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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If x > 0, b > a, and 2x + 5 < 3x + 1,
[#permalink]
Updated on: 09 Jun 2018, 08:47
Updated on: 09 Jun 2018, 08:47
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Question Stats:
33% (01:58) correct
66% (02:09) wrong based on 15 sessions
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If x > 0, b > a, and 2x + 5 < 3x + 1, then which of the following COULD be a value of x?
i) 4.39
ii) 7.17
iii) 9.27
A) i and ii only
B) ii and iii only
C) i and iii only
D) iii only
E) i, ii and iii
Originally posted by GreenlightTestPrep on 07 Jun 2018, 05:35.
Last edited by Carcass on 09 Jun 2018, 08:47, edited 1 time in total.
Last edited by Carcass on 09 Jun 2018, 08:47, edited 1 time in total.
Added the timer
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If x > 0, b > a, and 2x + 5 < 3x + 1,
[#permalink]
09 Jun 2018, 05:20
09 Jun 2018, 05:20
1
GreenlightTestPrep wrote:
If x > 0, b > a, and 2x + 5 < 3x + 1, then which of the following COULD be a value of x?
i) 4.39
ii) 7.17
iii) 9.27
A) i and ii only
B) ii and iii only
C) i and iii only
D) iii only
E) i, ii and iii
Here's a useful triangle property:
So, if b > a, then we know that 3x + 1 < 4x - 8
We're also told that 2x + 5 < 3x + 1
So, we can create the following 3-part inequality: 2x + 5 < 3x + 1 < 4x - 8
Subtract 2x from all 3 sides: 5 < x + 1 < 2x - 8
When we examine 5 < x + 1, we can conclude that 4 < x. So, x is greater than 4
When we examine x + 1 < 2x - 8, we can conclude that 9 < x. So, x is greater than 9
So, we know that x is greater than 4 AND x is greater than 9
So, it MUST be the case that x is greater than 9
Check the statements.....
Only statement iii works.
Answer: D
Cheers,
Brent
gmatclubot
Re: If x > 0, b > a, and 2x + 5 < 3x + 1, [#permalink]
09 Jun 2018, 05:20
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