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If a<b and c<b, is abc<a?

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If a<b and c<b, is abc<a?  [#permalink]

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New post 15 Apr 2018, 10:19
2
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (02:25) correct 38% (02:15) wrong based on 40 sessions

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If a<b and c<b, is abc<a?

1. ab < 0
2. ac > 0

Source: Experts Global

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Re: If a<b and c<b, is abc<a?  [#permalink]

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New post 15 Apr 2018, 10:39
pushpitkc wrote:
If a<b and c<b, is abc<a?

1. ab < 0
2. ac > 0

Source: Experts Global


IMO C. I will approach this question with cases approach!

Q: Is abc < a ?

based on sign of a we can simplify the question further... which will help us in the cases.

Case 1) a > 0 : then the Q : bc < 1 ? ( cancelling a )

Case 2) a < 0 : then the Q : bc > -1 ? ( cancelling negative a )

1. ab < 0

Case 1) not possible here as b > a : so if a > 0 ... b also > 0 and ab < 0 not possible.

Case 2) a < 0. If ab < 0 then b > 0. Our Q for case 2 is : Is bc > -1 ? we do not know if c > 0 or not.

Hence 1. is Insufficient!

2. ac > 0

Case 1) a > 0 . If ac > 0 then c > 0. Since we know b > c... hence b > 0. Q : is bc < 1 ? could be either yes or no depending on magnitude of b & c.

Case 2) a < 0. If ac > 0 then c < 0. Since we know b > c ... b could be positive or negative. Our Q: is bc > -1? could be yes or no depending on if b is positive or negative.

Hence 2. is insufficient.

(1) + (2) together...

Case 1) a > 0 is not possible as ab < 0. and b > a.

Case 2) a must be less than zero. Our question is : Is bc > -1?

ac > 0 hence c > 0 & b > c hence b >0

hence bc > 0.

Hence the answer to the question in this case is "YES" always. This means both 1. and 2. together are sufficient to answer.

Hence Option (C) is our answer.

Best,
Gladi
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Re: If a<b and c<b, is abc<a?  [#permalink]

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New post 29 Apr 2018, 05:16
I)ab<0
B>a given so b +be a-ve
But don't know abt C so insufficient

II) ac>0
Both a and c can be +be or -ve.also no info about b
So insufficient

Combining both
a>0 from 1) so c>0 from 2)
So got sign of a b and c

C is answer

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Re: If a<b and c<b, is abc<a? &nbs [#permalink] 29 Apr 2018, 05:16
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