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# a < b < c < d < 0

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a < b < c < d < 0 [#permalink]  30 Aug 2018, 16:30
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Question Stats:

72% (00:44) correct 27% (00:34) wrong based on 29 sessions
$$a < b < c < d < 0$$

 Quantity A Quantity B $$a - d$$ $$bc$$

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Sandy
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Re: a < b < c < d < 0 [#permalink]  01 Sep 2018, 10:01
sandy wrote:
$$a < b < c < d < 0$$

 Quantity A Quantity B $$a - d$$ $$bc$$

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

As per the condition we are told that all -ve numbers

Let's take $$a < b < c < d < 0$$ as $$-100 < -10 < -3 < -1 < 0$$

When we sum two numbers obviously we'll get -ve sum for the above case.

When we multiply two -ve digits we get +ve result.

So Col B > Col A.

B.
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Re: a < b < c < d < 0 [#permalink]  02 Sep 2018, 03:02
Given a < b < c < d < 0 that all the four numbers a, b, c, d are negative

A) a - d will be less than 0 because a is a bigger negative than d and so a - d will always result in a negative

B) bc => this is product of two negative numbers, which will always be positive

Hence quantity B is bigger, Answer choice B
Re: a < b < c < d < 0   [#permalink] 02 Sep 2018, 03:02
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