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# Is 10 percent of k less than 20 percent of n ? 1) k is less

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Senior Manager
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Is 10 percent of k less than 20 percent of n ? 1) k is less [#permalink]

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20 Jul 2006, 15:33
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Is 10 percent of k less than 20 percent of n ?
1) k is less than n.
2) 1 percent of k is less than 5 percent of n
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20 Jul 2006, 16:35
A nice way to do this question is to draw a quick graph, and you see that together they are sufficient, but separately neither is.
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20 Jul 2006, 20:25
Nice tricky question!

If you want a numerical solution: (I don't know how to draw the graph).

C it is.

A is not sufficient.
example, k=40, n=50 0,1k=4 <0,2n=10.
But: k=-50, n=-40 => 0,1k = -5 > 0,2n=-8.

B is not sufficient, too.

Both statements together are sufficient.
3 cases:
a) n, k > 0 (obvious)
b) k<0, n> 0 (obvious)
c) n, k < 0

k<n (by (1))
=> -k>-n (multiplying with -1)
Using (2)
0,01k < 0,05 n or equivalent k < 5n
Hence: -k > -5n (multiplying with -1)
-k > -2n must also be true because -5n > -2n (because -n > 0)
Senior Manager
Joined: 22 May 2006
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Location: Rancho Palos Verdes
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20 Jul 2006, 23:00
OA is C.

IME

Is 0.1k < 0.2n?

stmt1. k<n
only when k<n<0 => 0.1k > 0.2n : insuff.

stmt2. 0.01k < 0.05n
since k<n or k>n : insuff.

thust stmt1+stmt2
tells that 0 < k < n.

thus, sufficient. C.
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21 Jul 2006, 00:02
0.10k<0.20n reduces n > k/2, if we plot k and n on the x and y axes, we get the region above the line though the origin with slope 1/2. It's really quite fast, and this technique can come in handy!
21 Jul 2006, 00:02
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