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MGMAT Advanced Quant - Ch 4, In Action Q 13

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by leighton » Mon Mar 19, 2012 4:40 am
Is (a-k)/(b-k)>(a+k)/(b+k)?

(1) a>b>k

(2) k>0

OA:C

I know that with inequalities with variable expressions, if you don't know the sign of the variables, you must consider both a positive and negative case.

The answer explanation considers the following cases:

1) If (b-k)/(b+k)>0
2) If (b-k)/(b+k)<0

I actually don't understand why they've done this or their explanation in general. Can someone please help?

Thanks.
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Source: — Data Sufficiency |
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by Jim@StratusPrep » Mon Mar 19, 2012 6:24 am
1) will change if the numbers are positive or negative.

2) nothing about a or b

Logic goes like this when we know the order and all are positive: as a>b, the fraction a/b will always be greater than 1. If you think of averages when adding or subtracting k/k this problem gets easier. Adding 1/1, 2/2, 3/3, etc... Will always decrease the new "average" of the two fractions. 5/4 (1.25) becomes 6/5 (1.2), 10/9 (.1111) becomes 11/10 (.1). This will always persist. The opposite happens when you subtract.

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by killer1387 » Mon Mar 19, 2012 10:47 am
leighton wrote:Is (a-k)/(b-k)>(a+k)/(b+k)?

(1) a>b>k

(2) k>0

OA:C

I know that with inequalities with variable expressions, if you don't know the sign of the variables, you must consider both a positive and negative case.

The answer explanation considers the following cases:

1) If (b-k)/(b+k)>0
2) If (b-k)/(b+k)<0

I actually don't understand why they've done this or their explanation in general. Can someone please help?

Thanks.
(a-k)/(b-k)>(a+k)/(b+k)
=> k(a-b)/((b-k)(b+k))>0

now 1)
not no for the case of negative k
insuff

2) insuff

1+2

suff
hence
C
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