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Is a > (3b)/2, where a and b are real numbers? (1) b < a/2 (2) a and

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Is a > (3b)/2, where a and b are real numbers? (1) b < a/2 (2) a and  [#permalink]

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New post 23 Jul 2018, 21:11
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Question Stats:

41% (01:41) correct 59% (01:25) wrong based on 49 sessions

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Is a > (3b)/2, where a and b are real numbers?

(1) b < a/2

(2) a and b are non zero integers.
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Re: Is a > (3b)/2, where a and b are real numbers? (1) b < a/2 (2) a and  [#permalink]

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New post 23 Jul 2018, 22:22
To find whether a > \(\frac{3b}{2}\)

Statement 1

b < \(\frac{a}{2}\)

=> \(\frac{a}{2}\) > b

=> a > 2b

=> a > \(\frac{4b}{2}\)

=> a > \(\frac{3b}{2}\)

Statement 1 is sufficient

Statement 2

a and b are non zero integers

=> a can be 1 and b can be -1 => a > \(\frac{3b}{2}\)

=> a can be -1 and b can be 1 => a < \(\frac{3b}{2}\)

Statement 2 is not sufficient

Hence option A
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Re: Is a > (3b)/2, where a and b are real numbers? (1) b < a/2 (2) a and  [#permalink]

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New post 30 Jul 2018, 09:21
amanvermagmat wrote:
Is a > (3b)/2, where a and b are real numbers?

(1) b < a/2

(2) a and b are non zero integers.



Is a > 3b/2
or is 2a > 3b ?

1) b < a/2
or 2b < a
(a,b) = (3,1) --> 2a > 3b (yes)
(a,b) = (-1/2 , -1/3) --> 2a = 3b (no)
insuff.

2) a,b are integers (<>0)
insuf

1) + 2)
(a,b) = (3,1) --> 2a > 3b (yes)
(a,b) = (-3,-2) --> 2a = 3b (no)
insuff.
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Re: Is a > (3b)/2, where a and b are real numbers? (1) b < a/2 (2) a and &nbs [#permalink] 30 Jul 2018, 09:21
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