 < 1
(If | N | < 1, then N lies between −1 and 1.)
Adding 1 to every side of the inequality ⇒ 0 < | x − 1 | < 2
Considering the LHS of equation,
| x − 1 | > 0 ⇒ x ≠ 1
(Absolute value of any number is always positive as long as the value within modulus is not zero.)
Considering the RHS of equation,
| x − 1 | < 2
⇒ −2 < x − 1 < 2
(If | N | < 2, then N lies between −2 and 2.)
⇒ −1 < x < 3
Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.
Considering statement 2:
(x + 1)(x − 3) < 0
LHS of the inequality (x + 1)(x − 3) is equal to zero if x = −1 or 3. Hence, for any value of x between −1 and 3, the LHS is negative. This can be verified by picking any value of x between −1 and 3, say x = 0. Then x + 1 is a positive value and x − 3 is a negative value. Hence, the product of x + 1 and x − 3 is negative.
Hence, statement 2 ⇒ −1 < x < 3
Since we are getting a definite answer from above statement , statement 2 itself is sufficient to provide the answer.
Answer: D.
I think, I deserve a Kudus:D Lol
As we know | x | = x if x > 0, and | x | = −x if x < 0.
Considering statement 1:
| | x − 1 | − 1 | < 1
⇒ −1 < | x − 1 | − 1 < 1
(If | N | < 1, then N lies between −1 and 1.)
Adding 1 to every side of the inequality ⇒ 0 < | x − 1 | < 2
Considering the LHS of equation,
| x − 1 | > 0 ⇒ x ≠ 1
(Absolute value of any number is always positive as long as the value within modulus is not zero.)
Considering the RHS of equation,
| x − 1 | < 2
⇒ −2 < x − 1 < 2
(If | N | < 2, then N lies between −2 and 2.)
⇒ −1 < x < 3
Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.
Considering statement 2:
(x + 1)(x − 3) < 0
LHS of the inequality (x + 1)(x − 3) is equal to zero if x = −1 or 3. Hence, for any value of x between −1 and 3, the LHS is negative. This can be verified by picking any value of x between −1 and 3, say x = 0. Then x + 1 is a positive value and x − 3 is a negative value. Hence, the product of x + 1 and x − 3 is negative.
Hence, statement 2 ⇒ −1 < x < 3
Since we are getting a definite answer from above statement , statement 2 itself is sufficient to provide the answer.
Answer: D.
I think, I deserve a Kudus:D Lol
General Discussion
02 Aug 2018, 03:07
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[quote="gsingh0711"]Explanation:
As we know | x | = x if x > 0, and | x | = −x if x < 0.
Considering statement 1:
| | x − 1 | − 1 | < 1
⇒ −1 < | x − 1 | − 1 < 1
(If | N | < 1, then N lies between −1 and 1.)
Adding 1 to every side of the inequality ⇒ 0 < | x − 1 | < 2
Considering the LHS of equation,
| x − 1 | > 0 ⇒ x ≠ 1
(Absolute value of any number is always positive as long as the value within modulus is not zero.)
| x − 1 | < 2
⇒ −2 < x − 1 < 2
(If | N | < 2, then N lies between −2 and 2.)
⇒ −1 < x < 3
Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.
If X is not equal to 1,how come x lies between -1 and 3?
As we know | x | = x if x > 0, and | x | = −x if x < 0.
Considering statement 1:
| | x − 1 | − 1 | < 1
⇒ −1 < | x − 1 | − 1 < 1
(If | N | < 1, then N lies between −1 and 1.)
Adding 1 to every side of the inequality ⇒ 0 < | x − 1 | < 2
Considering the LHS of equation,
| x − 1 | > 0 ⇒ x ≠ 1
(Absolute value of any number is always positive as long as the value within modulus is not zero.)
| x − 1 | < 2
⇒ −2 < x − 1 < 2
(If | N | < 2, then N lies between −2 and 2.)
⇒ −1 < x < 3
Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.
If X is not equal to 1,how come x lies between -1 and 3?
