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17 Nov 2008, 17:57
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Pls explain your answer.
Pls explain your answer.
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17 Nov 2008, 18:05
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Not sure if someone has an elegant solution to it - i'd be interested in seeing one. But for test taking purposes, here's how I would tackle the problem.
Looking at the first data:
Knowing that a=b does not provide sufficient information. When b is negative, then 1/a > 1/(b+1) whereas when b is positive then 1/a < 1/(b+1)
Looking at the second data:
Knowing b>0 alone does not provide sufficient information. a could be any number so 1/a could be greater or less than 1/(b+1).
Looking at the two data combined:
If a is equal to b and b is a positive number (thus a is a positive number) then we have sufficient information to answer to question. if we sub in a for b, the question becomes is 1/a < 1/(a+1) where a is positive? The answer is intuitively no.
Thus, 1 and 2 combined is enough to answer the question but neither alone is sufficient.
Looking at the first data:
Knowing that a=b does not provide sufficient information. When b is negative, then 1/a > 1/(b+1) whereas when b is positive then 1/a < 1/(b+1)
Looking at the second data:
Knowing b>0 alone does not provide sufficient information. a could be any number so 1/a could be greater or less than 1/(b+1).
Looking at the two data combined:
If a is equal to b and b is a positive number (thus a is a positive number) then we have sufficient information to answer to question. if we sub in a for b, the question becomes is 1/a < 1/(a+1) where a is positive? The answer is intuitively no.
Thus, 1 and 2 combined is enough to answer the question but neither alone is sufficient.
17 Nov 2008, 18:06
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1/a < 1/(b+1) <=> 1/a - 1/(b+1) < 0
<=> (b+1-a)/[a(b+1] < 0
(1) a = b
Q <=> 1/[a(a+1]<0
<=> a(a+1) < 0 ==> insuff
if -1<a<0 => 1/a < 1/(b+1)
if a not in (-1,0) => 1/a > 1/(b+1)
(2) b > 0 insuff
(1) & (2)
a = b > 0 => 1/a > 1/(b+1) suff
Answer C
<=> (b+1-a)/[a(b+1] < 0
(1) a = b
Q <=> 1/[a(a+1]<0
<=> a(a+1) < 0 ==> insuff
if -1<a<0 => 1/a < 1/(b+1)
if a not in (-1,0) => 1/a > 1/(b+1)
(2) b > 0 insuff
(1) & (2)
a = b > 0 => 1/a > 1/(b+1) suff
Answer C
