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# Is 1/(a + b) > b - a?

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Is 1/(a + b) > b - a?  [#permalink]

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23 Mar 2017, 06:07
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65% (hard)

Question Stats:

42% (02:09) correct 58% (01:48) wrong based on 46 sessions

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Is 1/(a + b) > b - a?

(1) a + b > 0
(2) b^2 – a^2 > 1

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Re: Is 1/(a + b) > b - a?  [#permalink]

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23 Mar 2017, 10:14
False. Because a+b>0, 1/(a+b) will ALWAYS be less than 1. b^2-a^2>1 can be square rooted to b-a>1. This means b-a>1/(a+b). Always.

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Re: Is 1/(a + b) > b - a?  [#permalink]

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23 Mar 2017, 10:31
C.

Statement 1

a+b > 0

Let a=b=1 => (1/2 > 0) => Yes
Lets a=0 & b=1 => (1 > 1 ) => No

Thus, the statement is insufficient.

Statement 2

b^2 - a^2 > 1

Let a=1 & b=5 => (1/6 > 4) => No
Let a=0 & b=-2 => (1/-2 > -2) => Yes

Thus, statement 2 is also insufficient.

Combining both the statements:

We know that b has to be the bigger value since b^2 > 1 + a^2 ( which is always a +ve value )
Thus, a+b will also have be +ve ( since we cannot have a negative value for b )

Therefore, we can consider the above example where a=1 & b=5 we get a definite NO.
Hence, the answer is C.

Hope this reasoning is correct!
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Is 1/(a + b) > b - a?  [#permalink]

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Updated on: 23 Mar 2017, 13:42
Original statement:

Is 1/(a+b) > b - a ?

This can be re-arranged as 1 > (b - a)(b + a) if (a + b) is positive, or 1 < (b - a)(b + a) if (a + b) is negative.

Or, is 1 > (b^2 - a^2) when (a + b) is positive, or is 1 < (b^2 - a^2) if (a + b) is negative.

I:

(a + b) is positive. We therefore know we need to answer the inequality 1 > (b^2 - a^2). But we don't know anything about a, b, or (b^2 - a^2). Insufficient.

II:

b^2 - a^2 > 1.

On first glance, this looks like it satisfies the re-arrangement of the original statement. However, we do not know if (a + b) is positive or negative, so we do not know if the answer to the question is Yes or No. If (a + b) is positive, the answer is no, If (a + b) is negative, the answer is yes. Insufficient.

Combine:

We know that (a + b) is positive, and that b^2 - a^2 > 1.

From the original statement, we are asking if 1 > b^2 - a^2 if (a + b) is positive.

Since (a + b) is positive, b^2 - a^2 > 1, and the answer is No. Sufficient.

Originally posted by LeveragedTiger on 23 Mar 2017, 12:37.
Last edited by LeveragedTiger on 23 Mar 2017, 13:42, edited 2 times in total.
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Re: Is 1/(a + b) > b - a?  [#permalink]

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23 Mar 2017, 12:41
1
Hi Bunuel,

I somehow got this ques correct, but is there a better way to solve this rather than putting the values?
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Re: Is 1/(a + b) > b - a?  [#permalink]

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23 Mar 2017, 22:53
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Re: Is 1/(a + b) > b - a?  [#permalink]

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24 Mar 2017, 05:07
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Top Contributor
1
Hi,

The following points will hold you in good stead while solving questions on inequalities.

1. Always break down the question stem
2. Keep the right hand side of the inequality as 0 (This helps in an analysis and will save you the tedious task of plugging in values)

The question here is 'Is 1/(a + b) > b - a?'. Let us break down the question stem by keeping the right hand side 0 and simplifying.

1/(a + b) - (b - a) > 0
Simplifying the left hand side we get
(1 - (b - a)(b + a))/(a + b) > 0 ------> (1 - (b^2 - a^2)/(a + b)) > 0

So now the question can be rephrased as 'Is (1 - (b^2 - a^2)/(a + b)) > 0'

Statement 1 : a + b > 0

Here the denominator a + b in (1 - (b^2 - a^2)/(a + b)) is positive, but we have no information whether the numerator is positive or negative. So it is possible for us to get a YES and a NO. Insufficient.

Statement 2 : b^2 – a^2 > 1

The numerator 1 - (b^2 - a^2) will always be negative since b^2 - a^2 is greater than 1, but we have no information about the denominator a + b. Insufficient.

Combining 1 and 2 :

We know that the numerator is negative and the denominator is positive, so the entire term (1 - (b^2 - a^2)/(a + b)) will always be negative. This gives us a definite NO. Sufficient.

Hope this helps!

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Re: Is 1/(a + b) > b - a?   [#permalink] 24 Mar 2017, 05:07
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