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09 Apr 2020, 06:33
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:52) correct
34%
(01:56)
wrong
based on 346
sessions
History
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What is the largest positive integer n such that \(\frac{n^2 + 7n + 12}{n^2−n−12}\) is also positive integer?
A. 6
B. 8
C. 10
D. 12
E. 16
Are You Up For the Challenge: 700 Level Questions
A. 6
B. 8
C. 10
D. 12
E. 16
Are You Up For the Challenge: 700 Level Questions
09 Apr 2020, 07:08
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Bunuel
Two ways..
(I) Simplify and then substitute choices...
\(\frac{n^2 + 7n + 12}{n^2−n−12}\)..
\(=\frac{(n+4)(n+3)}{(n-4)(n+3)}\)..
\(=\frac{n+4}{n-4}\)..
As we are looking for the largest value, start from E..
E) n=16....\(\frac{n+4}{n-4}=\frac{16+4}{16-4}=\frac{20}{12}\)..Not an integer
D) n=12....\(\frac{n+4}{n-4}=\frac{12+4}{12-4}=\frac{16}{8}=2\)....YES
(II) Substitute the choices..
Again start from the largest
E) n=16.....\(\frac{n^2 + 7n + 12}{n^2−n−12}\)=\(\frac{16^2 + 7*16 + 12}{16^2−16−12}=\frac{256+112+12}{256-28}=\frac{380}{228}\)...NO
D) n=12.....\(\frac{n^2 + 7n + 12}{n^2−n−12}\)=\(\frac{12^2 + 7*12 + 12}{12^2−12−12}=\frac{12(12+7+1)}{12(12-2)}=\frac{12*20}{12*10}=2\).....YES
D
ArunSharma12
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09 Apr 2020, 07:25
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Bunuel
\(\frac{n^2 + 7n + 12}{n^2−n−12}\) = \(\frac{(n+4)(n+3)}{(n-4)(n+3)}\)
\(\frac{(n+4)}{(n-4)}\) > 0
starting from option E, \(\frac{(16+4)}{(16-4)}\)=\(\frac{(20)}{(12)}\); not an integer
\(\frac{(12+4)}{(12-4)}\)=\(\frac{(16)}{(8)}\)=2; this is positive integer
Ans: D
