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Source: — Data Sufficiency |

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by Anurag@Gurome » Tue Jun 19, 2012 6:03 am
hey_thr67 wrote:Is 1/(a-b)<(b-a)?

1. a < b
2. 1 < |a-b|
Statement 1: As a < b, (a - b) < 0
Hence, 1/(a - b) is negative but (b - a) is positive.
Hence, 1/(a - b) is always less than (b - a)

Sufficient

Statement 2: Consider the following two examples,
a = 3, b = 1 --> |a - b| > 1 ---> 1/(a - b) > (b - a) --> NO
a = 1, b = 3 --> |a - b| > 1 ---> 1/(a - b) < (b - a) --> YES

Not sufficient

The correct answer is A.
Last edited by Anurag@Gurome on Tue Jun 19, 2012 6:11 am, edited 1 time in total.
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by Anurag@Gurome » Tue Jun 19, 2012 6:11 am
hey_thr67 wrote:What is wrong with rephrasing the question to (b-a)^2 > 1
Because we cannot conclude that from the given condition.

Let us assume that 1/(a - b) is less than (b - a).

Now, for a < b,
  • (a - b) < 0 and (b - a) > 0
    So, 1/(a - b) < (b - a)
    --> (b - a)(a - b) < 1 ............. Change of inequality sign as we are multiplying with a negative number
    --> -(a - b)(a - b) < 1
    --> -(a - b)² < 1

    Now, (a - b)² is a non-negative quantity. Hence, -(a - b)² is always less than or equal to zero. Therefore, -(a - b)² will be always less than 1.

    Hence, our initial assumption that 1/(a - b) < (b - a) is true for a < b.
And, for a > b,
  • (a - b) > 0 and (b - a) < 0
    So, 1/(a - b) < (b - a)
    --> (b - a)(a - b) > 1 ............. No change of inequality sign as we are multiplying with a positive number
    --> -(a - b)(a - b) > 1
    --> -(a - b)² > 1

    Now, (a - b)² is a non-negative quantity. Hence, -(a - b)² is always less than or equal to zero. Therefore, -(a - b)² cannot be greater than 1.

    Hence, our initial assumption that 1/(a - b) < (b - a) must be false for a > b.
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