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If n is an integer, is \(\frac{(100-n)}{n}\) an integer?
(1) \(n > 4\)
(2) \(n^2 = 25\)
(1) \(n > 4\)
(2) \(n^2 = 25\)
04 Feb 2018, 04:36
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broall
If n is an integer, is (100-n)/n an integer?
(1) n > 4
(2) n^2 = 25
(1) n > 4
(2) n^2 = 25
1. If n is greater than 4, it could be 5 or 6
If n=5, \(\frac{100-n}{n} = \frac{95}{5} = 19\)
If n=6, \(\frac{100-n}{n} = \frac{94}{6} = \frac{47}{3}\) which is not an integer (Insufficient)
2. If n^2 = 25, n can be 5 or -5
If n=5, \(\frac{100-n}{n} = \frac{95}{5} = 19\)
If n=-5, \(\frac{100-n}{n} = \frac{105}{-5} = -21\) (Sufficient - Option B)
04 Feb 2018, 23:10
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broall
If n is an integer, is (100-n)/n an integer?
(1) n > 4
(2) n^2 = 25
(1) n > 4
(2) n^2 = 25
(100-n)/n can also be written as = 100/n - n/n = 100/n - 1
1 is an integer, so the above will be an integer only if 100 is divisible by n.
(1) n > 4, but we dont know whether n is a factor of 100 or not. If n=5 or say 10, then this will be an integer. But if n = 6 or 7 say, then not an integer.
So not sufficient.
(2) n^2 = 25 means that n = 5 or n = -5.
100/5 will give an integer (20) and 100/-5 will also give an integer (-20). In either case, 100/n - 1 will be an integer. So this is sufficient.
Hence B answer
