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If x > 0, b > a, and 2x + 5 < 3x + 1, then
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07 Jun 2018, 05:33
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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
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07 Jun 2018, 05:47
GMATPrepNow 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 *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)< (4x8)(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
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07 Jun 2018, 06:15
GMATPrepNow 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 *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
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08 Jun 2018, 04:01
Given: x>0, b>a, 2x+5 < 3x+1 Approach: As b>a, therefore 4x8 > 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
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08 Jun 2018, 08:53
GMATPrepNow 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 *kudos for all correct solutions I. InequalitiesTriangle 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\(4x8>3x+1\) \(x>9\)(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
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09 Jun 2018, 05:21
GMATPrepNow 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 *kudos for all correct solutions 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 3part 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 9Check 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
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09 Jun 2018, 06:15
GMATPrepNow 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 *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. 4x8 > 3x+1 x>9 Only iii fits 9.27 Answer D.
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