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If x > 0, b > a, and 2x + 5 < 3x + 1, then

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If x > 0, b > a, and 2x + 5 < 3x + 1, then [#permalink]

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New post 07 Jun 2018, 06:33
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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

*kudos for all correct solutions

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Re: If x > 0, b > a, and 2x + 5 < 3x + 1, then [#permalink]

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New post 07 Jun 2018, 06:47
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GMATPrepNow wrote:
Image

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

*kudos for all correct solutions


Given b>a-----(1)
and (2x+5) < (3x+1)
or, c < a-------(2) (Corresponding angles)

from (1) & (2), we have

c<a<b
Or, (2x+5) < (3x+1)< (4x-8)------------(3)

Now solving the above inequality, we have

x>4-------------(4)
and x>9-------------(5)

from (4) and (5), we have

x>9-----------(6)

Hence value of x could be iii) 9.27

Ans option (D)
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If x > 0, b > a, and 2x + 5 < 3x + 1, then [#permalink]

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New post 07 Jun 2018, 07:15
1
GMATPrepNow wrote:
Image

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

*kudos for all correct solutions


Since b > a, the side opposite b(4x - 8) must be greater than the side opposite to a(3x + 1)

i) \(4.39 -> 4*4.39 - 8 = 9.56 < 3*4.39 + 1 =14.17\) -> Side opposite angle a is bigger
ii) \(7.17 -> 4*7.17 - 8 = 20.68 < 3*7.17 + 1 =22.51\) -> Side opposite angle a is bigger
iii) \(9.27 -> 4*9.27 - 8 = 29.08 > 3*9.27 + 1 = 28.81\) -> Side opposite angle b is bigger
(To save time, you can approximate the value of 4.39 as 4.4, 7.17 as 7.2, and 9.27 as 9.3)

Therefore, Option D(iii only) can be a value for x from the 3 options given.
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Re: If x > 0, b > a, and 2x + 5 < 3x + 1, then [#permalink]

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New post 08 Jun 2018, 05:01
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Given: x>0, b>a, 2x+5 < 3x+1

Approach: As b>a, therefore 4x-8 > 3x+1
or, x>9...................................................1

we know that, 2x+5 < 3x+1
solving this, we get x>4 ............................2

From 1, we know that x>9
So, option D is correct.
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If x > 0, b > a, and 2x + 5 < 3x + 1, then [#permalink]

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New post 08 Jun 2018, 09:53
1
GMATPrepNow wrote:
Image

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

*kudos for all correct solutions

I. Inequalities
Triangle Side/Angle relationship: Larger angles' opposite sides are longer than
smaller angles' opposite sides

Although we can combine inequalities, calculating them separately
is just as quick and possibly less confusing.

(1) Inequality 1, given: b > a

The side opposite b > side opposite a

\(4x-8>3x+1\)
\(x>9\)


(2) Inequality 2, given:
\(2x + 5 < 3x + 1\)
\(x>4\)


Inequality 1's lower limit in the range is 9
x MUST be greater than 9

That condition includes and satisfies Inequality 2

Any number greater than 9 is also greater than 4
By contrast, if x is between 4 and 9 (x >4 has lower limit 4),
then x is NOT greater than 9. Not valid.

x must be > 9

Only option iii) 9.27 works

Answer D

II. Inequalities - Number line

Range of solutions for x > 9 [in which (9) = not including 9]

<--0----------------------------(9)-------------------->

Range of solutions for x > 4

<--0--------(4)--------------------------------------->

The lower limit, 9, from the range in Inequality 1, makes the second range true.

x > 9, by definition, means x > 4. A number greater than 9 is also greater than 4

But 4 cannot be the lower limit for the range of solutions

That is, x cannot lie between 4 and 9.

If x lies between 4 and 9, then x is not greater than 9

<--0--------(4)---------------(9)------------------>

The range of solutions for x > 9 satisfies both conditions
The range of solutions for x > 4 does not satisfy the condition that x > 9

x > 9, is the only possibility, and
only option iii) 9.27, is > 9

Answer D
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Re: If x > 0, b > a, and 2x + 5 < 3x + 1, then [#permalink]

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New post 09 Jun 2018, 06:21
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GMATPrepNow wrote:
Image

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

*kudos for all correct solutions


Here's a useful triangle property:
Image

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
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Re: If x > 0, b > a, and 2x + 5 < 3x + 1, then [#permalink]

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New post 09 Jun 2018, 07:15
1
GMATPrepNow wrote:
Image

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

*kudos for all correct solutions


x>0
b > a
2x + 5 < 3x + 1 => x > 4
All three options fit.
Let us check the other given parameters
b>a means side opposite angle b > side opposite angle a
i.e. 4x-8 > 3x+1
x>9
Only iii fits 9.27
Answer D.
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Re: If x > 0, b > a, and 2x + 5 < 3x + 1, then   [#permalink] 09 Jun 2018, 07:15
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