If a, b, c, and d are positive integers (a − b)/(c − d)

Question

If a, b, c, and d are positive integers is  \frac{a − b}{c − d} <−1  ?

(1) |a – b| > |c – d|

(2) c > b > d > a

GMATinsight · Accepted Answer

Question :  \frac{a − b}{c − d} <−1  ?

Statement 1 : |a – b| > |c – d|

i.e. Absolute value of a-b is greater than absolute value of c-d

i.e. for the fraction  \frac{a − b}{c − d}  the absolute value of numerator is greater than absolute value of denominator

BUT the value of  \frac{a − b}{c − d}  may be Less than -1 or Greater than +1   hence

NOT SUFFICIENT

Statement 2 : c > b > d > a

i,.e. we can deduce that  \frac{a − b}{c − d}  will be  negative  cause a-b is negative and c-d is positive

Still we can't compare absolute values of a-b and c-d hence result may be between 0 and -1 or less than -1 hence

NOT SUFFICIENT

Combining the two statements

We can now deduce that  \frac{a − b}{c − d}  MUST be Less than -1 as it's negative as explained in statement 2

SUFFICIENT

ANswer: Option C

chetan2u · Answer

Let us analyse \frac{a − b}{c − d} <−1 . The above will hold true when both of the below are true. (A) Both a-b and c-d should have opposite sign. One positive and other negative. Why? \frac{a-b}{c-d} is negative, so only possibilities are +/- or -/+. -/- and +/+ are both positive. (B) Absolute value of a-b is greater than that of c-d, that is, |a-b|>|c-d|. Why? Now, \frac{a − b}{c − d} <−1 Multiply by -1, so \frac{ b-a}{c-d} >1 . This will be possible only when numerator is more than denominator, ignoring the sign. Both of the above are true, the answer is yes. If we prove otherwise, the answer is NO. (1) |a – b| > |c – d| (B) is true but nothing about (A) Insufficient (2) c > b > d > a c-d>0 and b-a>0 or a-b<0. Thus both have opposite sign. (A) is true but nothing about (B). Insufficient Combined Both (A) and (B) are true. Hence, the answer is yes. Sufficient C

TestPrepUnlimited · Answer

We cannot move anything in the fraction as we don't know the signs of a - b or c - d.

Statement 1:

We can get  \frac{|a - b|}{|c - d|} > 1 , which is  |\frac{a - b}{c - d}| > 1 . So we may have  \frac{a - b}{c - d}  being bigger than 1, or being less than -1. Insufficient.

Statement 2:

We now know we have top negative and bottom positive. The result must be negative, but we do not know anything about the distances. Then the magnitude of  \frac{a - b}{c - d}  can be anything in the negative numbers. Insufficient.

Combined:

Knowing the fraction is negative, we can say  \frac{a - b}{c - d} < -1  from statement 1. Sufficient.

Ans: C

KARISHMA315 · Answer

Hi  chetan2u  can you please elaborate by both are required -

(A) Both a-b and c-d should have opposite sign. One positive and other negative. 
(B) Absolute value of a-b is greater than that of c-d, that is, |a-b|>|c-d|.

Thank You.

chetan2u · Answer

Please check the earlier post again. I have added some details. 
Please ask if any further queries.

PyjamaScientist · Answer

chetan2u  Sir, I did this via below method. Does this make the cut?

Posted from my mobile device

KARISHMA315 · Answer

Let us analyse \frac{a − b}{c − d} <−1 . The above will hold true when both of the below are true. (A) Both a-b and c-d should have opposite sign. One positive and other negative. (B) Absolute value of a-b is greater than that of c-d, that is, |a-b|>|c-d|. Both of the above are true, the answer is yes. If we prove otherwise, the answer is NO. (1) |a – b| > |c – d| (B) is true but nothing about (A) Insufficient (2) c > b > d > a c-d>0 and b-a>0 or a-b<0. Thus both have opposite sign. (A) is true but nothing about (B). Insufficient Combined Both (A) and (B) are true. Hence, the answer is yes. Sufficient chetan2u Thank you the explanation is just brilliant. May I pls ask one last point , is it practise that can help with such questions and do you possibly have a collection of such questions or what was your thinking to come to the point (B) Absolute value of a-b is greater than that of c-d, that is, |a-b|>|c-d|. Many thanks again

chetan2u · Answer

Hi, In inequality, >0 or >1 would give you certain relations between the numerator and the denominator. Say \frac{a}{b}>0 ...This will tell us about the relative signs and nothing about values. Here, a and b will have same sign, but it could be a>b or b>a or a=b But, if you are given \frac{a}{b}>1 ...This will tell us about both the relative signs and the relative values. Here, a and b will have same sign, and |a|>|b|, because |numerator|>|denominator| will give you a value greater than 1. So, either a>b>0 or 0

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If a, b, c, and d are positive integers (a − b)/(c − d) < −1 ?

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