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Bunuel
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|2a| + |6a + b| = 24, what is the value of 4a + b?

Case1: a>0; (6a+b) >0. 2a +6a+b = 24, or, 8a+b = 24.
Case2: a>0; (6a+b) <0. 2a -6a-b = 24, or, 4a+b =- 24.
Case3: a<0; (6a+b) >0. -2a +6a+b = 24, or, 4a+b = 24.
Case4: a<0; (6a+b) <0. -2a -6a-b = 24, or, 8a+b =- 24.

Stat1: 4a + b > 0
So, case 3 will be selected. Sufficient

Stat2: 8a + b > 0
So, case 1 will be selected, but we cannot get 4a+b value. Not sufficient

So, I think A. :)
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|2a| + |6a + b| = 24, what is the value of 4a + b?

(1) 4a + b > 0
(2) 8a + b > 0

Bunuel

Hi, Can you explain why OE is E.
Upon evaluating the equation we can find that 4a+b=+/-24 and by using statement 1 we get one unique solution to the question which is 4a+b=24.
Isn't Statement 1 sufficient to answer the question?

Thanks
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umangshah
|2a| + |6a + b| = 24, what is the value of 4a + b?

(1) 4a + b > 0
(2) 8a + b > 0

Bunuel

Hi, Can you explain why OE is E.
Upon evaluating the equation we can find that 4a+b=+/-24 and by using statement 1 we get one unique solution to the question which is 4a+b=24.
Isn't Statement 1 sufficient to answer the question?

Thanks

Not sure how you got that 4a + b is either 24 or -24 bur the OA is E. For instance, if a = 1 and b = 16, then 4a + b ≠ 24.
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TarunKumar1234
|2a| + |6a + b| = 24, what is the value of 4a + b?

Case1: a>0; (6a+b) >0. 2a +6a+b = 24, or, 8a+b = 24.
Case2: a>0; (6a+b) <0. 2a -6a-b = 24, or, 4a+b =- 24.
Case3: a<0; (6a+b) >0. -2a +6a+b = 24, or, 4a+b = 24.
Case4: a<0; (6a+b) <0. -2a -6a-b = 24, or, 8a+b =- 24.

Stat1: 4a + b > 0
So, case 3 will be selected. Sufficient

Stat2: 8a + b > 0
So, case 1 will be selected, but we cannot get 4a+b value. Not sufficient

So, I think A. :)


TarunKumar1234, Statement 1 does not imply that a<0 and (6a+b)>0. Case 1 could also potentially be true because 4a+b>0 could also occur when a>0 and (6a+b) >0. For e.g., if a=2 and b=8, Case 1 will be satisfied but 4a+b will also be greater than 0

umangshah hope this helps you as well!
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TarunKumar1234
|2a| + |6a + b| = 24, what is the value of 4a + b?

Case1: a>0; (6a+b) >0. 2a +6a+b = 24, or, 8a+b = 24.
Case2: a>0; (6a+b) <0. 2a -6a-b = 24, or, 4a+b =- 24.
Case3: a<0; (6a+b) >0. -2a +6a+b = 24, or, 4a+b = 24.
Case4: a<0; (6a+b) <0. -2a -6a-b = 24, or, 8a+b =- 24.

Stat1: 4a + b > 0
So, case 3 will be selected. Sufficient

Stat2: 8a + b > 0
So, case 1 will be selected, but we cannot get 4a+b value. Not sufficient

So, I think A. :)




I followed the same approach, then why answer E is right?
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Bunuel - If you create 4 scenarios and simply the equations you will get 4a+b=24 or -24 for (+,-) and (-,+) case.

So if you get an out right value of 24 after keeping (a), should the answer be A.

If no, can you please explain how to solve this.
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