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19 Nov 2009, 05:49
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Not quite satisfied with the OE for Statement 2. From picking numbers you can sort of figure it out but is there another method?
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19 Nov 2009, 07:22
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Question Stem : Is \(\frac{(\frac{a}{b})}{c}\) an integer?
Condition given is a, b and c are positive distinct integers.
I will just give the reasoning for St. (2) since that is what you have a problem with.
St. (2) : a = b + c
Substituting this in the question stem we get : Is \(\frac{b+c}{bc}\) an integer?
It can be further reduced to : Is \(\frac{1}{c} + \frac{1}{b}\) an integer?
Now we can have two cases :
Case 1 : When either b or c is = 1
In this case, the minimum value for the other will be 2. Therefore the maximum value of \(\frac{1}{c} + \frac{1}{b}\) will be 1.5.
Also, since \(\frac{1}{c}\) or \(\frac{1}{b}\) can never be 0, the value of \(\frac{1}{c} + \frac{1}{b}\) will always be greater than 1. Hence it can never be an integer.
Case 2 : When a and b are > 1
In this case, the minimum values that and b can take will be 2 and 3. Therefore the maximum value of \(\frac{1}{c} + \frac{1}{b}\) will be \(\frac{1}{2} + \frac{1}{3}\) = \(\frac{5}{6}\)
Also, since \(\frac{1}{c} + \frac{1}{b}\) can never be 0, the values for \(\frac{1}{c} + \frac{1}{b}\) will be greater than 0 but less than equal to \(\frac{5}{6}\). Hence it can never be an integer.
Since both cases in St. (2) tell us that \(\frac{(\frac{a}{b})}{c}\) can never be an integer, St. (2) is sufficient.
Answer : B
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