If x 0, b a, and 2x + 5

Question

https://i.imgur.com/jteIDVs.png

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

PKN · Answer

Given b>a-----(1) and (2x+5) < (3x+1) or, c < a-------(2) (Corresponding angles) from (1) & (2), we have c4-------------(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)

pushpitkc · Answer

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.

PP777 · Answer

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.

generis · Answer

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 <--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

BrentGMATPrepNow · Answer

Here's a useful triangle property:
 https://i.imgur.com/dLnB7G2.png

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

souvonik2k · Answer

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.

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

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