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14 Feb 2021, 22:30
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D
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Difficulty:
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Question Stats:
44% (01:49) correct
56%
(02:04)
wrong
based on 132
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Is the positive integer x greater than 3?
(1) \(|x-5|=|x-13|\)
(2) \(x^2+10x-35>10(x-1)\)
(1) \(|x-5|=|x-13|\)
(2) \(x^2+10x-35>10(x-1)\)
16 Feb 2021, 05:10
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Statement 1: |x - 5| = |x - 13|
If both LHS and RHS are positive or both negative, we cannot get a value for x
If LHS is +ve and RHS -ve, we get x - 5 = -(x - 13). Solving we get x = 4
If LHS is -ve and the RHS +ve then we have -(x - 5) = x - 13. Solving we get x = 4
We have a definite answer that x is > 3.
Therefore Statement 1 Alone is Sufficient. Answer options could be A or D
Statement 2: \(x^2 + 10x - 35 > 10(x - 1)\)
\(x^2 + 10x - 35 > 10x - 10\)
\(x^2 > 25\)
Since x is a positive integer, we again get a definite answer that x > 3
Therefore Statement 2 Alone is Sufficient.
Option D
Arun Kumar
If both LHS and RHS are positive or both negative, we cannot get a value for x
If LHS is +ve and RHS -ve, we get x - 5 = -(x - 13). Solving we get x = 4
If LHS is -ve and the RHS +ve then we have -(x - 5) = x - 13. Solving we get x = 4
We have a definite answer that x is > 3.
Therefore Statement 1 Alone is Sufficient. Answer options could be A or D
Statement 2: \(x^2 + 10x - 35 > 10(x - 1)\)
\(x^2 + 10x - 35 > 10x - 10\)
\(x^2 > 25\)
Since x is a positive integer, we again get a definite answer that x > 3
Therefore Statement 2 Alone is Sufficient.
Option D
Arun Kumar
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Joined: 12 Aug 2020
Last visit: 15 Apr 2023
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Given Kudos: 130
Location: India
Schools: HEC MiM "23 (A) HEC Sept'22 Intake (A)
GMAT 1: 700 Q49 V35
GPA: 3.75
17 Feb 2021, 07:28
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Given x>0
1. |x−5|=|x−13|
If we considered LHS and RHS positive then x will get cancelled. (x-5=x-13).
So we consider one side as negative and the other positive.
We get,
x-5 = -(x-13)
x-5 = 13-x
2x = 18
x=9
Sufficient.
2.\(x^2+10x−35>10(x−1)\)
\(x^2+10x-35>10x-10\)
\(x^2-25>0\)
\(x^2>25\)
\(x>5\) (x cannot be lesser than -5 as x is positive)
Sufficient.
Answer. D
1. |x−5|=|x−13|
If we considered LHS and RHS positive then x will get cancelled. (x-5=x-13).
So we consider one side as negative and the other positive.
We get,
x-5 = -(x-13)
x-5 = 13-x
2x = 18
x=9
Sufficient.
2.\(x^2+10x−35>10(x−1)\)
\(x^2+10x-35>10x-10\)
\(x^2-25>0\)
\(x^2>25\)
\(x>5\) (x cannot be lesser than -5 as x is positive)
Sufficient.
Answer. D
