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# Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b|

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Bunuel
SOLUTION

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

(1) a < b --> we can rewrite this as: $$a-b<0$$ so LHS is negative, also we can rewrite it as: $$b-a>0$$ so RHS is positive --> negative<positive. Sufficient.

(2) 1 < |a - b| --> if $$a-b=2$$ (or which is the same $$b-a=-2$$) then LHS>0 and RHS<0 and in this case the answer will be NO if $$a-b=-2$$ (or which is the same $$b-a=2$$) then LHS<0 and RHS>0 and in this case the answer will be YES. Not sufficient.

for the second statement;

1 < |a - b| --> 1< a-b or a-b < -1

AND

a-b < -1 --> 1< b-a therefore LHS will be negative and RHS will be positive -->negative<positive. Sufficient.

what is wrong with my approach ?
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Bunuel
SOLUTION

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

(1) a < b --> we can rewrite this as: $$a-b<0$$ so LHS is negative, also we can rewrite it as: $$b-a>0$$ so RHS is positive --> negative<positive. Sufficient.

(2) 1 < |a - b| --> if $$a-b=2$$ (or which is the same $$b-a=-2$$) then LHS>0 and RHS<0 and in this case the answer will be NO if $$a-b=-2$$ (or which is the same $$b-a=2$$) then LHS<0 and RHS>0 and in this case the answer will be YES. Not sufficient.

for the second statement;

1 < |a - b| --> 1< a-b or a-b < -1

AND

a-b < -1 --> 1< b-a therefore LHS will be negative and RHS will be positive -->negative<positive. Sufficient.

what is wrong with my approach ?

From $$1 < |a - b|$$ we can have two cases:

A. $$1 < a-b$$ --> $$(\frac{1}{a - b}=positive) > (b - a=negative)$$ --> answer YES.

B. $$a-b<-1$$ --> $$(\frac{1}{a - b}=negative) < (b - a=positive)$$ --> answer NO.

Hope it's clear.
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Re: Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b| [#permalink]
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I did it like this:

We can reduce the given statement as:

1/(a-b)>(b-a) -> Multiplying both sides by (a-b), we get: 1>(a-b)(b-a)

Taking negative out from RHS: 1<(a-b)(a-b), which is what we have to prove.

Now, Statement 1: a<b -> a-b<0. Squaring both sides: (a-b)(a-b)<0. Hence, the answer would be No. Thus, sufficient.

Statement 2: 1<|a-b| -> 1<(a-b) or 1>(a-b). Thus insufficient.

Please do let me know if this is a good way to proceed with such questions.

Thanks.
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I did it like this:

We can reduce the given statement as:

1/(a-b)>(b-a) -> Multiplying both sides by (a-b), we get: 1>(a-b)(b-a)

Taking negative out from RHS: 1<(a-b)(a-b), which is what we have to prove.

Now, Statement 1: a<b -> a-b<0. Squaring both sides: (a-b)(a-b)<0. Hence, the answer would be No. Thus, sufficient.

Statement 2: 1<|a-b| -> 1<(a-b) or 1>(a-b). Thus insufficient.

Please do let me know if this is a good way to proceed with such questions.

Thanks.

Unfortunately, most of it is wrong.

We cannot multiply 1/(a-b)>(b-a) by a-b, because we don't know its sign.
If a-b is positive, then we would have 1 > (b-a)(a-b);
If a-b is negative, then we would have 1 < (b-a)(a-b): flip the sign when multiplying by negative value.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

Next, you cannot square a-b<0 and write (a-b)^2<0. This is obviously wrong: the square of a number cannot be less than zero.

We can only raise both parts of an inequality to an even power if we know that both parts of the inequality are non-negative (the same for taking an even root of both sides of an inequality).

Hope this helps.
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Bunuel
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

(1) a < b
(2) 1 < |a - b|

Data Sufficiency
Question: 120
Category: Arithmetic; Algebra Arithmetic operations; Inequalities
Page: 161
Difficulty: 650

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

(1) a < b
(2) 1 < |a - b|

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

$$\frac{1+(a-b)^2}{(a-b)}$$ > 0 ? we can see that numerator is always +ive . all we need to know is a>b?

(1) a < b ; then $$\frac{1+(a-b)^2}{(a-b)}$$ < 0 . Sufficient.

(2) 1 < |a - b|
case 1 : a-b < -1 or a<b-1 --> a<b
case 2: a-b>1 or a> b+1------> a>b
not sufficient.

Ans : A
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Going through the solutions posted above, I realized that this question is a good illustration of the importance of first analyzing the question statement before moving to St. 1 and 2 in a DS question.

Most students dived straight into Statement 1 and then tried to draw inferences from St. 1 to determine whether the inequality given in the question statement was true or not.

Imagine if you had done this instead before going to St. 1:

$$\frac{1}{a-b} < b-a$$

Case 1: a - b is positive (that is, a > b)

Multiplying both sides of an inequality with with the positive number (a-b) will not change the sign of inequality.

So, the question simplifies to: Is 1 < (b-a)(a-b)

Now, b - a will be negative.

So, the question simplifies to: Is $$1 < -(a-b)^2$$ . . . (1)

$$(a-b)^2$$, being a square term, will be >0 (Note: a - b cannot be equal to zero because then the fraction given in the question: 1/a-b becomes undefined)

So, $$-(a-b)^2$$ will be < 0

Therefore, the question simplifies to: Is 1 < (a negative number?)

Case 2: a - b is negative (that is, a < b)

Multiplying both sides of an inequality with the negative number (a-b) will change the sign of inequality.

So, the question simplifies to: Is 1 > (b-a)(a-b)

Now, b - a will be positive.

So, the question simplifies to: Is $$1 > -(a-b)^2$$ . . . (2)

Again, by the same logic as above, we see that the question simplifies to: Is 1 > (a negative number?)

Thus, from the question statement itself, we've inferred that:

If a > b, the answer to the question asked is NO
If a < b, the answer to the question asked is YES

Thus, the only thing we need to find now is whether a > b or a < b.

Please note how we are going to Statement 1 now with a much simpler 'To Find' task now.

One look at St. 1 and we know that it will be sufficient.

One look at St. 2 and we know that it doesn't give us a clear idea of which is greater between a and b, and so, is not sufficient.

To sum up this discussion, spending time on analyzing the question statement before going to the two statements usually simplifies DS questions a good deal.

Hope this helped!

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Re: Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b| [#permalink]
[quote="Bunuel"]SOLUTION

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

(1) a < b --> we can rewrite this as: $$a-b<0$$ so LHS is negative, also we can rewrite it as: $$b-a>0$$ so RHS is positive --> negative<positive. Sufficient.

(2) 1 < |a - b| --> if $$a-b=2$$ (or which is the same $$b-a=-2$$) then LHS>0 and RHS<0 and in this case the answer will be YES if $$a-b=-2$$ (or which is the same $$b-a=2$$) then LHS<0 and RHS>0 and in this case the answer will be NO. Not sufficient.

Dear,

Please Clarify me Rewrite like this one.I thought rewite variable without considering its Sign is Wrong
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Bunuel
SOLUTION

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

(1) a < b --> we can rewrite this as: $$a-b<0$$ so LHS is negative, also we can rewrite it as: $$b-a>0$$ so RHS is positive --> negative<positive. Sufficient.

(2) 1 < |a - b| --> if $$a-b=2$$ (or which is the same $$b-a=-2$$) then LHS>0 and RHS<0 and in this case the answer will be YES if $$a-b=-2$$ (or which is the same $$b-a=2$$) then LHS<0 and RHS>0 and in this case the answer will be NO. Not sufficient.

Dear,

Please Clarify me Rewrite like this one.I thought rewite variable without considering its Sign is Wrong

We cannot multiply an inequality by a variable if don't know its sign but we can add/subtract a value to both sides. To get a-b<0 from a < b we are subtracting b from both sides and to get b - a > 0 we are subtracting a from both sides.
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Re: Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b| [#permalink]
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Let us look at this question differently here:

While solving DS inequality questions, the best approach is to always breakdown the question stem if possible. To breakdown the question stem, there are three hygiene factors, that if followed will simplify your analysis of the question.

1. Always keep the RHS of the inequality as 0
2. Simplify the LHS to a product or division of values (product and division of terms are easier to analyze)
3. Always try and maintain even powered terms (as the sign of them will always be 0 or positive)

Let us breakdown the question stem here:

1/(a - b) > b - a
(1/(a - b)) - (b - a) > 0

Since we have an option to get a squared term, before we take the LCM let us change b - a to a - b.

(1/(a - b)) + (a - b) > 0 ----> (1 + (a - b)^2)/(a - b) > 0

Now since we have deconstructed the question stem to a division we just need to worry about the signs of the numerator and denominator.

The RHS here is > 0, so we the numerator and denominator have to have the same sign. The numerator however will always be positive, since 1 + (a - b)^2 will always be positive. So we just need need the denominator to also be positive.

The entire question stem can now be rephrased as 'Is a - b > 0'?

Statement 1 : a - b < 0

This gives us a definite NO. So sufficient.

Statement 2 : 1 < |a - b|

|a - b| > 1. This only tells us that a - b > 1 or a - b < -1. So a - b can be both positive or negative. Insufficient.

Hope this helps!

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Re: Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b| [#permalink]
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Bunuel
Is 1/(a - b) > b - a ?

(1) a < b
(2) 1 < |a - b|

The Official Guide For GMAT® Quantitative Review, 2ND Edition

Asked: Is 1/(a - b) > b - a ?
1/(a-b) + (a-b) >0
(a-b)^2 + 1 / (a-b) >0
Since numerator >0
Q. a -b >0

(1) a < b
a-b<0
a-b is NOT >0
SUFFICIENT

(2) 1 < |a - b|
a-b > 1 or a-b < -1
NOT SUFFICIENT

IMO A
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Re: Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b| [#permalink]
Bunuel
SOLUTION

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

(1) a < b --> we can rewrite this as: $$a-b<0$$ so LHS is negative, also we can rewrite it as: $$b-a>0$$ so RHS is positive --> negative<positive. Sufficient.

(2) 1 < |a - b| --> if $$a-b=2$$ (or which is the same $$b-a=-2$$) then LHS>0 and RHS<0 and in this case the answer will be YES if $$a-b=-2$$ (or which is the same $$b-a=2$$) then LHS<0 and RHS>0 and in this case the answer will be NO. Not sufficient.

statement 1 says a<b so if we substitute values for a and b.
example a = 2 and b = 3
then from original equation we get
1/(2-3) > 3-2 ie 1/-1 > 1 ie -1>1 how is that possible? And why is statment 1 sufficient then
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Re: Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b| [#permalink]
siddharth287
Bunuel
SOLUTION

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

(1) a < b --> we can rewrite this as: $$a-b<0$$ so LHS is negative, also we can rewrite it as: $$b-a>0$$ so RHS is positive --> negative<positive. Sufficient.

(2) 1 < |a - b| --> if $$a-b=2$$ (or which is the same $$b-a=-2$$) then LHS>0 and RHS<0 and in this case the answer will be YES if $$a-b=-2$$ (or which is the same $$b-a=2$$) then LHS<0 and RHS>0 and in this case the answer will be NO. Not sufficient.

statement 1 says a<b so if we substitute values for a and b.
example a = 2 and b = 3
then from original equation we get
1/(2-3) > 3-2 ie 1/-1 > 1 ie -1>1 how is that possible? And why is statment 1 sufficient then

The question asks whether 1/(a - b) > b - a. If we can give a definite YES answer to the question (YES, 1/(a - b) IS greater than b - a) OR a definite NO answer to the question (No, 1/(a - b) is NOT greater than b - a), then the statement is sufficient (recall that a definite NO answer to a DS question is also sufficient).

Now, since a < b, then a - b < 0 and b - a > 0, thus 1/(a - b) = 1/negative = negative, while b - a = positive, so 1/(a - b) < b - a, which means that the answer to the question is NO, 1/(a - b) is NOT greater than b - a. That's why the first statement is sufficient.

Check the links below for DS Strategies and Tactics

For other subject check Ultimate GMAT Quantitative Megathread.

Hope it helps.
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Is 1/(a - b) > b - a ? (1) a < b (2) 1 < |a - b| [#permalink]
Bunuel
Is 1/(a - b) > b - a ?

(1) a < b
(2) 1 < |a - b|

The question asks is 1/(a - b) > b - a ?

We can tell that a-b and b-a have opposite signs and if we suppose that 1/(a-b)> b-a then a-b must be positive.
The question becomes is a>b?

Statement 1 tells us that a<b, so the answer to our question is a definte NO.
Statement 1 alone is sufficient.

Statement 2 tells us that |a - b|>1 which means that the distance between a and b on the number line is greater than 1 but we don't know which is to the right of which. a can be to the right of b, the same way b can be to the right of a (both the scenarios are possible).
Therefore, statement 2 is not sufficient.

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